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\title{Σημειώσεις Διαφορικές Εξισώσεις} | |
\date{2\textsuperscript{ο} εξάμηνο, 2016} | |
\author{\textlatin{\url{https://github.com/kongr45gpen/ece-notes}}} | |
\newtcbtheorem[number within=section]{theorem}{Θεώρημα}% | |
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\newtcbtheorem[number within=section]{exercise}{Άσκηση}% | |
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\begin{document} | |
\maketitle | |
\tableofcontents | |
\newpage | |
\part{Σεβαστιάδης} | |
Χρήστος Σεβαστιάδης | |
\begin{defn*}{Διαφορική εξίσωση} | |
Μια εξίσωση που αποτελείται από μια συνάρτηση και τις παραγώγους της | |
\end{defn*} | |
\paragraph{\textlatin{Langrange's}} | |
\(x',x'',x''',x^{(4)},\dots\) | |
\paragraph{\textlatin{Newton's}} | |
\(\dot{x}, \ddot{x}, \dddot{x}\) | |
\paragraph{\textlatin{Leibniz'}} | |
\(\od{x}{t}, \od[2]{x}{t}, \od[3]{x}{t}\) | |
\paragraph{} | |
\textit{π.χ.} | |
\[ | |
x(t) \od[2]{x(t)}{t} + 2 \od{x(t)}{t}=x(t)\sin (t) | |
\] | |
\begin{defn}{Τάξη}{} | |
\textbf{Τάξη} ονομάζεται ο μεγαλύτερος \emph{βαθμός} παραγώγου που εμφανίζεται στην εξίσωση | |
\end{defn} | |
\begin{defn}{Βαθμός}{} | |
\textbf{Βαθμός} ονομάζεται η μεγαλύτερη \emph{δύναμη} παραγώγου που εμφανίζεται στην εξίσωση | |
\end{defn} | |
\section{Διαφορική εξίσωση 1ης τάξης} | |
\begin{defn*}{} | |
\[ | |
\od{x}{t}=f(t,x) | |
\] | |
\end{defn*} | |
\paragraph{} | |
\subsection{Χωριζόμενες διαφορικές εξισώσεις} | |
Τυπική μορφή: | |
\[f(t,x) = \frac{-M(t,x)}{N(t,x)} = \od{x}{t} | |
\implies \underbrace{N(t,x)}_{N(x)} \dif x + \underbrace{M(t,x)}_{M(t)} \dif t = 0 \] | |
Αν δηλαδή τα \(N(t,x), \ M(t,x)\) εξαρτώνται μόνο από τα \(x\) και \(t\) αντίστοιχα, η εξίσωση ονομάζεται \textbf{χωριζόμενη}, και το αποτέλεσμά της μπορεί να βρεθεί με ολοκληρώματα: | |
\[ | |
\int N(x) \dif x + \int M(t) \dif t = c | |
\] | |
\begin{exercise*}{2.1} | |
\[x \dif x - t^2 \dif t = 0 \] | |
\tcblower | |
\[ N(x) = x, \quad M(t)=-t^2 \] | |
\begin{align*} | |
&\int x \dif x + \int (-t^2) \dif t = c \implies \\ | |
&\frac{1}{2} x ^ 2 - \frac{1}{3} t^3 = c \implies \\ | |
&x = \pm \sqrt{\frac{2}{3} t^3 + 2c} \implies \\ | |
&x= \pm \sqrt{\frac{2}{3} t^3 + \kappa } \\ | |
\text{ με } \kappa =2c | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{2.2} | |
\[x' = x^2t^3 \] | |
\tcblower | |
\begin{align*} | |
&\implies \od{x}{t}=x^2t^3 \\ | |
&\implies \frac{1}{x^2}\dif x - t^3 \dif t = 0 \\ | |
&\implies \int \frac{1}{x^2}\dif x + \int (-t^3) \dif t = c \\ | |
&\implies - \frac{1}{x} - \frac{t^4}{4} = c \\ | |
&\implies - \frac{1}{x} = c + \frac{t^4}{4} \\ | |
&\implies - \frac{4}{x} = 4c + t^4 \\ | |
&\implies x = \frac{-4}{t^4 + \kappa}, \quad \text{με } \kappa = 4c | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{2.3} | |
\[x' = \frac{t+1}{x^4+1} \] | |
\tcblower | |
\begin{align*} | |
&\implies \od{x}{t}=\frac{t+1}{x^4+1} \\ | |
&\implies (x^4+1) \dif x + (-t-1) \dif t = 0 \\ | |
&\implies \int (x^4+1)\dif x + \int (-t-1) \dif t = c \\ | |
&\implies \frac{x^5}{5} + x - \frac{t^2}{2} -t = c \\ | |
\end{align*} | |
\end{exercise*} | |
Παρατηρούμε ότι, χωρίς αρχική συνθήκη, βρίσκουμε γενικές λύσεις ως αποτέλεσμα. Με τη χρήση μιας αρχικής συνθήκης, μπορούμε να βρούμε και την ειδική λύση της εξίσωσης. | |
\begin{exercise*}{2.4} | |
\[e^t \dif t - x \dif x = 0;\quad x(0)=1 \leftarrow \text{αρχική συνθήκη} \] | |
\tcblower | |
\begin{align*} | |
&\implies \int x \dif x + \int (-e^t) \dif t = c \\ | |
&\implies \frac{x^2}{2} - e^t = c \\ | |
&\implies x^2 = 2e^t + 2c \\ | |
&\implies x^2 = 2e^t+ \kappa, \quad \text{με } \kappa = 2c | |
\end{align*} | |
Όμως \(x(0) = 1\), άρα: | |
\[ | |
\begin{cases} | |
x^2 = 2e^t+ \kappa \\ | |
x(0) = 1 | |
\end{cases} | |
\implies x(0)^2 = 2e^0 + \kappa \implies \boxed{\kappa = -1} | |
\] | |
Επομένως τελικά: | |
\[ | |
x^2=2e^t -1 \implies x = \pm \sqrt{2e^t-1} \implies \boxed{x=\sqrt{2e^t-1}} | |
\] | |
Η αρχική συνθήκη πράγματι επαληθεύει το αποτέλεσμα \(x\). Πρέπει όμως και \(x \in \mathbb R, \ 2e^t-1 \geq 0\). | |
Από τη διαφορική εξίσωση έχουμε \(x' = \frac{e^t}{x}\), άρα πρέπει \(2e^t - 1 > 0 \implies \boxed{t > \ln \frac{1}{2}} \). | |
\end{exercise*} | |
\[ | |
\int_{x_0}^x N(x) \dif x + \int_{t_0}^t M(t) \dif t = 0, \quad x(t_0)=x_0 | |
\] | |
\begin{exercise*}{2.5} | |
\[x \cos x \dif x + (1-6t^5) \dif t = 0; \quad t(\pi)=0 \] | |
\tcblower | |
\(x_0 = \pi,\ t_0 = 0\) | |
\begin{align*} | |
&\implies \int_\pi^x x \cos x \dif x + | |
\int_0^t (1-6t^5) \dif t = 0 \\ | |
&\implies | |
\left. x \sin x \right|_\pi^x | |
+ \left. \cos x \right|_\pi^x + \left. (t-t^6) \right|_0^t = 0 \\ | |
&\implies x \sin x + \cos x + 1 + t - t ^ 6 \\ | |
&\implies \boxed{ x \sin x + \cos x + 1 = t - t^6 } | |
\end{align*} | |
\end{exercise*} | |
\subsection{Ομοιογενείς} | |
\[ | |
f(t,x)= \frac{-M(t,x)}{N(t,x)} | |
\] | |
\begin{defn}{}{} | |
Αν \(\forall a \in \mathbb R: f(at,ax) = f(t,x)\), λέμε ότι η εξίσωση είναι \textbf{ομοιογενής}. | |
\end{defn} | |
\begin{theorem*}{} | |
Αν μια εξίσωση είναι ομοιογενής, μπορούμε να την λύσουμε μειώνοντάς/μετατρέποντάς την σε χωριζόμενη, εφαρμόζοντας το μαθηματικό κόλπο που ονομάζεται "αντικατάσταση μεταβλητής", δηλαδή, όπου \(u\) συνάρτηση: | |
\[ | |
\boxed{ | |
x=ut \implies \od{x}{t} = \od{u}{t}t+u | |
}\] | |
\end{theorem*} | |
\begin{exercise*}{2.6} | |
\[x' = \frac{x+t}{t} \] | |
\tcblower | |
\(\implies \od{x}{t} = \frac{x+t}{t}\), μη χωριζόμενη. | |
\[f(t,x) = \od{x}{t},\quad f(at,ax)= \frac{ax+at}{at} = \frac{x+t}{t} \text{ ομοιογενής} | |
\] | |
Θέτω \(x=ut, \ \od{x}{t}=\od{u}{t}t+u\), άρα η διαφορική εξίσωση γίνεται: | |
\begin{align*} | |
&\od{u}{t}t+u=\frac{ut+t}{t} \\ \implies& | |
\od{u}{t}t+u=u+1 \\ \implies& | |
t\od{u}{t} = 1 \\ \implies& | |
\frac{1}{t} \dif t - \dif u = 0 \text{ χωριζόμενη} \\ \implies& | |
\int \frac{1}{t} \dif t + \int (-1) \dif u = c \\ \implies& | |
\ln \abs t - u = c \\ \implies& | |
u = \ln \abs t -c \text{ με } c = - \ln \abs \kappa \\ \implies& | |
\boxed {u = \ln \abs{\kappa t}} \\ \implies& | |
\frac{x}{t} = \ln \abs{\kappa t} \implies | |
\boxed {x = t \ln \abs{\kappa t}} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{2.7} | |
\[x' = \frac{2x^4+t^4}{tx^3} \] | |
\tcblower | |
\(\implies \od{x}{t} = \frac{2x^4+t^4}{tx^3}\), μη χωριζόμενη. | |
\[f(t,x) = \od{x}{t},\quad f(at,ax)= \frac{2(ax)^4+(at)^4}{(at)(ax)^3} = \frac{a^4 2x^4 + a^4t^4}{a^4tx^3} = \frac{2x^4+t^4}{tx^3} \text{ ομοιογενής} | |
\] | |
Θέτω \(x=ut, \ \od{x}{t}=\od{u}{t}t+u\), άρα η διαφορική εξίσωση γίνεται: | |
\begin{align*} | |
& | |
\od{u}{t}t+u=\frac{2(ut)^4+t^4}{t(ut)^3} \\ \implies& | |
\od{u}{t}t+u=\frac{2u^4 \cancel{t^4}+\cancel{t^4}}{u^3 \cancel{t^4}} \\ \implies& | |
\od{u}{t}t+u=\frac{2u^4+1}{u^3} \\ \implies& | |
\od{u}{t}t=\frac{2u^4+1}{u^3}-u=\frac{u^4+1}{u^3} \\ \implies& | |
\frac{u^3}{u^4+1} \dif u - \frac{1}{t} \dif t = 0 \text{ χωριζόμενη} \\ \implies& | |
\int \frac{u^3}{u^4+1} \dif u + \int \frac{-1}{t} \dif t = c \\ \implies& | |
\frac{1}{4} \ln (u^4+1) - \ln \abs t = c \\ \implies& | |
\boxed{u^4+1 = (\kappa t)^4} \text{ με } c = \ln \abs x \\ | |
& x = ut \implies u = \frac{x}{t} \implies \left( \frac{x}{t} \right) ^ 4 | |
+ 1 = (\kappa t )^4 \\ \implies& | |
\frac{x^4}{t^4}+1 = \kappa ^4 t^4 \\ \implies & | |
\boxed{x^4 = c_1t^8 - t^4} \text{ με } c_1= \kappa ^4 | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{2.8} | |
\[x' = \frac{t^2+x^2}{tx}; x(1) = -2 \] | |
\tcblower | |
\(\implies \od{x}{t} = \frac{t^2+x^2}{tx}\), μη χωριζόμενη. | |
\[f(t,x) = \od{x}{t},\quad f(at,ax)= \frac{(at)^2+(ax)^2}{(at)(ax)} = \frac{\cancel{a^2}t^2+\cancel{a^2}x^2}{\cancel{a^2}tx} = \frac{t^2+x^2}{tx} \text{ ομοιογενής} | |
\] | |
Θέτω \(x=ut, \ \od{x}{t}=\od{u}{t}t+u\), άρα η διαφορική εξίσωση γίνεται: | |
\begin{align*} | |
& | |
\od{u}{t}t+u=\frac{t^2+(ut)^2}{t(ut)} \\ \implies& | |
\od{u}{t}t+u=\frac{\cancel{t^2}+\cancel{t^2}u^2}{\cancel{t^2}u} \\ \implies& | |
\od{u}{t}t+u=\frac{1+u^2}{u} \\ \implies& | |
\od{u}{t}t=\frac{1+\cancel{u^2}-\cancel{u^2}}{u}=\frac{1}{u} \\ \implies& | |
u \dif u - \frac{1}{t} \dif t = 0 \text{ χωριζόμενη} \\ \implies& | |
\int u \dif u + \int \frac{-1}{t} \dif t = c \\ \implies& | |
\frac{u^2}{2} - \ln \abs t = c \\ \implies& | |
u^2=2 \ln \abs t + 2c \\ \implies & | |
\boxed{u^2 = \ln t^2 + \kappa} \text{ με } \kappa = 2c \\ | |
& x = ut \implies u = \frac{x}{t} \implies \frac{x^2}{t^2} = \ln t^2 + \kappa \\ \implies& | |
\boxed{x^2=t^2 \ln t^2 + \kappa t ^2} | |
\end{align*} | |
Επειδή \(x(1)=2\), έχουμε: | |
\[ | |
(-2)^2=1^2 \ln 1^2 + \kappa 1 ^2 \implies 4 = 0 + \kappa \implies | |
\boxed{ \kappa = 4} | |
\] | |
Επομένως τελικά: | |
\[ | |
x^2=t^2 \ln t ^ 2 + 4t^2 \implies | |
\boxed{x = - \sqrt{t^2 \ln t^2 +4t^2}} | |
\] | |
\end{exercise*} | |
\subsection{Ακριβείς} | |
\begin{defn*}{} | |
Όταν: | |
\[ | |
\pd{M(t,x)}{x} = \pd{N(t,x)}{t} | |
\] | |
τότε η εξίσωση λέγεται ακριβής ή πλήρης. | |
Υπάρχει \(dF(t,x) = N(t,x) \dif x + M(t,x) \dif t\) | |
με Γενική Λύση \(F(t,x) = c\). | |
\end{defn*}{} | |
\begin{exercise*}{2.16} | |
\[ | |
(t + \sin x) \dif t + (t \cos x - 2x) \dif x = 0 | |
\] | |
\tcblower | |
\begin{align*} | |
\underbrace{(t + \sin x) \dif t}_{M(t,x)\dif t} + | |
\underbrace{(t \cos x - 2x) \dif x}_{N(t,x)\dif x} = 0 | |
\end{align*} | |
Δοκιμή: | |
\[ | |
\begin{cases} | |
M(t,x)&=t+\sin x\\ | |
N(t,x)&=t \cos x - 2x | |
\end{cases} | |
\implies | |
\pd{M(t,x)}{x} = \cos x | |
= \pd{N(t,x)}{t} = \cos x | |
\] | |
Άρα η ΔΕ είναι ακριβής, επομένως υπάρχει \(F(t,x)\) τέτοια ώστε: | |
\begin{align*} | |
&dF = N(t,x) \dif x + M(t,x) \dif t \\ | |
&\attnboxed{ | |
dF = \frac{\partial F}{\partial x} \dif x + \frac{\partial F}{\partial t} \dif t | |
} \leftarrow \text{ ολικό διαφορικό της } F | |
\end{align*} | |
\begin{gather*} | |
\pd{F(t,x)}{x}=N(t,x), \quad \pd{F(t,x)}{t}=M(t,x) | |
\xRightarrow{\text{ολοκλήρωση ως προς }t} \\ \implies | |
\cancelto{F(t,x)}{\int \frac{\partial F(t,x)}{\partial t} \dif t} | |
= \int (t+\sin x) \dif t \implies \\ \implies | |
\boxed{F(t,x) = | |
\frac{1}{2} t^2 + t \sin x + \overbrace{h(x)}^{\mathclap{\text{ολοκληρωτική σταθερά}}} | |
\hspace{27pt} | |
} | |
\end{gather*} | |
Έχουμε: | |
\begin{align*} | |
&\pd{F(t,x)}{x} = t \cos x + h'(x)\\ \implies& | |
t \cos x - 2x = t \cos x + h'(x)\\ \implies& | |
h'(x) = -2x \\ \implies& | |
\int h'(x) \dif x = \int (-2x) \dif x \\ \implies& | |
\boxed{h(x) = -x^2 + c_1} | |
\end{align*} | |
Επομένως: | |
\begin{align*} | |
F(t,x) &= \frac{1}{2}t^2 + t \sin x - x^2 + c_1 = c | |
\xRightarrow{c_2=c-c_1} \\ &\implies | |
\boxed{ | |
\frac{1}{2}t^2+t \sin x - x^2 = c_2 | |
} \text{ Γενική λύση} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{2.17} | |
\[ | |
\od{x}{t} = \frac{2+xe^{tx}}{2x-te^{tx}} | |
\] | |
\tcblower | |
\[ | |
\od{x}{t} = \frac{2+xe^{tx}}{2x-te^{tx}} | |
\xRightarrow{\text{διαφορική μορφή}} | |
\underbracket{(2+xe^{tx})}_{M(t,x)=2+xe^{tx}} \dif t | |
+ | |
\underbracket{(te^{tx}-2x)}_{N(t,x)=te^{tx}-2x} \dif x | |
= 0 | |
\] | |
Δοκιμή: | |
\[ | |
\pd{M(t,x)}{x}=e^{tx}+xte^{tx} | |
= \pd{N(t,x)}{t}=xte^{tx}+e^{tx} | |
\] | |
συνεπώς είναι ακριβής, οπότε υπάρχει \(F(t,x)\), με \(dF = M(t,x)\dif t + N(t,x) \dif x\), με λύση \(F(t,x) = c\). | |
\[ | |
\text{\small Ολικό διαφορικό} \rightarrow | |
dF = \frac{\partial F}{\partial x} \dif x + \frac{\partial F}{\partial t} \dif t | |
\] | |
Άρα: | |
\begin{align*} | |
\pd{F(t,x)}{x} = N(t,x), \quad | |
\pd{F(t,x)}{t} &= M(t,x) = 2+xe^{tx} \xRightarrow{\text{ολοκλήρωση ως προς }t} | |
\\ \implies | |
\int | |
\pd{F(t,x)}{t} \dif t &= | |
\int \left( 2+ xe^{tx} \right) \dif t | |
\implies \\ \implies | |
F(t,x) &= 2t +e ^ {tx} + h(x) | |
\end{align*} | |
\begin{align*} | |
\text{Παραγώγιση ως προς } x \rightarrow | |
\pd{F(t,x)}{x} = te^{tx} + h'(x) | |
\implies& | |
te^{tx}+h'(x)=te^{tx}-2x \implies \\ \implies& | |
h'(x) = -2x \implies \\ \implies& | |
h(x) = \int (-2x) \dif x \implies \\ \implies& | |
h(x) = -x^2+c_1 | |
