5 Wave Equation

Chapter 5
Wave Equation

The wave equation describes phenomena which propagate with finite speed through space time. The example of sound and electrodynamic (light) waves motivated the investigation of this equation in $n = 3$ , though it is also a useful model of vibrating strings and drums in $n = 1$ and $n = 2$ respectively. Later these methods were generalised to non-linear hyperbolic equations in order to describe gravitational waves.

In this final chapter we consider the homogeneous and inhomogeneous wave equation on open subsets of $ℝ^{n} \times ℝ$ for $n \leq 3$ . In particular we study the Cauchy problem for $t > 0$

\begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}} - △ u & = f & on & (x, t) \in ℝ^{n} \times ℝ^{+} & with \\ u (x, 0) & = g (x) & and & \frac{∂𝑢}{∂𝑡} (x, 0) = h (x) . \end{array}

The wave equation is a linear second order PDE. The coefficient matrix for the second derivatives has one positive and $n$ negative eigenvalues and is neither definite nor semi-definite. In the second chapter we introduced this differential equation as the simplest hyperbolic differential equation. The general theory of hyperbolic equations is distinctly different to that of elliptic and parabolic equations.

We see for the Cauchy problem that we have given not only the value of $u$ on the initial boundary but also its normal derivative. The intuition is that if you choose a point $(x_{0}, 0)$ then $△ u (x_{0}, 0) = △ g (x_{0})$ . Thus $\partial_{t}^{2} u (x_{0}, 0)$ can be determined from the PDE but not $\partial_{t} u (x_{0}, 0)$ . The simple example of the linear functions $u (x, t) = 𝑎𝑡 + b$ show that these two values are indeed independent. Conversely, for smooth functions $f, g, h$ these initial conditions are sufficient to determine all derivatives on $u$ at $(x_{0}, 0)$ . For example

\begin{array}{l} \partial_{t}^{3} u (x, 0) & = △ \partial_{t} u (x, 0) + \partial_{t}^{3} f (x, 0) = △ h (x) + \partial_{t}^{3} f (x, 0), \\ \partial_{t}^{4} u (x, 0) & = △ \partial_{t}^{2} u (x, 0) + \partial_{t}^{4} f (x, 0) = △^{2} g (x) + △ f (x, 0) + \partial_{t}^{4} f (x, 0) . \end{array}

This discussion may remind you of Definition 1.6 of characteristic and non-characteristic curves. Let’s make a brief detour to see how the method of characteristics can be generalised to the wave equation for $n = 1$ . Consider a path $(x (s), t (s))$ in the domain. Let us consider how the three functions $u$ , $v = \partial_{t} u$ , $w = \partial_{x} u$ behave along such a curve. We use a dot for derivative with respect to $s$ . By the chain rule $\dot{u} = 𝑣ṫ + 𝑤ẋ$ . The derivative for $v$ and $w$ are similar to one another.

\dot{v} = \partial_{t} 𝑣ṫ + \partial_{x} 𝑣ẋ, ẇ = \partial_{t} 𝑤ṫ + \partial_{x} 𝑤ẋ .

We need to relate these in such a way that we remove the direct dependence on $x$ and $t$ . The equality of partial derivatives implies $\partial_{x} v = \partial_{t} w$ and from the wave equation we have $\partial_{t} v - \partial_{x} w = 0$ . Substitution shows us that

\dot{v} = \partial_{t} 𝑣ṫ + \partial_{x} 𝑣ẋ, ẇ = \partial_{x} 𝑣ṫ + \partial_{t} 𝑣ẋ .

So we can equate these two expressions if $ẋ = ṫ = 1$ or $ẋ = - ṫ = - 1$ . Thus there are two characteristics through every point. Unlike for crossing characteristics in first order systems, this is not necessarily a problem. On the characteristic $x - t = c$ we have the system of ODEs

\dot{u} = v + w, \dot{v} - ẇ = 0 .

And on the characteristic $x + t = c$ we have

\dot{ũ} = ṽ - \tilde{w}, \dot{ṽ} + \dot{\tilde{w}} = 0 .

The tildes indicate that these functions are on different curves We see that both systems are underdetermined (three unknowns, two equations) so there is the possibility that they can be made to agree everywhere. The method of characteristics for higher order PDEs leads to the celebrated theorem of Cauchy and Kowalevski (also spelt Kovalevskaya) on the existence of PDEs with analytic coefficients. We do not pursue this line of inquiry further, nor shall we use Fourier analysis to solve the wave equation, though both methods work well. Instead we will use a classical method that links back to the first chapter. Hopefully the above digression has provided some deeper insight as to why the classical method works.

5.1 D’Alembert’s Formula

First we solve the Cauchy problem in one dimension (of space). We may factorise the wave operator (also called D’Alembert’s operator)

\frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}} = (\frac{\partial}{∂𝑡} + \frac{\partial}{∂𝑥}) (\frac{\partial}{∂𝑡} - \frac{\partial}{∂𝑥}) = (\frac{\partial}{∂𝑡} - \frac{\partial}{∂𝑥}) (\frac{\partial}{∂𝑡} + \frac{\partial}{∂𝑥}) .

If $u$ solves the homogeneous wave equation, then $v (x, t) = (\frac{\partial}{∂𝑡} - \frac{\partial}{∂𝑥}) u (x, t)$ solves $\frac{∂𝑣}{∂𝑡} + \frac{∂𝑣}{∂𝑥} = 0$ . This is the transport equation with constant coefficient with the unique solution

v (x, t) = a (x - t) with a (x) = v (x, 0) .

So the solution $u (x, t)$ of the wave equation solves the first order linear PDE

\frac{∂𝑢}{∂𝑡} - \frac{∂𝑢}{∂𝑥} = a (x - t) .

