Why does the characteristics method not work for boundary value problems?

Question

Let us consider, for example (all though the question applies more generally) the 1D wave equation $$ u_{tt} = c^2 u_{xx}.$$ This has characteristics $x-ct = const$ and $x+ct=const$ which is readily observable by factoring the differential operator as $$\frac{\partial^2}{\partial t^2} - c^2\frac{\partial^2}{\partial x^2} = (\frac{\partial}{\partial t} - c\frac{\partial}{\partial x})(\frac{\partial}{\partial t} + c\frac{\partial}{\partial x}).$$ Thus, making the variable transformation $\eta = x-ct$ and $\xi=x+ct$ yields $$ u_{\eta \xi} = 0 $$ from which we conclude that $$ u =F(x-ct) + G(x+ct) \tag{1}\label{1}$$ for some sufficiently smooth functions $F$, $G$.

If we consider the IVP: $u(x,0) = f(x)$ and $u_t(x,0) = g(x)$ we derive d'Alambert's solution from (1). However, this is the solution for a boundaryless domain and is incapable of capturing any sort of boundary values. e.g. a clamped boundary $u(0,t)=u(L,t)=0$. Yet, this is a well posed problem and solvable via separation of variables. So we can only conclude that (1) is not the general solution. What went wrong?

My guess is that in making the variable transformation and integrating we implicitly assume a certain global regularity which the boundary value problem does not satisfy. Any clarification or expansion on this?

Answer 1

To get the analogue of (1) in a more general case, e.g for an inhomogeneous boundary problem, you can put $$v = \left(\frac{\partial}{\partial t} + c\frac{\partial}{\partial x}\right)u$$ so that $v$ solves the transport equation $$\left(\frac{\partial}{\partial t} - c\frac{\partial}{\partial x}\right)v = 0.$$ You can solve that equation explicitly with the caracteristic method. Then, you can solve in the same way $$\left(\frac{\partial}{\partial t} + c\frac{\partial}{\partial x}\right)u = v$$ to find $u$ in terms of $v$.

Answer 2

If $u$ is a solution of the boundary value problem, odd reflections on the axes $x=0$, $x=L$ and in continuation at $x=kL$ for all $k\in \Bbb Z$, generates a $2L$-periodic function. Due to the odd symmetry at the reflection axes, $u_{xx}=0$ there and $u_{tt}=0$ due to the boundary conditions, so the wave equation is satisfied on $\Bbb R$.

Thus the functions $F,G$ for the solution formula exist. The boundary conditions applied to the solution formula give $F(ct)+G(-ct)=0$ and $F(L+ct)+G(L-ct)=0$. Combining both gives $$ G(-s)=-F(s)=-F(L+(s-L))=G(L-(s-L))=G(2L-s), $$ so $G$ is a $2L$-periodic function and $$u(x,t)=G(x+ct)-G(-x+ct).$$

Evaluating the initial conditions gives \begin{align} f(x)&=G(x)-G(-x)\\ \tfrac1c g(x)&=G'(x)-G'(-x)\\ \tfrac1c\int_0^x g(s)\,ds&=G(x)+G(-x) \end{align} so that in combination $$ G(x)=\begin{cases} \frac12\Bigl(f(x)+\tfrac1c\int_0^x g(s)\,ds\Bigr), & x\in[0,L],\\ \frac12\Bigl(-f(-x)+\tfrac1c\int_0^{-x} g(s)\,ds\Bigr), & x\in[-L,0]. \end{cases} $$ and $2L$-periodic continued outside $[-L,L]$.