\end{align*} | |
Άρα τελικά: | |
\begin{align*} | |
&F(t,x) = 2t+e^{tx}-x^2+c_1 \\ | |
\implies& 2t+e^{tx}-x^2+c_1 =c \\ | |
\implies& \boxed{2t+e^{tx} -x^2 = c_2, \qquad c_2=c-c_1} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{2.19} | |
\[ | |
\left( 2x^2t-2x^3 \right) \dif t + | |
\left( 4x^3-6x^2t+2xt^2 \right) \dif x = 0 | |
\] | |
\tcblower | |
\[ | |
\underbrace{\left( 2x^2t-2x^3 \right)}_{M(t,x)=2x^2t-2x^3} \dif t + | |
\underbrace{\left( 4x^3-6x^2t+2xt^2 \right)}_{N(t,x)=4x^3-6x^2t+2xt^2} \dif x = 0 | |
\] | |
\( | |
\pd{M(t,x)}{x}=4xt-6x^2= | |
\pd{N(t,x)}{t}=0-6x^2+4xt | |
\), ΔΕ ακριβής, οπότε υπάρχει \(F(t,x)\) με \(\dif F(t,x) = M(t,x) \dif t + N(t,x) \dif x\) με λύση \(F(t,x) = c\). | |
\[ | |
\dif F(t,x) = \pd{F(t,x)}{t} \dif t + \pd{F(t,x)}{x} \dif x | |
\] | |
\begin{align*} | |
\pd{F(t,x)}{x} = N(t,x),& \quad | |
\pd{F(t,x)}{t} = M(t,x) = 2x^2t-2x^3 &\implies \\ \implies& | |
\int \pd{F(t,x)}{t} \dif t | |
= | |
\int (2x^2t-2x^3) \dif t | |
&\implies \\ \implies& | |
F(t,x) = x^2t^2-2x^3t+h(x) & | |
\end{align*} | |
\begin{align*} | |
\pd{F(t,x)}{x}&=2xt^2-6x^2t+h'(x) \implies \\ \implies | |
\cancel{2xt^2}-\cancel{6x^2t}+h'(x)&=4x^3-6x^2t+\cancel{2x+2} \implies \\ \implies | |
h'(x)&=4x^3 \xRightarrow{\text{ολοκλ.}} h(x)=x^4+c_1 | |
\end{align*} | |
Άρα: | |
\begin{align*} | |
F(t,x) &= x^2t^2-2x^3t+x^4+c_1 \implies \\ \implies | |
x^2t^2-2^3t+x^4+c_1 &= c \implies \\ \implies | |
x^2t^2-2x^3t+x^4 = c-c_1 \implies \\ \implies | |
\begin{cases} | |
(x^2-xt)^2&= c_2\\ | |
c_2 &=c-c_1 | |
\end{cases} | |
&\implies \\ | |
\xRightarrow{c_3=\pm \sqrt{c_2}} x^2-xt &= c_3 \xRightarrow[\frac{-b \pm \sqrt{b^2-4ac}}{2a}]{ax^2+bx+c=0} | |
\\ \implies | |
\boxed{x = \frac{t \pm \sqrt{t^2+4c_3}}{2},\qquad c_3 = \pm \sqrt{c_2}} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{2.20} | |
\[ | |
2tx | |
\dif t | |
+ (1+t^2) | |
\dif x | |
= 0 | |
; \quad | |
x(2) = -5 | |
\] | |
\tcblower | |
\[ | |
\underbrace{2tx}_{\mathclap{M(t,x)}} | |
\dif t | |
+ \underbrace{(1+t^2)}_{N(t,x)} | |
\dif x | |
= 0 | |
; \quad | |
x(2) = -5 | |
\] | |
\begin{equation} \label{eq:e1} | |
M(t,x)=2tx,\quad N(t,x)=1+t^2 | |
\end{equation} | |
\(F(t,x)\), με \( \dif F(t,x) = \pd{F}{x} \dif x + \pd{F}{t} \dif t \). | |
\(\dif F(t,x) = N(t,x) \dif x + M(t,x) \dif t\) | |
\begin{equation} \label{eq:e2} | |
\pd{F(t,x)}{x} = N(t,x) | |
\end{equation} | |
\begin{align*} | |
\pd{F(t,x)}{t} = M(t,x) = 2tx \implies \\ \implies | |
\int \pd{F(t,x)}{t} \dif t | |
= \int (2tx) \dif t \implies | |
\end{align*} | |
\begin{equation} \label{eq:e3} | |
\implies F(t,x) = t^2x + h(x) | |
\end{equation} | |
\begin{align*} | |
\begin{cases} | |
\pd{F(t,x)}{x} = t^2+h'(x) \\ | |
\eqref{eq:e2},\ \eqref{eq:e1} | |
\end{cases} | |
\implies \\ \implies | |
t^2+h'(x)=1+t^2 \implies h'(x)=1 \implies \\ \implies | |
\begin{cases} | |
h(x)=x+c_1 \\ | |
\eqref{eq:e3} | |
\end{cases} | |
\implies | |
\begin{cases} | |
F(t,x)=t^2x \\ | |
(4) | |
\end{cases} | |
\implies | |
t^2+x+c_1 | |
\\ | |
\implies | |
t^2x+x=c_2 (c_2=c-c_1) \implies | |
x = \frac{c_2}{t^2+1} | |
\implies | |
(x(2)=5) | |
5 = \frac{c_2}{2^2+1} \implies x = \frac{-25}{t^2+1} | |
\end{align*} | |
\begin{equation} \label{eq:ef} | |
\implies F(t,x)=t^2x+x+c_1 | |
\end{equation} | |
\end{exercise*} | |
\subsection{\textlatin{Overview}} | |
\subsubsection{Συνήθεις Διαφορικές Εξισώσεις (ΣΔΕ - \textlatin{Ordinary Differential Equations})} | |
\begin{defn}{}{} | |
Εμπλέκουν: | |
\begin{itemize} | |
\item μία ανεξάρτητη μεταβλητή (π.χ. \(t,\ x\)) | |
\item μια εξαρτημένη και τις παραγώγους της (π.χ. \(i,y,u\)) | |
\end{itemize} | |
\[ | |
F(t,x,x',\dots,x^{(n)}) = 0 | |
\] | |
\end{defn}{}{} | |
\paragraph{Μη συνήθεις} | |
είναι οι Μερικές Διαφορικές Εξισσώεις \textlatin{(Partial Differential Equations - PDE)} που εμπλέκουν: | |
\begin{itemize} | |
\item πολλές ανεξάρτητες μεταβλητές (π.χ. \(x,y,z\)) | |
\item μία εξαρτημένη μεταβλητή και τις μερικές παραγώγους της | |
\end{itemize} | |
\subsubsection{1\textsuperscript{ης} τάξης ΔΕ} | |
\begin{defn}{}{} | |
όταν | |
\[ | |
x' = \frac{\dif x}{\dif t} = f(t,x) | |
\] | |
\begin{defn}{Τυπικής μορφής}{} | |
\[ | |
f(t,x) = \frac{-M(t,x)}{N(t,x)} | |
\] | |
\paragraph{Διαφορική μορφή} | |
\[N(t,x) \dif x + M(t,x) \dif t = 0 \] | |
\begin{defn}{Χωριζόμενη}{} | |
όταν | |
\[ | |
\begin{cases} | |
N(t,x) &= N(x) \\ | |
M(t,x) &= M(t) | |
\end{cases} | |
\] | |
τότε | |
\[ | |
N(x) \dif x + M(t) \dif t = 0 | |
\] | |
με λύση | |
\[ | |
\int N(x) \dif x + \int M(t) \dif t = c | |
\] | |
ή | |
\[ | |
\int_{x_0}^x N(x) \dif x + \int_{t_0}^t M(t) \dif = 0 | |
\] | |
\end{defn} | |
\begin{defn}{Ομογενής - Ομοιογενής}{} | |
όταν \(\forall a \in \mathbb R\) | |
\[ | |
F(at,ax)=f(t,x) | |
\] | |
τότε θέτω \(x = ut\), άρα \(\od{x}{t}=\od{u}{t}t+u\) | |
\end{defn} | |
\end{defn} | |
\end{defn} | |
\section{} | |
\subsection{ΔΕ 1\textsuperscript{ης} τάξης} | |
ΤΜ (Τυπική μορφή): \(x' = \od{x}{t} = f(t,x)\) | |
\boxed{ | |
\text{ΔΜ (Διαφορική μορφή): } N(t,x)\dif x + M(t,x) \dif t = 0 | |
} | |
Ακριβής: \( | |
\pd{M(t,x)}{x}=\pd{N(t,x)}{t} \rightarrow \dif F(t,x) | |
\) | |
\begin{gather*} | |
\dif F(t,x) = N(t,x) \dif x + M(t,x) \dif t | |
\\ F(t,x)=c | |
\end{gather*} | |
\begin{gather*} | |
\underbrace{G(t,x)}_{\mathclap{\text{Μπορεί να υπάρχει τέτοια συνάρτηση}}} \cdot | |
\left( | |
N(t,x)\dif x + M(t,x) \dif t | |
\right) = 0 | |
\end{gather*} | |
\begin{exercise*}{2.23 Ολοκληρωτικός παράγοντας, επίλυση μέσω ελέγχου} | |
\begin{gather*} | |
x\dif t - t\dif x = 0\\ | |
M(t,x)=x,\ N(t,x)=t\\ | |
\pd{M(t,x)}{x}=1,\ \od{N(t,x)}{t} = -1 \text{ δεν είναι ακριβής.} \\ | |
\overbrace{G(t,x)}^{\mathclap{\text{υποφήφιος}}} = -\frac{1}{t^2} \\ | |
-\frac{1}{t^2} (x\dif t - t\dif x) = 0 \implies | |
-\frac{x}{t^2} \dif t + \frac{\frac{1}{t}}{t^2} \dif x = 0 \\ | |
M(t,x) = - \frac{x}{t^2}m\ N(t,x) = \frac{1}{t}\\ | |
\pd{M(t,x)}{x}=-\frac{1}{t^2}=\overbrace{\pd{N(t,x)}{t}}^{\mathclap{\text{ακριβής}}} | |
=-\frac{1}{t^2} | |
\end{gather*} | |
\[\text{Αν }\boxed{\frac{1}{N} \left( \pd{M}{x} - \pd{N}{t} \right) \equiv g(t) \implies G = e^{\int g(t) \dif t}}\]. | |
Με διαφορά μερικών παραγώγων: | |
\[ | |
\boxed{\text{Αν } | |
\frac{1}{N} \left( \pd{M}{x} - \pd{N}{t} \right) = h(x) \implies | |
G = e^{\int g(t) \dif x} | |
} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{2.25} | |
\[ | |
x^2\dif t + tx\dif x = 0,\ | |
M(t,x)=x^2,\ | |
N(t,x) = tx | |
\] | |
\tcblower | |
\begin{gather*} | |
\pd{M(t,x)}{x}=2x\neq \pd{N(t,x)}{t}=x \text{ όχι ακριβής} \\ | |
\frac{1}{M} \left( | |
\pd{M}{x}-\pd{N}{t} \right) = \frac{1}{x^2}(2x-x)=\frac{1}{x} = h(x)\\ | |
G(t,x)=e^{-\int h(x)\dif x} = e^{-\int \frac{1}{x} \dif x} = e^{-\ln x} = \frac{1}{x} | |
\end{gather*} | |
\begin{gather*} | |
\frac{1}{x}(x^2\dif t + tx\dif x) = 0 \implies x\dif t + t\dif x = 0\\ | |
\pd{M}{x}=1=\pd{N}{t}=1 \text{ ακριβής} | |
\end{gather*} | |
\subparagraph{Μορφή των όρων $N$,$M$} | |
αν \(M=xf(tx)\) \textbf{και} \(N=tg(tx)\), τότε: | |
\[ | |
G(t,x) = \frac{1}{tM-xN} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{2.26} | |
\[ | |
x'=\frac{tx^2-x}{t} | |
\] | |
\tcblower | |
\begin{gather*} | |
\implies \od{x}{t}=\frac{tx^2-x}{t} \implies | |
t\dif x-(tx^2-x)\dif t = 0 \implies | |
x(1-tx)\dif t)+t\dif x = 0\\ | |
M(t,x)=x\cdot (1-tx) \implies \pd{M(t,x)}{x}=1-2tx \neq \pd{N(t,x)}{t} = 1 \text{ όχι ακριβής} | |
\end{gather*} | |
αλλά:\(M=xf(t)\) \textbf{και} \(N=tg(tx)\). | |
Επομένως: | |
\[ | |
G(t,x)=\frac{1}{tM-xN}=\frac{1}{tx(1-tx)-xt}=\frac{1}{-t^2x^2}=-\frac{1}{(tx)^2} | |
\] | |
Είναι: | |
\begin{align*} | |
-\frac{1}{(tx)^2} | |
\left( | |
x(1-tx)\dif t + t\dif x | |
\right) = 0 \implies \\ | |
\frac{tx-1}{t^2x}\dif t - \frac{1}{tx^2}\dif x = 0 | |
\end{align*} | |
και συνεχίζω με τη μέθοδο της ακριβούς. | |
\end{exercise*} | |
\section{Θεωρία των Λύσεων} | |
Μορφή ΔΕ $n$\textsuperscript{ης} τάξης: | |
\[ | |
b_n(t)\cdot x^{(n)} + b_{n-1}(t)\cdot x^{(n-1)} + \dots + b_2(t)x'' + b_1(t)x' + b_0(t)x = g(t) | |
\] | |
όπου: \(g(t), b_j(t) \quad (j=1,2,\dots,n)\) εξαρτώνται αποκλειστικά από το $t$. | |
Αν \(g(t)\equiv 0\), τότε η ΔΕ είναι ομογενής (ΟΜ - \textlatin{homogenous}). | |
Αν \(g(t) \neq 0\), τότε η ΔΕ είναι μη ομογενής (ΜΟ - \textlatin{non-homogenous}). | |
Όταν όλοι οι συντελεστές \(b_j(t)\) είναι σταθερές, τότε ΔΕΣΣ (σταθερών συντελεστών). | |
Όταν ένας τουλάχιστον \(b_j(t)\) δεν είναι σταθερά, ΔΕΜΣ (μεταβλητών συντελεστών). | |
\begin{theorem}{}{} | |
ΔΕ $n$ τάξης με $n$ ΑΣ (αρχικές συνθήκες): | |
\[ x(t_0) = c_0, x'(t_0) = c_1, x''(t_0)=c_2,\dots,\ x^{(n-1)}(t_0)=c_{n-1}\] | |
ΔΕ \(b_n(t)x^{(n)}+b_{n-1}(t)x^{(n-1)}+\dots+b_2(t)x''+b_1(t)x'+b_o(t)x=g(t)\) | |
Αν \(g(t)\) και \(b_i\) συνεχείς σε διάστημα $\phi$ που περιλαμβάνει το \(t_0\) και \(b_n(t) \neq 0\) στο \(\phi\), τότε το πρόβλημα εχει μία μοναδική λύση (ορισμένη στο \(\phi\)). | |
\end{theorem} | |
Διαιρώ με \(b_n(t)\) και έχω: | |
\[ | |
x^{(n)}+a_{n-1}(t)\cdot x^{(n-1)}+\dots+a_2(t)x''+a_1(t)x'+a_0(t)x=\phi(t) | |
\] | |
\begin{align*} | |
a'_j(t) &= \frac{b_j(t)}{b_n(t)} \quad (j=1,2,\dots,n-1) \\ | |
\phi(t) &= \frac{g(t)}{b_n(t)} | |
\end{align*} | |
\paragraph{Διαφορικός τελεστής \(\mathbf L(x)\)} | |
\begin{align*} | |
\mathbf L(x) &\equiv x^{(n)} +a_{n-1}(t)x^{(n-1)}+\dots+a_2(t)\cdot x''+a_1(t)\cdot x' + a_0(t)\cdot x | |
\\ | |
\mathbf L(x) &= \phi(t) \quad \text{ΜΟ ΓΡ ΔΕ $n$\textsuperscript{ης} τάξης}\\ | |
\mathbf{L}(x) &=0 \quad \text{ΟΜ} | |
\end{align*} | |
\begin{defn}{}{} | |
Το σύνολο \( \left\lbrace x_1(t),x_2(t),\dots,x_n(t) \right\rbrace \) είναι ΓΕ (γραμμικά εξαρτημένο) για ένα διάστημα $\Delta$ ότανν υπάρχουν συντελεστές όχι όλοι μηδενικοί τέτοιοι ώστε: | |
\[ | |
c_1x_1(t)+c_2x_2(t)+\dots+c_nx_n(t) \equiv 0 \quad \Delta | |
\] | |
\end{defn} | |
\begin{theorem}{}{} | |
Έστω η ομογενής $n$-οστής τάξης γραμμική διαφορική εξίσωση \(\mathbf{L}(x) = 0\). | |
Αν \(x_1(t),\ x_2(t),\dots,\ x_n(t)\) είναι λύσεις, τότε και ο γραμμικός τους συνδυασμός είναι γενική λύση της ομογενούς: | |
\[ | |
x(t) = c_1x_1(t)+c_2x_2(t)+\dots+c_nx_n(t) \quad\text{ΓΛ (Γενική Λύση)} | |
\] | |
\end{theorem} | |
\begin{theorem}{Βροσκιανή}{} | |
\[\text{Ορίζουσα } W(x_1,x_2,\dots,x_n)\] | |
\[\begin{cases} | |
W \neq 0 \text{ έστω σε ένα σημείο } \in \Delta &\rightarrow \text{ΓΑ (Γραμμικά Ανεξάρτητες)} \\ | |
W \equiv 0 \textbf{ και } \text{κάθε συνάρτηση είναι λύση της ίδιας ΔΕ} &\rightarrow \text{ΓΕ (Γραμμικά Εξαρτημένες)} | |
\end{cases} | |
\] | |
\end{theorem} | |
\begin{theorem}{}{} | |
\[ | |
\overbrace{\mathbf{L}(x)}^{\mathclap{\text{Ομογενής ΔΕ}}} = \phi(t) | |
\] | |
Έστω \( | |
\begin{cases} | |
x_n(t) & \text{ΓΛ της ΟΜ (Ομογενούς)} \\ | |
x_p(t) & \text{ΕΛ της ΜΟ (Μη ομογενούς)} | |
\end{cases} | |
\). | |
Τότε είναι ΓΛ ΜΟ (Γενική Λύση Μη Ομογενούς) η: | |
\[ | |
x(t) = x_n(t)+x_p(t) | |
\] | |
\end{theorem} | |
\begin{exercise*}{3.2} | |
\[ | |
\left\lbrace 1-t,\ 1+t,\ 1-3t | |
\right\rbrace | |
\] | |
\tcblower | |
\begin{align*} | |
W(1-t,\ 1+t,\ 1-3t) &= | |
\left| | |
\begin{matrix} | |
1-t & 1+t &1-3t\\ | |
\od{(1-t)}{t} & \od{(1+t)}{t} & \od{(1-3t)}{t} \\ | |
\od[2]{(1-t)}{t} & \od[2]{(1+t)}{t} & \od[2]{(1-3t)}{t} | |
\end{matrix} | |
\right| | |
\\ | |
&= \left| | |
\begin{matrix} | |
1-t&1+t&1-3t\\-1&1&-3\\0&0&0 | |
\end{matrix} \right| | |
= 0 | |
\end{align*} | |
\paragraph{(β)} | |
\begin{gather*} | |
c_1(1-t)+c_2(1+t)+c_3(1-3t)=0\\ | |
\underbrace{(c_1+c_2-3c_3)}_0t+\underbrace{(c_1+c_2+c_3)}_0 \equiv 0 | |
\end{gather*} | |
\[ | |
\begin{cases} | |
-c_1+c_2-3c_3 &= 0\\ | |
c_1+c_2+c_3 &=0 | |
\end{cases} | |
\implies | |
\begin{cases} | |
c_1&=-2c_3c_2\\ | |
c_2&=c_3\\ | |
c_3 &\text{ αυθαίρετη σταθερά} | |
\end{cases} | |
\implies | |
\begin{cases} | |
c_3&=1\\c_1&=-2\\c_2&=1 | |
\end{cases} | |
\implies\text{ΓΕ} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{3.3} | |
Βρείτε την Βροσκιανή: | |
\[ | |
\left\lbrace t,\ t^2,\ t^3 \right\rbrace | |
\] | |
\tcblower | |
\begin{align*} | |