This is an inhomogeneous transport equation with constant coefficients with the solution

\begin{array}{l} u (x, t) & = b (x + t) + \int_{0}^{t} a (x + (t - s) - s) d s = b (x + t) + \frac{1}{2} \int_{x - t}^{x + t} a (y) 𝑑𝑦 \end{array}

with $b (x) = u (x, 0)$ . The initial values $u (x, 0) = g (x)$ and $\frac{∂𝑢}{∂𝑡} (x, 0) = h (x)$ yields

\begin{array}{l} b (x) & = g (x) and & a (x) & = v (x, 0) = \frac{∂𝑢}{∂𝑡} (x, 0) - \frac{∂𝑢}{∂𝑥} (x, 0) = h (x) - g^{'} (x) . \end{array}

If we insert this in our solutions, then we obtain

u (x, t) = g (x + t) + \frac{1}{2} \int_{x - t}^{x + t} (h (y) - g^{'} (y)) 𝑑𝑦

Hence the solution of the initial value problem of the wave equation is given by

u (x, t) = \frac{1}{2} (g (x + t) + g (x - t)) + \frac{1}{2} \int_{x - t}^{x + t} h (y) 𝑑𝑦 .

Moreover, this must be the unique solution, since the transport equation has a unique solution. In summary

Theorem 5.1 (D’Alembert’s Formula). If $g : ℝ \to ℝ$ is twice continuously differentiable and $h : ℝ \to ℝ$ continuously differentiable, then

u (x, t) = \frac{1}{2} (g (x + t) + g (x - t)) + \frac{1}{2} \int_{x - t}^{x + t} h (y) 𝑑𝑦

is a twice continuously differentiable function on $ℝ \times ℝ_{0}^{+}$ that is the unique solution of the Cauchy problem of the homogeneous wave equation.

First an observation on the regularity. If solution is $k$ -times differentiable, if $g$ and $H$ are $k$ times differentiable, or equivalently if $g$ is $k$ times differentiable and $h$ is $(k - 1)$ times differentiable. So the regularity of the solution does not improve with time, as it does for solutions of the heat equation.

We interpret the fact that the value of the solution at $(x, t)$ depends only on the values of $g$ at $x \pm t$ and the values of $h$ at points in the interval $[x - t, x + t]$ as a bound of $1$ on the speed of propagation, since the trajectories from these points to $(x, t)$ propagate with speed not larger than $1$ . A stronger statement is possible. Using an antiderivative of $h$ then we can write

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

Conversely, every function of this form is a solution of the wave equation if $F$ and $G$ are twice differentiable (Exercise). Hence the value $u (x, t)$ of the solution at $(x, t)$ depends only on the values of $F$ and $G$ at $x \pm t$ and the propagation speed is exactly $1$ . We call this the decomposition into forward and backward travelling waves.

5.2 Solution on the half-line

While we are mainly interested in the Cauchy problem on $ℝ^{n}$ , in one dimension it is straightforward to reflect and derive the solution on the half-line. We will need this solution later in the chapter. Stated precisely, we solve the following problem.

\begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{2} u}{\partial x^{2}} & = 0 for (x, t) \in ℝ^{+} \times ℝ^{+}, & u (0, t) & = 0 & for & t \in ℝ_{0}^{+}, \\ u (x, 0) & = g (x) and & \frac{∂𝑢}{∂𝑡} (x, 0) & = h (x) & for & x \in ℝ^{+} . \end{array}

The trick is to extend the functions $g$ and $h$ to odd functions on the whole space $ℝ \times ℝ_{0}^{+}$ by a reflection:

\begin{array}{l} \tilde{g} (x) & = {\begin{matrix} g (x) & for x \geq 0, \\ - g (- x) & for x \leq 0, \end{matrix} & \tilde{h} (x) & = {\begin{matrix} h (x) & for x \geq 0, \\ - h (- x) & for x \leq 0 . \end{matrix} \end{array}

For any solution $ũ$ of the initial value problem

\begin{array}{l} \frac{\partial^{2} ũ}{\partial t^{2}} - \frac{\partial^{2} ũ}{\partial x^{2}} & = 0 & for & (x, t) \in ℝ \times ℝ^{+}, \\ ũ (x, 0) & = \tilde{g} (x) and & \frac{∂ũ}{∂𝑡} (x, 0) & = \tilde{h} (x) & for & x \in ℝ, \end{array}

the function $(x, t) \mapsto - ũ (- x, t)$ is also solution. Due to the uniqueness of the solution both solutions coincide: $ũ (- x, t) = - ũ (x, t)$ . By this argument we conclude that $ũ$ is an odd function and $ũ (0, t) = 0$ . Hence this solution restricts to give a solution of the half-line problem.

Conversely, if we take a solution to the half line problem, one can check that its reflected extension solves the Cauchy problem on $ℝ^{n}$ . The important point is the check the first and second derivatives of the reflection exist at $x = 0$ , but this is guaranteed by the fact that $u$ vanishes there. Therefore there is a bijection between solutions of the two problems and in particular the solution on the half-line is unique.

Explicitly the solution on the half-line is given by

u (x, t) = {\begin{matrix} \frac{1}{2} (g (x + t) + g (x - t) + \int_{x - t}^{x + t} h (y) 𝑑𝑦) & for 0 \leq t \leq x \\ \frac{1}{2} (g (t + x) - g (t - x) + \int_{t - x}^{t + x} h (y) 𝑑𝑦) & for 0 \leq x \leq t . \end{matrix}

Note that the waves propagating towards the boundary at $x = 0$ are reflected at the boundary and propagate back.

