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I've the following Poisson's equation:

$$\frac{\partial^2u(x, y)}{ \partial x^2} + \frac{\partial^2 u(x, y)}{ \partial y^2} = 6xy - 6xy^2 - 2x^3 = f(x, y) $$

what I'm trying to find are the original $u(x, y)$ function. So, what I'm trying to do right now is integrating the function $f(x, y)$, but I've never integrated a partial derivative.

I've seen here and here that integrating a partial derivative is similar to derivate a "normal function".

If I start integrating this as I know and if I don't do mistakes

$$\int 6xy - 6xy^2 - 2x^3 dx = \int 6xy dx - \int 6xy^2 dx - \int 2x^3 dx $$

Now, I didn't understand if I also need to consider $y$ here as a constant, but if yes, this would proceed as:

$$\int 6xy dx - \int 6xy^2 dx - \int2 x^3 dx$$

$$ 3x^2 y + C_1 - 3x^2 y^2 + C_2 - \frac{x^4}{2} + C_3 $$

I should integrate again, since we had initially a second order partial derivative... Also I integrated only with respect to $x$, so I should do the same for $y$..

Now this is not that useful but I've the following boundary conditions for the Poisson's equation:

$$u(0,y) = u(x,0) = u(x,1) = 0$$

$$u(1,y) = y(1−y)$$

Can I use this somehow to retrieve $u(x, y)$?

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  • When you integrate a function with respect to 'x' that has both 'x's and 'y's in it instead of getting a constant of integration 'C' you get a function 'h(y)'. So your integral will be of the form 'F(x,y) + H(y)'. However I do not believe that this will work here because your second partials are not with respect to the same variables. I believe that you can solve some of these equations with green's functions but they're a pain. – Zaros Nov 16 '16 at 22:21
  • Looks dubious as to the success. Take a look, instead, at Green's functions: https://en.wikipedia.org/wiki/Green's_function – avs Nov 16 '16 at 22:27

1 Answers1

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I found the same Poisson's equation/problem in this article:

http://farside.ph.utexas.edu/teaching/329/lectures/node71.html

Apparently $$u(x, y) = y(1 - y) x^3$$

Lets check that's real true, given the conditions in my question. Lets first check the boundary conditions.

  1. $u(0, y) = y(1 - y) * 0^3 = 0$

  2. $u(x, 0) = 0(1 - 0) x^3 = 0*1 * x^3 = 0$

  3. $u(x, 1) = 1 *(1 - 1) * x^3 = 1 * 0 * x^3 = 0$

  4. $u(1, y) = y(1 - y) * 1 = y(1 - y)$

So the boundary conditions are satisfied.

Now lets check the second order partial derivatives of $u(x, y)$ with both respect to $x$ and $y$.

$$u_x = \frac{\partial u(x, y)}{\partial x} = y(1 - y) 3x^2$$

then

$$u_{xx} = \frac{\partial^2 u(x, y)}{\partial x^2} = \frac{\partial u_x}{\partial x} = y(1 - y) 6x$$

Now lets take with respect to $y$

$$u_y = \frac{\partial u(x, y)}{\partial y} = x^3 - 2yx^3$$

$$u_{yy} = \frac{\partial^2 u(x, y)}{\partial y^2} = \frac{\partial u_y}{\partial x} = - 2x^3$$

Now lets verify that $\frac{\partial^2u(x, y)}{ \partial x^2} + \frac{\partial^2 u(x, y)}{ \partial y^2} = f(x, y) $.

So

$$6yx - 6xy^2 + (-2x^3) = 6yx - 6xy^2 -2x^3 = f(x, y)$$