Wave equation

The PDE

$$u_{tt}-c^2{u}_{xx}=0$$

is called the wave equation. If initial conditions

$$u(x,0)=f(x),u_t(x,0)=g(x)$$

are specified then the solution is uniquely determined.

Homogeneous solution

The solution is

$$\frac{1}{2}[f(x+ct)+f(x-ct)]+\frac{1}{2c}{\int }_{x-ct}^{x+ct}g(w)\mathrm{d}w.$$

This is called the d’Alembert solution.

Here is how to find it. Notice that

$$u_{tt}-c^2{u}_{xx}=\left(\frac{\partial }{\partial t}-c\frac{\partial }{\partial x}\right)\left(\frac{\partial }{\partial t}+c\frac{\partial }{\partial x}\right)u=0.$$

Inhomogeneous case

Let us see how to solve the inhomogeneous equation

$$u_{tt}-c^2{u}_{xx}=f,u(x,0)=f(x),u_t(x,0)=g(x).$$

This can be accomplished by solving the following PDEs:

$$u_{tt}-c^2{u}_{xx}=0,u(x,0)=f(x),u_t(x,0)=g(x),$$

and

$$u_{tt}-c^2{u}_{xx}=f,u(x,0)=0,u_t(x,0)=0.$$

If $u_1$ solves the first PDE and $u_2$ the second, then $u_1+u_2$ solves the original equation since the sum $u_1+u_2$ satisfies the original initial conditions and differentiating it gives us what we want.

The homogeneous equation is easy to solve (see above). We focus on solving the inhomogeneous equation below

Solving with change of coordinates

We can change to new coordinates $\xi =x+ct,\eta =x-ct$ to solve this. Let $\Phi$ change coordinates from $(x,t)$ to $(\xi ,\eta )$, and define $U=u\circ {\Phi }^{-1}$. Then $u=U\circ \Phi$ so

$$\begin{array}{rl}{u}_x& =\frac{\partial U}{\partial \xi }\frac{\partial \xi }{\partial x}+\frac{\partial U}{\partial \eta }\frac{\partial \eta }{\partial x}=\frac{\partial U}{\partial \xi }+\frac{\partial U}{\partial \eta },\\ u_t& =\frac{\partial U}{\partial \xi }\frac{\partial \xi }{\partial t}+\frac{\partial U}{\partial \eta }\frac{\partial \eta }{\partial t}=c\left(\frac{\partial U}{\partial \xi }-\frac{\partial U}{\partial \eta }\right).\end{array}$$

In Strauss PDEs they do this

$$\frac{\partial }{\partial x}=\frac{\partial }{\partial \xi }+\frac{\partial }{\partial \eta },\frac{\partial }{\partial t}=c\left(\frac{\partial }{\partial \xi }-\frac{\partial }{\partial \eta }\right)$$

We apply this to the operator mentioned above to see that

$$\begin{array}{rl}\frac{\partial }{\partial t}+c\frac{\partial }{\partial x}& =c\left(\frac{\partial }{\partial \xi }-\frac{\partial }{\partial \eta }\right)+c\left(\frac{\partial }{\partial \xi }+\frac{\partial }{\partial \eta }\right)=2c\frac{\partial }{\partial \xi }\\ \frac{\partial }{\partial t}-c\frac{\partial }{\partial x}& =c\left(\frac{\partial }{\partial \xi }-\frac{\partial }{\partial \eta }\right)-c\left(\frac{\partial }{\partial \xi }+\frac{\partial }{\partial \eta }\right)=-2c\frac{\partial }{\partial \eta }.\end{array}$$

Hence

$$u_{tt}-c^2{u}_{xx}=-4c^2{U}_{\xi \eta }=f\circ {\Phi }^{-1}.$$

Our new coordinates change the initial condition $u(x,0)=0$ to be $U(\xi ,\xi )=0$, since at $t=0$ we have $\xi =\eta =x$. (Here $U$ is the desired solution transformed to $(\xi ,\eta )$ coordinates). The initial condition $u_t(x,0)=0$ is transformed to $U_{\xi }(\xi ,\xi )=U_{\eta }(\xi ,\xi )$. So on the diagonal of the $(\xi ,\eta )$ plane, we see that the $U$ derivatives are 0.

We now proceed to solve this. Let $g=-(f\circ {\Phi }^{-1})/4c^2$ (technically $f$ should be $f\circ {\Phi }^{-1}$), so we have $U_{\xi \eta }=g$. Integrate $g$ from $\xi$ to $\eta$ to get

$$U_{\xi }(\xi ,\eta )-U_{\xi }(\xi ,\xi )={\int }_{\xi }^{\eta }g(\xi ,{\eta }^{\prime })\mathrm{d}{\eta }^{\prime }.$$

Since $U_{\xi }(\xi ,\xi )=0$ we have $U_{\xi }(\xi ,\eta )={\int }_{\xi }^{\eta }g(\xi ,{\eta }^{\prime })\mathrm{d}{\eta }^{\prime }=-{\int }_{\eta }^{\xi }g(\xi ,{\eta }^{\prime })\mathrm{d}{\eta }^{\prime }.$Note that the bounds of integration are deliberately chosen because we know $U_{\xi }(\xi ,\xi )=0$. Now integrate again in $\xi$ to get $$ U(\xi_0, \eta_0) - U(\eta_0, \eta_0) = \int_{\eta_0}^{\xi_0} U_\xi(\xi, \eta_0) \mathrm d \xi.