5.3 Spherical Means of the Wave Equation

When we studied the Laplace equation, we saw that the spherical means of its solutions (harmonic functions) did not depend on the radius of the sphere. The spherical means of solutions of the wave equation do depend on the radius of the sphere, but in a controlled way. In fact they obey a PDE! This PDE is similar to the one-dimensional wave equation. This opens an avenue to solve the initial value problem of the wave equation in any odd dimension, though for this course we will stick to $n = 3$ . We define for all $x \in ℝ^{n}, t \geq 0, r > 0$ the spatial-spherical mean

𝒮 [u] (x, r, t) : = \frac{1}{n ω_{n} r^{n - 1}} \int_{∂𝐵 (x, r)} u (y, t) d σ (y) .

Here $t$ is treated as an additional parameter and not integrated. With this understanding we reuse the same notation for the spherical means. For brevity we define $U (x, r, t) = 𝒮 [u] (x, r, t)$ , $G (x, r) = 𝒮 [g] (x, r)$ , and $H (x, r) = 𝒮 [h] (x, r)$ .

Lemma 5.2. If $u \in C^{m} (ℝ^{n} \times ℝ_{0}^{+})$ is a $m$ -times continuously differentiable solution of the initial value problem (with continuous partial derivatives of order $\leq m$ on $ℝ^{n} \times ℝ_{0}^{+}$ ) of the Cauchy problem of the homogeneous wave equation. The spherical mean $U (x, r, t)$ for fixed $x \in ℝ^{n}$ is an $m$ -times differentiable function on $(r, t) \in ℝ^{+} \times ℝ^{+}$ , which solves the following initial value problem of the Euler-Poisson-Darboux Equation (with continuous partial derivatives of order $\leq m$ ):

\frac{\partial^{2} U}{\partial t^{2}} (x, r, t) - \frac{\partial^{2} U}{\partial r^{2}} (x, r, t) - \frac{n - 1}{r} \frac{∂𝑈}{∂𝑟} (x, r, t) = 0 on (r, t) \in ℝ^{+} \times ℝ^{+}

U (x, r, 0) = G (x, r) and \frac{∂𝑈}{∂𝑡} (x, r, 0) = H (x, r)

Proof. By a substitution the domain of the integral becomes independent of $t$ and $r$ :

U (x, r, t) = \frac{1}{n ω_{n}} \int_{∂𝐵 (0, 1)} u (𝑟𝑦 + x, t) d σ (y) .

Hence we may calculate the derivative

\begin{array}{l} \frac{∂𝑈}{∂𝑟} (x, r, t) & = \frac{1}{n ω_{n}} \int_{∂𝐵 (0, 1)} \nabla u (x + 𝑟𝑦, t) \cdot y d σ (y) \\ = \frac{r}{n ω_{n}} \int_{B (0, 1)} △ u (x + 𝑟𝑦, t) dn y = \frac{r}{n ω_{n} r^{n}} \int_{B (x, r)} △ u (y, t) dn y . \end{array}

In the limit at $r ↓ 0$ , we can recognise the last expression as $\frac{r}{n}$ multiplied with the ball mean of $△ u$ . The mean has a limit $△ u (x, r)$ , which implies $\lim_{r \to 0} \frac{∂𝑈}{∂𝑟} (x, r, t) = 0$ . Differentiating further we get

\begin{array}{l} \frac{\partial^{2} U}{\partial r^{2}} (x, r, t) & = \frac{\partial}{∂𝑟} (\frac{1}{n ω_{n} r^{n - 1}} \int_{0}^{r} \int_{∂𝐵 (x, s)} △ u (y, t) dn y d s) \\ = \frac{1 - n}{n ω_{n} r^{n}} \int_{B (x, r)} △ u (y, t) dn y + \frac{1}{n ω_{n} r^{n - 1}} \int_{∂𝐵 (x, r)} △ u (y, t) d σ (y) \\ = \frac{1 - n}{r} \frac{∂𝑈}{∂𝑟} (x, r, t) + 𝒮 [△ u] (x, r, t) . \end{array}

Finally, we use the wave equation to change this last term.

𝒮 [△ u] (x, r, t) = 𝒮 [\partial_{t}^{2} u] (x, r, t) = \partial_{t}^{2} 𝒮 [u] (x, r, t) = \frac{\partial^{2} U}{\partial t^{2}} (x, r, t) .

□

5.4 Solution in Dimension 3

We shall see that for odd dimensions the spherical means of solutions of the wave equation can be transformed into solutions of the one-dimensional wave equation, but not for even dimensions. For this reason we shall next solve the initial value problem of the wave equation in three dimensions. In this section we consider for any $x \in ℝ^{3}$ the following initial value problem for the spherical means of a solution of the wave equation:

\frac{\partial^{2} U}{\partial t^{2}} - \frac{\partial^{2} U}{\partial r^{2}} - \frac{2}{r} \frac{∂𝑈}{∂𝑟} = 0 on (x, r, t) \in {x} \times ℝ^{+} \times ℝ^{+}

U = G and \frac{∂𝑈}{∂𝑡} = H on (x, r, t) \in {x} \times ℝ^{+} \times {0} .

The substitution $Ũ = 𝑟𝑈$ transforms the above into the following:

\begin{array}{l} \frac{\partial^{2} Ũ}{\partial t^{2}} - \frac{\partial^{2} Ũ}{\partial r^{2}} & = 0 on (x, r, t) \in {x} \times ℝ^{+} \times ℝ^{+}, & Ũ (x, 0, t) & = 0 for t \in ℝ_{0}^{+}, \\ Ũ (x, r, 0) & = \tilde{G} (x, r) = 𝑟𝐺 (x, r) and \frac{∂Ũ}{∂𝑡} (x, r, 0) = & \tilde{H} (x, r) & = 𝑟𝐻 (x, r) for r \in ℝ^{+} . \end{array}

We solved this initial value problem in the Section 5.2. The solution is

Ũ (x, r, t) = \frac{1}{2} (\tilde{G} (x, r + t) - \tilde{G} (x, t - r)) + \frac{1}{2} \int_{- r + t}^{r + t} \tilde{H} (x, s) d s for 0 \leq r \leq t .