W(t,t^2,t^3)&=\left| | |
\begin{matrix} | |
t&t^2&t^3\\ \od{(t)}{t}& \od{(t^2)}{t} & \od{(t^3)}{t}\\ | |
\od[2]{(t)}{t}& \od[2]{(t^2)}{t} & \od[2]{(t^3)}{t} | |
\end{matrix}\right| | |
\\&= | |
\left| | |
\begin{matrix} | |
t&t^2&t^3\\ | |
1&2t&3t^2\\ | |
0&2&6t | |
\end{matrix} | |
\right|=2t^3 | |
\end{align*} | |
\[ | |
(-\infty,\infty),\ t=3,\ W=54 \neq 0 \implies \text{ΓΑ, θαυμάσια!} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{3.4} | |
\[ | |
\left\lbrace t^3, \left|t^3\right| \right\rbrace \quad[-1,1] | |
\] | |
\begin{gather*} | |
c_1t^3+c_2\left|t^3\right|\equiv 0\\ | |
\left|t^3\right|=t^3,\ t\geq0 \quad \big/\quad \left|t^3\right|=-t^3,\ t\leq0 \\ | |
\begin{cases} | |
c_1t^3+ c_2t^3&\equiv0 \quad t \geq 0 \\ | |
c_1t^3- c_2t^3&\equiv0 \quad t < 0 | |
\end{cases} | |
\implies c_1=c_2=0 \text{ ΓΑ}\\ | |
\od{|t^3|}{t} = \begin{cases} | |
3t^2&\quad\text{ αν } t>0\\ | |
0&\quad\text{ αν } t=0\\ | |
-3t^2&\quad\text{ αν } t<0 | |
\end{cases}\implies | |
\begin{cases} | |
\text{για } t>0:\ &W\left(t^3,|t^3| \right) = \left| | |
\begin{matrix} | |
t^3&t^3\\3t^2&3t^2 | |
\end{matrix} | |
\right|\equiv 0 \\ | |
\text{για } t=0:\ &W\left(t^3,|t^3| \right) = 0 \\ | |
\text{για } t<0:\ &W\left(t^3,|t^3| \right) = \left| | |
\begin{matrix} | |
t^3&-t^3\\3t^2&-3t^2 | |
\end{matrix} | |
\right|\equiv 0 | |
\end{cases} | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{3.5} | |
\begin{gather*} | |
x''-2x'+x=0\\ | |
e^t,\ te^t \text{ Λύσεις} | |
\end{gather*} | |
Ο γραμμικός συνδυασμός \(X=c_1e^t+c_2te^t\) είναι λύση της εξίσωσης? | |
\tcblower | |
\begin{align*} | |
W(e^t,te^t) &= | |
\left| | |
\begin{matrix} | |
e^t&te^t\\ | |
e^t&e^t+te^t | |
\end{matrix} | |
\right|=e^{2t}\nequiv 0 | |
\end{align*} | |
Άρα οι εξισώσεις είναι γραμμικά ανεξάρτητες, άρα, επειδή είναι λύσεις της διαφορικής, ο γραμμικός συνδυασμός τους είναι γενική λύση. | |
\paragraph{} | |
Μη ομογενής: \(x''-2x'+x=e^{3t}\) | |
Ειδική λύση: \(\frac{1}{4}e^{3t} \rightarrow x_p = \frac{1}{4}e^{3t}\) | |
Γενική λύση μη ομογενούς: \(\underbrace{x(t)}_{\mathclap{\text{ΜΟ}}} =\underbrace{x_h(t)}_{\mathclap{\text{ΟΜ}}}+\underbrace{x_p(t)}_{\mathclap{\text{ΜΟ}}}\) | |
Άρα: | |
\begin{align*} | |
x(t)=c_1e^t+c_2te^t+\frac{1}{4}e^{3t} | |
\end{align*} | |
\end{exercise*} | |
\section{} | |
\subsection{ΓΡ/ΔΕ/1\textsuperscript{ης}} | |
\begin{itemize} | |
\item \( | |
\od{x}{t}+p(t)x=q(t) | |
\)\\ή \(\underbrace{f(t,x)}_{\mathclap{f(t,x)=\od{x}{t}}}=q(t)-p(t)x\) | |
Τότε ΟΠ (Ολοκληρωτικός Παράγοντας) \(G(t)=e^{\int p(t) \dif x}\). Πολλαπλασιάζοντας με τον ολοκληρωτικό παράγωντα παίρνουμε: | |
\[ | |
G(t)\od{x}{t}+G(t)p(t)x=G(t)q(t) | |
\] | |
ή | |
\[ | |
\od{(Gt)}{t} = Gq(t) | |
\] | |
που είναι μια ακριβής διαφορική εξίσωση. | |
Λύση: | |
\[ | |
x(t)=e^{-\int p(t)\dif t} | |
\left( | |
\int e^{\int p(t)\dif t} | |
q(t)\dif t + c | |
\right) | |
\] | |
Αν τα \(p(t)=a\) και \(q(t)=b\) είναι σταθερά: | |
\[ | |
\od{x}{t}+ax=b,\ | |
x(t)=e^{-at} \left( \frac{b}{a} e^{at}+c\right) = \frac{b}{a}+ce^{-at} | |
\] | |
\end{itemize} | |
\subsubsection{\textlatin{Bernoulli}} | |
\[ | |
\od{x}{t}+p(t)x = q(t)x^n,\quad\text{με } n\neq1,0 | |
\] | |
Αντικατάσταση μεταβλητών: \(u=x^{1-n}\rightarrow x,\ x'\) | |
\begin{exercise*}{4.1} | |
ΓΡ/ΔΕ/1\textsuperscript{ης} | |
\[ | |
x'-3x=6 | |
\] | |
\tcblower | |
\begin{gather*} | |
\od{x}{t}+ax=b\\ | |
a=-3,\ b=6\\ | |
x(t)=\frac{b}{a}+ce^{-at}=\frac{6}{-3}+ce^{3t} | |
\implies x(t)=ce^{3t}-2 | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{4.2} | |
ΓΡ/ΔΕ/1\textsuperscript{ης} | |
\[ | |
\od{x}{t}-2tx=t | |
\] | |
\tcblower | |
\begin{gather*} | |
\od{x}{}+p(t)x=q(t)\implies\begin{cases} | |
p(t)&=-2t\\q(t)=t | |
\end{cases}\\ | |
\text{ΟΠ } G(t) = e^{\int p(t)\dif t}=e^{\int(-2t)\dif t}\\ | |
\int(-2t)\dif t = -t^2 \text{ άρα } G(t)=e^{-t^2}\\ | |
e^{-t^2}\od{x}{t}-2te^{-t^2}x=te^{-t^2} \implies | |
\od{}{t}\left(xe^{-t^2}\right)=te^{-t^2}\implies | |
\int \od{}{t}\left(xe^{-t^2}\right) \dif t=\int te^{-t^2} \dif t \implies | |
\\ \implies | |
xe^{-t^2}=-\frac{1}{2}e^{-t^2}+c \implies | |
\boxed{x=ce^{t^2}-\frac{1}{2}} | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{4.3} | |
\[ | |
x' + \left( \frac{4}{t}\right)x=t^4 | |
\] | |
\tcblower | |
\begin{gather*} | |
x'+p(t)x=t^4\qquad p(t)=\frac{4}{t},\ q(t)=t^4\\ | |
G(t)=e^{\int p(t)\dif t}=e^{\int \frac{4}{t}\dif t} | |
=e^{4\ln|t|}=e^{\ln t^4}=t^4\\ | |
t^4\od{x}{t}+t^4\left(\frac{4}{t}\right)x=t^4\cdot t^4 \implies | |
\od{x}{t}\left(t^4x\right)=t^8\implies | |
\\ \implies | |
\int\od{}{t}\left(t^4x\right) \dif t = \int t^8\dif t\implies | |
t^4x=\frac{1}{9}t^9+c\implies | |
\boxed{x=\frac{1}{9}t^5+\frac{c}{t^4}} | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{4.4} | |
\begin{gather*} | |
x'+x=\sin t\\ | |
x(\pi)=1 | |
\end{gather*} | |
\tcblower | |
\begin{gather*} | |
\od{x}{t}+p(t)x=q(t)\qquad p(t)=1,\ q(t)=\sin t\\ | |
\end{gather*} | |
\begin{gather*} | |
G(t)=e^{\int p(t)\dif t}=e^{\int 1 \dif t}=e^t | |
\end{gather*} | |
\begin{gather*} | |
e^t\left(x'+x\right)=e^t\sin t\implies\\ | |
\implies\int\od{}{t}\left(e^tx\right)=\int e^t\sin t\dif t \implies\\ | |
\implies e^tx=\frac{e^t}{2}\left( \sin t -\cos t \right)+c \implies\\ | |
\implies\boxed{ x(t)=ce^{-t}+\frac{1}{2}\sin t-\frac{1}{2}\cos t} \\ | |
1=ce^{-\pi}+\frac{1}{2}\sin\pi - \frac{1}{2}\cos\pi \implies c =e^\pi \\ | |
\text{ΕΛ}\quad x(t)= \frac{1}{2}e^\pi e^{-t}+\frac{1}{2}\sin t-\frac{1}{2}\cos t | |
\implies \\ \implies | |
x(t)=\frac{1}{2}\left( | |
e^{\pi-t}+\sin t -\cos t | |
\right) | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{4.6} | |
\[ | |
\od{z}{x}-xz=-x;\quad z(0)=-4 | |
\] | |
\tcblower | |
\[ | |
p(x)=-x,\ q(x)=-x | |
\] | |
\[ | |
G(x) = e^{\int p(x)\dif x}= e^{\int (-x)\dif x}=e^{-\frac{x^2}{2}} | |
\] | |
\begin{align*} | |
e^{-\frac{x^2}{2}}\left(\od{z}{x}-xz\right)=e^{-\frac{x^2}{2}}(-x) &\implies\\ | |
\od{}{x}\left(e^{-\frac{x^2}{2}}z\right)=e^{-\frac{x^2}{2}}z &\implies\\ | |
\int\od{}{x}\left(e^{-\frac{x^2}{2}}z\right)\dif x=\int\left(e^{-\frac{x^2}{2}}x\right)\dif x &\implies\\ | |
e^{-\frac{x^2}{2}}z=e^{-\frac{x^2}{2}}+c &\implies\\ | |
\boxed{z=ce^{\frac{x^2}{2}}+1} &\text{ ΓΛ}\\ | |
-4=ce^{\frac{0^2}{2}}+1\implies c=-5 &\implies \boxed{z(x)=-5e^\frac{x^2}{2} +1} \text{ ΕΛ} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{4.7} | |
\[ | |
z'-\frac{2}{x}z=\frac{2}{3}x^4 | |
\] | |
\tcblower | |
\begin{gather*} | |
p(x)=-\frac{2}{x}\\ | |
G(x)=e^{\int\left(\frac{-2}{x}\right)}=e^{-2\ln|x|}=e^{\ln x^{-2}}\\ | |
G(x)=x^{-2}\\ | |
x^{-2}(z'-\frac{2}{x}z)=\frac{2}{3}x^4x^{-2}\implies\dots\implies z(x)=cx^2+g^2x^5 | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{4.10} | |
\[ | |
y'+xy=xy^2 | |
\] | |
\tcblower | |
\textlatin{Bernoulli} | |
\begin{align*} | |
u=y^{1-n}=y^{1-2}=y^-1 \implies \\ | |
u=\frac{1}{y} \implies y=\frac{1}{u} \text{ και } y' = -\frac{1}{u^2}\od{u}{x}=-\frac{1}{u^2}u' \\ | |
-\frac{1}{u^2}\od{u}{x}+x\frac{1}{u}=x\left(\frac{1}{u}\right)^2\implies u'-xu=-x\\ | |
G(x)=e^{\int (-x)\dif x}=e^{-\frac{x^2}{2}},\ | |
e^{-\frac{x^2}{2}}u'-e^{-\frac{x^2}{2}}xu=e^{-\frac{x^2}{2}}x \implies \\ | |
\boxed{ | |
\od{}{x}\left(ue^{-\frac{x^2}{2}}\right) =-xe^{-\frac{x^2}{2}} | |
}\\ | |
\int \od{}{x} \left(ue^{-\frac{x^2}{2}}\right) \dif x = | |
\int \left(-xe^{-\frac{x^2}{2}}\right) \dif x \implies \\ | |
ue^{-\frac{x^2}{2}} + c \implies \boxed{u=ce^{\frac{x^2}{2}}+1} \text{ ΓΛ} | |
\frac{1}{y}=ce^{\frac{x^2}{2}}+1 \implies \boxed{y(x)=\frac{1}{ce^\frac{x^2}{2}+1}} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{4.11} | |
\[ | |
y' - \frac{3}{x}y = x^4y^\frac{1}{3} | |
\] | |
\tcblower | |
\begin{gather*} | |
n=\frac{1}{3},\ u=y^{1-n}=y^{1-\frac{1}{3}}=y^\frac{2}{3}\implies y =u^\frac{3}{2} | |
\\ | |
\implies y'=\frac{3}{2}u^\frac{1}{2}u' | |
\\ | |
\intertext{Άρα η διαφορική εξίσωση γίνεται} | |
\frac{3}{2}u^\frac{1}{2}u'-\frac{3}{x}u^\frac{3}{2}=x^4\left(u^\frac{3}{2}\right)^\frac{1}{3} \implies | |
\\ | |
\frac{3}{2}u'u-\frac{3}{x}u^2=x^4u\implies\\ | |
\frac{3}{2}u'-\frac{3}{x}u=x^4\implies\\ | |
u'-\frac{2}{x} u =\frac{2}{3}x^4 \\ | |
\od{u}{t}+p(t)u=q(t)\\ | |
G(x)=e^{\int\left(-\frac{2}{x}\right)\dif x}=e^{-2\ln|x|}=-\frac{1}{x^2}\\ | |
\left(\frac{1}{x^2}\right)u'+\left(\frac{1}{x^2}\right)\left(-\frac{2}{x}\right)u=\frac{1}{x^2}\frac{2}{3}x^4 \implies\\ | |
\od{}{x}(x^{-2}u)=\frac{2}{3}x^2 \xRightarrow{\text{ολοκλ.}}\\ | |
\int \od{}{x}(x^{-2}u)\dif x = \int\frac{2}{3}x^2\dif x \implies x^{-2}u=\frac{2}{9}x^3+c \implies \\ | |
\boxed{ | |
u(x)=cx^2+\frac{2}{9}x^5 } \implies \\ | |
y^\frac{2}{3}=cx^2+\frac{2}{9}x^5\implies \\ | |
\boxed{ | |
y = \pm \left( | |
cx^2+\frac{2}{9}x^5 | |
\right)^\frac{3}{2} | |
} | |
\end{gather*} | |
\end{exercise*} | |
\subsection{ΟΜ/ΓΡ/ΔΕ/2,$n$\textsuperscript{ης}/ΣΣ} | |
Ομογενείς Γραμμικές Διαφορικές Εξισώσεις 2\textsuperscript{ης} και $n$\textsuperscript{ης} τάξης με Σταθερούς Συντελεστές | |
\subsubsection{2\textsuperscript{ης} τάξης} | |
\[ | |
\od[2]{x}{t}+a_1\od{x}{t}+a_0x=0 | |
\] | |
\paragraph{Χαρακτηριστική εξίσωση \emph{ΧΕ}} | |
\[ | |
\lambda^2+a_1\lambda+a_0 = 0 \xRightarrow{\text{παραγοντοποιείται}}\left(\lambda-\lambda_1\right)\left(\lambda-\lambda_2\right)=0 | |
\] | |
\paragraph{(1) \(\lambda_1 \neq \lambda_2\)} | |
\[ | |
\text{ΓΛ} \quad \boxed{ x(t)=c_1e^{\lambda_1t}+c_2e^{\lambda_2t}} | |
\] | |
\paragraph{(2) \(\lambda_1 = a+ib\) και \(\lambda_2=a-ib\)} | |
\begin{align*} | |
\text{ΓΛ} \quad x(t)&=c_1e^{(a+ib)t}+c_2e^{(a-ib)t} \\ | |
\Aboxed{x(t) &= \kappa_1 e^{at}\cos bt + \kappa_2 e^{at} \sin bt}\\ | |
&\left.\begin{array}{r l} | |
\kappa_1&=c_1+c_2\\ | |
\kappa_2&=i(c_1-c_2) | |
\end{array} | |
\middle|\left( | |
\begin{array}{l} | |
\text{λύσεις στο }\mathbb R | |
\\ | |
k_1,k_2 \in \mathbb R \\ | |
c_1,c_2 \rightarrow \text{συζυγείς} | |
\end{array} | |
\right)\right. | |
\end{align*} | |
\paragraph{(3) \(\lambda_1=\lambda_2\) διπλή} | |
\[ | |
\text{ΓΛ}\quad \boxed{ x(t)=c_1e^{\lambda_1t}+c_2te^{\lambda_1t}} | |
\] | |
\subsubsection{$n$\textsuperscript{ης} τάξης} | |
\[ | |
\od[n]{x}{t}+a_{n-1}\od[n-1]{x}{t} | |
+\dots+ | |
a_1\od{x}{t}+a_0x=0 | |
\] | |
\paragraph{Χαρακτηριστική εξίσωση} | |
\[ | |
\lambda^n+a_{n-1}\lambda^{n-1}+\dots+a_1\lambda+a_0=0 | |
\] | |
με λύσεις \(\lambda_1,\lambda_2,\dots,\lambda_n\) | |
\paragraph{(1) Λύσεις \( \in \mathbb R\) διακριτές} | |
\[ | |
\boxed{ | |
x(t)=c_1e^{\lambda_1t}+c_2e^{\lambda_2t}+\dots+c_ne^{\lambda_nt} | |
} | |
\] | |
\paragraph{(2) Μερικές \( \mathbb R \) διακριτές, μερικές \( \mathbb C\) συζυγείς} | |
Ομοίως. | |
\paragraph{(3) \(\lambda_\kappa \) πολλαπλότητας $p$} | |
Δηλαδή \(\left(\lambda-\lambda_\kappa\right)^p\) παράγοντας της ΧΕ αλλά όχι \(n(\lambda-\lambda_\kappa)^{p+1}\). | |
\(p\) ΓΑ λύσεις | |
\[ | |
e^{\lambda_\kappa t},\ te^{\lambda_\kappa t},\ t^2e^{\lambda_\kappa t},\ \dots,\ | |
t^{p-1}e^{\lambda_\kappa t} | |
\] | |
\begin{exercise*}{5.1} | |
\[ | |
y''-y'-2y=0 | |
\] | |
\tcblower | |
\[ | |
\lambda^2-\lambda-2=0 \implies | |
(\lambda+1)(\lambda-2)=0 \implies | |
\lambda_1=-1,\ \lambda_2=2 | |
\] | |
Άρα: | |
\begin{align*} | |
\boxed{y(t)=c_1e^{-t}+c_2e^{2t}} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{5.4} | |
\[ | |
\ddot y +10\dot y+21y=0 | |
\] | |
\tcblower | |
ΧΕ \(\lambda^2+10\lambda+21=0 \implies (\lambda+3)(\lambda+7) \implies | |
\boxed{ | |
c_1e^{-3t}+c_2e^{-7t} | |
} | |
\) | |
\end{exercise*} | |
\begin{exercise*}{5.11} | |
\[ | |
y''-8y'+16y=0 | |
\] | |
\tcblower | |
ΧΕ \( | |
\lambda^2-8\lambda+16=0 \implies (\lambda-4)^2=0 \implies \lambda_\kappa=4 | |
\) διπλή | |
\[ | |
\boxed{ | |
y(x)=c_1e^{4x}+c_2xe^{4x} | |
} | |
\] | |
\end{exercise*} | |
\begin{exercise*}[breakable]{5.5} | |
\[RLC,\ \text{σειρά},\ R=10\Omega,\ C=10^{-2} \mathrm F,\ L = \frac{1}{2}H,\ v = 12V\] | |
Αρχικά κανένα ρεύμα, κανένα φορτίο, τάση εφαρμόζεται για $t=0$. | |
Να βρεθεί \(i\) για $t$ μετά το $0$. | |
\tcblower | |
\begin{circuitikz} \draw | |
(0,0) to[european voltage source=$v$] (0,2) | |
