Solve the following equation: -4u + 3u_{xx} + u_{yy} + sin(2x) = 0, on a square 0 < x, y < π subject to zero boundary conditions: u = 0 on top, bottom, left and right sides.

Steps to Solve

1. Assume a Solution Form

Since the boundary conditions are $u(0, y) = u(\pi, y) = u(x, 0) = u(x, \pi) = 0$ and there is a $sin(2x)$ term, we assume a solution of the form:

$u(x, y) = X(x)Y(y)$.

2. Consider the form of $X(x)$

Given the $sin(2x)$ term in the equation and the boundary conditions $u(0, y) = u(\pi, y) = 0$, we can deduce that $X(x)$ should be proportional to $sin(2x)$. So, we assume: $u(x, y) = f(y)sin(2x)$

3. Substitute into the PDE

Substitute $u(x, y) = f(y)sin(2x)$ into the given PDE: $-4u + 3u_{xx} + u_{yy} + sin(2x) = 0$

$-4f(y)sin(2x) + 3f(y)(-4)sin(2x) + f''(y)sin(2x) + sin(2x) = 0$

4. Simplify the equation

Divide the entire equation by $sin(2x)$ (since $sin(2x)$ is not always zero in the given domain): $-4f(y) - 12f(y) + f''(y) + 1 = 0$ $f''(y) - 16f(y) + 1 = 0$ $f''(y) - 16f(y) = -1$

5. Solve the homogeneous equation

Solve the homogeneous equation $f''(y) - 16f(y) = 0$. The characteristic equation is $r^2 - 16 = 0$, which gives $r = \pm 4$. The homogeneous solution is $f_h(y) = c_1e^{4y} + c_2e^{-4y}$.

6. Find a particular solution

For a particular solution, let $f_p(y) = A$. Then $f_p''(y) = 0$. Substituting into $f''(y) - 16f(y) = -1$, we get $0 - 16A = -1$, so $A = \frac{1}{16}$. Thus, $f_p(y) = \frac{1}{16}$.

7. General solution

The general solution is $f(y) = f_h(y) + f_p(y) = c_1e^{4y} + c_2e^{-4y} + \frac{1}{16}$.

8. Apply boundary conditions

We know $u(x, 0) = 0$ and $u(x, \pi) = 0$, so $f(0) = 0$ and $f(\pi) = 0$. $f(0) = c_1 + c_2 + \frac{1}{16} = 0$ $f(\pi) = c_1e^{4\pi} + c_2e^{-4\pi} + \frac{1}{16} = 0$

From the first equation, $c_2 = -c_1 - \frac{1}{16}$. Substituting into the second equation: $c_1e^{4\pi} + (-c_1 - \frac{1}{16})e^{-4\pi} + \frac{1}{16} = 0$ $c_1(e^{4\pi} - e^{-4\pi}) = \frac{1}{16}e^{-4\pi} - \frac{1}{16}$ $c_1(e^{4\pi} - e^{-4\pi}) = \frac{1}{16}(e^{-4\pi} - 1)$ $c_1 = \frac{e^{-4\pi} - 1}{16(e^{4\pi} - e^{-4\pi})} = \frac{e^{-4\pi} - 1}{16(e^{4\pi} - e^{-4\pi})} \cdot \frac{e^{4\pi}}{e^{4\pi}} = \frac{1 - e^{4\pi}}{16(e^{8\pi} - 1)}$ $c_1 = \frac{-(e^{4\pi} - 1)}{16(e^{8\pi} - 1)} = \frac{-(e^{4\pi} - 1)}{16(e^{4\pi} - 1)(e^{4\pi} + 1)} = -\frac{1}{16(e^{4\pi} + 1)}$

$c_2 = -c_1 - \frac{1}{16} = \frac{1}{16(e^{4\pi} + 1)} - \frac{1}{16} = \frac{1 - (e^{4\pi} + 1)}{16(e^{4\pi} + 1)} = \frac{-e^{4\pi}}{16(e^{4\pi} + 1)}$

$f(y) = -\frac{e^{4y}}{16(e^{4\pi} + 1)} - \frac{e^{-4\pi}e^{-4y}}{16(e^{4\pi} + 1)} + \frac{1}{16}$ = $\frac{1}{16} - \frac{e^{4y} + e^{4(\pi - y)}}{16(e^{4\pi} + 1)}$

$f(y) = \frac{e^{4\pi} + 1 - e^{4y} - e^{4(\pi - y)}}{16(e^{4\pi} + 1)}$

9. Final Solution

$u(x, y) = f(y)sin(2x)$ $u(x, y) = \frac{e^{4\pi} + 1 - e^{4y} - e^{4(\pi - y)}}{16(e^{4\pi} + 1)}sin(2x)$ Which can also be written as: $u(x,y) = \frac{1 - \cosh(4y - 2\pi)}{8(1 + \cosh(4\pi))}sin(2x)$