
# 4.3: Inhomogeneous Equations

Here we consider the initial value problem

\begin{eqnarray}
\label{inh1}
\Box u&=&w(x,t)\ \ \mbox{on}\ x\in\mathbb{R}^n,\ t\in\mathbb{R}^1\\
\label{inh2}
u(x,0)&=&f(x)\\
\label{inh3}
u_t(x,0)&=&g(x),
\end{eqnarray}

where $$\Box u:=u_{tt}-c^2\triangle u$$. We assume $$f\in C^3$$, $$g\in C^2$$ and $$w\in C^1$$, which are given.

Set $$u=u_1+u_2$$, where $$u_1$$ is a solution of problem (\ref{inh1})-(\ref{inh3}) with $$w:=0$$ and $$u_2$$ is the solution where $$f=0$$ and $$g=0$$ in (\ref{inh1})-(\ref{inh3}). Since we have explicit solutions $$u_1$$ in the cases $$n=1$$, $$n=2$$ and $$n=3$$, it remains to solve

\begin{eqnarray}
\label{duhu2gl}
\Box u&=&w(x,t)\ \ \mbox{on}\ x\in\mathbb{R}^n,\ t\in\mathbb{R}^1\\
\label{duhu2in1}
u(x,0)&=&0\\
\label{duhu2in2}
u_t(x,0)&=&0.
\end{eqnarray}

The following method is called Duhamel's principle which can be considered as a generalization of the method of variations of constants in the theory of ordinary differential equations.

To solve this problem, we make the ansatz

\label{duh1}
u(x,t)=\int_0^t\ v(x,t,s)\ ds,

where $$v$$ is a function satisfying

\label{duh2}
\Box v=0\ \ \mbox{for all}\ s

and

\label{duh3}
v(x,s,s)=0.

From ansatz (\ref{duh1}) and assumption (\ref{duh3}) we get

\begin{eqnarray}
u_t&=&v(x,t,t)+\int_0^t\ v_t(x,t,s)\ ds,\nonumber\\
\label{duh4}
&=&\int_0^t\ v_t(x,t,s).
\end{eqnarray}

It follows $$u_t(x,0)=0$$. The initial condition $$u(x,t)=0$$ is satisfied because of the ansatz (\ref{duh1}). From (\ref{duh4}) and ansatz (\ref{duh1}) we see that

\begin{eqnarray*}
u_{tt}&=&v_t(x,t,t)+\int_0^t\ v_{tt}(x,t,s)\ ds,\\
\triangle_x u&=&\int_0^t\ \triangle_x v(x,t,s)\ ds.
\end{eqnarray*}

Therefore, since $$u$$ is an ansatz for (\ref{duhu2gl})-(\ref{duhu2in2}),

\begin{eqnarray*}
u_{tt}-c^2\triangle_x u&=&v_t(x,t,t)+\int_0^t(\Box v)(x,t,s)\ ds\\
&=&w(x,t).
\end{eqnarray*}

Thus necessarily $$v_t(x,t,t)=w(x,t)$$, see (\ref{duh2}). We have seen that the ansatz provides a solution of (\ref{duhu2gl})-(\ref{duhu2in2}) if for all $$s$$

\label{duh5}
\Box v=0,\ \ v(x,s,s)=0,\ \ v_t(x,s,s)=w(x,s).

Let $$v^*(x,t,s)$$ be a solution of

\label{duh6}
\Box v=0,\ \ v(x,0,s)=0,\ \ v_t(x,0,s)=w(x,s),

then

$$v(x,t,s):=v^*(x,t-s,s)$$

is a solution of (\ref{duh5}).
In the case $$n=3$$, where $$v^*$$ is given by, see Theorem 4.2,

$$v^*(x,t,s)=\frac{1}{4\pi c^2 t}\int_{\partial B_{ct}(x)}\ w(\xi,s)\ dS_\xi.$$

Then

\begin{eqnarray*}
v(x,t,s)&=&v^*(x,t-s,s)\\
&=&\frac{1}{4\pi c^2 (t-s)}\int_{\partial B_{c(t-s)}(x)}\ w(\xi,s)\ dS_\xi.
\end{eqnarray*}

from ansatz (\ref{duh1}) it follows

\begin{eqnarray*}
u(x,t)&=&\int_0^t\ v(x,t,s)\ ds\\
&=&\frac{1}{4\pi c^2}\int_0^t\ \int_{\partial B_{c(t-s)}(x)}\ \frac{w(\xi,s)}{t-s}\ dS_\xi ds.
\end{eqnarray*}

Changing variables by $$\tau=c(t-s)$$ yields

\begin{eqnarray*}
u(x,t)&=&\frac{1}{4\pi c^2}\int_0^{ct}\ \int_{\partial B_{\tau}(x)}\ \frac{w(\xi,t-\tau/c)}{\tau}\ dS_\xi d\tau\\
&=&\frac{1}{4\pi c^2} \int_{ B_{ct}(x)}\ \frac{w(\xi,t-r/c)}{r}\ d\xi,
\end{eqnarray*}

where $$r=|x-\xi|$$.

Formulas for the cases $$n=1$$ and $$n=2$$ follow from formulas for the associated homogeneous equation with inhomogeneous initial values for these cases.

Theorem 4.4. The solution of

$$\Box u=w(x,t),\ \ u(x,0)=0,\ \ u_t(x,0)=0,$$

where $$w\in C^1$$, is given by:

Case $$n=3$$:

$$u(x,t)=\frac{1}{4\pi c^2} \int_{ B_{ct}(x)}\ \frac{w(\xi,t-r/c)}{r}\ d\xi,$$

where $$r=|x-\xi|$$, $$x=(x_1,x_2,x_3)$$, $$\xi=(\xi_1,\xi_2,\xi_3)$$.

Case $$n=2$$:

$$u(x,t)=\frac{1}{4\pi c}\int_0^t\ \left( \int_{ B_{c(t-\tau)}(x)}\ \frac{w(\xi,\tau)}{\sqrt{c^2(t-\tau)^2-r^2}}\ d\xi\right)\ d\tau,$$

$$x=(x_1,x_2)$$, $$\xi=(\xi_1,\xi_2)$$.

Case $$n=1$$:

$$u(x,t)=\frac{1}{2c}\int_0^t\ \left(\int_{x-c(t-\tau)}^{x+c(t-\tau)}\ w(\xi,\tau)\ d\xi\right)\ d\tau.$$

Remark. The integrand on the right in formula for $$n=3$$ is called retarded potential. The integrand is taken not at $$t$$, it is taken at an earlier time $$t-r/c$$.

### Contributors

• Integrated by Justin Marshall.