to[R=$R$] (2,2) | |
to[L=$L$] (4,2) | |
to[C=$C$] (6,2) -- (6,0) -- (0,0); | |
\end{circuitikz} | |
\begin{align*} | |
\begin{cases} | |
Ri+L\od{i}{t}+\frac{1}{C}q-v=0\\ | |
i=\od{q}{t} | |
\end{cases}\xRightarrow[t]{\text{διαφ.}}& | |
R\od{i}{t}+L\od[2]{i}{t}+\frac{1}{C}i=\od{v}{t}\\ | |
\implies & | |
\od[2]{i}{t}+\frac{R}{L}\od{i}{t}+\frac{1}{LC}i=\frac{1}{L}\od{u}{t}\\ | |
\implies & | |
\od[2]{i}{t}+\frac{10}{ ^1/_2}\od{i}{t}+\frac{1}{ ^1/_2\left(10^{-2}\right)}i = | |
\frac{1}{ ^1/_2}\cancelto{0}{\od{}{t}\big(12\big)} | |
\\ \implies & | |
\od[2]{i}{t}+20\od{i}{t}+200i = 0 | |
\\ \text{ΧΕ } \implies & \lambda^2+20\lambda+200=0 \quad | |
\lambda_{1,2} = \frac{-20\pm\sqrt{20^2-4(200)}}{2} | |
\\ \implies & | |
\lambda_1=-10+10j,\ \lambda_2=-10-10j | |
\\ \text{ΓΛ } \rightarrow & | |
\boxed{ | |
i(t)=c_1e^{(-10+10j)t}+c_2e^{(-10-10j)t} | |
} \quad \mathsmaller{c_1,c_2 \in \mathbb C} | |
\\ \implies & | |
i(t) = e^{-10t}\left( | |
\kappa_1\cos10t+\kappa_2\sin10t | |
\right) | |
\end{align*} | |
ΑΣ \(\boxed{i(0)=0},\ q(0)=0\). Ψάχνω \(\left.\od{i}{t}\right|_{t=0}\). | |
\begin{align*} | |
Ri+L\od{i}{t}+\frac{1}{C}q-v=0 | |
\\ | |
R(0)+L\left.\od{i}{t}\right|_{t=0}+\frac{1}{C}(0)-12 = 0 \implies \\ | |
\left.\od{i}{t}\right|_{t=0}=\frac{1}{L}v-\frac{1}{LC}\cancelto{0}{q}-\frac{R}{L}\cancelto{0}{i} = \\ | |
=\frac{1}{ ^1/_2}12=24\implies\boxed{ | |
\left.\od{i}{t}\right|_{t=0}=24 | |
} | |
\end{align*} | |
\begin{align*} | |
\od{i(t)}{t}&=-10e^{-10t}\left( | |
\kappa_1\cos10t+\kappa_2\sin10t | |
\right)+ | |
e^{-10t}\left( | |
-10\kappa_1\sin10t+10\kappa_2\cos10t | |
\right)\\ | |
\cancelto{0}{i(0)}&=\cancelto{1}{e^{\cancelto{0}{-10(0)}}}\left( | |
\kappa_1 \cancelto{1}{\cos10(0)} + \kappa_2 \cancelto{1}{\sin10(0)} | |
\right) \implies \boxed{ | |
\kappa_1=0 | |
}\\ | |
\cancelto{24}{\left.\od{i}{t}\right|_{t=0}} &= | |
-10e^{-10(0)}\left( | |
(0)\cos10(0)+\kappa_2\sin10(0) | |
\right)+e^{-10(0)}\left(-10(0)\cos10(0)+10\kappa_2\cos10(0)\right) | |
\implies | |
\\ &\implies | |
\boxed{\kappa_2=\frac{12}{5}} | |
\end{align*} | |
Άρα: | |
\[ | |
\boxed{ | |
\text{ΕΛ} \quad | |
\begin{array}{l} | |
i(t) = e^{-10t} \frac{12}{5}\sin10t\\ | |
\qquad t>0 | |
\end{array} | |
} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{5.16} | |
\begin{gather*} | |
y'''-6y''+11y'-6y=0\\ | |
y(\pi),\ y'(\pi)=0,\ y''(\pi)=1 | |
\end{gather*} | |
\tcblower | |
ΧΕ \(\lambda^3-6\lambda^2+11\lambda-6=0\) | |
\[ | |
\begin{array}{rrrr|l} | |
1&-6&11&-6 & 1\\ | |
\downarrow & 1 & -5 & 6 & \\ | |
1 & -5 & 6 & 0 | |
\end{array} | |
\] | |
\[ | |
\lambda ^3-6\lambda ^2+11\lambda -6=(\lambda -1)(\lambda ^2-5\lambda +6) | |
\] | |
Άρα $\lambda_1=1,\lambda _2=2,\lambda _3=3$. | |
\[ | |
\begin{matrix} | |
\text{ΓΛ:} \quad & c_1e^t+c_2e^{2t}+c_3e^{3t}\\ | |
\text{ΕΛ:} \quad & y'_n = c_1e^t+2c_2e^{2t}+3c_3e^{3t},\ y''_h = c_1e^t+4c_2e^{2t}+9c_3e^{3t} | |
\end{matrix} | |
\] | |
\[ | |
\left. | |
\begin{matrix} | |
\cancelto{0}{y_h(\pi)}=c_1e^\pi +c_2e^{2\pi}+c_3e^{3\pi}\\ | |
\cancelto{0}{y'_h(\pi)}=c_1e^\pi +2c_2e^{2\pi}+3c_3e^{3\pi}\\ | |
\cancelto{1}{y''_h(\pi)} = c_1e^\pi+4c_2e^{2\pi}+9c_3e^{3\pi} | |
\end{matrix} | |
\right\rbrace | |
\implies | |
\begin{matrix} | |
c_1 = \frac{1}{2}e^{-\pi}\\ | |
c_2 = -e^{-2\pi}\\ | |
c_3 = \frac{1}{2}e^{-3\pi} | |
\end{matrix} | |
\] | |
ΕΛ: \(y_p = \frac{1}{2}e^{-\pi}e^t-e^{-2\pi}e^{2t}+\frac{1}{2}e^{-3\pi}e^{3t}\) | |
\end{exercise*} | |
\begin{exercise*}{5.15} | |
\[ | |
y''' - 6y'' + 2y' + 36y = 0 | |
\] | |
\tcblower | |
ΧΕ: \( | |
\lambda^3-6\lambda ^2+2\lambda +36=0 | |
\) | |
\(\lambda _1=-2\) | |
\[ | |
\begin{array}{r|llll} | |
& 1 & -6 & 2 & 36 \\ | |
-2 & & -2 & 16 & -36 \\ | |
& 1 & -8 & 18 & 0 | |
\end{array} | |
\] | |
\( | |
(\lambda +2)(\lambda ^2-8\lambda +18)=0 \implies | |
\lambda _1 = -2,\ \lambda _2=4+i\sqrt{2},\ \lambda_3 = 4-i\sqrt{2} | |
\) | |
ΓΛ: \( | |
c_1e^{-2t}+c_2e^{(4+i\sqrt{2})t}+c_3e^{(4-i\sqrt{2})t} | |
\) | |
\end{exercise*} | |
\begin{exercise*}{5.21} | |
\[ | |
y^{(4)}+8y'''+24y''+32y'+16y=0 | |
\] | |
\tcblower | |
ΧΕ: \( | |
\lambda ^4+8\lambda ^3+24\lambda ^2+32\lambda +16=0 | |
\) | |
\( | |
\lambda _1 = -2 | |
\) | |
\[ | |
\begin{array}{rrrrr|l} | |
1&8&24&32&16&-2\\ | |
\hline | |
&-2&-12&-24&-16\\ | |
\hline | |
1&6&12&8&0 | |
\end{array} | |
\] | |
\( | |
(\lambda +2)(\lambda ^3+6\lambda ^2+12\lambda +8)=0 | |
\) | |
\[ | |
\begin{array}{rrrr|l} | |
1&6&12&8&-2 \\ | |
\hline | |
&-2&-8&-8&\\ | |
\cline{1-4} | |
1&4&4&0& | |
\end{array} | |
\] | |
Άρα \( | |
(\lambda +2)^2(\lambda ^2+4\lambda +4)=(\lambda +2)^2(\lambda +2)^2=(\lambda +2)^4 | |
\). Άρα ρίζες: $\lambda =-2$ τετραπλή. | |
\[ | |
y_h = c_1e^{-2t}+c_2te^{-2t}+c_3t^2e^{-2t}+c_4t^3e^{-2t} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{5.22} | |
\[ | |
\od[5]{P}{t} | |
-\od[4]{P}{t} | |
-2\od[3]{P}{t} | |
+2\od[2]{P}{t} | |
+\od{P}{t}-P=0 | |
\] | |
\tcblower | |
ΧΕ \( | |
\lambda^5-\lambda^4-2\lambda^3+2\lambda^2+\lambda-1=0 | |
\) | |
Τριπλή \(\lambda_{1,2,3}=1\), διπλή \(\lambda_{4,5}=-1\) | |
ΓΛ \( | |
P_h = c_1e^t+c_2te^t+c_3t^2e^t+c_4e^{-t}+c_5te^{-t} | |
\) | |
\end{exercise*} | |
\begin{exercise*}{5.14} | |
Μάζα 2 \textlatin{kg} ανάρτηση ελατήριο στ. ελαστικότητας 10 \(\mathrm{N/m}\) ηρεμία. | |
Μετά κίνηση αρχ. ταχύτητα 150 \(\mathrm{km/s}\). | |
\begin{enumerate} | |
\item Έκφραση της κίνησης της μάζας, χωρίς απώλειες | |
\item κυκλική συχνότητα, φυσική συχνότητα, περίοδος | |
\end{enumerate} | |
\tcblower | |
\begin{center} | |
\begin{tikzpicture} | |
\draw[decoration={aspect=0.3, segment length=1mm, amplitude=1mm, coil},decorate] | |
(1.5, 0) -- (1.5,1); | |
\draw (1.7, 0.5) node {$\kappa$}; | |
\draw (2.9, 0.6) node {$a$}; | |
\draw (2,1.25) node {$m$}; | |
\draw (2.5, 0) -- (2.5, 0.5); | |
\draw (2.4, 0.5) -- (2.6, 0.5); | |
\draw (2.4, 0.5) -- (2.4, 0.7); | |
\draw (2.6, 0.5) -- (2.6, 0.7); | |
\draw (2.5, 0.6) -- (2.5, 1); | |
\draw (1, 1) rectangle (3,1.5); | |
\fill[pattern=north west lines] (0,-0.5) rectangle (4,0); | |
\draw[thick] (0,0) -- (4,0); | |
\end{tikzpicture} | |
\hspace{20pt} | |
\begin{tikzpicture} | |
\draw[decoration={aspect=0.3, segment length=2mm, amplitude=1mm, coil},decorate] | |
(0,1.5) -- (2,1.5) node[midway,above] {$\kappa$}; | |
\draw[,opacity=.4] (2.5,-0.7) -- (2.5,2.5); | |
\draw[->,opacity=.4] (2.4,-0.6) -- (5,-0.6); | |
\draw (1, 0.65) node {$a$}; | |
\draw (2.5,1.3) node {$m$}; | |
\draw (0, 1) -- (1, 1); | |
\draw (1.1, 1.1) -- (1.1, 0.9); | |
\draw (1.1, 1.1) -- (0.9, 1.1); | |
\draw (1.1, 0.9) -- (0.9, 0.9); | |
\draw (1.1, 1) -- (2, 1); | |
\draw (2, 2) rectangle (3,0.6); | |
\fill[pattern=north west lines] (0,2.1) rectangle (-0.5,0.3) rectangle (4, -0.2); | |
\draw[thick] (0,2.1) -- (0,0.3) -- (4,0.3); | |
\end{tikzpicture} | |
\end{center} | |
\paragraph{(α)} | |
\begin{align*} | |
m\ddot{x} &= -\kappa x - a \dot{x} +F(t) | |
\intertext{ή} | |
\ddot{x}+\frac{a}{m}\dot{x}+\frac{\kappa}{m} &= \frac{F(t)}{m} | |
\intertext{ΑΣ} | |
\dot{x}(0) &= 150\ \mathrm{ ^{cm}/_s}\\ | |
x(0) &= 0\\ | |
\ddot{x}+5x&=0 | |
\intertext{ΓΡ/ΟΜ/ΔΕ/2τ/ΣΣ} | |
\lambda^2+5 = 0 \implies \lambda_{1,2} = \pm i\sqrt{5}\\ | |
\Aboxed{ | |
\text{ΓΛ} \quad x(t)&=c_1\cos\sqrt{5}t+c_2\sin\sqrt{5}t | |
}\\ | |
x'(t)&=-c_1\sqrt{5}\sin\sqrt{5}t+c_2\sqrt{5}\cos\sqrt{5}t\\ | |
x(0)&=0 \implies 0 =\cancelto{1}{c_1\cos\sqrt{5}}(0)+c_2\cancelto{0}{\sin\sqrt{5}(0)} \implies 0 =c_1\\ | |
\dot{x}(0) &= 1.5 \implies 1.5 = -c_1\sqrt{5}\sin\sqrt{5}(0)+c_2\sqrt{5}\cos\sqrt{5}(0)\implies c_2=\frac{1.5}{\sqrt{5}} | |
\end{align*} | |
\paragraph{(β)} | |
\begin{gather*} | |
\omega = \sqrt{5}\ \mathrm{rad/s}\\ | |
f=\frac{\omega}{2\pi}=\frac{\sqrt{5}}{2\pi}\; \mathrm{Hz}\\ | |
T=\frac{1}{f} | |
\end{gather*} | |
\end{exercise*} | |
\subsection{ΓΛ ΜΟ/ΔΕ} | |
\[ | |
x=\underbrace{x_n}_{\text{ΓΛ ΟΜ}}+\overbrace{x_p}^{???} | |
\] | |
\[ | |
\mathbf{L}(x) = \phi(t) | |
\] | |
\subsection*{Μέθοδος απροσδιόριστων τελεστών} | |
\( | |
\phi(t) | |
\) και όλες οι παράγωγοί της \( | |
\left\lbrace x_1,x_2,\dots,x_n \right\rbrace | |
\) | |
\paragraph{Αρχικοποίηση} | |
\[ | |
x_p(t) = \underbrace{A_1x_1(t)+A_2x_2(t) + \dots + A_nx_n(t)}_{\text{αυθαίρετοι}} | |
\] | |
\subparagraph{Περίπτωση 1} | |
\begin{align*} | |
\phi(t)&=p_n(t)\\ | |
\tilde x_p &= A_nt^n+A_{n-1}t^{n-1}+\dots+A_1t+A_0\quad A_j (j=0,\dots,n) | |
\end{align*} | |
\subparagraph{Περίπτωση 2} | |
\begin{align*} | |
\phi(t)&=\kappa e^{at}\\ | |
\tilde x_p &= Ae^{at} | |
\end{align*} | |
\subparagraph{Περίπτωση 3} | |
\begin{align*} | |
\phi(t) &= \kappa_1\sin\beta t + \kappa_2\cos\beta t\\ | |
\tilde x_p &= A\sin\beta t + B\cos\beta t | |
\end{align*} | |
\subparagraph{Γενίκευση} | |
\( | |
\begin{matrix} | |
\phi(t) \text {συνδυασμός περιπτώσεων.} \\ | |
\rightarrow \tilde x_p \text{ αντίστοιχος συνδυασμός} | |
\end{matrix} | |
\) | |
\paragraph{Τροποποίηση} | |
Όταν η \(\widetilde{x}\) έχει κοινό όρο με τη \(x_h\), τότε πολλαπλασιάζουμε με \(t^m\) την \(\widetilde{x_p}\) ώστε να μην υπάρχει κοινός όρος. | |
\begin{exercise*}{5.24} | |
\[y''-y'-2y=4x^2\] | |
\tcblower | |
βλ. 5.1 | |
\[ | |
y_h = c_1e^{-x}+c_2e^{2x} | |
\] | |
\begin{gather*} | |
\phi(x)=4x^2\\ | |
\boxed{y_p = A_2x^2+A_1x+A_0}\\ | |
y'_p = 2A_2x+1_1\\ | |
y''_p = 2A_2 | |
\end{gather*} | |
Αντικαθιστώντας στη διαφορική μας εξίσωση έχουμε: | |
\begin{align*} | |
2A_2-(2A_2x+A_1)-2(A_2x^2+A_1x+A_0)&=4x^2 \implies \\ | |
(-2A_2)x^2+(-2A_2-2A_1)x+(2A_2-A_1-2A_0)&=4x^2+(0)x+0 \implies \\ | |
\begin{cases} | |
-2A_2 &= 4\\ | |
-2A_2-2A_1 &= 0 \\ | |
2A_2-A_1-2A_0 &=0 | |
\end{cases} | |
&\implies | |
\begin{cases} | |
A_2 &= -2 \\ | |
A_1 &= 2 \\ | |
A_0 &= -3 | |
\end{cases} \implies \\ | |
\Aboxed{y_p &= -2x^2+2x-3} \\ | |
\text{ΓΛ ΜΟ} | |
\quad | |
y = y_h+y_p &= c_1e^{-x}+c_2e^{2x}-2x^2+2x-3 | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{5.25} | |
\[ | |
y''-y'-2y = e^{3x} | |
\] | |
\tcblower | |
5.1 ΟΜ | |
\(y_h = c_1e^{-x}+c_2e^{2x}\) | |
\(\phi(x)=e^{3x}, \quad y_p(x) = Ae^{3x} | |
\) | |
\( | |
y'_p=3Ae^{3x}, \quad y''_p = 9Ae^{3x} | |
\) | |
\( | |
9Ae^{3x}-3Ae^{3x}-2Ae^{3x}=e^{3x} \implies 4Ae^{3x}=e^{3x} \implies 4A=1 \implies \boxed{A = \frac{1}{4}} | |
\) | |
ΕΛ ΜΟ \(y_p = \frac{1}{4} e^{3x}\) | |
ΓΛ ΜΟ \(y(x) = c_1e^{-x}+c_2e^{2x}+\frac{1}{4}e^{3x} \) | |
\end{exercise*} | |
\begin{exercise*}{5.25} | |
\[ | |
y'' -y' -2y = \sin 2x | |
\] | |
\tcblower | |
5.1 OM \(y_h = c_1e^{-x} + c_2e^{2x}\) | |
\(\phi(x) = \sin 2x \) | |
\( | |
y_p = A\sin 2x + B \cos 2x | |
\) | |
\(y'_p=2A\cos 2x-2B\sin2x \) | |
\(y''_p=-4A\sin2x-4B\cos2x\) | |
\begin{align*} | |
(-4A\sin2x-4B\cos2x) - (2A\cos2x-2B\sin2x)-2(A\sin2x+B\cos2x) &= \sin2x \implies \\ | |
(-6A+2B)\sin2x+(-6B-2A)\cos2x &= (1)\sin2x+(0)\cos2x \implies \\ | |
\begin{cases} | |
-6A-2B &=1 \\ | |
-2A-6B &= 0 | |
\end{cases} | |
\implies \begin{cases} | |
A &= -\frac{3}{20} \\ | |
B &= \frac{1}{20} | |
\end{cases} \implies \text{ΕΛ ΜΟ } \Aboxed{ y_p &= -\frac{3}{20}\sin2x+\frac{1}{20}\cos 2x} | |
\end{align*} | |
\[ | |
\text{ΓΛ ΜΟ } \boxed{ | |
y = c_1e^{-x} + c_2e^{2} - \frac{3}{20}\sin 2x + \frac{1}{20}\cos 2x | |
} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{5.30} | |
\[ | |
y'''-6y''+11y'-6y = 2xe^{-x} | |
\] | |
\tcblower | |
5.16 ΟΜ \(y_h = c_1e^x +c_2e^{2x}+c_3e^{3x} \) | |
\( | |
\phi(x) = 2xe^{-x} \quad \phi(x)=e^{ax}p_n(x), \ a=-1,\ p_n(x)=2x | |
\) | |
\( | |
y_p = e^{-x}(A_1x+A_0) \implies \boxed{y_p = A_1xe^{-x}+A_0e^{-x}} | |
\) | |
\( | |
y'_p = -A_1xe^{-x} + A_1e^{-x}-A_0e^{-x} | |
\) | |
\( | |
y''_p= A_1xe^{-x} -2A_1e^{-x} + A_0e^{-x} | |
\) | |
\( | |
y'''_p = -A_1xe^{x} + 3A_1e^{-x} - A_0e^{-x} | |
\) | |
\begin{align*} | |
-24A_1xe^{-x} + (26A_1-24A_0)e^{-x} &= 2xe^{-x} + (0)e^{-x} \implies \\ | |
\begin{cases} | |
-24A_1 &= 2 \\ | |
26A_1-24A_0 &=0 | |
\end{cases} | |
\implies \begin{cases} | |
A_1 &= -\frac{1}{12} \\ | |
A_0 &= -\frac{13}{144} | |
\end{cases} | |
\text{ΕΛ ΜΟ } y_p &= -\frac{1}{12}xe^-x-\frac{13}{144}e^{-x}\\ | |
