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I'm struggling to find an exact solution to the 2D wave equation, $u_{tt} = c^2(u_{xx} + u_{yy})$ with the following boundary and initial conditions:

The left edge of the square is equal to 1, while the top, bottom and right are 0. $u_x(0,y,t) =1$ $u_y(x,0,t) = u_y (x,1,t)= u_x(1,y,0)=0$

Initial conditions $u(x,y,0) = u_t (x,y,0)=0 $.

By separation of variables, let U(x,y,t) = F(x,y)G(t), and going through the usual process letting $-v^2$ be the constant that lets the two functions equal to each other.

This reduces to two equations requiring to be solved $G'' + \lambda^2 G = 0 $ therefore $G(t) = A\cos(\lambda t) + B\sin(\lambda t)$ where $\lambda = cv$. And $F_{xx} + F_{yy} + v^2 F = 0$ (2D Helmholtz equation). However when I apply the conditions all i get is zero, which shouldn't be the case.

jake rowbottom
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  • If $F(x,y) = 0$ and $G(t)=0$, then $u_x(0,y,t)=F_x(0,y)G(t)\neq 1$, so you are applying the boundary conditions wrong. Can you show us your work, so we can point out the mistake? – Vasily Mitch Jan 22 '19 at 17:27

1 Answers1

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I first tried to solve this problem by finding a time-independent solution that satisfies that Laplace equation with the corresponding boundary conditions. But as it turns out, no such solution exists.

What you actually have to do is to separate the solution into $u(x,y,t) = w(x,y,t) + v(x,y,t)$, where $w$ is some particular function that satisfies the boundary conditions, and $v$ is homogeneous on the boundary. There are many options, but for simplicity, we pick a $w$ wuch that $w_y=0$ and $w_x=1-x$, so

$$ w(x,y,t) = -\frac{(1-x)^2}{2} $$

Plugigng this into the original equation, we have

\begin{align} v_{tt} - c^2(v_{xx} + v_{yy}) &= -c^2 \\ v_y(x,0,t) = v_y(x,1,t) &= 0 \\ v_x(0,y,t) = v_x(1,y,t) &= 0 \\ v(x,y,0) &= \frac{(1-x)^2}{2} \\ v_t(x,y,0) &= 0 \end{align}

We can find $v$ by decomposing into a triple product: $v(x,y,t) = T(t)X(x)Y(y)$, such that $X(x)$ and $Y(y)$ are eigenfunctions of the Laplacian operators, each satisfying Neumann boundary conditions. You can find them by solving

\begin{align} X'' + \lambda^2 X &= 0, \quad X'(0) = X'(1) = 0 \\ Y'' + \mu^2 Y &= 0, \quad Y'(0) = Y'(1) = 0 \end{align}

which gives

$$ X_n(x) = \cos(n\pi x), \quad Y_m(y) = \cos(m\pi y) $$

However, notice that the solution does not need to depend on $y$, so we're left with

$$ v(x,y,t) = T_0(t) + \sum_{n=1}^\infty T_n(t)\cos(n\pi x) $$

Plugging this into the equation:

$$ T_0'' + \sum_{n=1}^\infty [T_n'' + c^2n^2\pi^2 T_n]\cos(n\pi x) = -c^2 $$

$$ \implies \begin{cases} T_0'' = -c^2 \\ T_n'' + c^2n^2\pi^2 T_n = 0 &&, n \ge 1 \end{cases} $$

The last step is to break down the initial condition into its Fourier series

$$ v(x,y,0) = \frac{(1-x)^2}{2} = T_0(0) + \sum_{n=1}^\infty T_n(t)\cos(n\pi x) $$

$$ \implies \begin{cases} T_0(0) = \int_0^1 \frac{(1-x)^2}{2}\ dx \\ T_n(0) = \dfrac{\int_0^1 \frac{(1-x)^2}{2}\cos(n\pi x)\ dx}{\int_0^1 \cos^2 (n\pi x)\ dx} &&, n \ge 1 \end{cases} $$

and $T_n'(0)=0, \forall n$.

Solve the remaining IVPs for $T_n(t)$

You may also review this method which uses the characteristic lines to form a recursive solution. It's less work if you understand the concepts.

Dylan
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