Question

Let \( k \in \mathbb{R} \) and \( D = \{(r, \theta) : 0<r<2, 0<\theta<\pi\ \). Let \( u(r, \theta) \) be the solution of the following boundary value problem \[ \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0, \quad (r, \theta) \in D, \] \[ u(r, 0) = u(r, \pi) = 0, \quad 0 \leq r \leq 2, \] \[ u(2, \theta) = k \sin(2\theta), \quad 0<\theta<\pi. \] If \( u\left(\frac{1}{4}, \frac{\pi}{4}\right) = 2 \), then the value of \( k \) is equal to ………. (round off to TWO decimal places).

Hint:

For separable PDEs, solve each part independently and apply boundary conditions carefully to find the constants.

Answer 1

Step 1: Solving the boundary value problem. The given PDE is separable, so let \( u(r, \theta) = R(r)\Theta(\theta) \). Substituting into the PDE and separating variables, we get: \[ \frac{r^2 R''(r) + rR'(r)}{R(r)} = -\frac{\Theta''(\theta)}{\Theta(\theta)} = \lambda. \] Step 2: Solving for \( \Theta(\theta) \). The angular part satisfies \( \Theta''(\theta) + \lambda \Theta(\theta) = 0 \) with boundary conditions \( \Theta(0) = \Theta(\pi) = 0 \). The solution is: \[ \Theta(\theta) = \sin(2\theta), \quad \lambda = 4. \] Step 3: Solving for \( R(r) \). The radial part satisfies: \[ r^2 R''(r) + rR'(r) - 4R(r) = 0. \] The general solution is: \[ R(r) = C_1 r^2 + C_2 r^{-2}. \] Step 4: Applying boundary conditions. Using \( u(2, \theta) = k \sin(2\theta) \), we determine \( C_1 = k/4 \) and \( C_2 = 0 \), so: \[ u(r, \theta) = \frac{k}{4} r^2 \sin(2\theta). \] Step 5: Calculating \( k \). Using \( u\left(\frac{1}{4}, \frac{\pi}{4}\right) = 2 \): \[ 2 = \frac{k}{4} \left(\frac{1}{4}\right)^2 \sin\left(2 \cdot \frac{\pi}{4}\right), \] \[ 2 = \frac{k}{4} \cdot \frac{1}{16} \cdot 1, \quad k = 8. \] Step 6: Conclusion. The value of \( k \) is \( {8} \).