\text{ΓΛ ΜΟ } \Aboxed{ y &= c_1e^x+c_2e^{2x}+c_3e^{3x}-\frac{1}{12}xe^{-x}-\frac{13}{144}e^{-x} } | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{5.31} | |
\[ | |
y''=9x^2+2x-1 | |
\] | |
ΟΜ ΔΕ \( | |
y'' = 0, \boxed{y_n=c_1x+c_0} | |
\) | |
\( | |
\phi(x) = 9x^2+2x-1 \quad \underbrace{y_p}_{\mathclap{\cdot\; x^m}} = A_2x^2+A_1x+A_0 | |
\) | |
\paragraph{Τροποποίηση} | |
\(\boxed{ | |
y_p = A_2x^4+A_1x^3+A_0x^2 | |
} \) | |
\(y'_p = \dots,\quad y''_p = \dots\) | |
\begin{align*} | |
12A_2x^2 + 6 A_1x+2A_0 = 9x^2+2x -1 \xRightarrow{\dots} \begin{cases} | |
A_2 &= \frac{3}{4}\\ | |
A_1 &= \frac{1}{3} \\ | |
A_0 &= -\frac{1}{2} | |
\end{cases} \implies | |
\frac{3}{4}x^4+\frac{1}{3}x^3-\frac{1}{2}x^2 | |
\end{align*} | |
\[ | |
\text{ΓΛ ΜΟ } | |
y=c_1x+c_0+\frac{3}{4}x^4+\frac{1}{3}x^3-\frac{1}{2}x^2 | |
\] | |
\end{exercise*} | |
\begin{exercise*}{5.36} | |
\(RLC\) κύκλωμα σε σειρά, \(R = 180\; \Omega,\ C = \frac{1}{280}\; \mathrm F,\ L=20\; \mathrm H\) | |
Εφαρμόζεται τάση \(v(t)=10\sin t\), καμία αρχική φόρτιση και αρχικό ρεύμα \(1\; \mathrm A\) για \(t = 0\), οπότε εφαρμόζεται η τάση. | |
Να βρεθεί το φορτίο στον πυκνωτή. | |
\tcblower | |
\begin{center} | |
\begin{circuitikz} \draw | |
(0,0) to[vsourcesin=$v(t)$] (0,2) | |
to[R=$R$] (2,2) | |
to[L=$L$] (4,2) | |
to[C=$C$] (6,2) -- (6,0) -- (0,0); | |
\end{circuitikz} | |
\end{center} | |
\begin{gather*} | |
Ri + \frac{1}{C}q + L\od{i}{t} -v = 0\\ | |
i = \od{q}{t} = q \implies \od{i}{t} = \od[2]{q}{t} = \ddot{q} | |
\implies R \dot q + \frac{1}{C}q + L \ddot q = v \\ | |
\implies \boxed{\ddot q + \frac{R}{L} \dot q + \frac{1}{LC} q = \frac{1}{L}v} \quad \text{ΜΟ/ΓΡ/ΔΕ/2-τ/ΣΣ} \\ | |
q(0)=0,\ i(0)=1\implies \dot q (0) = 1\\ | |
\ddot q + \frac{180}{20}\dot q + \frac{1}{20\left( ^1/_{180}\right)}q = \frac{10}{20}\sin t \implies \\ | |
\boxed{ | |
\ddot q + 9\dot q + 14q = \frac{1}{2}\sin t | |
} \quad \text{ΔΕ}\\ | |
\text{ΓΛ ΟΜ} \quad \ddot q + 9\dot q + 14q = 0 \\ | |
\text{ΧΕ} \quad \lambda^2 +9\lambda +14=0\\ | |
\lambda_1=-2,\lambda_2=-7\\ | |
\text{ΓΛ ΟΜ} \quad \boxed{q_h =c_1e^{-2t}+c_2e^{-7t}} \\ | |
\text{ΓΛ ΜΟ} \quad q = q_h+q+p\\ | |
\phi(t) = \frac{1}{2} \sin t\\ | |
q_p = A\sin t + B\cos t\\ | |
\dot q_p = A\cos t - B\sin t\\ | |
\ddot q_p = -A\sin t - B\cos t\\ | |
-A\sin t - B\cos t + 9 A\cos t - 9B\sin t + 14A\sin t+ 14B\cos t = \frac{1}{2}\sin t \\ | |
\implies (-A-9B+14A)\sin t + (-B+9A+14B)\cos t = \left(\frac{1}{2}\right)\sin t + \big(0\big)\cos t \\ | |
\begin{cases} | |
13A-9B&=\frac{1}{2}\\ | |
9A+13B&=0 | |
\end{cases} \implies \begin{cases} | |
A &= \frac{13}{500} \\ | |
B &= -\frac{9}{500} | |
\end{cases} | |
\\ | |
\text{ΕΛ ΜΟ} \quad q_p = \frac{13}{500}\sin t -\frac{9}{500}\cos t\\ | |
\text{ΓΛ ΜΟ} \quad q = c_1e^{-2t}+c_2e^{-7t}+\frac{13}{500}\sin t - \frac{9}{500}\cos t\\ | |
\cdots \implies \begin{cases} | |
c_1 &= \frac{110}{500} \\ | |
c_2 &= -\frac{101}{500} | |
\end{cases} \xrightarrow{\text{ΕΛΜΟ}} q = \frac{1}{500} (110e^{-2t}-101e^{-7t}+13\sin t-9\cos t) | |
\end{gather*} | |
\end{exercise*} | |
\subsection{ΓΡ/ΔΕ/ΜΣ: Μέθοδος μεταβολής των παραμέτρων} | |
\begin{align*} | |
\begin{matrix} | |
\text{ΜΟ}\\\text{ΔΕ}\end{matrix}\quad P_0(x)y'' + P_1(x)y'+P_2(x)y &= F(X) \\ | |
\begin{matrix} | |
\text{ΟΜ}\\\text{ΔΕ}\end{matrix}\quad P_0(x)y'' + P_1(x)y'+P_2(x)y &=0 \\ | |
\text{Σύνολο λύσεων ΟΜ/ΔΕ } \left\lbrace y_1,y_2 \right\rbrace & | |
\end{align*} | |
\begin{align*} | |
\text{ΓΛ ΜΟ}\quad y = & y_p + c_1y_1+c_2y_2=y_p+y_h\\ | |
& y_p \text{ ΕΛ ΜΟ ΔΕ} | |
\end{align*} | |
\(P_0,P_1,P_2,F\) συνεχείς \((a,b)\) | |
\(P_0 \) χωρίς μηδενικά | |
\paragraph{\textlatin{a})} | |
\( | |
y_p = u_1y_1+u_2y_2 | |
\) | |
\paragraph{\textlatin{b})} | |
\( | |
\begin{cases} | |
u'_1y_1+u'_2y_2 &=0\\ | |
u'_1y'_1+u'_2y'_2 &= \frac{F}{P_0} | |
\end{cases} | |
\) | |
\paragraph{\textlatin{c)}} | |
\( | |
u'_1,u'_2 | |
\) | |
\paragraph{\textlatin{d)}} | |
υπολογίζουμε \(u_1,u_2\), με ολοκλήρωση (ολ. σταθ \(\rightarrow 0\)) | |
\paragraph{\textlatin{e)}} | |
αντικατάσταση \(u_1,u_2\) στην \(y_p\) | |
\begin{exercise*}{6.1} | |
\[ | |
x^2y'' -2xy' -2y = x^{\frac{9}{2}} | |
\] | |
\( | |
y_h = c_1x+c_2x^2 = c_1y_1 + c_2y_2 \text{ ΓΛ ΟΜ} | |
\) | |
\tcblower | |
ΕΛ | |
\paragraph{\textlatin{a})} | |
\begin{align*} | |
y_p &= u_1y_1 + u_2y_2 \\ | |
&= u_1x + u_2x^2 | |
\end{align*} | |
\paragraph{\textlatin{b})}\[ | |
\begin{cases} | |
u_1'x+u_2'x^2 &=0 \\ | |
u_1' + 2u_2'x &= \frac{x^{\left( ^9/_2\right)}}{x^2} | |
\end{cases} \impliedby \begin{cases} | |
u_1'y_1+u_2'y_2 &= 0\\ | |
u_1'y_1'+u_2'y_2' &= \frac{F}{P_1} | |
\end{cases} | |
\] | |
\paragraph{\textlatin{c})} | |
\begin{align*} | |
u_1' &= -u_2' x\\ | |
u_2'x &= x ^ {^5/_2} \implies u_2' = x^{^3/_2}, u_1' = -x^{^5/_2} | |
\end{align*} | |
\paragraph{\textlatin{d}) ολοκλήρωση} | |
\[ | |
u_1=-\frac{2}{7}x^{^7/_2},\ u_2=\frac{2}{5}x^{^5/_2} | |
\] | |
\paragraph{\textlatin{e})} | |
\[ | |
y_p = -\frac{2}{7}x^{^7/_2}x + \frac{2}{5}x^{^5/_2}x^2 = \frac{4}{35}x^{^9/_2} | |
\] | |
\[ | |
\boxed{ | |
y=y_h+y_p=c_1+c_2x^2+\frac{4}{35}x^{^9/_2} | |
} \quad \text{ΓΛ ΜΟ} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{12.4} | |
\(y''-3y'+4y=0, \quad y(0)=1,\ y'(0)=5\) | |
\tcblower | |
\[ | |
\mathscr L \left\lbrace y'' \right\rbrace - 3\mathscr L\left\lbrace y' \right\rbrace | |
+4\mathscr L\left\lbrace y \right\rbrace = \mathscr L\left\lbrace 0 \right\rbrace | |
\] | |
\( \left[s^2Y(s)-Y(s)-5\right] -3\left[sY(s)-1 \right] + 4Y(s) = 0\) \\ | |
\( \implies Y(s) = \frac{s+2}{s^2-3s+4} \) | |
Παρονομαστής \( s^2-3s+4 = (s^2-3s)+4 = \left[ s^2-3s+\left( | |
\frac{-3}{2} | |
\right)^2 \right] + | |
\left[ 4-\left( | |
\frac{-3}{2} | |
\right)^2 \right] = \left( | |
s-\frac{3}{2} | |
\right)^2 + \left(\frac{\sqrt{7}}{2}\right)^2 | |
\) | |
Αριθμητής: \( s+2 = \left(s-\frac{3}{2}\right) + \frac{7}{2} | |
= \left(s-\frac{3}{2}\right)+ \sqrt{7}\frac{\sqrt{7}}{2} | |
\) | |
Άρα \begin{align*} | |
Y(s) &= \frac{(s-^3/_2)+\sqrt{7}\frac{\sqrt{7}}{2}}{(s-^3/_2)^2+\left(\frac{\sqrt{7}}{2}\right)^2} | |
\\ &= | |
\frac{s-^3/_2}{(s- ^3/_2)^2+\left(\frac{\sqrt{7}}{2}\right)}+\sqrt{7}\frac{\frac{\sqrt{7}}{2}}{(s-^3/_2)^2+\left(\frac{\sqrt{7}}{2}\right)} \\ | |
\implies y(x) &= \mathscr L^{-1} \left\lbrace Y(s) \right\rbrace = \\ | |
&= \mathscr L^{-1}\left\lbrace | |
\frac{s-^3/_2}{(s- ^3/_2)^2+\left(\frac{\sqrt{7}}{2}\right)} | |
\right\rbrace + \sqrt{7} \mathscr L^{-1} \left\lbrace | |
\frac{\frac{\sqrt{7}}{2}}{(s-^3/_2)^2+\left(\frac{\sqrt{7}}{2}\right)} | |
\right\rbrace \\ | |
\implies y(x) &= e^{\frac{3}{2}x}\cos\left(\frac{\sqrt{7}}{2}x\right) | |
+\sqrt{7}e^{\frac{3}{2}x}\sin\left(\frac{\sqrt{7}}{2}x\right) | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{12.5} | |
\( y''-4y=2e^{3t},\quad y(0)=1,\ y'(0)=-1 \) | |
\tcblower | |
\( \mathscr L \left\lbrace y' \right\rbrace -4\mathscr L \left\lbrace y \right\rbrace = 2 \mathscr L\left\lbrace e^{3t} \right\rbrace \implies \) | |
\( \implies \left[ | |
s^2Y(s)-s+1 | |
\right] + \left[-4Y(s)\right] = \frac{2}{s-3} \implies \) | |
\( (s^2-4)Y(s) = \frac{2}{s-3}+s-1=\frac{2+(s-1)(s-3)}{s-3} \xRightarrow{s^2-4=(s-2)(s+2)} \implies Y(s) = \frac{2+(s-1)(s-3)}{(s-2)(s+2)(s-3)} \) | |
\begin{attnbox}{} | |
\[ | |
\frac{1}{(s-a)^m} \rightarrow \frac{A_1}{(s-a)}+\frac{A_2}{(s-a)^2}+\dots + \frac{A_m}{(s-a)^m} | |
\] | |
\[ | |
\frac{1}{(s^2+bs+c)^p} \rightarrow | |
\frac{B_1s+C_1}{s^2+bs+c} + \frac{B_2s+C_2}{(s^2+bs+c)^2} + \dots + \frac{B_ps+C_p}{(s^2+bs+c)^p} | |
\] | |
\end{attnbox} | |
\( \implies Y(s) = \frac{A}{s-2}+\frac{B}{s+2} + \frac{c}{s-3} \implies \) | |
\( \implies \frac{s^2-4s+5}{(s-2)(s+2)(s-3)} = \frac{A(s+2)(s+3)+B(s-2)(s-3)+C(s-2)(s+2)}{(s-2)(s+2)(s-3)} \implies \) | |
\( s^2-4s+5 = As^2-3As+2As-6A +Bs^2-3Bs-2Bs+6B+Cs^2+2Cs-2Cs-4C \) | |
\( s^2-4s+5 = (A+B+C)s^2 - (-A-5B)s + (-6A+6B-4C) \implies \begin{cases} | |
A+B+C &= 1 \\ | |
-A-5B &= 4 \\ | |
-6A +6B -4C &= 5 | |
\end{cases} \xRightarrow{\dots} \begin{cases} | |
A &= -\frac{1}{4} \\ | |
B &= \frac{17}{30} \\ | |
C &= \frac{3}{5} | |
\end{cases} \) | |
\( Y(s) = -\frac{1}{4}\frac{1}{s-2} + \frac{17}{20} \frac{1}{s+2} + \frac{2}{5}\frac{1}{s-3} \) | |
\[ \implies \boxed{ y(x) = -\frac{1}{4}e^{2x}+\frac{17}{20}e^{-2x}+\frac{2}{5}e^{3x}} \] | |
\end{exercise*} | |
\begin{exercise*}{12.6} | |
\[ | |
y''+3y'+2y=6e^t,\quad y(0) = 1,\ y'(0) = -1 | |
\] | |
\tcblower | |
\( \left[s^2Y(s)-s+1\right] +3\left[sY(s)-1\right]+2Y(s) = 6\frac{1}{s-1} | |
\implies \) \\ | |
\( (s^2+3s+2)Y(s) = \frac{6}{s-1}+(s-1)+3 = \frac{6+(s-1)(s+2)}{s-1} \) \\ | |
Εφόσον \( s^2+3s+2=(s+1)(s+2) \) | |
\( \implies Y(s) = \frac{6+(s-1)(s+2)}{(s-1)(s+2)(s+1)} | |
= \frac{1}{s-1}+\frac{2}{s+2}-\frac{2}{s+1} | |
\implies \boxed{ y(t) = e^t+2e^{-2t}-2e^{-t} } | |
\) | |
\end{exercise*} | |
\begin{attnbox}{Προσοχή} | |
Σημαντικός ο ύπνος για τις εξετάσεις | |
\end{attnbox} | |
\begin{exercise*}{} | |
\[y''-2y'-3y=10\cos t, | |
\qquad y(0) = 2,\ y'(0)=7 | |
\] | |
\tcblower | |
\begin{alignat*}{3} | |
& \left[s^2Y(s)-2s-7\right]-2\left[sY(s)-2\right]-3Y(s)=\frac{10s}{s^2 + 1} && \implies \\ | |
\implies & (\underbrace{s^2-2s-3}_{\mathclap{(s-3)(s-1)}})Y(s) = \frac{10s}{s^2+1} + (7+2s)-4 = \frac{10s}{s^2+1}+(2s+3) && \implies \\ | |
\implies & Y(s) = \frac{10s}{(s-3)(s+1)(s^2+1)} + \frac{2s+3}{(s-3)(s+1)} && | |
\end{alignat*} | |
\[ | |
\frac{2s+3}{(s-3)(s+1)} = \frac{9}{4}\frac{1}{s-3}-\frac{1}{4}\frac{1}{s+1} \xleftrightarrow{\mathscr L^{-1}} \frac{9}{4}e^{3t}-\frac{1}{4}e^{-t} | |
\] | |
\[ | |
\frac{10s}{(s-3)(s+1)(s^2+1)} = \frac{A}{s-3} + \frac{B}{s+1} + \frac{Cs+D}{s^2+1} | |
\] | |
όπου \( \left(A(s+1)-B(s-3)\right) (s^2+1)+ (Cs+D)(s-3)(s+1)=10s \) | |
Θέτω \( | |
\begin{array}{l} | |
s = 3 \rightarrow \left(A(3+1) + 0\right)(3^2+1)+(\quad)(0)(\quad)=10\cdot 3\implies 40A=30 \\ | |
s = 1 \rightarrow -8B=-10 \\ | |
s=0 \rightarrow A-3B-3D = 0 \\ | |
\text{εξ. συντ } \rightarrow A+B+C = 0 | |
\end{array} \implies \begin{cases} | |
A &= \frac{3}{4} \\ | |
B &= \frac{5}{4} \\ | |
C &= -2 \\ | |
D &= -1 | |
\end{cases} | |
\) | |
\begin{align*} | |
\frac{10s}{(s-3)(s+1)(s^2+1)} &= \frac{3}{4}\frac{1}{s-3} + \frac{5}{4}\frac{1}{s+1}-\frac{2s+1}{s^2+1} \\ | |
&\xleftrightarrow{\mathscr L^{-1}}\frac{3}{4}e^{3t}+\frac{5}{4}e^{-t}-2\cos t - \sin t \\ | |
\implies & \boxed{ | |
y(t) = -\sin t -2\cos t + 3e^{3t}+e^{-t} | |
} | |
\end{align*} | |
\end{exercise*} | |
\begin{attnbox}{} | |
\[ | |
e^{ax}\cos bx \leftrightarrow \frac{s-a}{(s-a)^2+b^2} | |
\] | |
\tcblower | |
\[ | |
e^{ax}\sin bx \leftrightarrow \frac{b}{(s-a)^2+b^2} | |
\] | |
\end{attnbox} | |
\begin{exercise*}{12.8} | |
\[ | |
y''+4y = 8\sin 2t + 9\cos t,\quad y(0)=1,\ y'(0) = 0 | |
\] | |
\tcblower | |
\begin{alignat*}{3} | |
& \left[s^2Y(s)-1s-0\right] + 4Y(s) = \frac{16}{s^2+4}+\frac{9s}{s^2+1} && \implies \\ | |
\implies & Y(s) = \frac{16}{(s^2+4)^2}+\frac{9s}{(s^2+1)(s^2+4)} + \frac{s}{s^2+4} && | |
\end{alignat*} | |
Γνωρίζουμε ότι \( t\cos(at) \leftrightarrow \frac{s^2-a^2}{(s^2+a^2)^2} \implies | |
t\cos 2t \leftrightarrow \frac{s^2-4}{(s^2+4)^2} = \frac{s^2+4}{(s^2+4)^2} - \frac{8}{(s^2+4)^2} | |
= \frac{1}{s^2+4}-\frac{8}{(s^2+4)^2} | |
\). | |
Άρα \( \frac{8}{(s^2+4)^2} = \frac{1}{s^2+4} - \mathscr L \left\lbrace t\cos 2t \right\rbrace | |
\implies \frac{16}{(s^2+4)^2} \xleftrightarrow{\mathscr L^{-1}}\sin 2t-2t\cos 2t | |
\) | |
Αντικαθιστώντας \( x=s^2 \) στον \( \frac{9}{(x+4)(x+1)} = \frac{3}{x+1}-\frac{3}{x+4} \) και πολλαπλασιάζοντας με \( s \), μάς δίνει \( \frac{9s}{(s^2+4)(s^2+1)} = \frac{3s}{s^2+1} -\frac{3s}{s^2+4} \xleftrightarrow{\mathscr L^{-1}}3\cos t-3\cos 2t \). | |
\( \frac{s}{s^2+4} \xleftrightarrow{\mathscr L^{-1}} \cos 2t \) | |
\[ | |
\boxed{ | |
y(t) = -(2t+2)\cos 2t + \sin 2t +3\cos t | |
} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{12.9} | |
\[ | |
y''-2y'+2y=2t,\quad y(0) = 2, \ y'(0) = -7 | |
\] | |
\tcblower | |
\begin{align*} | |
& \left[s^2Y(s)-2s+7\right]+2\left[sY(s)-2\right]+2Y(s) = \frac{2}{s^2} \implies \\ | |
\implies & (s^2+2s+2)Y(s) = \frac{2}{s^2}+(2s-7)+4=\frac{2}{s^2}+2s -3 | |
