Derivation, vibrating string, vibrating membranes

The wave equation with several space variables is

$$\frac{\partial^2 u}{\partial t^2} - \Delta u = 0,\qquad \Delta u{\stackrel{\textrm{def}}=}\frac{\partial^2 u}{\partial x_1^2} + \cdots + \frac{\partial^2 u}{\partial x_n^2}$$

or more generally

$$\frac{\partial^2 u}{\partial t^2} - c^2\Delta u=0,$$

where $c>0$ is a positive real number called the wave speed.

The unknown is a function $u:\mathbb R\times\mathbb R^n \to\mathbb R$.

The equation appears in physics in the following contexts

• vibrating string ($n=1$, slides for the derivation)
• vibrating membrane ($n=2$)
• propagation of light waves, or of sound waves ($n=3$)
• more …

D’Alembert’s solution ($n=1$)

Changing coordinates.

Suppose that $u(x,t)$ satisfies

$$u_{tt}-u_{xx}=0 \text { for all }(x, t).\tag{1}$$

We introduce new coordinates $(p,q)$ related to our original space-time coordinates $(x, t)$ by

$$p=x+t,\qquad q=x-t$$

If we know the $(p,q)$ coordinates of a point then we can recover the $(x, t)$ coordinates of the point from

$$x=\frac{p+q}{2}, \qquad t = \frac{p-q}{2}.$$

Consider the quantity $u$ expressed as a function of the $(p,q)$ coordinates, i.e. we consider the new function

$$v(p,q) \stackrel{\sf def}{=} u\Bigl(\frac{p+q}{2}, \frac{p-q}{2}\Bigr).$$

and ask what equation it satsifies. The computation is easiest if we first write $u(x, t)$ in terms of $v$, i.e.

$$u(x, t) = v(x+t, x-t).$$

We can substitute this in the wave equation and apply the chain rule several time. We compute the derivatives one by one. The first derivative is

$$\begin{aligned} u_{t} &= \frac{\partial v(\overbrace{x+t}^{=p}, \overbrace{x-t}^{=q})}{\partial t}\\ &=v_p(x+t, x-t) \frac{\partial (x+t)}{\partial t} +v_q(x+t, x-t)\frac{\partial (x-t)}{\partial t}\\ &=v_p(x+t, x-t) - v_q(x+t, x-t) \end{aligned}$$

Differentiate again:

$$\begin{aligned} u_{tt} &= \frac{\partial \{v_p(x+t, x-t) - v_q(x+t, x-t)\}}{\partial t}\\ &=v_{pp}-v_{pq}-v_{qp}+v_{qq}\quad \text{at }(x+t, x-t)\\ &=v_{pp}-2v_{pq}+v_{qq}\quad \text{at }(x+t, x-t) \end{aligned}$$

Similarly, the second $x$ derivative is

$$u_{xx} = v_{pp}+2v_{pq}+v_{qq} \quad \text{at }(x+t, x-t)$$

Therefore

$$u_{tt}-u_{xx}=v_{pp}-2v_{pq}+v_{qq}-\bigl(v_{pp}+2v_{pq}+v_{qq}\bigr) =-4v_{pq}$$

We see that if $u$ is a $C^2$ function, then $u$ satisfies the wave equation if and only if $v$ satisfies the equation

$$\frac{\partial^2 v}{\partial p\partial q} = 0 \text{ for all }(p,q).\tag{2}$$

The point of choosing the coordinates $p, q$ is that the new equation (2) is easier to solve than the original wave equation (1).

Theorem.

(a) If $u:\R^2\to\R$ is a $C^2$ solution of the wave equation $u_{tt}=u_{xx}$ then there exist $C^2$ functions $F, G:\R\to\R$ such that $u(x, t)=F(x+t)+G(x-t)$ for all $(x, t)\in\R^2$.
(b) If $F, G:\R\to\R$ are $C^2$ functions then $u(x, t)=F(x+t)+G(x-t)$ satisfies the wave equation.

