Question

Let \(u(x, t)\) be the solution of the non-homogeneous wave equation \[ \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial t^2} = \sin x \sin(2t), \quad 0<x<\pi, \ t>0 \] \[ u(x, 0) = 0, \text{ and } \frac{\partial u}{\partial t}(x, 0) = 0, \quad \text{for } 0 \le x \le \pi, \] \[ u(0, t) = 0, \quad u(\pi, t) = 0, \quad \text{for } t \ge 0. \] Then the value of \( u \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \) is ................. (rounded off to 2 decimal places).

Hint:

Pay close attention to the standard form of the wave equation \(u_{tt} - c^2 u_{xx} = F\). The given equation might need to be rearranged, which could introduce a negative sign in the forcing term, as it did in this problem. This is a common source of error.

Answer 1

Step 1: Understanding the Concept:
This problem involves solving a non-homogeneous 1D wave equation with specified boundary and initial conditions. The method of separation of variables, specifically using a sine series expansion, is suitable for this type of problem.
Step 2: Key Formula or Approach:
1. Rewrite the PDE in the standard form: \(u_{tt} - u_{xx} = F(x,t)\).
2. Assume a solution of the form \(u(x,t) = \sum_{n=1}^{\infty} u_n(t) \sin(nx)\), which automatically satisfies the boundary conditions \(u(0,t)=0\) and \(u(\pi,t)=0\).
3. Substitute this series into the PDE and the forcing term \(F(x,t)\) to obtain an ODE for each mode \(u_n(t)\).
4. Use the initial conditions \(u(x,0)=0\) and \(u_t(x,0)=0\) to find the initial conditions for each \(u_n(t)\).
5. Solve the ODE for \(u_n(t)\) with its initial conditions.
6. Construct the final solution \(u(x,t)\) and evaluate it at the given point.
Step 3: Detailed Explanation or Calculation:
The PDE is \(u_{xx} - u_{tt} = \sin x \sin(2t)\). Rearranging it gives: \[ u_{tt} - u_{xx} = -\sin x \sin(2t) \] So the forcing function is \(F(x,t) = -\sin x \sin(2t)\).
Let \(u(x,t) = \sum_{n=1}^{\infty} u_n(t) \sin(nx)\). Substitute this into the PDE: \[ \sum_{n=1}^{\infty} u_n''(t) \sin(nx) - \sum_{n=1}^{\infty} (-n^2) u_n(t) \sin(nx) = -\sin x \sin(2t) \] \[ \sum_{n=1}^{\infty} [u_n''(t) + n^2 u_n(t)] \sin(nx) = -\sin x \sin(2t) \] By comparing the coefficients of \(\sin(nx)\) on both sides (Fourier sine series expansion):
For n = 1: \(u_1''(t) + 1^2 u_1(t) = -\sin(2t)\).
For n>1: \(u_n''(t) + n^2 u_n(t) = 0\).
The initial conditions are:
\(u(x,0) = \sum u_n(0) \sin(nx) = 0 \implies u_n(0) = 0\) for all n.
\(u_t(x,0) = \sum u_n'(0) \sin(nx) = 0 \implies u_n'(0) = 0\) for all n.
For n>1, the solution to \(u_n'' + n^2 u_n = 0\) is \(u_n(t) = A\cos(nt) + B\sin(nt)\). With \(u_n(0)=0\) and \(u_n'(0)=0\), we get \(A=0\) and \(B=0\). So \(u_n(t)=0\) for \(n>1\).
For n = 1, we solve \(u_1'' + u_1 = -\sin(2t)\) with \(u_1(0)=0\) and \(u_1'(0)=0\).
The general solution is \(u_1(t) = u_h(t) + u_p(t)\).
The homogeneous solution is \(u_h(t) = A\cos(t) + B\sin(t)\).
For the particular solution, we try \(u_p(t) = C\sin(2t)\). Then \(u_p''(t) = -4C\sin(2t)\).
Substituting into the ODE: \(-4C\sin(2t) + C\sin(2t) = -\sin(2t) \implies -3C = -1 \implies C = 1/3\).
So, \(u_p(t) = \tfrac{1}{3}\sin(2t)\).
The general solution is \(u_1(t) = A\cos(t) + B\sin(t) + \tfrac{1}{3}\sin(2t)\). Apply initial conditions:
\(u_1(0) = A\cos(0) + B\sin(0) + \tfrac{1}{3}\sin(0) = A = 0\).
\(u_1'(t) = -A\sin(t) + B\cos(t) + \tfrac{2}{3}\cos(2t)\).
\(u_1'(0) = B\cos(0) + \tfrac{2}{3}\cos(0) = B + \tfrac{2}{3} = 0 \implies B = -2/3\).
So, \(u_1(t) = -\tfrac{2}{3}\sin(t) + \tfrac{1}{3}\sin(2t)\).
The full solution is \(u(x,t) = u_1(t)\sin(x)\):
\[ u(x,t) = \left( -\tfrac{2}{3}\sin(t) + \tfrac{1}{3}\sin(2t) \right) \sin(x) \] Step 4: Final Answer:
We need to find \(u(\pi/2, 3\pi/2)\).
\[ u\left(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right) = \left( -\tfrac{2}{3}\sin\left(\tfrac{3\pi}{2}\right) + \tfrac{1}{3}\sin\left(2 \cdot \tfrac{3\pi}{2}\right) \right) \sin\left(\tfrac{\pi}{2}\right) \] We know \(\sin(3\pi/2) = -1\), \(\sin(3\pi) = 0\), and \(\sin(\pi/2) = 1\).
\[ u\left(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right) = \left( -\tfrac{2}{3}(-1) + \tfrac{1}{3}(0) \right) \times (1) = \tfrac{2}{3} \] Rounding to 2 decimal places, \(2/3 \approx 0.67\).
Step 5: Why This is Correct:
The problem was correctly identified as a non-homogeneous wave equation solvable by Fourier series. The sign of the forcing term was handled correctly. The resulting ODE was solved with the correct initial conditions. The final evaluation is arithmetically correct and matches the provided answer range.