\end{align*} | |
Εφ' όσον \( s^2+2s+2 =(s+1)^2+1 \implies Y(s) = \frac{2}{s^2\left((s+1)^2+1\right)} | |
+\frac{2s-3}{(s+1)^2+1} | |
\) | |
\[ | |
\frac{2s-3}{(s+1)^2+1} = \frac{2(s+1)-5}{(s+1)^2+1}=2\frac{s+1}{(s+1)^2+1}-5\frac{1}{(s+1)^2+1} | |
\xleftrightarrow{\mathscr L^{-1}}2e^{-t}\cos t-5e^{-t}\sin t | |
\] | |
\[ | |
\frac{2}{s^2\left((s+1)^2+1\right)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{(s+1)^2+1} | |
\] | |
όπου \( (As+B)\left((s+1)^2+1\right) + s^2\left(C(s+1)+D\right) =2 \) \\ | |
\( \implies (A+C)s^3+(2A+B+C+D)s^2 + 2(A+B)s+2B = 2 \) \\ | |
\( \implies \begin{cases} | |
2B&= 2 \quad s = 0\\ | |
-A+B+D &= 2 \quad s = -1 \\ | |
A+C &= 0 \quad \text{εξ. συντ. } s^3 \\ | |
2A+B+C+D &=0 \quad \text{εξ. συντ. } s^2 | |
\end{cases} \) | |
Λύνοντας \( | |
\begin{cases} | |
A &= -1 \\ B &= 1 \\ C &= 1 \\ D &= 0 | |
\end{cases} | |
\). Συνεπώς \( \frac{2}{s^2\left((s+1)^2+1\right)} = \frac{1}{s}+\frac{1}{s^2}+ \frac{s+}{(s+1)^2+1} | |
\leftrightarrow -1+t+e^{-t}\cos t | |
\) | |
\end{exercise*} | |
\begin{exercise*}{12.10} | |
\[ | |
y''-4y'+5y = e^{-t}\left(\cos t +3\sin t \right) | |
\] | |
\( y(0) = 0,\ y'(0) = 4 \) | |
\tcblower | |
\( | |
\left[s^2Y(s)-0s-4\right]+4\left[sY(s)-0\right] + 5Y(s) = | |
\frac{s+1}{(s+1)^2+1}+\frac{3}{(s+1)^2+1} \implies | |
\) | |
\begin{tcolorbox}{} | |
\( \mathscr L\left\lbrace \od[n]{y}{x} \right\rbrace | |
= s^nY(s)-s^{n-1}y(0)-s^{n-2}y'(0)-\dots - sy^{(n-2)}-y^{(n-1)}(0) | |
\) | |
\end{tcolorbox} | |
\( | |
\implies \underbrace{(s^2+4s+5)}_{\mathclap{=(s+2)^2+1 }}Y(s)=\frac{s+4}{(s+1)^2+1}+4 \implies | |
\) | |
\( | |
Y(s) = \frac{s+4}{\left((s+1)^2+1\right)\left((s+2)^2+1\right)} | |
+\underbrace{\frac{4}{(s+2)^2+1}}_{\mathclap{4e^{-2t}\sin t}} | |
\implies | |
\) | |
\( \frac{s+4}{\left( (s+1)^2+1 \right)\left( (s+1)^2+1 \right)} | |
= \frac{A(s+1)+B}{(s+1)^2+1}+\frac{C(s+2)+D}{(s+2)^2+1} \implies | |
\) | |
\( | |
\implies | |
\left( | |
A(s+1)+B | |
\right)\left((s+2)^2+1\right)+\left(C(s+2)+D\right)\left((s+1)^2+1\right) = 4+s | |
\implies | |
\begin{cases} | |
5A+5B+4C+2D &= 4, \quad \text{για } s=0 \\ | |
2B+C+D &= 3, \quad \text{για } s=-1\\ | |
-A+B+2D&=2, \quad \text{για } s=-2\\ | |
A+C&=0 \quad \text{εξ. συντ του } s^3 | |
\end{cases} \implies \begin{cases} | |
A &= -1\\B&=1\\C&=1\\D&=0 | |
\end{cases} | |
\) | |
Άρα \(\cdots = \frac{-(s+1)+1}{(s+1)^2+1} + \frac{s+2}{(s+2)^2+1} =e^{-t}(-\cos t + \sin t) + e^{-2t}\cos t \) | |
Άρα \( \boxed{y(t) = e^{-t}(-\cos t + \sin t)+e^{-2t}(\cos t + 4\sin t)} \) | |
\end{exercise*} | |
\begin{exercise*}{12.11} | |
\[ | |
4y''+4y'+y=3\sin t + \cos t | |
\] | |
\( y(0) = 2,\ y'(0)=-1 \) | |
\tcblower | |
\begin{gather*} | |
4\left[s^2Y(s)-2s+1\right] + 4\left[sY(s)-2\right]+Y(s)=\frac{3}{s^2+1}+\frac{s}{s^2+1} \\ | |
\implies \underbrace{(4s^2+4s+1)}_{\mathclap{4\left(s+\frac{1}{2}\right)^2}}Y(s) = \frac{3+s}{s^2_1}+4(-1+2s)+8=\frac{3+s}{s^2+1}+8s+4 | |
\\ \implies | |
Y(s) = \frac{3+s}{4\left(s+^1/_2\right)(s^2+1)} + \frac{2}{s+^1/_2} | |
\\ \frac{3+s}{4\left(s+^1/_2\right)(s^2+1)} | |
= \frac{A}{s+^1/_2}+ \frac{B}{\left(s+^1/_2\right)^2}+\frac{Cs+D}{s^2+1} \\ | |
\implies \left(A(s+^1/_2)+B\right)+\left(s^2+1\right)+(Cs+D)\left(s+\frac{1}{2}\right)^2=\frac{3+s}{4} \\ \implies | |
\begin{cases} | |
\text{για } s = - \frac{1}{2} \rightarrow 10B &= 5 \\ | |
\text{για } s = 0 \rightarrow 2A +4B + D &= 3 \\ | |
\text{για } s = 1 \rightarrow 12A+8B+9C+9D &=4 \\ | |
\text{εξ. αντ. συντ. } s^3 \rightarrow A+C &= 0 | |
\end{cases} \implies \begin{cases} | |
A &= ^3/_5 \\ | |
B &= ^1/_2 \\ | |
C &= -^3/_5 \\ | |
D &= -^1/_5 | |
\end{cases} \\ | |
\implies \frac{3+s}{4\left(s+^1/_2\right)^2(s^2+1)} = \frac{3}{5} \frac{1}{s+^1/_2}+\frac{1}{2}\frac{1}{\left(s+^1/_2\right)^2}-\frac{1}{5}\frac{3s+1}{s^2+1} \leftrightarrow \frac{3}{5}e^{-\frac{t}{2}}+\frac{1}{2}te^{-\frac{t}{2}}-\frac{1}{5}\cdot (3\cos t +\sin t)\\ | |
\frac{2}{1+ ^1/_2} \leftrightarrow 2e^{-\frac{t}{2}} \implies | |
\boxed{y(t) = \frac{e^{- ^t/_2}}{10}}\left(5t+26\right)-\frac{1}{5}(3\cos t+\sin t) | |
\end{gather*} | |
\end{exercise*} | |
\section{Συστήματα Διαφορικών Εξισώσεων} | |
\begin{exercise*}{12.12} \[ | |
\begin{cases} | |
u'+u-v &=0 \\ | |
v'-u+v &= 2t | |
\end{cases} | |
\] | |
\begin{align*} | |
u(0) &= 1 \\ | |
v(0) &= 2 | |
\end{align*} | |
\tcblower | |
\begin{align*} | |
\mathscr L\left\lbrace u(x) \right\rbrace &= U(s) \\ | |
\mathscr L\left\lbrace v(x) \right\rbrace &= V(s) | |
\end{align*} | |
\begin{align*} | |
& \begin{cases} | |
\left[sU(s)-1\right]+U(s)-V(s) &=0 \\ | |
\left[sV(s)-2\right]-U(s)+V(s) &= \frac{2}{s} | |
\end{cases} \\ \implies& \begin{cases} | |
(s+1)U(s) - V(s) &= 1 \\ | |
-U(s) + (s+1)V(s) &= \frac{2(s+1)}{s} | |
\end{cases} \\ \implies & | |
\begin{cases} | |
U(s) &= \frac{s+1}{s^2} \\ V(s) &= \frac{2s+1}{s^2} | |
\end{cases} \\ \implies& | |
\begin{cases} | |
\mathscr L^{-1} \left\lbrace U(s) \right\rbrace = \mathscr L^{-1} \left\lbrace \frac{s+1}{s^2} \right\rbrace = \mathscr L^{-1} \left\lbrace \frac{1}{s} +\frac{1}{s^2}\right\rbrace \implies u(x) &= 1+x \\ | |
\mathscr L^{-1}\left\lbrace V(s) \right\rbrace = \mathscr L ^{-1}\left\lbrace \frac{2s+1}{s^2} \right\rbrace = \mathscr L ^{-1}\left\lbrace \frac{2}{s} + \frac{1}{s^2} \right\rbrace \implies v(x) &= 2+x | |
\end{cases} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{12.13} | |
\[ | |
y'+z=x,\ z'+4y=0,\ y(0)=1,\ z(0)=-1 | |
\] | |
\tcblower | |
\begin{align*} | |
&\mathscr L \left\lbrace y(x) \right\rbrace = Y(s),\ \mathscr L \left\lbrace z(x) \right\rbrace = Z(s) \\ | |
\implies & | |
\begin{cases} | |
\left[sY(s)-1\right]+Z(s) &= \frac{1}{s^2} \\ | |
\left[sZ(s)+1\right]+4Y(s) &=0 | |
\end{cases} | |
\\ \implies & | |
\begin{cases} | |
sY(s) + Z(s) &= \frac{s^2+1}{s^2} \\ | |
4Y(s) + sZ(s) &= -1 | |
\end{cases} | |
\\ \implies & | |
\begin{cases} | |
Y(s) &= \frac{s^2+s+1}{s(s^2-4)} \\ | |
Z(s) &= -\frac{s^3+4s^2+4}{s^2(s^2+4)} | |
\end{cases} | |
\end{align*} | |
\begin{align*} | |
\frac{s^2+s+1}{s(s^2-4)} = \frac{s^2+s+1}{s(s+2)(s-2)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s-2} \\ | |
\implies s^2+s+1 = (A+B+C)s^2+2(C-B)s-4A \\ | |
\implies \begin{cases} | |
A+B+C &= 1 \\ | |
2(C-B) &= 1 \\ | |
-4A &= 1 | |
\end{cases} \\ | |
\implies \begin{cases} | |
A &= - \frac{1}{4} \\ | |
B &= \frac{3}{8} \\ | |
C &= \frac{7}{8} | |
\end{cases} \\ | |
Y(s) = \frac{- ^1/_4}{s}+\frac{^3/_8}{s+2}+\frac{^7/_8}{s-2} \leftrightarrow \boxed{y(x) = -\frac{1}{4}+\frac{3}{8}e^{2x}+\frac{7}{8}e^{2x}} | |
\end{align*} | |
Από την 1\textsuperscript{η} σχέση, | |
\[ | |
\boxed{z = x+\frac{3}{4}e^{-2x}-\frac{7}{4}e^{2x}} | |
\] | |
\end{exercise*} | |
\begin{exercise*}{12.14}\[ | |
\left. | |
\begin{cases} | |
w'+y &= \sin x \\ | |
y'-z &= e^x \\ | |
z'+w+y &= 1 | |
\end{cases} | |
\middle| | |
\begin{array}{l} | |
w(0)=0\\y(0)=1\\z(0)=1 | |
\end{array} | |
\right. | |
\] | |
\tcblower | |
\begin{align*} | |
&\begin{cases} | |
\left[sW(s)-0\right]+Y(s) &= \frac{1}{s^2+1} \\ | |
\left[sY(s)-1\right]-Z(s) &= \frac{1}{s-1} \\ | |
\left[sZ(s)-1\right]+W(s)+Y(s) &= \frac{1}{s} | |
\end{cases} | |
\\ &\implies | |
\begin{cases} | |
sW(s)+Y(s)&= \frac{1}{s^2+1} \\ | |
sY(s)-Z(s) &= \frac{s}{s+1} \\ | |
W(s)+Y(s)+sZ(s) &= \frac{s+1}{s} | |
\end{cases} \\ &\implies \begin{cases} | |
W(s) &= \frac{-1}{s(s-1)} \\ | |
Y(s) &= \frac{s^2+s}{(s-1)(s^2+1)} \\ | |
Z(s) &= \frac{s}{s^2+1} | |
\end{cases} | |
\\ &\implies | |
\begin{cases} | |
w(x) &= 1-e^x \\ | |
y(x) &= e^x-\sin x\\ | |
z(x) &= \cos x | |
\end{cases} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{12.15} | |
\begin{align*} | |
y''+z+y &= 0 \\ | |
z'+y' &= 0 | |
\end{align*} | |
\begin{align*} | |
y(0) &= 0 \\ y'(0) &= 0 \\ z(0)&=1 | |
\end{align*} | |
\tcblower | |
\begin{align*} | |
\implies& \begin{cases} | |
\left[s^2Y(s)-(0)s-(0)\right] + Z(s) + Y(s) &= 0 \\ | |
\left[sZ(s)-1\right]+\left[sY(s)-0\right] &= 0 | |
\end{cases} | |
\\ \implies & \begin{cases} | |
(s^2+1)Y(s)+Z(s) &= 0\\ | |
Y(s) + Z(s) &= \frac{1}{s} | |
\end{cases} | |
\\ \implies & \begin{cases} | |
Y(s) &= -\frac{1}{s^3} \\ | |
Z(s) &= \frac{1}{s} + \frac{1}{s^3} | |
\end{cases} \\ \implies & \begin{cases} | |
y(x) &= -\frac{1}{2}x^2 \\ | |
z(x) &= 1 + \frac{1}{2}x^2 | |
\end{cases} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{12.16} | |
\[ | |
\begin{cases} | |
z''+y'&=\cos x \\ | |
y''-z&=\sin x | |
\end{cases} | |
\] | |
\begin{align*} | |
z(0) &= -1\\ | |
z'(0) &= -1 \\ | |
y(0) &= 1 \\ | |
y'(0) &= 0 | |
\end{align*} | |
\tcblower | |
\begin{align*} | |
& \begin{cases} | |
\left[s^2Z(s)+s+1\right]+\left[sY(s-1)\right] &= \frac{s}{s^2+1} \\ | |
\left[s^2Y(s)-s-0\right]-Z(s) &= \frac{1}{s^2+1} | |
\end{cases} \\ &\implies | |
\begin{cases} | |
s^2Z(s)+sY(s) &= -\frac{s^3}{s^2+1} \\ | |
-Z(s)+s^2Y(s) &= \frac{s^3+s+1}{s^2-1} | |
\end{cases} \\ | |
& \implies \begin{cases} | |
Z(s) &= -\frac{s+1}{s^2+1} \\ | |
Y(s) &= \frac{s}{s^2+1} | |
\end{cases} \\ | |
& \implies \begin{cases} | |
z(x) &= -\cos x-\sin x \\ | |
y(x) &= \cos x | |
\end{cases} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{12.17} | |
\[ | |
\begin{cases} | |
w''-y+2z &= 3e^{-x} \\ | |
-2w'+2y'+z &= 0 \\ | |
2w'-2y+z'+2z'' &= 0 | |
\end{cases} | |
\] | |
\begin{align*} | |
w(0)=1 \quad & y(0) = 2 \\ | |
w'(0)=1 \quad & z(0) = 2 \\ | |
& z'(0) = -2 | |
\end{align*} | |
\tcblower | |
\begin{align*} | |
& \begin{cases} | |
\left[s^2W(s)-s-1\right]-Y(s)+2Z(s) &= \frac{3}{s+1} \\ | |
-2\left[sW(s)-1\right]+2\left[sY(s)-2\right]+Z(s) &= 0 \\ | |
2\left[sW(s)-1\right]-2Y(s) + \left[sZs-2\right]+2\left[s^2Z(s)-2s+2\right] &= 0 | |
\end{cases} \\ \implies & | |
\begin{cases} | |
s^2W(s) - Y(s)+2Z(s) &= \frac{s^2+2s+4}{s+1} \\ | |
-2sW(s)+2sY(s)+Z(s) &= 2 \\ | |
2sW(s)-2Y(s)+(2s^2+s)Z(s) &= 4s | |
\end{cases} \\ \implies & | |
\begin{cases} | |
W(s) &= \frac{1}{s-1} \leftrightarrow e^x =w(x) \\ | |
Y(s) &= \frac{2s}{(s-1)(s+1)} = \frac{1}{s-1}+\frac{1}{s+1} \leftrightarrow e^x+e^{-x}=y(x) \\ | |
Z(s) &= \frac{2}{s+1} \leftrightarrow 2e^{-x}=z(x) | |
\end{cases} | |
\end{align*} | |
\end{exercise*} | |
\section{Επίλυση ΓΡ ΔΕ με γενικευμένες/ασυνεχείς συναρτήσεις ως πηγές} | |
\paragraph{Βηματική συνάρτηση \( u \)} | |
\[ | |
u(t)= \begin{cases} | |
0 \quad t<0 \\ | |
1 \quad t \geq0 | |
\end{cases} | |
\] | |
\paragraph{Μετατοπισμένη βηματική συνάρτηση} | |
\[ | |
u(t-c)=u_c(t) | |
\] | |
\[ | |
u_c(t) = \begin{cases} | |
0 \quad t < c \\ | |
1 \quad t \geq c | |
\end{cases} | |
\] | |
\paragraph{Συνάρτηση τινάγματος ή πάγκου \textlatin{(bench)}} | |
\begin{align*} | |
b(t) &= u(t-a)-u(t-b)\\ | |
b(t) &= \begin{cases} | |
0 \quad t <a \\ | |
1 \quad a \leq t \leq b \\ | |
0 \quad b \leq t | |
\end{cases} | |
\end{align*} | |
\begin{figure} | |
\centering | |
\begin{subfigure}[b]{0.31\textwidth} | |
\begin{tikzpicture}[scale=0.8] | |
\begin{axis}[% | |
gray | |
,xlabel=$t$ | |
,ylabel=$u(t)$ | |
,axis lines = center | |
,ymax=1.5 | |
,ytick={0,1} | |
,xtick={0} | |
] | |
\addplot+[const plot, no marks,very thick] coordinates {(-4,0) (0,1) (4,1)} node[above,pos=.57,black] {$u$}; | |
\end{axis} | |
\end{tikzpicture} | |
\caption{Βηματική συνάρτηση \(u\)} | |
\end{subfigure} | |
~ | |
\begin{subfigure}[b]{0.31\textwidth} | |
\begin{tikzpicture}[scale=0.8] | |
\begin{axis}[% | |
gray | |
,xlabel=$t$ | |
,axis lines = center | |
,ymax=1.5 | |
,ytick={0,1} | |
,xtick={0,1} | |
,xticklabels={$0$,$c$} | |
] | |
\addplot+[const plot, no marks,very thick] coordinates {(-4,0) (1,1) (4,1)} node[below,pos=.57,black] {$u_c$}; | |
\end{axis} | |
\end{tikzpicture} | |
\caption{Μετατοπισμένη βηματική \(u_c\)} | |
\end{subfigure} | |
~ | |
\begin{subfigure}[b]{0.31\textwidth} | |
\begin{tikzpicture}[scale=0.8] | |
\begin{axis}[% | |
gray | |
,xlabel=$t$ | |
,axis lines = center | |
,ymax=1.5 | |
,ytick={0,1} | |
,xtick={0,-1,1.5} | |
,xticklabels={$0$,$a$,$b$} | |
,xmin=-2.5 | |
,xmax=2.5 | |
] | |
