jkwiens.com: Answer: Solve Laplace’s Equation

Answer: Solve Laplace’s Equation

Monday, 17 of March , 2008 at 9:10 pm

Question: Find the solutions to Laplace’s Equation:

$\frac{\partial^2 u(x,y)}{\partial x^2} + \frac{\partial^2 u(x,y)}{\partial y^2} = 0$

Answer:
First assume that the solution to the PDE

$\frac{\partial^2 u(x,y)}{\partial x^2} + \frac{\partial^2 u(x,y)}{\partial y^2} = 0$

has the form

$u(x,y) = \phi(x) \xi (y)$

Therefore, we can reduce the equation to the following:

$\xi (y) \frac{\partial^2 \phi(x)}{\partial x^2} + \phi(x) \frac{\partial^2 \xi (y)}{\partial y^2} = 0$
$\frac{1}{\phi (x)} \frac{\partial^2 \phi(x)}{\partial x^2} = \frac{-1}{\xi (y)}\frac{\partial^2 \xi (y)}{\partial y^2}$

Since this equation is true for all x and y, therefore both sides of the equation must equal a constant.

$\frac{1}{\phi (x)} \frac{\partial^2 \phi(x)}{\partial x^2} = \frac{-1}{\xi (y)}\frac{\partial^2 \xi (y)}{\partial y^2} = \lambda$

This implies that we need to solve two ODEs.

$\frac{\partial^2 \phi(x)}{\partial x^2} =\lambda\phi (x)\frac{\partial^2 \xi (y)}{\partial y^2} = -\lambda\xi (y)$

Since the solutions to the two ODEs will be very similar, I will solve the ODE

$\frac{\partial^2 \Psi}{\partial r^2} =\kappa \Psi(r)$

and apply the results to the two ODEs.

There are 3 cases which we need to solve ( $\kappa > 0$ , $\kappa = 0$ , and $\kappa < 0$ ).

Case $\kappa = 0$
The ODE for this case would be

$\frac{\partial^2 \Psi}{\partial r^2} =0$

which has the solution

$\Psi = C_1 r + C_2$

Case $\kappa < 0$ and $\kappa > 0$
The ODE

$\frac{\partial^2 \Psi}{\partial r^2} =\kappa \Psi(r)$

which will have the solution

$\Psi = B_1 e^{\sqrt{\kappa}r} + B_2 e^{-\sqrt{\kappa}r}$

Please note that $\sqrt{\kappa}$ will be an imaginary number when $\kappa < 0$ .

Therefore, if we apply the above solution, we can find the functions that solve the ODEs

$\frac{\partial^2 \phi(x)}{\partial x^2} =\lambda\phi (x)\frac{\partial^2 \xi (y)}{\partial y^2} = -\lambda\xi (y)$

which would be

$\phi (x)= A_1 e^{i\sqrt{\lambda}x} + A_2 e^{-i\sqrt{\lambda}x}$

$\xi (y)= B_1 e^{\sqrt{\lambda}y} + B_2 e^{-\sqrt{\lambda}y}$ when $\lambda < 0$

$\phi (x)= A_1 e^{\sqrt{\lambda}x} + A_2 e^{-\sqrt{\lambda}x}$

$\xi (y)= B_1 e^{i\sqrt{\lambda}y} + B_2 e^{-i\sqrt{\lambda}y}$ when $\lambda > 0$

$\phi (x)= C_1 x + C_2$

$\xi (y)= D_1 y + D_2$ when $\lambda = 0$

Therefore, the solution would have the form

$u(x,y) = \begin{cases}(A_1 e^{i\sqrt{\lambda}x} + A_2 e^{-i\sqrt{\lambda}x})(B_1 e^{\sqrt{\lambda}y} + B_2 e^{-\sqrt{\lambda}y}) & for~\lambda < 0 \\ (A_1 e^{\sqrt{\lambda}x} + A_2 e^{-\sqrt{\lambda}x})(B_1 e^{i\sqrt{\lambda}y} + B_2 e^{-i\sqrt{\lambda}y}) & for~\lambda > 0 \\ (C_1 x + C_2)(D_1 y + D_2) & for~\lambda = 0\end{cases}$

Any superposition of the above equation will satisfy Laplace’s equation. In order to reduce this solution more, we would need to be given Boundary and Initial Conditions.

Category: Answers, Differential Equations

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