But this isn’t what we wanted. We wanted the solve the wave equation. Thus we must undo all the transforms and recover $u$ . The continuity of $u (x, t)$ implies

u (x, t) = \lim_{𝑟↓ 0} U (x, r, t) = \lim_{𝑟↓ 0} \frac{Ũ (x, r, t)}{r} .

We compute this limit for each part of the formula of $Ũ$ .

\begin{array}{l} \lim_{𝑟↓ 0} & \frac{1}{2 r} (\tilde{G} (x, r + t) - \tilde{G} (x, t - r)) = \lim_{𝑟↓ 0} \frac{1}{2} (\frac{\tilde{G} (x, t + r) - \tilde{G} (x, t)}{r} + \frac{\tilde{G} (x, t - r) - \tilde{G} (x, t)}{- r}) \\ = \frac{\partial \tilde{G} (x, t)}{∂𝑡} = \frac{\partial}{∂𝑡} (t 𝒮 [g] (x, t)) = 𝒮 [g] (x, t) + \frac{t}{4 π t^{2}} \int_{∂𝐵 (x, t)} \nabla g (y) \cdot N d σ (y) \end{array}

using Equation (3.4), and

\begin{array}{l} \lim_{𝑟↓ 0} & \frac{1}{2 r} \int_{- r + t}^{r + t} \tilde{H} (x, s) d s = \tilde{H} (x, t) = t 𝒮 [h] (x, t) \end{array}

Therefore we obtain for all $x \in ℝ^{3}, t > 0$

u (x, t) = \frac{1}{4 π t^{2}} \int_{∂𝐵 (x, t)} (𝑡h (y) + g (y)) d σ (y) + \frac{1}{4 π t^{2}} \int_{∂𝐵 (x, t)} \nabla_{y} g (y) \cdot (y - x) d σ (y)

using the fact that $𝑡𝑁 (y) = y - x$ for points $y \in ∂𝐵 (x, t)$ . The is Kirchhoff’s Formula for the solution of the initial value problem of the three dimensional wave equation. We see, like the one dimensional wave equation, that for the three dimensional wave equation the value at $(x, t)$ only depends on the values $(y, 0)$ for $y \in ∂𝐵 (x, t)$ . We again stylise this fact to mean that all waves travel at speed $1$ .

5.5 Solution in Dimension 2

In two dimensions the Euler-Poisson-Darboux equations cannot be transformed into the one-dimensional wave equation. We present another method, the method of descent, and transform the initial value problem of the two-dimensional wave equation into a special type of initial value problem of the three-dimensional wave equation: We choose initial values which depend only on the coordinates $x_{1}$ and $x_{2}$ and not on the coordinate $x_{3}$ . If $g, h$ are the initial values of the 2-dimensional problem, let

ḡ (x_{1}, x_{2}, x_{3}) = g (x_{1}, x_{2}), \bar{h} (x_{1}, x_{2}, x_{3}) = h (x_{1}, x_{2}) .

By the previous section, we know how to calculate the solution $ū (x, t)$ on $(x, t) \in ℝ^{3} \times ℝ^{+}$ of the initial value problem

\begin{array}{l} \frac{\partial^{2} ū (x, t)}{\partial t^{2}} - △ ū (x, t) & = 0 & for & (x, t) \in ℝ^{3} \times ℝ^{+} \\ ū (x, 0) & = ḡ (x) and & \frac{∂ū}{∂𝑡} (x, 0) & = \bar{h} (x) & for & x \in ℝ^{3} . \end{array}

We observe that if a function $f$ does not depend $x_{3}$ then the mean of that function over $∂𝐵 (x, r)$ also does not depend on $x_{3}$ :

\frac{\partial}{\partial x_{3}} 𝒮 [f] (x, r) = \frac{\partial}{\partial x_{3}} \frac{1}{n ω_{n} r^{n - 1}} \int_{∂𝐵 (0, r)} f (x + y) d σ (y) = \frac{1}{n ω_{n} r^{n - 1}} \int_{∂𝐵 (0, r)} 0 d σ (y) = 0 .

The solution $ū$ is given by Kirchhoff’s formula. The second expression in that formula is not quite a spherical mean, because the integrand also depends on $x$ . We need to check it directly

\begin{array}{l} \frac{\partial}{\partial x_{3}} \int_{∂𝐵 (x, t)} \nabla_{y} g (y) \cdot (y - x) d σ (y) & = \frac{\partial}{\partial x_{3}} \int_{∂𝐵 (0, t)} \nabla_{y} g (x + y) \cdot y d σ (y) = 0 . \end{array}

Together this shows that $ū$ does not depend on $x_{3}$ . If we define $u (x_{1}, x_{2}, t) = ū (x_{1}, x_{2}, 0, t)$ then

(\partial_{t}^{2} - △_{ℝ^{2}}) u = (\partial_{t}^{2} - △_{ℝ^{2}}) ū - \frac{\partial^{2} ū}{\partial x_{3}^{2}} = (\partial_{t}^{2} - △_{ℝ^{3}}) ū = 0 .

Hence we have found a solution to the two dimensional wave equation. The initial conditions are clear. The choice of $x_{3} = 0$ is not important; $ū$ is constant in $x_{3}$ so any other choice gives the same function.

