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 $u(x, y)$?
I found the same Poisson's equation/problem in this article:
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.
$u(0, y) = y(1 - y) * 0^3 = 0$
$u(x, 0) = 0(1 - 0) x^3 = 0*1 * x^3 = 0$
$u(x, 1) = 1 *(1 - 1) * x^3 = 1 * 0 * x^3 = 0$
$u(1, y) = y(1 - y) * 1 = y(1 - y)$
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
$$6yx - 6xy^2 + (-2x^3) = 6yx - 6xy^2 -2x^3 = f(x, y)$$