Discussion
Let \(u(x,t)\) be the solution of the following initial-boundary value problem
\[ \begin{aligned} &\frac{\partial u}{\partial t} - \frac{\partial^{2} u}{\partial x^{2}} = 0, \quad x \in (0,\pi), \quad t \gt 0, \\ &u(0,t) = u(\pi,t) = 0, \quad u(x,0) = \sin 4x \cos 3x. \end{aligned} \]
Then, for each \(t \gt 0\), the value of \(u\left(\frac{\pi}{4}, t\right)\) is
Answer: A
The value is \(\dfrac{e^{-49t}}{2\sqrt{2}}(e^{48t} - 1)\).
Rewrite the initial data as sine modes, which satisfy the boundary conditions:
\[ \begin{aligned} \sin 4x\cos 3x &= \tfrac{1}{2}(\sin 7x + \sin x) \\ u(x,t) &= \tfrac{1}{2}(e^{-49t}\sin 7x + e^{-t}\sin x) \end{aligned} \]
since the mode \(\sin nx\) decays as \(e^{-n^{2}t}\). At \(x = \pi/4\), \(\sin(7\pi/4) = -\tfrac{1}{\sqrt{2}}\) and \(\sin(\pi/4) = \tfrac{1}{\sqrt{2}}\):
\[ \begin{aligned} u &= \frac{e^{-t} - e^{-49t}}{2\sqrt{2}} \\ &= \frac{e^{-49t}}{2\sqrt{2}}(e^{48t} - 1) \end{aligned} \]
The (1 - e48t) option has the sign of \(\sin(7\pi/4)\) wrong, and the \(4\sqrt{2}\) option halves the correct coefficient.
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