Discussion
Let u(x, t) be the solution of the wave equation
\[ \begin{aligned} &\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{2} u}{\partial x^{2}} = 0, \\ &0 \lt x \lt \pi,\ t \gt 0, \end{aligned} \]
with the initial conditions
\[ u(x,0) = \sin x + \sin 2x + \sin 3x, \]
\[ \frac{\partial u}{\partial t}(x,0) = 0, \; 0 \lt x \lt \pi \]
and the boundary conditions u(0, t) = u(\(\pi\), t) = 0, \(t \geq 0\). Then, the value of \(u\left(\frac{\pi}{2}, \pi\right)\) is
Answer: B
The value is 0.
With zero initial velocity and fixed ends on (0, \(\pi\)), each mode sin kx oscillates as cos kt, so by superposition
\[ u(x,t) = \sin x\cos t + \sin 2x\cos 2t + \sin 3x\cos 3t \]
At x = \(\pi/2\), t = \(\pi\):
\[ \begin{aligned} u &= (1)(-1) + (0)(1) + (-1)(-1) \\ &= -1 + 0 + 1 = 0 \end{aligned} \]
The answer 1 comes from forgetting that \(\sin(3\pi/2) = -1\), and \(\pm 1/2\) has no source in this solution. D'Alembert's formula, u = [g(x + t) + g(x - t)]/2 with g extended as an odd \(2\pi\)-periodic function, gives the same 0.
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