Ross Ogilvie      2nd December, 2024
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42.

Method of Descent

In this exercise we will apply the method of descent to solve the wave equation on ${ℝ}^2$ for a particular set of initial conditions. The idea is to help you understand the key ideas and notation of the method. It is a combination of results from Sections 5.1–5.

Consider the wave equation on ${ℝ}^2$ with initial conditions

$$\begin{array}{llll}\hfill & {\partial }_t^{2}u-\mathrm{\Delta }u=0\text{ on }(x,t)\in {ℝ}^2\times (0,\infty ),& \hfill & \\ \hfill & u(x,0)=g(x)={\chi }_{[0,\infty )}(x_1),{\partial }_{t}u(x,0)=h(x)=0.& \hfill & \end{array}$$

(a)

Suppose $u$ is a solution of the wave equation for $n=2$. Why does $(x_1,x_2,x_3,t)\mapsto u(x_1,x_2,t)$ solve the wave equation on ${ℝ}^3$? (note, the Laplacians are different in different dimensions). (1 point)

(b)

Conversely, prove that a solution $ū$ to the 3-dimensional wave equation that does not depend on $x_3$ gives a solution to the 2-dimensional wave equation. (1 point)

(c)

By (a) and (b), we now must solve a wave equation on ${ℝ}^3$. The key to solving the 3-dimensional wave equation is to consider the (spatial-)spherical means

$$U(x,t,r)=\frac{1}{4\pi r^2}{\int \; <!--\; nolimits\; -->}_{\mathit{\partial 𝐵}(x,r)}ū(z,t)\mathit{𝑑𝜎}(z),$$

and likewise let $G$ and $H$ be the spherical means of $ḡ$ and $\overline{h}$ respectively. Explain why $ḡ(x_1,x_2,x_3)={\chi }_{[0,\infty )}(x_1)$ and $\overline{h}=0$ (or give the definition of bar). Show that

$$G(x,r)=\{\begin{array}{cc}0\hfill & \text{for }{x}_1\le -r\hfill \\ \frac{1}{2}\frac{x_1+r}{r}\hfill & \text{for }|x_1|\le r\hfill \\ 1\hfill & \text{for }r\le x_1\hfill \end{array}\text{and }H(x,r)=0.$$

You may use the following geometric fact: for $-R<a<b<R$, the surface area of the part of the sphere $\mathit{\partial 𝐵}(0,R)$ with $a<x_1<b$ is $2\mathit{𝜋𝑅}(b-a)$. (4 points)

(d)

We know by Lemma 5.2 that $U$ obeys the Euler-Poisson-Darboux equation. Let $Ũ(x,t,r):=\mathit{𝑟𝑈}(x,t,r)$. Show that $Ũ$ obeys the following PDE $$\begin{array}{llll}\hfill & {\partial }_t^2Ũ-{\partial }_r^2Ũ=0\text{ on }(t,r)\in [0,\infty )\times [0,\infty ),& \hfill & \\ \hfill & Ũ(x,0,r)=\mathit{𝑟𝐺}(x,r),{\partial }_tŨ(x,0,r)=\mathit{𝑟𝐻}(x,r).& \hfill & \end{array}$$

Note that there are no $x$-derivatives in this PDE, so we can think of it as a family of PDEs in the variables $r,t$ parametrised by $x$. (2 points)

(e)

Thus we see that $Ũ$ obeys the 1-dimensional wave equation on the half-line $r\in [0,\infty )$. This is solved by a trick using reflection, and the formula is at the end of Section 5.2. We only need the solution for small $r$, so it is enough to consider the case $0\le r\le t$. In this case, show

$$Ũ(x,t,r)=\{\begin{array}{cc}0\hfill & \text{for }{x}_1\le -(t+r)\hfill \\ \frac{1}{4}(x_1+t+r)\hfill & \text{for }-(t+r)\le x_1\le -(t-r)\hfill \\ \frac{1}{2}r\hfill & \text{for }|x_1|\le t-r\hfill \\ \frac{1}{4}(x_1-t+3r)\hfill & \text{for }t-r\le x_1\le t+r\hfill \\ r\hfill & \text{for }{x}_1\ge t+r.\hfill \end{array}$$

(4 points)

(f)

Recover $ū$ from $Ũ$ using a certain property of spherical means. (3 points)

Observe that $ū$ does not depend on $x_3$. So by part (b) we have a solution to the 2-dimensional wave equation:

$$u(x_1,x_2,t)=\{\begin{array}{cc}0\hfill & \text{for }{x}_1<-t\hfill \\ 0.25\hfill & \text{for }{x}_1=-t\hfill \\ 0.5\hfill & \text{for }-t<x_1<t\hfill \\ 0.75\hfill & \text{for }{x}_1=t\hfill \\ 1\hfill & \text{for }{x}_1>t.\hfill \end{array}$$

This solution has jump discontinuities, but this is unsurprising since the initial conditions also had them.

43.

Wave energy modes

Let us work with $n=1$ for simplicity. If we apply the Fourier transform to the wave equation, we arrive at

$$\frac{{\partial }^2û}{\partial t^2}+4{\pi }^2{k}^2û=0.$$

This leads us to define the spectral energy density

$$2E(k,t):={\left|\frac{\mathit{\partial û}}{\mathit{\partial 𝑡}}\right|}^2+4{\pi }^2{k}^2{\left|û\right|}^2.$$

(a)

Through calculation, show that $E$ is constant in $t$. Note that $û$ is a complex valued function and it is important to respect complex conjugation: $|g{|}^2=\mathit{𝑔ḡ}$. (1 point)

(b)

Consider a complex valued function $g$ and its conjugate $h=ḡ$. Prove that $ĥ(k)=\overline{ĝ(-k)}$.
(1 point)

(c)

Using an equation on page 73 for the script, establish Plancherel’s theorem (1 point)

$${\int \; <!--\; nolimits\; -->}_{ℝ}|h{|}^2\mathit{𝑑𝑥}={\int \; <!--\; nolimits\; -->}_{ℝ}|ĥ{|}^2\mathit{𝑑𝑘}.$$

(d)

Use this to show that the energy of a wave is constant. (2 points)