Let’s try to use Kirchhoff’s formula but remove any mention of $x_{3}$ . We use the notation $\bar{x} = (x_{1}, x_{2})$ when $x = (x_{1}, x_{2}, x_{3}) \in ℝ^{3}$ . We need to integrate over spheres. The height function $γ (z) = \sqrt{r^{2} - | z - \bar{x} |^{2}}$ on the two-dimensional ball $z \in B (\bar{x}, r)$ yields by the formula $Ψ (z) = (z, \pm γ (z))$ a parametrisations of both hemispheres of the boundary of the three-dimensional ball $B ((\bar{x}, 0), r)$ by the two-dimensional ball $B (\bar{x}, r)$ . The two hemispheres do not cover $∂𝐵 ((\bar{x}, 0), r)$ completely, but the missing equator is one-dimensional and has measure zero with respect to $d σ (y)$ . We have already made some calculations for parametrisations that are graphs after Lemma 2.8, and using those formulas here gives

\sqrt{\det {(Ψ^{'} (z))}^{T} Ψ (z)} = \sqrt{1 + {(\nabla γ (z))}^{2}} = \frac{r}{\sqrt{r^{2} - | z - \bar{x} |^{2}}} .

By the definition of integration over a submanifold:

\begin{matrix} \int_{∂𝐵 ((\bar{x}, 0), r)} ḡ (y) d σ (y) = 2 \int_{B (\bar{x}, r)} g (z) \sqrt{1 + {(\nabla γ (z))}^{2}} d^{2} z = 2 r \int_{B (\bar{x}, r)} \frac{g (z)}{\sqrt{r^{2} - | z - \bar{x} |^{2}}} d^{2} z . \end{matrix}

This gives finally the following formula for $u (\bar{x}, t)$ on $(\bar{x}, t) \in ℝ^{2} \times ℝ^{+}$ :

\begin{array}{l} u (\bar{x}, t) & = \frac{1}{4 π t^{2}} \int_{∂𝐵 ((\bar{x}, 0), r)} (t \bar{h} (y) + ḡ (y) + \nabla_{y} ḡ (y) \cdot (y - x)) d σ (y) \\ = \frac{1}{2 𝜋𝑡} \int_{B (\bar{x}, t)} \frac{𝑡h (z) + g (z) + \nabla g (z) \cdot (z - \bar{x})}{\sqrt{t^{2} - | z - \bar{x} |^{2}}} d^{2} z . \end{array}

This formula also carries the name Poisson’s formula. It shows that in two dimensions the propagation speed is bounded by $1$ .

This method of deriving the solution of the initial value problem in a lower dimension by transforming the initial value problem into an initial value problem in the higher dimensional space, is called the method of descent. Here the initial values do not depend on some of the coordinates of the higher dimensional space. Ponder this: can we obtain the solution of the one-dimensional wave equation by this method of descent from Poisson’s formula?

5.6 Inhomogeneous Wave Equation

We have seen in exercises how Duhamel’s principle can use the solution of the initial value problem of a homogeneous time-evolution equation to solve the inhomogeneous equation. It also applies to the wave equation, after we put it into the appropriate form: a first order linear ODE on the function space consisting of pairs of functions on $x \in ℝ^{n}$ :

\frac{d}{𝑑𝑡} (\begin{matrix} u (\cdot, t) \\ \frac{∂𝑢}{∂𝑡} (\cdot, t) \end{matrix}) = (\begin{matrix} 0 & 1 \\ △ & 0 \end{matrix}) (\begin{matrix} u (\cdot, t) \\ \frac{∂𝑢}{∂𝑡} (\cdot, t) \end{matrix}) + (\begin{matrix} 0 \\ f (\cdot, t) \end{matrix}) .

In accordance with the principle we try calculate the special solution of the inhomogeneous wave equation

\begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}} - △ u & = f & for & (x, t) \in ℝ^{n} \times ℝ^{+} \\ u (x, 0) & = 0 and & \frac{∂𝑢}{∂𝑡} (x, 0) & = 0 & for & x \in ℝ^{n} \end{array}

via the family of solutions of the homogeneous wave equation whose initial values is given by the inhomogeneity. Suppose $u (x, t, s)$ solves

\begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}} - △ u & = 0 & for & (x, t) \in ℝ^{n} \times (s, \infty) \\ u (x, s, s) & = 0 and & \frac{∂𝑢}{∂𝑡} (x, s, s) & = f (x, s) & for & x \in ℝ^{n}, \end{array}

for any $s \in ℝ^{+}$ , then $u (x, t) = \int_{0}^{t} u (x, t, s) d s$ solves the former inhomogeneous wave equation since

\begin{matrix} \frac{\partial^{2} u}{\partial t^{2}} (x, t) = \frac{\partial}{∂𝑡} (u (x, t, t) + \int_{0}^{t} \frac{∂𝑢}{∂𝑡} (x, t, s) d s) = \frac{\partial}{∂𝑡} \int_{0}^{t} \frac{∂𝑢}{∂𝑡} (x, t, s) d s = \\ = \frac{∂𝑢}{∂𝑡} (x, t, t) + \int_{0}^{t} \frac{\partial^{2} u}{\partial t^{2}} (x, t, s) d s = f (x, t) + \int_{0}^{t} △ u (x, t, s) d s = f (x, t) + △ u (x, t) . \end{matrix}

Consequently the initial value problem of the inhomogeneous wave equation

\begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}} - △ u & = f & for & (x, t) \in ℝ^{n} \times ℝ^{+} \\ u (x, 0) & = g (x) and & \frac{∂𝑢}{∂𝑡} (x, 0) & = h (x) & for & x \in ℝ^{n} \end{array}

is the sum of the former special solution with trivial initial value and the solution of the corresponding homogeneous initial value problem.

