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I have difficulties solving the following exercise: Consider the IBVP on the half line $(0,\infty)$ (with $T \in (0,\infty)$:

$u_{xx}-u_{tt}=0$, on $(0,T)\times(0,\infty)$
$u(0,x)=u_{0}(x), x>0$, $u_t(0,x)=v_{0}(x), x>0$ and finally the boundary condition: $u(t,0)=h(t), t\in(0,T)$, where $u_0 \in C^2([0,\infty))$,$v_0 \in C^1([0,\infty))$, $h \in C^2([0,T))$.

Question: Find a formula for solutions of the preceding IBVP, and show uniqueness of its solutions $u \in C^2([0,T)×[0,\infty))$. In addition, provide compatibility conditions for $u_0, v_0, h$ which ensure the existence of such a solution u, and determine the largest open superset of $[0,T)×[0,\infty)$ to which u can be extended as the unique $C^2$ solution of the 1-dimensional wave equation.

The answer is supposed to be quite similar to D'Alamberts formula. I tried to derive a formula in the same fashion but it didn't really work. Compatibility conditions: Probably $u_0(0)=h(0)$ and $v_0(0)=h'(0)$. Uniqueness at least is quite clear, but that is it.

Thanks in advance

Dylan
  • 16,575

1 Answers1

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Follow the method in this similar question. First note that the initial conditions and the boundary conditions are separated by the characteristic line $x=t$, so we find the solution in 2 different regions

$$ u(x,t) = \begin{cases} u_1(x,t), & 0 \le x < t \\ u_2(x,t), & x > t \end{cases} $$

For $x > t$, the solution is given by d'Alembert's formula

$$ u_2(x,t) = \frac{u_0(x+t)+u_0(x-t)}{2} + \frac{1}{2}\int_{x-t}^{x+t} v_0(s)\ ds $$

For $0 < x < t$, we find a solution of the form

$$ u_1(x,t) = F(t+x) + G(t-x) $$

The given B.C. gives

$$ u_1(0,t) = F(t) + G(t) = h(t) $$

And the continuity condition at $x=t$ gives

$$ u_1(t,t) = F(2t) + G(0) = u_2(t,t) = \frac{u_0(2t)+u_0(0)}{2} + \frac{1}{2}\int_0^{2t}v_0(s)\ ds $$

Solving for $F$ and $G$ gives

\begin{align} F(z) &= -G(0) + \frac{u_0(z)+u_0(0)}{2} + \frac{1}{2}\int_0^z v_0(s)\ ds \\ G(z) &= G(0) + h(z) - \frac{u_0(z)+u_0(0)}{2} - \frac{1}{2}\int_0^z v_0(s)\ ds \end{align}

So finally

$$ \implies u_1(x,t) = h(t-x) + \frac{u_0(t+x)-u_0(t-x)}{2} + \frac{1}{2}\int_{t-x}^{t+x} v_0(s)\ ds $$

If the boundary function $h(t)$ has compact support, then the solution in $0<x<t$ is additionally bounded above by the characteristic line $x = t - T$, so the domain of $u_1$ is $\{x > 0, x < t < x + T\}$

Dylan
  • 16,575