\addplot+[const plot, no marks,very thick] coordinates {(-4,0) (-1,1) (1.5,0) (4,0)} node[above,pos=.35,black] {$b$}; | |
\end{axis} | |
\end{tikzpicture} | |
\caption{Συνάρτηση πάγκου \(b\)} | |
\end{subfigure} | |
\end{figure} | |
\begin{align*} | |
u(t-c) & \leftrightarrow e^{-cs}\frac{1}{s}\\ | |
u(t-c)f(t-c) &\leftrightarrow e^{-cs}F(s) \\ | |
e^{ct}f(t) &\leftrightarrow F(s-c) | |
\end{align*} | |
\begin{exercise*}{13.1} | |
\[ | |
y'+2y=u(t-4) | |
\] | |
\( y(0)=3 \) | |
\tcblower | |
\( [sY(s)-3]+2Y(s) = e^{-4s}\frac{1}{s} \) | |
\( \implies Y(s) = e^{-4s}\frac{1}{s(s+2)}+\frac{3}{s+2} \) | |
\begin{align*} | |
\frac{3}{s+2} &\leftrightarrow 3e^{-2t}\\ | |
\frac{1}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2} = \frac{A(s+2)+Bs}{s(s+2)} = \frac{A(+b)s+(2A)}{s(s+2)} \\ | |
\implies \begin{cases} | |
A+B&=0\\2A&=1 | |
\end{cases} \implies \begin{cases} | |
A &= ^1/_2 \\ B&= - ^1/_2 | |
\end{cases} \\ | |
\frac{1}{s(s+2)} = \frac{1}{2} \left[\frac{1}{s}-\frac{1}{s+2}\right] | |
\\ \implies e^{-4s}\frac{1}{s(s+2)} = \frac{1}{2} \left( | |
e^{-4s}\frac{1}{s}-e^{-4s}\frac{1}{s}-e^{-4s}\frac{1}{s+2} | |
\right) &\leftrightarrow \frac{1}{2}\left(u(t-4)-u(t-4)e^{-2(t-4)}\right) | |
\\ | |
\implies \boxed{ | |
y(t) = \frac{1}{2}u(t-4)\left(1-e^{-2(t-4)}\right)+3e^{-2t} | |
} | |
\end{align*} | |
\end{exercise*} | |
\begin{exercise*}{13.2} | |
\( y''+y'+\frac{5}{4}y=b(t) \) | |
\( y(0)=0,\ y'(0)=0 \) | |
\( b(t) = \begin{cases} | |
1 \quad 0 \leq t \leq \pi \\ 0 \quad t \geq \pi | |
\end{cases} \) | |
\tcblower | |
\begin{centering} | |
\begin{tikzpicture} | |
\begin{axis}[% | |
% gray | |
,xlabel=$t$ | |
% ,ylabel=$u(t)$ | |
,axis lines = center | |
,ymax=1.5 | |
% ,axis x line = bottom,axis y line = left | |
,ytick={0,1} | |
,xtick={0,3.14} | |
,xticklabels={$0$,$\pi$} | |
,xmin=-1 | |
,xmax=7 | |
] | |
\addplot+[const plot, no marks,very thick,color=magenta] coordinates {(-1,1) (3.14,0) (7,0)} node[midway,anchor=east,black] {$b(t)$}; | |
\end{axis} | |
\end{tikzpicture} | |
~ | |
\begin{tikzpicture} | |
\begin{axis}[% | |
% gray | |
,xlabel=$t$ | |
% ,ylabel=$u(t)$ | |
,axis lines = center | |
,ymax=1.5 | |
% ,axis x line = bottom,axis y line = left | |
,ytick={0,1} | |
,xtick={0,3.14} | |
,xticklabels={$0$,$\pi$} | |
,xmin=-1 | |
,xmax=7 | |
% ,ymax=1.2 % or enlarge y limits=upper | |
% ,xmin=-1.2 | |
] | |
%\addplot+[const plot, no marks,thick,opacity=1] coordinates {(-1,0) (3.14,1) (7,1)} node[midway,anchor=east,black] {$b(t)$}; | |
\addplot+[const plot, no marks,very thick,opacity=0.75] coordinates {(-1,0) (0,1) (7,1)} node[pos=0.3,anchor=south,black] {$\mathrm u(t)$}; | |
\addplot+[const plot, no marks,very thick,color=red,opacity=0.75] coordinates {(-1,0) (3.14,1) (7,1)} node[right,pos=0.3,black] {$\mathrm u(t-\pi)$}; | |
\addplot+[const plot, no marks,very thick,color=orange,opacity=0] coordinates {(-1,1) (3.14,0) (7,0)} node[midway,anchor=east,black] {$b(t)$}; | |
%\addplot+[const plot, blue, no marks,very thick] coordinates {(-4,0) (1,1) (4,1)} node[below,pos=.57,black] {$u_c$}; | |
%\addplot+[const plot, no marks, thick] coordinates {(0,0) (1,0.25) (2,0.4) (3,0.5) (4,1) (4.49,1)} node[below=1.15cm,pos=.76,black] {$F_y$}; | |
\end{axis} | |
\end{tikzpicture} | |
\end{centering} | |
\begin{gather*} | |
b(t) = u(t) - u(t-\pi) \implies \mathscr L \left\lbrace b(t) \right\rbrace = \mathscr L \left\lbrace u(t) \right\rbrace - \mathscr L \left\lbrace u(t-\pi) \right\rbrace \\ | |
\implies B(s) = \frac{1}{s} - e^{-\pi s}\frac{1}{s}\\ | |
\mathscr \rightarrow \left[s^2Y(s)-(0)s-(0)\right]+\left[sY(s)-(0)\right] + \frac{5}{4}Y(s) = \frac{1}{s}-e^{-\pi s} \frac{1}{s} \\ | |
\implies Y(s) = \left(1-e^{-\pi s}\right) \frac{1}{s(s^2+s+^5/_4)}\\ | |
H(s) = \frac{1}{s(s^2+s+^5/_4)} = \frac{A}{s}+\frac{Bs+C}{s^2+s+^5/_4} \\ | |
\implies \begin{cases} | |
A = \frac{4}{5},\ B= -\frac{4}{5},\ C = -\frac{4}{5} | |
\end{cases}\\ | |
\implies H(s) = \frac{4}{5} \left( | |
\frac{1}{s}-\frac{s+1}{s^2+s+^5/_4}\right) \\ | |
s^2+s+^5/_4 = \left[s^2+2\left(\frac{1}{2}\right)s+\frac{1}{4}\right]-\frac{1}{4}+\frac{5}{4} = \left(s+\frac{1}{2}\right)^2+1 \\ | |
s+1 = s +\frac{1}{2}+\frac{1}{2} = \left(s+\frac{1}{2}\right) + \frac{1}{2}\\ | |
\implies H(s) = \frac{4}{5} \left[ | |
\frac{1}{s}-\frac{s+^1/_2}{\left(s+^1/_2\right)^2+1}-\frac{1}{2}\frac{1}{\left(s+^1/_2\right)^2} \right] \\ | |
\implies Y(s) = \left(1-e^{-\pi s}\right)H(s) \implies Y(s) = H(s) - e^{-\pi s} H(s) \\ | |
\mathscr L^{-1} \left\lbrace H(s) \right\rbrace = \frac{4}{5} \left( | |
1-e^{-^t/_2}\cos t - \frac{1}{2}e^{-^t/_2}\sin t | |
\right) = h(t) \\ | |
\mathscr L^{-1} \left\lbrace e^{-\pi s}H(s)\right\rbrace = \mathrm u(t-\pi)h(t-\pi) = \mathrm u_\pi(t)h(t-\pi) | |
\\ y(t) = h(t)-\mathrm u(t-\pi)h(t-\pi) | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{13.3} | |
\( y''+y'+\frac{5}{4}y=g(t) \) | |
\( y(0) = 0,\ y'(0) = 0 \) | |
\( g(t) = \begin{cases} | |
\sin (t) \quad& 0 \leq t < \pi \\ | |
0 & t \geq \pi | |
\end{cases} \) | |
\tcblower | |
\begin{centering} | |
\begin{tikzpicture} | |
\begin{axis}[% | |
% gray | |
,xlabel=$t$ | |
% ,ylabel=$u(t)$ | |
,axis lines = center | |
,ymax=1.5 | |
,ymin=-1.5 | |
% ,axis x line = bottom,axis y line = left | |
,ytick={0,1} | |
,xtick={0,3} | |
% ,xticklabels={$0$,$\pi$} | |
,xmin=-1 | |
,xmax=7 | |
% ,ymax=1.2 % or enlarge y limits=upper | |
% ,xmin=-1.2 | |
] | |
%\addplot+[const plot, no marks,thick,opacity=1] coordinates {(-1,0) (3.14,1) (7,1)} node[midway,anchor=east,black] {$b(t)$}; | |
\addplot[samples=200,very thick,color=orange,opacity=1,domain=0:3*pi] {cos(deg(x))}; | |
\addplot+[const plot, no marks,very thick,opacity=0.75,color=blue] coordinates {(-1,0) (0,1) (7,1)} node[pos=0.3,anchor=south,black] {$\mathrm u(t)$}; | |
\addplot+[const plot, no marks,very thick,color=red,opacity=0.75] coordinates {(-1,0) (3,1) (7,1)} node[right,pos=0.3,black] {$\mathrm u(t-3)$}; | |
%\addplot+[const plot, blue, no marks,very thick] coordinates {(-4,0) (1,1) (4,1)} node[below,pos=.57,black] {$u_c$}; | |
%\addplot+[const plot, no marks, thick] coordinates {(0,0) (1,0.25) (2,0.4) (3,0.5) (4,1) (4.49,1)} node[below=1.15cm,pos=.76,black] {$F_y$}; | |
\end{axis} | |
\end{tikzpicture} | |
~ | |
\begin{tikzpicture} | |
\begin{axis}[% | |
% gray | |
,xlabel=$t$ | |
% ,ylabel=$u(t)$ | |
,axis lines = center | |
,ymax=1.5 | |
,ymin=-1.5 | |
% ,axis x line = bottom,axis y line = left | |
,ytick={0,1} | |
,xtick={0,3} | |
% ,xticklabels={$0$,$\pi$} | |
,xmin=-1 | |
,xmax=7 | |
% ,ymax=1.2 % or enlarge y limits=upper | |
% ,xmin=-1.2 | |
] | |
\addplot[samples=200,very thick,color=orange,opacity=1,domain=0:3] {cos(deg(x))}; | |
\addplot+[const plot, no marks,very thick,opacity=1,color=orange] coordinates {(3,-1) (3,0) (6.8,0)} node[midway,above,black] {$g(t)$}; | |
\end{axis} | |
\end{tikzpicture} | |
\end{centering} | |
\begin{gather*} | |
g(t) = \left[u(t)-u(t-\pi)\right]\sin t \\ | |
\implies g(t) = u(t)\sin t +u(t-\pi)\sin(t-\pi)\\ | |
\leftrightarrow \frac{1}{s^2+1}+e^{-\pi s}\frac{1}{s^2+1} | |
\\ | |
Y(s) \overset{\text{13.2}}{=}\left(1+e^{-\pi s}\right)\frac{1}{(s^2+s+^5/_4)\left(s^2+1\right)} | |
\\ | |
H(s) = \frac{1}{(s^2+s+^5/_4)\left(s^2+1\right)} = \frac{As+B}{s^2+s+^5/_4} + \frac{Cs+D}{s^2+1} | |
\\ | |
\implies \begin{cases} | |
A &= \frac{16}{17} \\ | |
B &= \frac{12}{17} \\ | |
C &= - \frac{16}{17} \\ | |
D &= \frac{4}{17} | |
\end{cases} \implies H(s) = \frac{4}{17} \left[ | |
\frac{4s+3}{s^2+s+^5/_4}+ \frac{-4s+}{s^2+1} | |
\right] | |
\\ | |
s^2+s+^5/_4 = \left(s+\frac{1}{2}\right)^2 + 1 \\ | |
4s+3 = 4\left(s+\frac{1}{2}-\frac{1}{2}\right)+3 = 4\left(s+\frac{1}{2}\right)+1 | |
\\ | |
H(s) = \frac{4}{17} \left[ | |
4+\frac{s+^1/_2}{\left(s+^1/_2\right)^2+1}+\frac{1}{\left(s+\frac{1}{2}\right)^2+1}-4\frac{s}{s^2+1}+\frac{1}{s^2+1} | |
\right] \leftrightarrow \\ | |
\leftrightarrow \frac{4}{17} \left[ | |
4e^{-\frac{t}{2}}\cos t + e^{-\frac{t}{2}}\sin t - 4\cos t +\sin t =h(t) | |
\right] \\ | |
Y(s) = | |
\left(1+e^{-\pi s}\right)H(s) = H(s) + e^{-\pi s}H(s) \leftrightarrow h(t) + \mathrm u(t-\pi)h(t-\pi) | |
\end{gather*} | |
\end{exercise*} | |
\subsection{Συνάρτηση δέλτα - \textlatin{Dirac}} | |
\begin{align*} | |
\mathrm{\delta}(t) =& \lim_{n\to \infty}\mathrm \delta_n(t),\ t \in \mathbb R, & \delta_n(t) = n \left[ | |
\mathrm u(t) - u\left(t-\frac{1}{n}\right) | |
\right] \\ | |
& | |
\left\lbrace \mathrm \delta_n \right\rbrace_{n=1}^\infty | |
& \mathrm{\delta}_n(t) = \begin{cases} | |
0 \quad& t\leq0 \\ n \quad& 0 \leq t < \frac{1}{n} \\ 0 \quad& t \geq \frac{1}{n} | |
\end{cases} \\ & & \mathrm \delta(t) = \begin{cases} | |
0 \quad& \text{για } t \neq 0\\ | |
\infty \quad& \text{για } t = 0 | |
\end{cases} | |
\end{align*} | |
\subsubsection{Συνάρτηση κρουστικής απόκρισης} | |
στο \( c \geq 0 \) \( \boxed{\mathbf L(y) = y''+a_1y'+a_0y_0} \) | |
με \( \mathbf L(y_\delta) = \delta (t-c) \) | |
\( \delta(t-c) \leftrightarrow e^{-cs} \quad c \geq 0 \) | |
\begin{exercise*}{13.4} | |
Να βρεθεί η συνάρτηση της κρουστικής απόκρισης του τελεστή \( \mathbf L \) | |
\[ | |
\mathbf L (y) = y''+2y'+2y | |
\] | |
\( t=0 \) | |
\tcblower | |
\( y_\delta(0) = 0,\ y_\delta'(0) = 0 \) | |
\begin{gather*} | |
\xrightarrow{\mathscr L} | |
\left[ | |
s^2Y_\delta(s) - (0)s-(0) | |
\right] + 2 \left[sY_\delta(s)-(0) \right] +2Y_\delta(s) = 1 | |
\\ \implies Y_\delta(s) = \frac{1}{s^2+2s+2} = \frac{1}{(s+1)^2+1} \implies \boxed{y_\delta(t) = e^{-t}\sin t} | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{13.5} | |
\( t = c\geq0 \) | |
\[ | |
L(y) = y''+2y'+2y | |
\] | |
\begin{gather*} | |
y_\delta \ \text{Α.Τ.} y''_\delta + 2y_\delta' + 2y_\delta = \delta(t-c) | |
\\ y_\delta(0) = 0, y_\delta'(0) = 0, c \geq 0 \\ | |
\xrightarrow{\mathscr L} | |
s^2Y_\delta(s)+2sY_\delta(s)+2Y_\delta(s) = e^{-cs} | |
\\ \implies Y_\delta(s) = \frac{e^{-cs}}{s^2+2s+2} = \frac{e^{-cs}}{(s+1)+1} | |
\\ \implies \boxed{ | |
y_\delta(t) = \mathrm u(t-c)e^{-(t-c)}\sin (t-c) | |
} | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{13.6} | |
\[ | |
y''-y=-20\delta(t-3), \quad y(0) = 1,\ y'(0) = 0 | |
\] | |
\tcblower | |
\begin{gather*} | |
\left[s^2Y(s)-s-0\right]-Y(s)=-20e^{-3s} \\ | |
\implies Y(s) = \frac{s}{s^2-1}-20e^{-3s}\frac{1}{s^2-1} \leftrightarrow \\ | |
\leftrightarrow \cosh(t)-20\mathrm u(t-3)\sinh(t-3) | |
\end{gather*} | |
\end{exercise*} | |
\begin{exercise*}{13.7} | |
\[ | |
y''+4y = \delta(t-\pi)-\delta(t-2\pi); \quad y(0) = 0,\ y'(0)=0 | |
\] | |
\tcblower | |
\begin{gather*} | |
\left[s^2Y(s)\right]+4Y(s) = e^{-\pi s}-e^{-2\pi s} | |
\\ | |
\implies Y(s) = \frac{e^{-\pi s}}{s^2+4} - \frac{e^{-2\pi s}}{s^2+4} = \frac{e^{-\pi s}}{2}\frac{2}{s^2+4}-\frac{e^{-2\pi s}}{2}\frac{2}{s^2+4}\\ | |
\leftrightarrow \frac{1}{2}\mathrm u(t-\pi) \sin \left[2(t-\pi)\right]-\frac{1}{2}\mathrm u(t-2\pi)\sin\left[2(t-2\pi)\right]=y(t) | |
\\ | |
y(t) = \frac{1}{2} \left[ | |
\mathrm u(t-\pi)-\mathrm u(t-2\pi) | |
\right] \sin 2t | |
\end{gather*} | |
\end{exercise*} | |
\newpage | |
\part{Κεχαγιάς: Ολοκληρωτικοί μετασχηματισμοί} | |
\textlatin{(Fourier, Laplace)} | |
Τετάρτη 17:00-18:30 | |
\section{Κεφάλαιο 7: Εισαγωγή στην ανάλυση του \textlatin{Fourier}} | |
\begin{circuitikz} \draw | |
(0,0) to[american voltage source=$V$] (0,4) | |
to[R=$R$] (4,4) | |
to[C=$C$] (4,0) -- (0,0); | |
\end{circuitikz} | |
H συμπεριφορά του κυκλώματος μπορεί να περιγραφεί με μια διαφορική εξίσωση. | |
\(Q(t)\): Το φορτίο του πυκνωτή σε χρονική στιγμή \(t\) | |
\begin{align*} | |
v_1 &= R\cdot i(t) = \frac{\dif Q}{\dif t}\\ | |
v_2 &= \frac{Q(t)}{C} \\ | |
&\boxed{v_1 + v_2 = V(t) \implies \frac{\dif Q}{\dif t} + \frac{Q(t)}{RC} = \frac{1}{R}V(t), \quad \text{με αρχική συνθήκη }Q(0)=0} | |
\end{align*} | |
Θα προσπαθήσω να λύσω την εξίσωση για τρεις περιπτώσεις: | |
\pgfplotsset{width=6.5cm} | |
\paragraph{} | |
\begin{tikzpicture} | |
\begin{axis}[ | |
xlabel=$t$, | |
ylabel={$V(t)$}, | |
axis lines=middle, | |
xtick=\empty, | |
ytick=\empty, | |
title={$V(t)=V_0$} | |
] | |
\addplot[blue,thick,domain=0:5] {5}; | |