Proof of (a). If $u$ satisfies the wave equation then we have shown that $v(p,q)$ defined above satisfies

$$\frac{\partial^2 v}{\partial q\partial p} = \frac{\partial v_p}{\partial q} = 0.$$

Hence $v_p$ does not depend on $q$, i.e. $v_p(p,q)$ is a function of $p$ only, i.e.  there is a function $f:\R\to\R$ such that

$$\frac{\partial v}{\partial p} = f(p)$$

We can now integrate this. Let $F:\R\to\R$ be an antiderivative of $f$. Then there is a function $G:\R\to\R$ such that

$$v(p,q) = \int f(p)\; dp + G(q) = F(p)+G(q)$$

Substitute $p=x+t$, $q=x-t$ and we get

$$u(x,t) = F(x-t)+G(x+t)$$

If we are given the function $u:\R^2\to\R$ then we can find the functions $F$ and $G$ up to a constant from the formulas

$$u(s/2, s/2) = F(s)+G(0), \qquad u(s/2, -s/2) = F(0) + G(s).$$

We find these formulas by setting $x=t=s/2$ (for $F$) or $x=s/2, t=-s/2$ (for $G$) in $u(x, t)=F(x+t)+G(x-t)$.
If $u$ is a $C^2$ function then it follows from $F(s)=u(s/2,s/2)-G(0)$ that

$$F'(s) = \tfrac12 u_x(\tfrac s2, \tfrac s2)+\tfrac12 u_t(\tfrac s2, \tfrac s2),\qquad F''(s) = \tfrac14 u_{xx}(\tfrac s2, \tfrac s2)+ \tfrac12 u_{xt}(\tfrac s2, \tfrac s2)+\tfrac14 u_{tt}(\tfrac s2, \tfrac s2)$$

Therefore $F$ also has two derivatives, and they are continuous.

Proof of (b). If $F, G$ are given functions, and if $u(x, t) = F(x-t)+G(x+t)$, then

$$u_{t}(x, t) = F'(x+t)-G'(x-t), \qquad u_{x}(x, t) = F'(x+t)+G'(x-t)$$

and hence

$$u_{tt}(x, t) = F''(x+t)+G''(x-t), \qquad u_{xx}(x, t) = F''(x+t)+G''(x-t)$$

This implies that $u_{tt}=u_{xx}$, so $u$ satisfies the wave equation. $\hspace{1in}$ Q.E.D.

Theorem — solution in terms of the initial values and velocities. If $u$ is a $C^2$ solution of the wave equation, then

$$u(x,t) = \frac{u(x-t,0)+u(x+t,0)}{2} + \frac12 \int_{x-t}^{x+t}u_t(\xi,0)\mathrm d\xi$$

Conversely, if $U, \dot U:\R\to\R$ are two functions of which $U$ is $C^2$ and $\dot U$ is $C^1$, then

$$u(x,t) = \frac{U(x-t)+U(x+t)}{2} + \frac12 \int_{x-t}^{x+t}\dot U(\xi)\mathrm d\xi$$

is a solution of the wave equation that satisfies

$$u(x, 0)= U(x),\qquad u_t(x, 0)=\dot U(x)\qquad \text{ for all }x\in\R.$$

Proof. If $u$ is a solution then it is of the form $u(x,t)=F(x+t)+G(x-t)$ for certain functions $F$ and $G$. To find these functions we note that

$$u_t(x, t) = F'(x+t)-G'(x-t).$$

Set $t=0$:

$$u(x,0) = F(x)+G(x), \quad u_t(x,0)=F'(x)-G'(x)$$

Differentiate the first equation with respect to $x$,

$$u_x(x,0) = F'(x)+G'(x), \quad u_t(x,0)=F'(x)-G'(x)$$

which we solve for $F'$ and $G'$:

$$F'(x) = \frac{u_x(x,0)+u_t(x,0)}{2},\qquad G'(x) = \frac{u_x(x,0)-u_t(x,0)}{2}$$

The fundamental theorem of calculus implies

$$\begin{aligned} F(x) &= F(0)+\frac12 \int_0^x \bigl\{u_x(\xi,0)+u_t(\xi,0)\bigr\}\,d\xi\\ &=F(0) + \frac12 u(x, 0) - \frac12u(0,0) + \frac12 \int_0^x u_t(\xi, 0)\,d\xi \end{aligned}$$