Finally we investigate how the present determines the past. The wave equations is invariant with respect to time translation and reversal $t \mapsto T - t$ . However, this transformation replaces $\frac{∂𝑢}{∂𝑡}$ by $- \frac{∂𝑢}{∂𝑡}$ . Therefore the values $u (x, t)$ of the solution of the final value problem

\begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}} - △ u & = f & for & (x, t) \in ℝ^{n} \times ℝ^{-} \\ u (x, T) & = g (x) and & \frac{∂𝑢}{∂𝑡} (x, T) & = h (x) & for & x \in ℝ^{n} \end{array}

are given by the values $u (x, T - t)$ of the solution of the initial value problem with initial values $g$ and $- h$ and inhomogeneity $(x, t) \mapsto f (x, T - t)$ . This means that we can derive both the future and the past from the present. Both solutions fit together and form a solution $u (x, t)$ of the wave equation on $(x, t) \in ℝ^{n} \times ℝ$ which is completely determined by its values $u (x, 0)$ and $\frac{∂𝑢}{∂𝑡} (x, 0)$ on $x \in ℝ^{n}$ .

5.7 Energy Methods

Unlike elliptic and parabolic PDES, hyperbolic PDEs do not satisfy a maximum principle. The key idea of the maximum principle was the connection between the differential operator and the Hessian, to control where extrema can occur. The case that the Hessian is never definite is exactly the elliptic PDEs and their limiting cases as the parabolic PDEs (Theorems 3.13 and 4.11). Our calculations above prove that solutions of the Cauchy problem are unique, by virtue of reducing the problem to a transport equation. But is there a similar general principle that we can call upon to prove uniqueness directly? The class of techniques we are about to see go by the name “energy methods” due to their inspiration from physics.

Theorem 5.3 (Uniqueness of the solutions of the wave equation). Let $Ω \subset ℝ^{n}$ be a bounded domain. Then the following initial values problem of the wave equation

\begin{array}{l} \frac{\partial^{2} u}{\partial t^{2}} - △ u & = f & on & Ω \times (0, T) \\ u (x, t) & = g (x, t) & on & Ω \times {t = 0} & and on & \partial Ω \times (0, T) \\ \frac{∂𝑢}{∂𝑡} (x, 0) & = h (x) & on & Ω \times {t = 0} \end{array}

has a unique solution in $C^{2} (Ω \times (0, T))$ with continuous extensions of $\partial^{α} u$ to $\bar{Ω} \times [0, T]$ for $| α | \leq 2$ .

Proof. The difference of two solutions solves the analogous homogeneous initial value problem with $f = g = h = 0$ . For such a solution we define the energy as

e (t) = \frac{1}{2} \int_{Ω} ({(\frac{∂𝑢}{∂𝑡} (x, t))}^{2} + {| \nabla u (x, t) |}^{2}) dn x .

Then we calculate

\begin{array}{l} \frac{𝑑𝑒}{𝑑𝑡} (t) & = \int_{Ω} (\frac{\partial^{2} u}{\partial t^{2}} (x, t) \frac{∂𝑢}{∂𝑡} (x, t) + \frac{\partial \nabla u}{∂𝑡} u (x, t) \nabla u (x, t)) dn x \\ = \int_{Ω} \frac{∂𝑢}{∂𝑡} (x, t) (\frac{\partial^{2} u}{\partial t^{2}} (x, t) - △ u (x, t)) dn x = 0 . \end{array}

Here we applied once the divergence theorem to the vector field $\frac{∂𝑢}{∂𝑡} \nabla u$ which vanishes at $\partial Ω \times [0, T]$ together with $u$ and $\frac{∂𝑢}{∂𝑡}$ . Initially the energy is zero $e (0) = 0$ . Since the energy starts as zero, it stays zero for all positive times $t > 0$ . This shows that $u$ is constant and moreover it vanishes on $Ω \times [0, T)$ since it vanishes initially. □

The proof gives the same conclusion if we assume that the normal derivative $\nabla u (x, t) \cdot N (x, t)$ is given on $\partial Ω \times [0, T]$ instead of the values of $u (x, t)$ .

We give a related proof that the length of the speed of propagation is bounded by $1$ .

Theorem 5.4. If $u$ is any solution of the homogeneous wave equation obeying $u = \frac{∂𝑢}{∂𝑡} = 0$ on $B (x_{0}, t_{0})$ for $t = 0$ , then $u$ vanishes on the cone ${(x, t) ∣ | x - x_{0} | \leq t_{0} - t, t > 0}$ .

Proof. Again we calculate the time derivative of the energy

\begin{array}{l} e (t) & = \frac{1}{2} \int_{B (x_{0}, t_{0} - t)} ({(\frac{∂𝑢}{∂𝑡} (x, t))}^{2} + {(\nabla u (x, t))}^{2}) dn x as \\ \frac{𝑑𝑒}{𝑑𝑡} (t) & = \frac{1}{2} \frac{d}{𝑑𝑡} \int_{0}^{t_{0} - t} \int_{∂𝐵 (x_{0}, s)} ({(\frac{∂𝑢}{∂𝑡} (x, t))}^{2} + {(\nabla u (x, t))}^{2}) d σ (x) d s \\ = \int_{B (x_{0}, t_{0} - t)} (\frac{\partial^{2} u}{\partial t^{2}} (x, t) \frac{∂𝑢}{∂𝑡} (x, t) + \frac{\partial \nabla u}{∂𝑡} (x, t) \nabla u (x, t)) dn x \\ - \frac{1}{2} \int_{∂𝐵 (x_{0}, t_{0} - t)} ({(\frac{∂𝑢}{∂𝑡} (x, t))}^{2} + {(\nabla u (x, t))}^{2}) d σ (x) \\ = \int_{B (x_{0}, t_{0} - t)} \frac{∂𝑢}{∂𝑡} (x, t) (\frac{\partial^{2} u}{\partial t^{2}} (x, t) - △ u (x, t)) dn x \\ + \int_{∂𝐵 (x_{0}, t_{0} - t)} (\frac{∂𝑢}{∂𝑡} (x, t) \nabla u (x, t) \cdot N (x, t) - \frac{1}{2} {(\frac{∂𝑢}{∂𝑡} (x, t))}^{2} - \frac{1}{2} {(\nabla u (x, t))}^{2}) d σ (x) \\ = \int_{∂𝐵 (x_{0}, t_{0} - t)} (\frac{∂𝑢}{∂𝑡} (x, t) \nabla u (x, t) \cdot N (x, t) - \frac{1}{2} {(\frac{∂𝑢}{∂𝑡} (x, t))}^{2} - \frac{1}{2} {(\nabla u (x, t))}^{2}) d σ (x) . \end{array}

Since the outer normal has length one we derive

\frac{∂𝑢}{∂𝑡} (x, t) \nabla u (x, t) \cdot N (x, t) \leq \frac{1}{2} {(\frac{∂𝑢}{∂𝑡} (x, t))}^{2} + \frac{1}{2} {(\nabla u (x, t))}^{2}

with $a = \nabla u (x, t)$ and $b = \frac{∂𝑢}{∂𝑡} (x, t) N (x, t)$ from the following inequality:

a \cdot b \leq a \cdot b + \frac{1}{2} (a - b) \cdot (a - b) = \frac{1}{2} a^{2} + \frac{1}{2} b^{2} .

(5.1)

So by $ė (t) \leq 0$ the energy is monotonically decreasing. Because the energy is non-negative and vanishes initially it stays zero for all positive times in $t \in [0, t_{0}]$ . This implies $u = 0$ on ${(x, t) ∣ | x - x_{0} | \leq t_{0} - t, t > 0}$ . □

The uniqueness of the Cauchy problem now follows as an easy consequence. Suppose, as is standard, that there are two solutions to a given Cauchy problem, and consider their difference. The above theorem now applies to this difference, and tells us that it is identically zero.

By the invariance with respect to time reversal we can also deduce the vanishing of $u$ on the cone ${(x, t) ∣ | x - x_{0} | < t_{0} + t, t < 0}$ from the vanishing of $u$ and $\frac{∂𝑢}{∂𝑡} = 0$ on $(x, t) \in B (x_{0}, t_{0}) \times {0}$ .

Finally, let us briefly tour the energy method for the Laplace and heat equations. Suppose that we have a solution $u$ to the equation $- △ u = 0$ . Consider what happens if we multiply this by $u$ and integrate:

0 = \int_{Ω} - △ u u dn x = \int_{Ω} | \nabla u |^{2} dn x .

By the fundamental lemma of the calculus of variations, we know that the integrand is zero, i.e. $u$ is constant. If we further know that $u$ is zero on the boundary, then we have that $u \equiv 0$ .

In analogy to the Laplace equation one can show the uniqueness for the heat equation. From intuition, the energy of a solution to the heat equation should be proportional to the total temperature, which we know is conserved. However we have seen that it is important to have positive functions, so that we can conclude that the function is zero if the integral is. Therefore we look at a simple positive quantity and define

e (t) = \int_{Ω} u^{2} (x, t) dn x .

If $u$ solves the homogeneous heat equation and vanishes at the boundary of $Ω$ , then this functional is monotonically decreasing with respect to time:

ė (t) = 2 \int_{Ω} u (x, t) \dot{u} (x, t) dn x = 2 \int_{Ω} u (x, t) △ u (x, t) dn x = - 2 \int_{Ω} {| \nabla u (x, t) |}^{2} dn x \leq 0 .

If $u (x, t)$ vanishes at $t = 0$ , and if $u (\cdot, t)$ and $\nabla u (\cdot, t)$ are square integrable for $t > 0$ , then $u$ vanishes identically since $\nabla u (\cdot, t)$ vanishes and $u (\cdot, t)$ is constant for $t > 0$ .

This idea is strong enough to show the uniqueness of the solution of the Dirichlet problem for all three of the second order equations in this course. In fact, Dirichlet’s insight was that the unique solution of Dirichlet’s Problem solves the following variational problem:

Dirichlet’s Principle 5.5. Let $Ω \subset ℝ^{n}$ be bounded and open and obey the assumptions of the Divergence Theorem. For continuous real functions $f$ on $\bar{Ω}$ and $g$ on $\partial Ω$ the solution $u$ of the Dirichlet Problem 3.14 is the minimizer of the following functional:

\begin{array}{l} I : {w \in C^{2} (\bar{Ω}) ∣ w |_{\partial Ω} = g} & \to ℝ, & w & \mapsto I (w) = \int_{Ω} (\frac{1}{2} \nabla w \cdot \nabla w - 𝑤𝑓) dn x . \end{array}

Proof. Let $u$ be a solution of the Dirichlet Problem and $w$ another function in the domain ${w \in C^{2} (\bar{Ω}) ∣ w |_{\partial Ω} = g}$ of $I$ . An integration by parts yields

0 = \int_{Ω} (- △ u - f) (u - w) dn x = \int_{Ω} (\nabla u \cdot \nabla (u - w) - f (u - w)) dn x .

We can rearrange this to give two expressions that are almost $I (u)$ and $I (w)$ . All that remains is to deal with the mixed term $\nabla u \cdot \nabla w$ using the Cauchy-Schwarz inequality (5.1).

\begin{array}{rcl} \int_{Ω} (\nabla u \cdot \nabla u - 𝑓𝑢) dn x & = & \int_{Ω} (\nabla u \cdot \nabla w - 𝑓𝑤) dn x \leq \\ \leq & \int_{Ω} \frac{1}{2} \nabla u \cdot \nabla u dn x + \int_{Ω} (\frac{1}{2} \nabla w \cdot \nabla w - 𝑓𝑤) dn x \end{array}

Moving the term across shows $I (u) \leq I (w)$ .

If, conversely, $u$ is a minimum, then all $v \in C^{2} (\bar{Ω})$ which vanish on $\partial Ω$ obey

\begin{array}{l} 0 = {\frac{d}{𝑑𝑡} I (u + 𝑡𝑣) |}_{t = 0} & = {\frac{d}{𝑑𝑡} (I (u) + t \int_{Ω} (\nabla u \cdot \nabla v - 𝑓𝑣) dn x + \frac{t^{2}}{2} \int_{Ω} \nabla v \cdot \nabla v dn x) |}_{t = 0} \\ = \int_{Ω} (\nabla u \cdot \nabla v - 𝑓𝑣) dn x = \int_{Ω} (- △ u - f) v dn x . \end{array}

The final integration by parts shows $- △ u = f$ on $Ω$ . □

This result naturally suggests a way of showing the existence of a solution to the Dirichlet problem for the Poisson equation, namely to show that there is a function that achieves this minimum. Dirichlet offered this as a proof of the existence, however he took for granted the existence of a function that attains the minimum.

The validity of Dirichlet’s proof was questioned by Weierstrass, who offered the following example of a functional that had an infimum but no minimum. We use this as an example of the subtlety that exists, even though it does not directly involve the Laplace equation. Consider the set of functions in $C^{1} ([- 1, 1])$ with the boundary condition $φ (- 1) = - 1$ and $φ (1) = 1$ . On this set, consider the functional $J (φ) = \int_{- 1}^{1} {(x ϕ^{'})}^{2} d x$ . Clearly we have $J \geq 0$ . On the other hand, consider the following family of functions

φ_{𝜀} (x) = {\begin{matrix} - 1 & if x < - 𝜀, \\ - 1 + 𝜀^{- 1} (x + 𝜀) & if | x | < 𝜀, \\ 1 & if x > 𝜀 \end{matrix}

Strictly speaking, these function are not continuously differentiable, so we should smooth the function at $x = \pm 𝜀$ . This will not affect the following calculation significantly. We compute

J (φ_{𝜀}) = \int_{- 𝜀}^{𝜀} {(x 𝜀^{- 1})}^{2} d x = 𝜀^{- 2} \frac{1}{3} x^{3} |_{- 𝜀}^{𝜀} = \frac{2}{3} 𝜀 .

This shows that the infimum of $J$ is exactly 0. However there is no function that achieves this value. If $J (φ) = 0$ then we must have $x φ^{'} = 0$ . This forces $φ^{'} (x) = 0$ for $x \in [- 1, 1] ∖ {0}$ and hence for all $x$ by continuity. But there is no constant function $φ$ that meets the boundary condition.

In order that we finish the course on a positive note, we will consider one final example. This deals with some of the subtleties of Dirichlet’s principle, which also also goes by the name “the direct method of the calculus of variations”. In the previous example, the issue is that limit of $φ_{𝜀}$ is not a continuous function (let alone a $C^{1}$ function). This is typical and one must choose a larger class of functions, usually a Sobolev space. In this example we define a notion of weak solution of the Laplace equation, namely we say that $u \in {w \in C (\bar{Ω}) ∣ w |_{\partial Ω} = g}$ on a bounded domain $Ω$ is a weak solution of the Laplace equation if for any test function $φ \in C_{0}^{\infty} (Ω)$ it obeys

\int_{Ω} u △ φ dn x = 0 .

This is precisely the statement that the distribution $F_{u}$ is harmonic, which explains the use of the term ‘weak’, but we will not use anything from Chapter 3. By integration by parts, a (twice continuously differentiable) harmonic function is a weak solution.

Suppose that we have a sequence of weak solutions $u_{k}$ that converge locally uniformly to a function $u$ . We can show that $u$ is also a weak solution. By local uniform convergence, we mean that every point in $Ω$ has a neighbourhood $V$ such that $∥ u - u_{k} ∥_{L^{\infty} (V)}$ tends to zero. Equivalently we can say that the restrictions $u_{k} |_{V}$ converge uniformly to $u |_{V}$ . This latter phrasing shows that $u$ is continuous, since it is the uniform limit of continuous functions. Clearly $u |_{\partial Ω} = g$ . It remains to show that $u$ has the necessary integral property.

\begin{array}{l} | \int_{Ω} u △ φ dn x | & = | \int_{Ω} u △ φ dn x - \lim_{k \to \infty} \int_{Ω} u_{k} △ φ dn x | = \lim_{k \to \infty} | \int_{Ω} (u - u_{k}) △ φ dn x | \\ \leq \lim_{k \to \infty} \int_{Ω} | u - u_{k} | | △ φ | dn x = \lim_{k \to \infty} \int_{supp φ} | u - u_{k} | | △ φ | dn x . \end{array}

The point of the last step, reducing the integration to the support of $φ$ , is that the support is compact. For each point of the support, choose a neighbourhood $V$ such that we have uniform convergence. Because of compactness, we can choose finitely many $V_{i}$ such that the support remains covered. Moreover the integrand is positive, so if we integrate certain parts of the domain multiple times, it only increases the value. Thus we continue with our calculation

\leq \lim_{k \to \infty} \sum_{i} \int_{V_{i}} | u - u_{k} | | △ φ | dn x \leq \lim_{k \to \infty} \sum_{i} ∥ u - u_{k} ∥_{L^{\infty} (V_{i})} \int_{V_{i}} | △ φ | dn x \to 0 .

This is only possible if $\int_{Ω} u △ φ dn x = 0$ . I hope that this gives you an idea of what goes into an existence proof using the energy method.

[next] [prev] [prev-tail] [front] [up]

Chapter 5Wave Equation