Solve PDE Uxx+Uyy=-2 with Boundary Conditions

[chwala]

Homework Statement

solve the inhomogenous pde below by considering the boundary conditions given

Relevant Equations

separation of variables.

ial

this is the question.

 

[chwala]

its very long since i solved such questions during my post graduate class, i am trying to see if i can remember...
i attempted to solve the homogenous part, i just want to know if i am on the right track...

am i thinking right or i am missing something...from literature i can see that this does not qualify to be solved by the approach i have used of trying to solve homogenous pde...breaking them into ode. Since we have $-2$ on the rhs then it becomes poisson equation?

 

[pasmith]

As with any linear problem the answer is "take something which satisfies the RHS and add something from the kernel".

So here: Find a function of (x,y) which satisfies \nabla^2 f = -2 and add solutions of Laplace's equation to satisfy the boundary condition. This is easier if your particular solution satisfies as much of the boundary condition as possible. Here I think you have less work if your function vanishes on x = 0 and x = 1.

Here you have u(x,0) = u(x,1) = u(1,-x) = -\sinh \pi \sin(\pi x).

Don't forget that you can use \cosh (kx) and \sinh(kx) in place of e^{\pm kx} and that \sinh(k(1 - x)) is a linear combination of these.

 

[chwala]

I should be able to look at this during december...i have had too much work on my desk... and at same time need to familiarise with literature on inhomegenous pde's...

 

[chwala]

* Am on this now...allow me to deal with it slowly...am now conversant with pde's. This is Laplace equation...steady-state equation (Equilibrium solution).

We shall solve this by separation of variables, we shall seek solutions of the form:

$U(x,y) = X(x) Y(y)$

For the Homogenous part we shall have;

$X^{''}(x) Y(y) + X(x) Y^{''} (y) = 0$

it follows that,

$\dfrac{X^{''}(x)}{X(x)} = -\dfrac{Y^{''}(y)}{Y(y)} =±μ^2$

case 1; using eigenvalue $[λ=μ^2]$

$X^{''} -μ^2X=0$
$Y^{''}+μ^2Y=0$

$X^{''} -μ^2X=0$

it follows that

$m^2-μ^2=0$ is our characteristic equation, solving we get;

$m^2=μ^2$

$⇒m=\sqrt{μ^2}=μ$ our solution would therefore be of the form;

$X=e^{μx}=A \cosh μx +B \sinh μx$

$X=A \cosh μx +B \sinh μx$

...will continue later.

 

[pasmith]

Solution

Spoiler

<br /> u(x,y) = x(1-x) - \sum_{n=1}^\infty Y_n(y) a_n \sin (n \pi x) + \sinh(\pi x)\sin(\pi y) <br /> - ( \sinh (\pi y) + \sinh (\pi(1 - y)) )\sin (\pi x) where <br /> Y_n(y) = \frac{\sinh(n\pi y) + \sinh(n\pi(1-y))}{\sinh (n \pi)}<br /> is chosen such that Y_n(0) = Y_n(1) = 1 and <br /> \sum_{n=1}^\infty a_n \sin (n \pi x) = x(1-x).