Similarly,

$$G(x) = G(0) + \frac12 u(x, 0) - \frac12u(0,0) - \frac12 \int_0^x u_t(\xi, 0)\,d\xi$$

It follows that the solution $u(x, t)$ is given by

$$\begin{aligned} u(x, t)&=F(x+t)+G(x-t)\\ &=\frac{u(x+t, 0)+u(x-t, 0)}{2}+\frac12\int_0^{x+t}u_t(\xi, 0)\,d\xi -\frac12\int_0^{x-t}u_t(\xi, 0)\,d\xi \\ &\qquad\qquad+F(0)+G(0)-u(0,0). \end{aligned}$$

We can combine the two integrals:

$$\int_0^{x+t}u_t(\xi, 0)\,d\xi -\int_0^{x-t}u_t(\xi, 0)\,d\xi = \int_0^{x+t}u_t(\xi, 0)\,d\xi +\int^0_{x-t}u_t(\xi, 0)\,d\xi = \int_{x-t}^{x+t}u_t(\xi, 0)\,d\xi.$$

The terms $F(0)+G(0)-u(0,0)$ cancel because

$$u(x,t)=F(x+t)+G(x-t) \implies u(0,0) = F(0)+G(0).$$

Therefore we end up with

$$u(x,t) = \frac{u(x-t,0)+u(x+t,0)}{2} + \frac12 \int_{x-t}^{x+t}u_t(\xi,0)\,d\xi$$

as claimed.

Weak and classical solutions

For many partial differential equations the naive notion of a “solution” is not satisfactory, as there may be functions that do not have enough derivatives to allow verification of the equation, but which one really would like to call solutions, for other reasons. This has led to the introduction of many theories of “generalized solution.” Here we will see one version of such a theory.

Definition. A classical solution of the wave equation is a $C^2$ function $u:\R^2\to\R$ that satisfies the equation, i.e. for which $u_{tt}=u_{xx}$ holds at all $(x, t)\in\R^2$.

A weak solution of the wave equation is a continuous function that satisfies

$$u(x+h, t)+u(x-h,t) = u(x, t+h)+u(x, t-h) \tag{$\star$}$$

for all $x, t\in\R$ and $h>0$.

The new notion of solution has the following features:

• checking if some function $u$ satisfies $(\star)$ does not require one to differentiate $u$
• Classical solutions, which we originally thought of as solutions, should still be solutions
• Weak solutions should have some of the same properties as classical solutions

Theorem. If $F, G:\R\to\R$ are continuous functions, then $u(x, t)=F(x+t)+G(x-t)$ is a weak solution to the Wave Equation.

Proof. For any $x, t\in\R$ and $h>0$ we have

$$\begin{aligned} u(x+h, t)+u(x-h, t) &= F(x+h+t) + G(x+h-t) + F(x-h+t) + G(x-h-t)\\ u(x, t+h)+u(x, t-h) &= F(x+t+h) + G(x-t-h) + F(x+t-h) + G(x-t+h). \end{aligned}$$

Therefore $(\star)$ holds. $\qquad$ Q.E.D

Theorem. Any classical solution of the wave equation is also a weak solution

Proof. Since $u$ is a $C^2$ solution of the wave equation there exist functions $F, G$ such that $u(x, t)=F(x+t)+G(x-t)$ for all $x, t$. The previous theorem then implies that $u$ is a weak solution$\qquad$Q.E.D

Theorem. If $u$ is a weak solution, and if $u$ is $C^2$, then $u$ is also a classical solution of the wave equation.

Proof. Suppose $u$ is a $C^2$ function satisfying $(\star)$. Then we differentiate $(\star)$ on both sides twice with respect to $h$ and set $h=0$

$$\begin{aligned} &&u(x+h, t)+u(x-h,t) &= u(x, t+h)+u(x, t-h) \\ \implies&&u_x(x+h, t)-u_x(x-h,t) &= u_t(x, t+h)-u_t(x, t-h) \\ \implies&&u_{xx}(x+h, t)+u_{xx}(x-h,t) &= u_{tt}(x, t+h)+u_{tt}(x, t-h) \\ \implies&&2u_{xx}(x, t) &= 2u_{tt}(x, t) \end{aligned}$$

Therefore $u$ satisfies the wave equation.$\qquad$Q.E.D

Theorem. If $u:\R^2\to\R$ is a weak solution of the Wave Equation, then there exist continuous function $F, G:\R\to\R$ such that

$$u(x,t)=F(x+t)+G(x-t)$$

for all $x, t$.

