The wave equation is the usual utt−c2uxx=0utt−c2uxx=0u_{tt} – c^2u_{xx} = 0. We are only considering 0≤x≤10 \le x \le 1 and 0≤t≤10 \le t \le 1. Here are the boundary conditions:

u(x,0)=f(x)u(x,1)=g(x)u(0,t)=h(t)u(1,t)=k(t)\begin{align}

u(x, 0) = f(x) \, \, \, \, \, \, \, \, \, \, & u(x, 1) = g(x) \\

u(0, t) = h(t) \, \, \, \, \, \, \, \, \, \, & u(1, t) = k(t)

\end{align}

In the above, f,g,h,kf, g, h, k are continuously differentiable. From the derivation of d’Alembert’s solution, we know that, to find the solution, we need the full set of the first-order derivative along one side of the square. For example:

ux(x,0)=f(x)ut(x,0)=???\begin{align}

u_x(x, 0) = f(x) \, \, \, \, \, \, \, \, \, \, u_t(x, 0) = ???

\end{align}

So potentially, there is no solution here.

On the other hand, given a set of boundary conditions, it is necessary for them to be compatible. For example:

u(x,y)=f(x,y)ux(x,y)=ϕ(x,y)→dudx=ux+uydydxut(x,y)=η(x,y)\begin{align}

u(x, y) & = f(x, y) \\

u_x(x, y) & = \phi(x, y) \, \, \, \, \, \, \, \, \, \, \rightarrow \, \, \, \, \, \, \, \, \, \, \frac{du}{dx} = u_x + u_y \frac{dy}{dx}\\

u_t(x, y) & = \eta(x, y) \\

\end{align}

So, from the above, we need something like fx(x,y)=ϕ(x,y)+η(x,y)dydxf_x(x, y) = \phi(x, y) + \eta(x, y)\frac{dy}{dx}. I would like to construct a counter-example that fails compatibility. Would you be able to help?

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Are you talking about 2D or 1D in space? And why is ut(x,0)≠0u_t(x,0) \ne 0?

– Biswajit Banerjee

Oct 21 at 1:00

This is 2D, and the question is what happens when you don’t have ut(x,0)u_t(x, 0).

– Andy Tam

Oct 21 at 3:40

Hint: Write the wave equation as a first order system. How many initial conditions and boundary conditions do you need for each PDE?

– Mattos

2 days ago

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