Transport with decay

Author: jhobbs

content/applied-math/differential-equations/partial-differential-equations/intro.md

@@ -28,15 +28,17 @@ $$ u(t,x) = f(x), \quad \text{where} \quad f(x) = u(0,x). $$
 
 Note that this solution is a function of the space variable $x$ alone. We only require that $f(x)$ be continuously differentiable (why?). This solution represents a stationary wave - it does not change in time. The initial profile stays frozen in place and the system remains in equilibrium (in the same meaning as equilibrium in dynamical systems - a fixed point?)
 
-# Transport Equation
+# Transport Equations
+
+## Basic Transport Equation
 
 The basic transport equation is
 
-$$ u_t + c u_x = 0. \tag{a} $$
+$$ u_t + c u_x = 0, \quad u(0, x) = f(x). \tag{a} $$
 
-We'll find its characteristic curves. First, let's parameterize $u(x,t)$ to get $v(t) = u(x(t), t).$ Now,
+We'll find its characteristic curves. First, let's parameterize $u(t,x)$ to get $v(t, x(t)) = u(t, x).$ Now,
 
-$$ \frac{dv}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial t} \frac{dt}{dt} = u_t + \frac{dx}{dt} u_x. \tag{b} $$
+$$ \frac{dv}{dt} = \frac{\partial u}{\partial t} \frac{dt}{dt} + \frac{\partial u}{\partial x} \frac{dx}{dt} = u_t + \frac{dx}{dt} u_x. \tag{b} $$
 
 Comparing (b) to (a), we see that if if $u_t + c u_x = 0,$ then $\frac{dx}{dt} = c.$ Now, we solve that ODE:
 
@@ -48,9 +50,9 @@ x & = ct + k \\
 x - ct & = k.
 \end{aligned} $$
 
-So, we see that $x - ct$ is constant, and if we let $t=0,$ we get $x_0 = k.$ So, our characteristic curve is $\xi{(x,t)} = x - ct.$
+So, we see that $x - ct$ is constant, and if we let $t=0,$ we get $x_0 = k.$ So, our characteristic curve is $\xi{(t,x)} = x - ct.$
 
-Now, we'll let perform a change of variables and let $\xi = x - ct.$ So we have
+Now, we'll let perform a change of variables (to a moving coordinate system that will create stations waves) and let $\xi{(t,x)} = x - ct.$ So we have
 
 $$ u(t,x) = v(t, \xi) = v(t, x - ct). $$
 
@@ -68,7 +70,7 @@ Substituting our terms from (c) into (a) gives us
 $$ \begin{aligned}
 v_t - c v_{\xi} + c v_{\xi} & = 0 \\
 v_t = \frac{dv}{dt} = 0.
-\end{aligned} $$
+\end{aligned} \tag{d} $$
 
 Note: we can go from $v_t to \frac{dv}{dt}$ because $v(t, x - ct)$ is only a function of $t,$ because $x - ct$ is constant.
 
@@ -80,6 +82,30 @@ v(t, \xi) - v(0, x_0) & = 0 \\
 v(t, \xi) & = v(0, x_0). \\
 \end{aligned} $$
 
-We were given that $u(0, x) = f(x),$ and $\frac{dv}{dt} = 0,$ i.e. $v$ doesn't change with time so $v(t, \xi) = v(t, x_0) = v(0, x_0) = f(x_0) = f(\xi) = f(x - ct).$
+We were given that $u(0, x) = f(x),$ and found that $\frac{dv}{dt} = 0,$ i.e. $v$ doesn't change with time so $v(t, \xi) = v(t, x_0) = v(0, x_0) = f(x_0) = f(\xi) = f(x - ct).$
+
+Since we defined $u(t,x) = v(t, \xi),$ our solution is $u(t,x) = f(x - ct) = f(\xi)$ for any $f \in C^1.$ This meany any reasonable function of $\xi$ will solve our PDE, i.e. $\xi^2 +1$ or $\cos{\xi}$ will produce a corresponding solution such as $(x-ct)^2 + 1$ or $\cos{(x - ct)}.$
+
+## Transport with Decay
+
+Let $a > 0$ be a positive constant, and $c$ an arbitrary constant. The homogeneous linear first-order partial differential equation
+
+$$ \frac{\partial u}{\partial t} + c \frac{\partial u }{\partial x} + a u = 0 $$
+
+models the transport of, for example, a radioactively decaying solute in a uniform fluid flow with wave speed $c,$ and the coefficient $a$ modeling the rate of decay.
+
+We can reuse the same characteristic as we used in the basic transport equation (since, I believe, it is determined only by the differential terms of the equation.) Then, following what we did in (c) and (d) above we get
+
+$$ \frac{dv}{dt} + av = 0. $$
+
+This is a first order linear differential equation. We find the integrating factor to be $e^{at},$ and the solution to be
+
+$$ e^{at} v(t, \xi) = C. \tag{e} $$
+
+If we let $t = 0, $ we get $v(0, \xi) = f(\xi) = C.$ Then if we divide both sides by $e^{at}$ we get
+
+$$ v(t, \xi) = f(\xi)e^{-at}, $$
+
+and since $u(t,x) = v(t, \xi)$ we have $u(t,x) = f(x - ct)e^{-at}$ as our solution.
+
 
-Since we defined $u(t,x) = v(t, \xi),$ so our solution is $u(t,x) = f(x - ct).$