# Thursday October 31

Today: Some topics in PDEs.

## The Heat / Diffusion Equation in the Plane

Setup: 
Let $\vector \in \RR^2$ be a plate, and consider it evolving over time $t$. 

![Image](figures/2019-10-31-11:29.png)\

So we have pairs $(x, t) \in \RR^2 \cross \RR_{\geq 0}$.

We have some initial distribution of heat on the plate, we want to know how it evolves over time.

This is modeled by the equation

\begin{align*}
\dd{u}{t} = \frac{1}{4\pi} \left( \dd{^2 u}{x_1^2} + \dd{^2u}{x_2^2} \right) \definedas \frac{1}{4\pi} \Delta u\\
u(x, 0) = f(x)
.\end{align*}


Consider a point and a small ball around that point. 
Then heat flow at any point $x_0$ is given by $\nabla_x u(x_0, t)$. 
Now think about the change in energy contained in this ball. 
We should have

\begin{align*}
\dd{}{t} \int_{B} u(x, t) ~dx &= \text{Flux across boundary} \\
&= \int_B \nabla \cdot \nabla_x u(x, t) ~dx \quad  \text{by Green's/Divergence theorem} \\
&\definedas \int_B \Delta_x u(x, t) ~dx,
.\end{align*}


which is the heat equation.

### Solution
We can use Fourier transforms to help solve these. 
Recall the identities:

- $\widehat{\dd{}{x_j} f}(\xi) = 2\pi i \xi_j \widehat{f}(\xi)$.
- $\widehat{\dd{^2}{x_j^2} f}(\xi) = (2\pi i \xi_j)^2\widehat{f}(\xi) = - 4\pi^2 \xi^2 \hat{f}(\xi)$.
- $\widehat{\Delta f}(\xi) = 4\pi^2 \abs{\xi}^2 \hat{f}(\xi)$.

If we take the Fourier transform in the $x$ variable, we get 
$$
\widehat{\dd{u}{t}} 
= \dd{}{t} \hat{u}(\xi, t) 
= -\pi \abs{\xi}^2 \hat{u}(\xi, t)
.$$

Then the boundary conditions become $\hat{u}(\xi, 0) = \xi{f}(\xi)$.
But note that this is now a first order ODE!

This is easy to solve, we get 
$$
\hat{u} (\xi, t) = c(\xi) e^{-\pi \abs{\xi}^2 t} = \hat{f}(\xi) e^{-\pi \abs{\xi}^2 t}
.$$

But then 
$$
e^{-\pi \abs{\xi}^2 t} = \hat{G} (\sqrt t \xi)
$$ 
where $G(x) = e^{-\pi \abs{x}^2}$.

We now have $\hat{u} = \hat{f} \hat{G} = \widehat{f\ast G}$, but **if the transforms are equal then the original functions are equal by the inversion formula**.

We thus obtain 
$$
u(x, t) = f \ast G_{\sqrt t}(x) \text{ where } G_{\sqrt t}(x) = \frac{1}{t^{n/2}} e^{-\pi \abs{x}^2/ t}
.$$

Note that $f \ast g \to f$ as $t \to 0$, which matches with the original boundary conditions, and $f \ast g \to 0$ as $t \to \infty$, which corresponds with heat dissipating.


## Dirichlet problem in the upper half-plane

Setup:

![Image](figures/2019-10-31-11:28.png)\

We want to solve

\begin{align*}
\Delta u  &= 0 \\
u(x, 0)   &= f(x)
.\end{align*}

### Solution

We'll use the same technique as the heat equation, and obtain
$$
\Delta u = 0 \implies -4\pi^2 \abs{\xi}^2 \hat{u} (\xi ,y) + \dd{^2}{y^2} \hat{u} (\xi, y) = 0
$$

But this is a homogeneous second order ODE, so we can look at the auxiliary polynomial. 
If we have distinct roots, the general solution is $c_1 e^{r_1x} + c_2 e^{r_2 x}$.

We thus obtain
$$
\hat{u}(\xi, y) = A(\xi) e^{-2\pi \abs{\xi} y} + B(\xi) e^{2\pi \abs{\xi} y}
$$

In particular, we can just take the first term, since the second term won't vanish at infinity.
We again find that $A(\xi) = \hat{f}(\xi)$ by checking initial conditions, so 
$$
\hat{u}(\xi, y) = \hat{f}(\xi) \hat{P}(y\xi) = \widehat{f \ast P_y}
\quad \text{ where } P(x) = \frac{1}{\pi} \frac{1}{1+x^2}
,$$ 
and
$f\ast P_y \to f$ as $y\to 0$ as desired.
.
## Wave Equation (Cauchy Problem in $\RR^n$)

Same situation as the heat equation, but now in $\RR^n \cross \RR_{\geq 0}$:

\begin{align*}
\dd{^2 u}{t^2} = \Delta_x u \\
u(x, 0) = f(x) \\
\dd{u}{t}(x, 0) = g(x)
.\end{align*}


This models something like plucking a string with initial shape $f$ and initial velocity $g$.

> Note that this now involves a *second* derivative!

### Solution

Using the same technique, we have

\begin{align*}
\dd{^2}{t^2} \hat{u}(\xi, t) &= -4\pi^2 \abs{\xi}^2 \hat{u} (\xi, t) \\
\hat{u}(\xi, 0) &= \hat{f}(\xi) \\
\dd{}{t} \hat{u}(\xi, 0) &= \hat{g}(\xi)
.\end{align*}


This is again 2nd order linear homogeneous, except there is now a complex conjugate pair of roots, so we get
$$
\hat{u}(\xi, t) = \hat{f}(\xi) \cos(2\pi \abs \xi t) + \frac{\hat{g}(\xi) \sin(2\pi \abs \xi t)}{2\pi \abs \xi}.
$$

Note that the derivative of the first term is exactly the second term, so we have

$$
u(x, t) = f\ast \dd{}{t} W_t(x) + g\ast W_t(x), \quad \hat{W_t}(\xi) = \frac{\sin(2\pi \abs \xi t)}{2\pi \abs \xi}.
$$

From the homework problems, we know:

- $n=1$ implies $\chi_{[-1, 1]} (x)$

- $n = 2$ implies $\frac{1}{\sqrt{1 - \abs{x}^2}} \chi_{-1, 1}(x)$

- $n=3$ implies we only get a measure, i.e. $w(x) = \sigma(x)$ where $\sigma$ is a surface measure on $S^2$.

- For $n > 3$, $W$ is a *distribution*.

Note that there is a solution given by D'Alembert,
$$
u(x, t) = \frac{1}{2} ( f(x +t) + f(x - t) ) + \frac{1}{2} \int_{x-t}^{x+t} g(y) ~dy
$$

Note the similarities -- the first term is a rough average, the second term is a more continuous average.

*Exercise*: 
Verify that these two solutions are equivalent.