an interesting solution approach for a wave equation geometrically

Question

Let $PQRS$ be a rectangle in first quadrant whose adjacent sides $PQ$ and $QR$ have slopes $1$ and $-1$ respectively. If $u(x,t)$ is a solution of $\displaystyle{\frac{\partial^2 u}{\partial t^2}-\frac{\partial^2 u}{\partial x^2}}=0$ and $u(P)=1, u(Q)=-\frac{1}{2}, u(R)=\frac{1}{2}$, then $u(S)$ equals

$(a)$ $2$

$(b)$ $1$

$(c)$ $\displaystyle\frac{1}{2}$

$(d)$ $\displaystyle-\frac{1}{2}$

I at first considered a rectangle in first quadrant as shown below :

Now this problem is really unconventional to me as no standard boundary conditions are provided in the problem, the boundary conditions are given only on the vertex of the curve. I know that Riemann-Volterra method is used for determining the solution of a PDE along any closed curve at any arbitrary point on it, but I don't know that method too well to apply at this particular problem. What may be the best possible way for solving this problem at hand? Thanks in advance.

Answer 1

All solutions of the $1$D wave equation are of the form

$$u(x,t) = f(x-t) + g(x+t)$$

Let's write the equations of each line with their standard form in $\mathbb{R}^2$

$$\begin{cases} PQ: & x-t = k_1 \\ QR: & x+t = k_2 \\ RS: & x-t = k_3 \\ SP: & x+t = k_4 \\ \end{cases} $$

where $k_i$ are all constants. Since each point is located at the intersection of two lines we have the following system of equations:

$$\begin{cases} u(P) = f(k_1) + g(k_4) = 1 \\ u(Q) = f(k_1) + g(k_2) = -\frac{1}{2} \\ u(R) = f(k_3) + g(k_2) = \frac{1}{2} \\ \end{cases} $$

Thus we get that

$$u(S) = f(k_3) + g(k_4) = f(k_3) + g(k_2) - g(k_2) - f(k_1) +f(k_1) + g(k_4)$$

$$= u(R) - u(Q) + u(P) = 2$$

Although since the question was multiple choice, there was a faster way of seeing the answer had to be $(a) \: 2$. The points of the rectangle all fall on lines of characteristics for the $1$D wave equation, yet the values on the corners were unequal. The value on the last corner could not share a value with any other corner.