Solve the following Laplace equation: -8u + 2u_{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.... Solve the following Laplace equation: -8u + 2u_{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. Find u(x, y). Read more

Steps to Solve

1. Assume a solution form

Since the boundary conditions are $u=0$ on $x=0, x=\pi, y=0,$ and $y=\pi$, we can assume a solution of the form $u(x, y) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} A_{nm} \sin(nx) \sin(my)$. However, due to the form of the forcing function $\sin(2x)$, we can try a simpler solution of the form $u(x, y) = X(x)Y(y)$. Furthermore, let's assume that the solution can be written as $u(x, y) = f(y) \sin(2x)$, as dictated by the forcing function.

2. Substitute the assumed solution into the PDE

Substituting $u(x, y) = f(y) \sin(2x)$ into the given PDE $-8u + 2u_{xx} + u_{yy} + \sin(2x) = 0$, we have: $$ -8f(y)\sin(2x) + 2f(y) \frac{\partial^2}{\partial x^2}\sin(2x) + \frac{\partial^2}{\partial y^2}f(y) \sin(2x) + \sin(2x) = 0 $$

3. Calculate the derivatives

Calculate the required derivatives: $\frac{\partial^2}{\partial x^2} \sin(2x) = -4\sin(2x)$, $\frac{\partial^2}{\partial y^2}f(y) = f''(y)$.

4. Substitute derivatives back into equation

Substituting these derivatives into the equation, we get: $$ -8f(y)\sin(2x) + 2f(y)(-4\sin(2x)) + f''(y)\sin(2x) + \sin(2x) = 0 $$ $$ -8f(y)\sin(2x) - 8f(y)\sin(2x) + f''(y)\sin(2x) + \sin(2x) = 0 $$ $$ f''(y)\sin(2x) - 16f(y)\sin(2x) + \sin(2x) = 0 $$

5. Simplify the equation Dividing by $\sin(2x)$ (since $\sin(2x) \neq 0$ for $0 < x < \pi$), we have: $$ f''(y) - 16f(y) + 1 = 0 $$ $$ f''(y) - 16f(y) = -1 $$

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

7. Find particular solution For a particular solution, we assume a constant solution $f_p(y) = A$. Then $f_p''(y) = 0$. Substituting into the non-homogeneous equation gives $0 - 16A = -1$, so $A = \frac{1}{16}$. Thus, $f_p(y) = \frac{1}{16}$.

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

9. Apply Boundary Conditions

Now, we apply the boundary conditions $u(x, 0) = 0$ and $u(x, \pi) = 0$, which implies $f(0) = 0$ and $f(\pi) = 0$. $f(0) = c_1e^{0} + c_2e^{0} + \frac{1}{16} = c_1 + c_2 + \frac{1}{16} = 0$. Thus, $c_1 + c_2 = -\frac{1}{16}$.

$f(\pi) = c_1e^{4\pi} + c_2e^{-4\pi} + \frac{1}{16} = 0$. Thus, $c_1e^{4\pi} + c_2e^{-4\pi} = -\frac{1}{16}$.

10. Solve for coefficients We have the system of equations: $c_1 + c_2 = -\frac{1}{16}$ $c_1e^{4\pi} + c_2e^{-4\pi} = -\frac{1}{16}$ From the first equation, $c_2 = -\frac{1}{16} - c_1$. Substituting into the second equation: $c_1e^{4\pi} + (-\frac{1}{16} - c_1)e^{-4\pi} = -\frac{1}{16}$ $c_1e^{4\pi} - \frac{1}{16}e^{-4\pi} - c_1e^{-4\pi} = -\frac{1}{16}$ $c_1(e^{4\pi} - e^{-4\pi}) = \frac{1}{16}e^{-4\pi} - \frac{1}{16}$ $c_1 = \frac{\frac{1}{16}(e^{-4\pi} - 1)}{e^{4\pi} - e^{-4\pi}} = \frac{e^{-4\pi} - 1}{16(e^{4\pi} - e^{-4\pi})} = \frac{e^{-4\pi} - 1}{16(e^{4\pi} - e^{-4\pi})} = \frac{e^{-4\pi} - 1}{16(e^{4\pi} - e^{-4\pi})} \cdot \frac{e^{4\pi}}{e^{4\pi}} = \frac{1 - e^{4\pi}}{16(e^{8\pi} - 1)}$ $c_1 = \frac{-(e^{4\pi} - 1)}{16(e^{8\pi} - 1)} = -\frac{e^{4\pi} - 1}{16(e^{4\pi} - 1)(e^{4\pi} + 1)} = -\frac{1}{16(e^{4\pi} + 1)}$.

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

11. Final Solution Therefore, $$ f(y) = -\frac{e^{4y}}{16(e^{4\pi} + 1)} - \frac{e^{-4y}e^{4\pi}}{16(e^{4\pi} + 1)} + \frac{1}{16} = \frac{1}{16} - \frac{e^{4y} + e^{4\pi-4y}}{16(e^{4\pi} + 1)} $$ $$ u(x, y) = f(y) \sin(2x) = \left( \frac{1}{16} - \frac{e^{4y} + e^{4\pi-4y}}{16(e^{4\pi} + 1)} \right) \sin(2x) $$ $$ u(x, y) = \frac{1}{16} \left(1 - \frac{e^{4y} + e^{4(\pi-y)}}{e^{4\pi} + 1}\right) \sin(2x) $$