\end{axis} | |
\end{tikzpicture} | |
\hskip 10pt | |
\begin{tikzpicture} | |
\begin{axis}[ | |
xlabel=$t$, | |
ylabel={$V(t)$}, | |
axis lines=middle, | |
xtick=\empty, | |
ytick=\empty, | |
xmin=0, | |
title={$V(t)=V_0\cdot sin(nt)$} | |
] | |
\addplot[blue,thick,samples=200] {5*sin(deg(3*x))}; | |
\end{axis} | |
\end{tikzpicture} | |
\hskip 10pt | |
\begin{tikzpicture} | |
\begin{axis}[ | |
xlabel=$t$, | |
ylabel={$V(t)$}, | |
axis lines=middle, | |
xtick=\empty, | |
ytick=\empty, | |
xmin=0, | |
xmax=8, | |
ymin=-1.4, | |
ymax=1.4, | |
title={$V(t)\text{ τετραγωνική συνάρτηση}$} | |
] | |
\addplot+[blue,thick,mark=none,const plot] | |
coordinates | |
{(0,1) (1,-1) (2,1) (3,-1) (4,1) (5,-1) (6,1) (7,-1)}; | |
\end{axis} | |
\end{tikzpicture} | |
\subsubsection{\(V(t) = V_0\)} | |
\[ | |
\frac{\dif x}{\dif t}+ax=b | |
\] | |
Θα εξετάσω τη γενική λύση \(x_0(t)\) της ομογενούς ΔΕ, και \\ | |
θα ψάξω μία ειδική λύση της μη ομογενούς ΔΕ. | |
Ομογενής: \(b=0 \implies \frac{\dif x}{\dif t} = -ax \implies x(t)=ce^{-at}.\)\\\(x(0)=0\implies c=0 \implies x_0(t)=0\). | |
Μη ομογενής: \(\frac{\dif x}{\dif t}+ax=b\). | |
\[ | |
x(t)=k \implies \frac{\dif x}{\dif t}+ak=b \implies k =\frac{b}{a} \implies x(t)=k=\frac{b}{a} | |
\] | |
\begin{theorem*}{} | |
%\paragraph{Θ.} | |
Η γενική λύση της μη ομογενούς είναι: | |
\[x(t) = x_h(t)+x_i(t) \] | |
\end{theorem*} | |
Άρα \[ | |
\begin{cases} | |
x(t) = ce^{-at} - \frac{b}{a} \\ | |
x(0) = 0 | |
\end{cases} | |
\implies 0=x(0)=c+\frac{b}{a} | |
\implies x(t) =\frac{b}{a}-\frac{b}{a}e^{-at} \text{ ή και } | |
x(t) =\frac{b}{a}(1-e^{-at}) | |
\] | |
\begin{tikzpicture} | |
\begin{axis}[ | |
xlabel=$t$, | |
ylabel={$x(t)$}, | |
axis lines=left, | |
xtick=\empty, | |
ytick=\empty | |
] | |
\addplot[blue,thick,domain=0:5] {1-e^(-x)}; | |
\end{axis} | |
\end{tikzpicture} | |
\[ | |
a=\frac{1}{RC}, \quad b= \frac{V_0}{R} | |
\] | |
\subsubsection{\(V(t)=V_0 \sin(nt)\)} | |
\[ | |
\frac{\dif x}{\dif t} + ax = b\sin (nt) | |
\] | |
Είναι \(x_h(t) = ce^{-at}\). | |
Υποθέτω \(x(t) = c_2 \sin (nt) + c_3 \cos (nt) \). Τότε \( \frac{\dif x}{\dif t} = nc_2 \cos (nt) - nc_3 \sin (nt) \): | |
\[ \frac{\dif x}{\dif t} +ax = | |
\left( ac_2 - nc_3 | |
\right) | |
\sin (nt) + | |
\left( ac_3 + nc_2 | |
\right) \cos (nt) | |
= b \sin (nt) \implies | |
\] | |
\[ | |
\implies | |
\begin{cases} | |
ac_2-nc_3&=b \\ | |
nc_2+ac_3&=0 | |
\end{cases} | |
\implies \cdots \implies | |
\begin{cases} | |
c_2 &= \frac{ab}{a^2+n^2} \\ | |
c_3 &= -\frac{bn}{a^2+n^2} | |
\end{cases} | |
\] | |
Θυμάμαι ότι \(x(t) = x_h(t)+x_i(t) = c_1e^{-at} + \frac{ab}{a^2+n^2} \sin (nt) - \frac{bn}{a^2+n^2} \cos (nt)\) και από το \(x(0)=0\) βρίσκω \(c_1 = \frac{bn}{a^2+n^2}\). | |
Άρα: | |
\[ | |
x(t) = \frac{bn}{a^2+n^2} + \frac{ab}{a^2+n^2} \sin (nt) - \frac{bn}{a^2+n^2} \cos (nt) | |
\] | |
Για το \(RC\) κύκλωμα, \(a=\frac{1}{RC} \leftarrow \) χρονική σταθερά κυκλώματος, \(b=\frac{V_0}{R}\), άρα: | |
\[ | |
Q(t) = \frac{V_0C^2Rn}{C^2R^2n^2+1}e^{-\frac{t}{RC}} + \frac{CV_0\sin (nt) - C^2RnV_0 \cos (nt)}{C^2R^2n^2+1} | |
\] | |
\begin{attnbox}{} | |
\begin{align*} | |
p \cos (\omega t) + q \sin (\omega t) = \\ | |
\sqrt{p^2+q^2} \left( \frac{p}{\sqrt{p^2+q^2}} \cos \omega t+ \frac{q}{\sqrt{p^2+q^2}} \sin \omega t \right) = \\ | |
\sqrt{p^2+q^2} \left ( \sin \phi \cos \omega + \cos \phi \sin \omega t \right) = \\ | |
\sqrt{p^2+q^2} \sin ( \omega t + \phi ), \quad \phi = \arctan \frac{p}{q} | |
\end{align*} | |
\end{attnbox} | |
Παρατηρούμε ότι ο πυκνωτής φορτίζει περισσότερο αν είναι μικρότερη η συχνότητα του εναλλασσόμενου ρεύματος. | |
\pgfplotsset{width=0.8\textwidth} | |
\begin{tikzpicture} | |
\begin{axis}[ | |
height=5cm, | |
xlabel=$t$, | |
ylabel={$Q(t)$}, | |
axis lines=left, | |
xtick=\empty, | |
ytick=\empty | |
] | |
\addplot[blue,samples=100,thick,domain=0:2*pi] {1.5*e^(-0.75*x)+0.1*sin(400*x)}; | |
\end{axis} | |
\end{tikzpicture} | |
\subsubsection{\(V(t) = \mathrm{square}(t)\)} | |
%\begin{tikzpicture}[domain=0:14] | |
%\draw (0,0) -- (0,4); | |
%\draw (0,4) -- (2,4); | |
%\draw (2,4) -- (2,0) node[anchor=west] {\(\pi\)}; | |
%\draw (2,0) -- (2,-4) -- (4,-4) -- (4,4) -- (6,4) -- (6,-4); | |
%\draw[color=blue] | |
%plot (\x,{exp(-\x) + 0.35*sin(5*\x r)}) | |
%node[right] {$f(x) = \sin x$}; | |
%\end{tikzpicture} | |
\[ | |
V(t)= \sum _{n = (1,3,5,\dots)} \frac{4}{n \pi} \sin (n t) = | |
\frac{4}{ \pi} \sin (n t) + \frac{4}{3 \pi} \sin (3 t) + | |
+ \frac{4}{5 \pi} \sin (5 t) \frac{4}{7 \pi} \sin (7 t) + \cdots | |
\] | |
Έτσι γίνεται η ανάλυση \textlatin{Fourier}, και αυτό θα το δούμε την επόμενη Τετάρτη, που θα πάμε στο Κεφάλαιο 8, που λέει σειρές \textlatin{Fourier}. | |
\[ | |
V_N(t) = \sum _{n = (1,3,5,\dots)}^N \frac{4}{n \pi} \sin (n t) | |
\] | |
\[ | |
V(t) = \sum _{n = (1,3,5,\dots)}^\infty \frac{4}{n \pi} \sin (n t) = \lim_{t \to \infty} V_N(t) | |
\] | |
\paragraph{} | |
Άρα: | |
\[ | |
\frac{\dif R}{\dif t} + \frac{1}{RC}Q(t) = \frac{V_0 \sin (nt)}{R} \implies | |
Q_n(t) = \frac{V_0C^2Rn}{C^2R^2n^2+1}e^{\frac{t}{RC}} + \frac{CV_0\sin (nt) - C^2RnV_0 \cos (nt)}{C^2R^2n^2+1} | |
\] | |
Οπότε αν: | |
\[ | |
\frac{\dif R}{\dif t} + \frac{1}{RC}Q(t) = \frac{4}{\pi} \frac{\sin (nt)}{R} \implies | |
Q_1(t) = \frac{4}{\pi} \left( \frac{C^2R}{C^2R^1+1} e^{-\frac{1}{RC}}+ \cdots \right) | |
\] | |
\[ | |
\frac{\dif R}{\dif t} + \frac{1}{RC}Q(t) = \frac{4}{3\pi} \frac{\sin (3t)}{R} \implies | |
Q_3(t) = \frac{4}{3\pi} \left( \frac{3C^2R}{9C^2R^1+1} e^{-\frac{1}{RC}}+ \cdots \right) | |
\] | |
\[ | |
\frac{\dif R}{\dif t} + \frac{1}{RC}Q(t) = \frac{4}{5\pi} \frac{\sin (5t)}{R} \implies | |
Q_5(t) = \cdots | |
\] | |
Άρα: | |
\[ | |
Q(t) = \sum_{n \in \lbrace1,3,5,\dots\rbrace}Q_n(t) | |
\] | |
Γιατί όμως, αν \(V_1(t) \rightarrow Q_1(t), \ V_2(t) \rightarrow Q_2(t)\), τότε \(k_1V_1+k_2V_2 = k_1Q_1 +k_2Q_2\) σε αυτό το κύκλωμα (αρχή επαλληλίας/γραμμικότητα)? | |
\section{Σειρές \textlatin{Fourier}} | |
\begin{defn*}{} | |
Μία συνάρτηση \(f(t)\) λέγεται \textbf{τμηματικά συνεχής} στο \([t_1,t_2]\) ανν μπορώ να διαμερίσω: | |
\[ [t_1,t_2] = [\tau_0,\tau_1] \cup [\tau_1,\tau_2] \cup \cdots \cup [\tau_{n-1},\tau_n] \] | |
όπου \(\tau_0 = t_1,\ \tau_n = t_2\), τέτοια ώστε \(f(t)\) συνεχής στο κάθε \((\tau_{i-1},\tau_i)\), και υπάρχουν \(\lim_{t \to \tau_i^-} f(t),\ \lim_{t \to \tau_1^+}\ f(t) \forall i\) | |
\end{defn*} | |
\textit{π.χ} | |
\begin{center} | |
\begin{tikzpicture}[scale=0.5] | |
\draw[->] (-5,0) -- (11,0); | |
\draw[->] (0,-5) -- (0,5); | |
\draw[dashed] (-pi,0) -- (-pi,-1); | |
\draw[dashed] (1*pi,-1) -- (1*pi,1); | |
\draw[dashed] (2*pi,-1) -- (2*pi,1); | |
\draw[dashed] (3*pi,0) -- (3*pi,1); | |
\draw[thick,blue] (-pi, -1) -- (0,-1); | |
\draw[thick,blue] (0,1) -- (pi,1); | |
\draw[thick,blue] (pi,-1) -- (2*pi,-1); | |
\draw[thick,blue] (2*pi,1) -- (3*pi,1); | |
\filldraw[blue] (-pi,-1) circle (2.5pt); | |
\filldraw[blue,fill=white] (0,-1) circle (2.5pt); | |
\filldraw[blue] (0,1) circle (2.5pt); | |
\filldraw[blue,fill=white] (pi,1) circle (2.5pt); | |
\filldraw[blue] (pi,-1) circle (2.5pt); | |
\filldraw[blue,fill=white] (2*pi,-1) circle (2.5pt); | |
\filldraw[blue] (2*pi,1) circle (2.5pt); | |
\filldraw[blue,fill=white] (3*pi,1) circle (2.5pt); | |
\filldraw[blue] (3*pi,0) circle (2.5pt); | |
\end{tikzpicture} | |
\end{center} | |
Η \(f(t)\) είναι τμηματικά συνεχής στο \([-\pi,3\pi]\), επειδή, για \(t_1=-\pi,t_2=3\pi\): | |
\[ | |
[-\pi,3\pi] = [-\pi,0] \cup [0,\pi] \cup [\pi,2\pi] \cup [2\pi,3\pi] | |
\] | |
Στα \( (-\pi,0),\ (0,\pi) ,\ (\pi,2\pi) ,\ (2\pi,3\pi)\) η \(f\) είναι συνεχής, και υπάρχουν τα αντίστοιχα πλευρικά όρια | |
άρα η \(f\) είναι τμηματικά συνεχής. | |
\subsubsection{Συνθήκες του \textlatin{Dirichlet}} | |
\begin{enumerate} | |
\item Η \(f(t)\) είναι ορισμένη στο \((-L,L)\) | |
\item Η \(f(t)\) είναι τμηματικά συνεχής στο \((-L,L)\) | |
\item Η \(f(t)\) είναι περιοδική με περίοδο \(2L\). | |
\end{enumerate} | |
\subsection*{} | |
\begin{theorem*}{} | |
Έστω \(f(t)\) η οποία ικανοποιεί τις συνθήκες \textlatin{Dirichlet} στο \((-L,L)\). Τότε: | |
\begin{enumerate} | |
\item | |
Για κάθε σημείο συνέχειας της \(f(t)\): | |
\[ | |
f(t) = \frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos \frac{n\pi t}{L} | |
+\sum_{n=1}^\infty b_n \sin \frac{n\pi t}{L} | |
\] | |
όπου: | |
\begin{align*} | |
a_n&=\frac{1}{L} \int_{-L}^L f(t) \cos \frac{n\pi t}{L} \dif t \\ | |
b_n&=\frac{1}{L} \int_{-L}^L f(t) \sin \frac{n\pi t}{L} \dif t | |
\end{align*} | |
\item Σε κάθε σημείο ασυνέχειας \(\tau\): | |
\[ | |
\frac{1}{2} \left( \lim_{t \to \tau^-}f(t) + \lim_{t \to \tau^+}f(t) \right) | |
= \frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos \frac{n\pi t}{L} | |
+\sum_{n=1}^\infty b_n \sin \frac{n\pi t}{L} | |
\] | |
\end{enumerate} | |
\end{theorem*} | |
\paragraph{Παρ.} | |
\(f(t) = \text{τετραγωνικός παλμός}\) | |
\subparagraph{Λύση} | |
Η \(f(t)\) ικανοποιεί τις Σ.\textlatin{D} με \(L=\pi\) | |
\begin{align*} | |
a_0 &= \frac{1}{\pi} \int_{-\pi}^\pi f(t) \dif t = \frac{1}{\pi}\int_{-\pi}^0(-1)\dif t+\frac{1}{\pi}\int_{-\pi}^0 1\dif t = -1 + 1 = 0 \\ | |
b_n &= \frac{1}{\pi} \int_{-\pi}^\pi f(t) \sin \frac{n \pi t}{\pi} \dif t | |
= \frac{2}{\pi} \int_0^\pi 1 \sin (nt) \dif t \\ | |
&= \frac{2}{\pi} \cdot \left( \frac{-\cos nt}{n} \right)_{t=0}^\pi | |
=\frac{2}{\pi} \cdot \left( \frac{1-\cos n\pi}{\pi} \right) = | |
\frac{2}{\pi} \left( \frac{1-(-1)^n}{n} \right) = \frac{2}{n\pi} \text{ για άρτια }n | |
\end{align*} | |
Άρα: | |
\begin{align*} | |
a_0=a_1=a_2=\cdots=0 \\ | |
b_1=\frac{4}{\pi}, \quad b_3=\frac{4}{3\pi} \\ | |
\quad b_2=0, \quad b_4 = 0, \dots | |
\end{align*} | |
%a_n &= \frac{1}{\pi} \int_{-\pi)^\pi \underbrace{f(t)}_{\text{άρτια}} \underbrace{\cos \frac{n \pi t}{\pi}}_{\text{άρτια}} | |
\paragraph{Απόδειξη} (Μερική) | |
Θα δεχτούμε ότι η \(f(t)\) γράφεται στη μορφή \(f(t) = \frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos \frac{n\pi t}{L} | |
+\sum_{n=1}^\infty b_n \sin \frac{n\pi t}{L}\), | |
και θα δείξουμε τους τύπους \( | |
a_n=\frac{1}{L} \int_{-L}^L f(t) \cos \frac{n\pi t}{L} \dif t ,\ | |
b_n=\frac{1}{L} \int_{-L}^L f(t) \sin \frac{n\pi t}{L} \dif t | |
\) | |
Έστω \(f(t)= \frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos \frac{n\pi t}{L} | |
+\sum_{n=1}^\infty b_n \sin \frac{n\pi t}{L}\). | |
Tότε: \[\int_{-L}^L f(t) \dif t = \int_{-L}^L f(t) \cdots t | |
= \int_{-L}^L \frac{a_0}{2} \dif t + | |
\int_{-L}^L a_1 \cos \frac{\pi t}{L} \dif t + | |
\int_{-L}^L a_2 \cos \frac{2\pi t}{L} + \cdots = a_0\cdot L + 0 + 0 + \dots | |
\] | |
Άρα: | |
\[ | |
a_0 = \frac{1}{L} \int_{-L}^L f(t) \dif t | |
\] | |
\subparagraph{Συνέχεια απόδειξης} | |
Υποθέτω ότι υπάρχει \textbf{κάποια} σειρά της μορφής *, θα δείξω ότι οι συντελεστές δίνονται από τους τύπους **. | |
ίό | |
Παρνω τυχόν \(m \in \mathbb N\) και εξετάζω το | |
\begin{align*} | |
\int_{-L}^L f(t) \cos \frac{m\pi t}{L} \dif t | |
&= \\ &= | |
\int_{-L}^L \left( \frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos \frac{n\pi t}{L} | |
+\sum_{n=1}^\infty b_n \sin \frac{n\pi t}{L}\right)\cos \frac{m\pi t}{L} \dif t | |
\\ &= | |
\underbrace{\int_{-L}^L \frac{a_0}{2}\cos \frac{m\pi t}{L} \dif t}_{ | |
=0\text{ ολοκληρώνω πάνω σε }m\text{ περιόδους} | |
} | |
+ \sum_{n=1}^\infty \int_{-L}^L a_n \cos \frac{n\pi t}{L} | |
\cos \frac{m \pi t}{L} \dif t | |
+ \underbrace{\sum_{n=1}^\infty \int_{-L}^L b_n \sin \frac{n\pi t}{L} | |
\cos \frac{m \pi t}{L} \dif t}_{ | |
=\frac{b_n}{2} \left( \int_{-L}^L\sin\frac{n\pi t + m\pi t}{L} \dif t + | |
\int_{-L}^L\sin\frac{n\pi t - m\pi t}{L} \dif t | |
\right) = 0 | |
} | |
\\ &= | |
\sum_{n=1}^\infty a_n \int_{-L}^L \cos\frac{n\pi t}{L} \cos\frac{m\pi t}{L } \dif t | |
\\ & \overset{\cos a \cdot \cos \beta= \frac{\cos(a+\beta)+\cos(a-\beta)}{2}}{=} | |
\sum_{n=1}^\infty \frac{a_n}{2} \int_{-L}^L \left( \cos \frac{(n+m)\pi t}{L} | |
+ \cos \frac{(n-m)\pi t}{L} | |
\right) \dif t | |
\\ &= | |
\sum_{n=1}^\infty \begin{cases} | |
0, \quad & n \neq m \\< |