# Solution Poisson Equation by Homotopy Perturbation Method

I need to solve the following Boundary Value problem

$$\frac{\partial^2w }{\partial x^2}+\frac{\partial^2w }{\partial y^2}=c$$

Boundary conditions are

$$w(x,h)=0$$

$$w\left(\frac{\pm h}{\sqrt3},y\right)=0$$

I am trying to solve it by Homotopy Perturbation Method, but unable to succeed to get desired answer which should be

$$w=\frac{c(y-h)(3x^2-y^2)}{4h}$$

• is $c$ constant? and what's the domain of the problem? – Dylan Feb 28 at 14:45
• Yes $c$ is contant,and I think it's in real domain. – Ubaid Ur Rehman Mar 2 at 10:45
• It can't be $\Bbb R^2$ since your boundaries are $x=\pm h/\sqrt{3}$ and $y=h$. Either it's $(-h/\sqrt{3},h/\sqrt{3}) \times (h,\infty)$ or $(-h/\sqrt{3},h/\sqrt{3}) \times (-\infty,h)$. I'm asking you which is it? – Dylan Mar 2 at 13:34
• its $(-h/sqrt(3), h/sqrt(3)) × (h,∞)$ – Ubaid Ur Rehman Mar 6 at 12:54

## 1 Answer

You might want to verify your given solution. I don't think it satisfies the equation.

First, let

$$w(x,y) = \frac{c}{6}(3x^2-h^2) + v(x,y)$$

then $$v(x,y)$$ is harmonic with boundary conditions

$$\begin{cases} v\left(\pm \frac{h}{\sqrt3},y\right) = 0 \\ v(x,h) = -\frac{c}{6}(3x^2-h^2) \end{cases}$$

The solution to this problem isn't unique. If we assume $$v(x,y\to\infty)$$ is bounded, then separation of variables gives

$$v(x,y) = \sum_{n=1}^\infty c_n \exp\left(-\frac{n\pi\sqrt 3}{2h}(y-h)\right) \sin\left(\frac{n\pi}{2h}\big(\sqrt3 x+h\big)\right)$$

where

$$c_n = \frac{\sqrt 3}{h}\int_{-h/\sqrt3}^{h/\sqrt3} v(x,h) \sin\left(\frac{n\pi}{2h}\big(\sqrt3 x+h\big)\right) \ dx$$