Proof. Define

$$F(x) = u\left(\frac x2, \frac x2\right), \qquad G(x) = u\left(\frac x2, -\frac x2\right) - u(0,0).$$

and consider the function

$$v(x, t) \stackrel{\sf def}= F(x+t)+G(x-t).$$

We will show that $u(x, t)=v(x, t)$ for all $x, t$. To do this we begin by verifying

• $u(t, t)=v(t, t)$ for all $t\in\R$
• $v$ is a continuous function
• $v$ also satisfies the condition $(\star)$

Let $h>0$ be given. Then by repeatedly using the condition $(\star)$ we conclude that

$$u((m-n)h, (m+n)h) = v((m-n)h, (m+n)h) \text{ for all }m, n\in\Z$$

If $(x, t)\in\R^2$ is a given point, then we can find sequences $m_k, n_k\in \Z$ such that $(m_k-n_k) 2^{-k}\to x$ and $(m_k+n_k)2^{-k}\to t$ as $k\to\infty$. Since $u$ and $v$ are continuous functions we then have

$$u(x, t) = \lim_{k\to\infty} u\left(\frac{m_k-n_k}{2^k}, \frac{m_k+n_k}{2^k}\right) =\lim_{k\to\infty} v\left(\frac{m_k-n_k}{2^k}, \frac{m_k+n_k}{2^k}\right) = v(x, t).$$

Q.E.D.

Problems

Suppose that $u(x,t)$ is a solution of

$$u_{tt} - 2u_{xt} - 15u_{xx} = 0 \tag{\dag}$$

For certain values of $c\in\mathbb R$ the function $u(x,t) = F(x-ct)$ is a solution of $(\dag)$ for any $F:\R\to\R$ that is $C^2$. Find all $c\in \mathbb R$ with this property.

Consider the PDE

$$u_{tt} - u_{xt} - 2u_{xx} = 0 \tag{\ddag}$$

and consider the coordinate transformation

$$\begin{aligned} r &= x-t \\ s&=x+2t \end{aligned} \iff \begin{aligned} x&=\tfrac13(2r+s)\\[2pt] \quad t&=\tfrac13(-r+s) \end{aligned}$$

1. Which differential equation does the function $v(r, s) = u\bigl(\frac{2r+s}{3}, \frac{-r+s}{3}\bigr)$ satisfy? (The computation is easier if you begin with $u(x, t) = v(x-t, x+2t)$ and substitute that in the equation for $u$).
2. Find the general solution for $(\ddag)$.

Differentiability of d’Alembert’s solution.

Suppose $f:\R\to\R$ is $C^2$ and $g:\R\to\R$ is $C^1$, and consider

$$u(x, t) = \frac{f(x+t)+f(x-t)}{2} + \frac12 \int_{x-t}^{x+t} g(\xi)\, d\xi.$$

1. Find $C^2$ functions $F. G:\R\to\R$ such that $u(x, t)=F(x+t)+G(x-t)$.
2. Show that $u$ is a $C^2$ function. (This function appears in our formulation of d’Alembert’s solution; we never checked that $u$ is twice differentiable, and this problem asks you to do that.)

The initial value problem for weak solutions.

Suppose $f:\R\to\R$ is $C^1$ and $g:\R\to\R$ is continuous, and again consider

$$u(x, t) = \frac{f(x+t)+f(x-t)}{2} + \frac12 \int_{x-t}^{x+t} g(\xi)\, d\xi.$$

1. Find $C^1$ functions $F. G:\R\to\R$ such that $u(x, t)=F(x+t)+G(x-t)$.
2. Show that $u(x, 0)=f(x)$ and $u_t(x, 0)=g(x)$.
3. Show that for every $C^1$ function $f:\R\to\R$ and every continuous $g:\R\to\R$ there is a weak solution to the wave equation with $u(x, 0)=f(x)$ and $u_t(x, 0)=g(x)$ for all $x\in\R$.