one single fourier transform thing

Author: jhobbs

content/applied-math/differential-equations/partial-differential-equations/fourier-transform.md

@@ -0,0 +1,37 @@
+---
+description: Fourier Transform 
+layout: page
+title: Fourier Transform
+---
+
+
+:::theorem
+$$\hat{\hat{f}}(\omega) = f(-\omega).$$
+
+::::proof
+First we just write the definition:
+$$ \hat{\hat{f}}(\omega) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}  \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) e^{- i k x} dx e^{-ik \omega} dk $$
+
+Now we move the outer exponential into the inner integral, which is allowed since it's constant in $x.$:
+
+$$ \hat{\hat{f}}(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}  \int_{-\infty}^{\infty} f(x) e^{- i k (x + \omega )} dx dk $$
+
+Now we swap the order if integration. We do this in calc 3... some justify it with Fubini's theorem, it's not clear exactly why this is allowed but this is how it goes (TODO: figure this out)
+
+$$ \hat{\hat{f}}(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}  \int_{-\infty}^{\infty} f(x) e^{- i k (x + \omega )} dk dx $$
+
+Now pull out $f(x)$ from the inner integral since it's a constant in $k:$
+
+$$ \hat{\hat{f}}(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} f(x) \int_{-\infty}^{\infty} e^{- i k (x + \omega )} dk dx $$
+
+The inner integral is just a delta function:
+
+$$ \hat{\hat{f}}(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} f(x) 2 \pi \delta(x + \omega). dx $$
+
+Now by, the sifting property of the delta function, this just reduces to
+
+$$ \hat{\hat{f}}(\omega) = f(-\omega). $$
+
+::::
+:::
+

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

@@ -36,9 +36,9 @@ The basic transport equation is
 
 $$ u_t + c u_x = 0, \quad u(0, x) = f(x). \tag{a} $$
 
-We'll find its characteristic curves. First, let's parameterize $u(t,x)$ to get $v(t, x(t)) = u(t, x).$ Now,
+We'll find its characteristic curves. First, let's parameterize $u(t,x)$ to get $h(t, x(t)) = u(t, x).$ Now,
 
-$$ \frac{dv}{dt} = \frac{\partial u}{\partial t} \frac{dt}{dt} + \frac{\partial u}{\partial x} \frac{dx}{dt} = u_t + \frac{dx}{dt} u_x. \tag{b} $$
+$$ \frac{dh}{dt} = \frac{\partial u}{\partial t} \frac{dt}{dt} + \frac{\partial u}{\partial x} \frac{dx}{dt} = u_t + \frac{dx}{dt} u_x. \tag{b} $$
 
 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:
 
@@ -72,7 +72,7 @@ v_t - c v_{\xi} + c v_{\xi} & = 0 \\
 v_t = \frac{dv}{dt} = 0.
 \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.
+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.
 
 Now we solve this ODE: