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# 4.5.3: Inhomogeneous Wave Equations

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Let $$\Omega\subset\mathbb{R}^n$$ be a bounded and sufficiently regular domain. In this section we consider the initial-boundary value problem

\begin{eqnarray}
\label{waveinh1}\tag{4.5.3.1}
u_{tt}&=&Lu+f(x,t)\ \ \mbox{in}\ \Omega\times\mathbb{R}^1\\
\label{waveinh2} \tag{4.5.3.2}
u(x,0)&=&\phi(x)\ \ x\in\overline{\Omega}\\
\label{waveinh3} \tag{4.5.3.3}
u_t(x,0)&=&\psi(x)\ \ x\in\overline{\Omega}\\
\label{waveinh4} \tag{4.5.3.4}
u(x,t)&=&0\ \ \mbox{for} \ x\in\partial\Omega\ \mbox{and}\ t\in\mathbb{R}^1,
\end{eqnarray}

where $$u=u(x,t)$$, $$x=(x_1,\ldots,x_n)$$, $$f,\ \phi,\ \psi$$ are given and $$L$$ is an elliptic differential operator. Examples for $$L$$ are:

1. $$L=\partial^2/\partial x^2$$, oscillating string.
2. $$L=\triangle_x$$, oscillating membrane.
3. $$Lu=\sum_{i,j=1}^n\frac{\partial}{\partial x_j}\left(a^{ij}(x)u_{x_i}\right),$$

where $$a^{ij}=a^{ji}$$ are given sufficiently regular functions defined on $$\overline{\Omega}$$. We assume $$L$$ is uniformly elliptic, that is, there is a constant $$\nu>0$$ such that

$$\sum_{i,j=1}^na^{ij}\zeta_i\zeta_j\ge\nu|\zeta|^2$$

for all $$x\in\Omega$$ and $$\zeta\in\mathbb{R}^n$$.

4. Let $$u=(u_1,\ldots,u_m)$$ and

$$Lu=\sum_{i,j=1}^n\frac{\partial}{\partial x_j}\left(A^{ij}(x)u_{x_i}\right),$$

where $$A^{ij}=A^{ji}$$ are given sufficiently regular $$(m\times m)$$-matrices on $$\overline{\Omega}$$. We assume that $$L$$ defines an elliptic system. An example for this case is the linear elasticity.

Consider the eigenvalue problem

\begin{eqnarray}
\label{osceigen1} \tag{4.5.3.5}
-Lv&=&\lambda v\ \ \mbox{in}\ \Omega\\
\label{osceigen2} \tag{4.5.3.6}
v&=&0\ \ \mbox{on}\ \partial\Omega.
\end{eqnarray}

Assume there are infinitely many eigenvalues

$$0<\lambda_1\le\lambda_2\le\ldots\ \to\infty$$

and a system of associated eigenfunctions $$v_1,\ v_2,\ldots$$ which is complete and orthonormal in $$L^2(\Omega)$$. This assumption is satisfied if $$\Omega$$ is bounded and if $$\partial\Omega$$ is sufficiently regular.

For the solution of (\ref{waveinh1})-(\ref{waveinh4}) we make the ansatz

\label{oscinh1} \tag{4.5.3.7}
u(x,t)=\sum_{k=1}^\infty v_k(x)w_k(t),

with functions $$w_k(t)$$ which will be determined later. It is assumed that all series are convergent and that following calculations make sense.

Let

\label{oscinh2} \tag{4.5.3.8}
f(x,t)=\sum_{k=1}^\infty c_k(t)v_k(x)

be Fourier's decomposition of $$f$$ with respect to the eigenfunctions $$v_k$$. We have

\label{oscinh3} \tag{4.5.3.9}
c_k(t)=\int_\Omega\ f(x,t)v_k(x)\ dx,

which follows from (\ref{oscinh2}) after multiplying with $$v_l(x)$$ and integrating over $$\Omega$$.

Set

$$\langle\phi,v_k\rangle=\int_\Omega\ \phi(x)v_k(x)\ dx,$$

then

\begin{eqnarray*}
\phi(x)&=&\sum_{k=1}^\infty\langle\phi,v_k\rangle v_k(x)\\
\psi(x)&=&\sum_{k=1}^\infty\langle\psi,v_k\rangle v_k(x)
\end{eqnarray*}

are Fourier's decomposition of $$\phi$$ and $$\psi$$, respectively.

In the following we will determine $$w_k(t)$$, which occurs in ansatz (\ref{oscinh1}), from the requirement that $$u=v_k(x)w_k(t)$$ is a solution of

$$u_{tt}=Lu+c_k(t)v_k(x)$$

and that the initial conditions

$$w_k(0)=\langle\phi,v_k\rangle,\ \ \ w_k'(0)=\langle\psi,v_k\rangle$$

are satisfied. From the above differential equation it follows

$$w_k''(t)=-\lambda_kw_k(t)+c_k(t).$$

Thus

\begin{eqnarray}
\label{oscinh4} \tag{4.5.3.10}
w_k(t)&=&a_k\cos(\sqrt{\lambda_k}t)+b_k\sin(\sqrt{\lambda_k}t)\\
&&+\frac{1}{\sqrt{\lambda_k}}\int_0^t\ c_k(\tau)\sin(\sqrt{\lambda_k}(t-\tau))\ d\tau,\nonumber
\end{eqnarray}

where

$$a_k=\langle\phi,v_k\rangle,\ \ \ b_k=\frac{1}{\sqrt{\lambda_k}}\langle\psi,v_k\rangle.$$

Summarizing, we have

Proposition 4.2. The (formal) solution of the initial-boundary value problem (\ref{waveinh1})-(\ref{waveinh4}) is given by

$u(x,t)=\sum_{k=1}^\infty v_k(x)w_k(t),$

where $$v_k$$ is a complete orthonormal system of eigenfunctions of (\ref{osceigen1}), (\ref{osceigen2}) and the functions $$w_k$$ are defined by (\ref{oscinh4}).

## The Resonance Phenomenon

Set in (\ref{waveinh1})-(\ref{waveinh4}) $$\phi=0$$, $$\psi=0$$ and assume that the external force $$f$$ is periodic and is given by

$$f(x,t)=A\sin(\omega t)v_n(x),$$

where $$A,\ \omega$$ are real constants and $$v_n$$ is one of the eigenfunctions of (\ref{osceigen1}), (\ref{osceigen2}). It follows

$$c_k(t)=\int_\Omega\ f(x,t)v_k(x)\ dx=A\delta_{nk}\sin(\omega t).$$

Then the solution of the initial value problem (\ref{waveinh1})-(\ref{waveinh4}) is
\begin{eqnarray*}
u(x,t)&=&\frac{Av_n(x)}{\sqrt{\lambda_n}}\int_0^t\ \sin(\omega\tau)\sin(\sqrt{\lambda_n}(t-\tau))\ d\tau\\
&=&Av_n(x)\frac{1}{\omega^2-\lambda_n}\left(\frac{\omega}{\sqrt{\lambda_n}}\sin(\sqrt{\lambda_k}t)-\sin(\omega t)\right),
\end{eqnarray*}
provided $$\omega\not=\sqrt{\lambda_n}$$. It follows

$$u(x,t)\to\frac{A}{2\sqrt{\lambda_n}}v_n(x)\left(\frac{\sin(\sqrt{\lambda_n}t)}{\sqrt{\lambda_n}}-t\cos(\sqrt{\lambda_n} t)\right)$$

if $$\omega\to\sqrt{\lambda_n}$$. The right hand side is also the solution of the initial-boundary value problem if $$\omega=\sqrt{\lambda_n}$$.

Consequently $$|u|$$ can be arbitrarily large at some points $$x$$ and at some times $$t$$ if $$\omega=\sqrt{\lambda_n}$$. The frequencies $$\sqrt{\lambda_n}$$ are called critical frequencies at which resonance occurs.

## A Uniqueness Result

The solution of the initial-boundary value problem (\ref{waveinh1})-(\ref{waveinh4}) is unique in the class $$C^2(\overline{\Omega}\times\mathbb{R}^1)$$.

Proof. Let $$u_1$$, $$u_2$$ are two solutions, then $$u=u_2-u_1$$ satisfies

\begin{eqnarray*}
u_{tt}&=&Lu\ \ \mbox{in}\ \Omega\times\mathbb{R}^1\\
u(x,0)&=&0\ \ x\in\overline{\Omega}\\
u_t(x,0)&=&0\ \ x\in\overline{\Omega}\\
u(x,t)&=&0\ \ \mbox{for} \ x\in\partial\Omega\ \mbox{and}\ t\in\mathbb{R}^n.
\end{eqnarray*}

As an example we consider Example 3 from above and set

$$E(t)=\int_\Omega\ (\sum_{i,j=1}^na^{ij}(x)u_{x_i}u_{x_j}+u_tu_t)\ dx.$$

Then
\begin{eqnarray*}
E'(t)&=&2\int_\Omega\ (\sum_{i,j=1}^na^{ij}(x)u_{x_i}u_{x_jt}+u_tu_{tt})\ dx\\
&=&2\int_{\partial\Omega}\ (\sum_{i,j=1}^na^{ij}(x)u_{x_i}u_tn_j)\ dS\\
&&+2\int_\Omega\ u_t(-Lu+u_tt)\ dx\\
&=&0.
\end{eqnarray*}

It follows $$E(t)=const.$$ From $$u_t(x,0)=0$$ and $$u(x,0)=0$$ we get $$E(0)=0$$. Consequently $$E(t)=0$$ for all $$t$$, which implies, since $$L$$ is elliptic, that $$u(x,t)=const.$$ on $$\overline{\Omega}\times\mathbb{R}^1$$. Finally, the homogeneous initial and boundary value conditions lead to $$u(x,t)=0$$ on $$\overline{\Omega}\times\mathbb{R}^1$$.

$$\Box$$

## Contributors

• Integrated by Justin Marshall.