Again the bounds of integration were chosen because $U(\eta_0, \eta_0) = 0$. We combine everything to see that $$ U(\xi_0, \eta_0) = - \int_{\eta_0}^{\xi_0} \int_{\eta_0}^\xi g(\xi, \eta') \mathrm{d} \eta' \mathrm d \xi.

We now analyze the region of integration. First notice that ${\eta }_0\le \xi \le {\xi }_0$ from the outer integral and that ${\eta }_0\le {\eta }^{\prime }\le \xi$ from the inner integral. So we have ${\eta }_0\le {\eta }^{\prime }\le \xi \le {\xi }_0$. The bounds give us a triangle ${\Delta }_{{\xi }_0,{\eta }_0}$ where the points $(\eta ,\xi )$ lie in (visualize in https://www.desmos.com/calculator/la96trwbe9). Thus

$$U({\xi }_0,{\eta }_0)=\frac{1}{4c^2}{\int }_{{\Delta }_{{\xi }_0,{\eta }_0}}(f\circ {\Phi }^{-1})(\eta ,\xi )\mathrm{d}\eta \mathrm{d}\xi .$$

By change of variables

$${\int }_{{\Delta }_{{\xi }_0,{\eta }_0}}(f\circ {\Phi }^{-1})(\eta ,\xi )\mathrm{d}\eta \mathrm{d}\xi ={\int }_{{\Phi }^{-1}[{\Delta }_{{\xi }_0,{\eta }_0}]}f|\det D\Phi |\mathrm{d}x\mathrm{d}t,$$

and the Jacobian determinant is $-2c$ so

$$u(x_0,t_0)=u(\Phi ({\eta }_0,{\xi }_0))=U({\xi }_0,{\eta }_0)=\frac{1}{2c}{\int }_{{\Phi }^{-1}[{\Delta }_{{\xi }_0,{\eta }_0}]}f\mathrm{d}x\mathrm{d}t$$

The region ${\Phi }^{-1}[{\Delta }_{{\xi }_0,{\eta }_0}]$ is the characteristic triangle with vertices $(x_0,t_0),(x_0-c{t}_0,0),(x_0+c{t}_0,0)$. (I’ll work this out eventually but it does make sense)

So overall, the integral is just

$$u(x_0,t_0)=\frac{1}{2c}{\int }_0^{t_0}{\int }_{x_0-c(t_0-t)}^{x_0+c(t_0-t)}f(x,t)\mathrm{d}x\mathrm{d}t.$$

See claude chat Superposition for inhomogeneous wave equations

Solving with Duhamel’s principle

Duhamel’s principle states that for the PDE

$$u_{tt}-c^2{u}_{xx}=f,u(x,0)=0,u_t(x,0)=0,$$

if for every $s\ge 0$ the function $u^{(s)}$ satisfies

$$u_{tt}^{(s)}-c^2{u}_{xx}^{(s)}=0(t>s),u^{(s)}(x,s)=0,u_t^{(s)}(x,s)=f(x,s),$$

then

$$u(x_0,t_0)={\int }_0^{t_0}{u}^{(s)}(x_0,t_0)\mathrm{d}s.$$

Proof: Define

$$G(x,t)={\int }_0^t{u}^{(s)}(x,t)\mathrm{d}s.$$

We will check that $G$ satisfies all the conditions of the original PDE; by the uniqueness of the solution $G=u$. Firstly, $G(x,0)=0$ trivially by plugging $t=0$ for the bounds of integration. Now,

$$G_t(x,0)=\frac{\partial }{\partial t}{\int }_0^t{u}^{(s)}(x,t)\mathrm{d}s.$$

We use Leibniz’s rule to rewrite this as

$$u^{(t)}(x,t)+{\int }_0^t\frac{\partial }{\partial t}{u}^{(s)}(x,t)\mathrm{d}s.$$

(Note: Stefanos never mentioned Leibniz’s rule in the lecture but that’s what he is using). Evaluate $G_t(x,0)=u^{(t)}(x,0)+0=0$, since $u^{(t)}(x,t)=0$ for all $t$. Finally, we directly check $G_{tt}-c^2{G}_{xx}$ which should directly give us the answer.

Now we need to find these functions $u^{(s)}$. The rough idea is that each $u^{(s)}$ is solving a homogeneous wave equation with initial conditions given at $t=s$. So we just need to change the coordinates back to $t=0$.

Fix $s$. Let $\tau =t-s$, set $v(x,\tau )=u^{(s)}(x,s+\tau )$. Then $v$ satisfies the wave equation PDE

$$v_{\tau \tau }-c^2{v}_{xx}=0,v(x,0)=0,v_{\tau }(x,0)=f(x,s).$$

The solution to this is

$$v(x,\tau )=\frac{1}{2c}{\int }_{x-c\tau }^{x+c\tau }f(w,s)\mathrm{d}w.$$

Undo the change of variables to get

$$u^{(s)}(x,t)=\frac{1}{2c}{\int }_{x-c(t-s)}^{x+c(t-s)}f(w,s)\mathrm{d}w.$$

See claude chat Duhamel’s principle for wave equation.