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[unreal dirge]

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[sullen drum]

Have not done any work -

[rustic bone]

You could try two things here

First of all, if you denote L = (Δ + 100²), and the rhs f(x, y)

you want to find the following

1. the solutions of Lu = 0
2. a particular solution to Lu = -f

You may have a good shot if you look at stationary solutions in the form u(x, y) = v(x)w(y)

[sullen drum]

This was the also what I thought

[sullen drum]

Somehow I can use this to do Separation of variables

onto a Sturm Louiville System

@rustic bone

[rustic bone]

I dunno what that is

but if you have a solution for it, use it

[rustic bone]

Here I think in 2) you can get away with it with a little guess work

but in 1) it's a lot more complicated

According to wikipedia you solve it with separation of variables yes

[sullen drum]

Can you write it out?

[rustic bone]

$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + k^2 u = 0$

[split flare]

Rion

[rustic bone]

as we substitute

$\frac{d^2 v}{d x^2}(x) w(y) + v(x) \frac{d^2 w}{d y^2}(y) + k^2 v(x)w(y) = 0$

[split flare]

Rion

[rustic bone]

Somehow this is just screaming periodic functions

I found some solutions, which are as follows:

$v(x) = A_1\cos\left( \frac{k}{\sqrt{2}} x + \phi_1 \right)$

and

$w(y) = A_2\cos\left( \frac{k}{\sqrt{2}} y + \phi_2 \right)$

Or anything in their span I guess

[split flare]

Rion

[rustic bone]

However I don't really know how to show that we do not have any other plausible solution

which may not be true anyway

yeah but that's absolutely rigged

it doesn't exactly show that there is no other solution

that is not a combination of those

And it doesn't help our cause anyway, because we want to get a homogeneous equation solution

But in a sense I suppose that if we know that the above v and w will help construct a solution

then we'd want to look into sums of those solutions I suppose

$\sum_{i=0}^{\infty} a_j \cos\left( \frac{k}{\sqrt{2}} x + \phi_{1j} \right) \cos\left( \frac{k}{\sqrt{2}} y + \phi_{2j} \right)$

[split flare]

Rion

[rustic bone]

@sullen drum

This page gives you insights on how to find solutions for Lu = 0

just copy that shit

it's close enough to what i had in mind

[sullen drum]

hmm

[rustic bone]

For the particular solution you may want to try u(x, y) = v(x)w(y) with v and w quadratics wrt x and y

and you want to identify the coefficients

by expanding the rhs too

maybe that will work

[sullen drum]

Not sure I still get it

[rustic bone]

Well you see that the rhs is some sort of polynomial with x and y

So a plausible idea would be to look for polynomial particular solutions as well

[rustic bone]

Consider:

$u(x, y) = Kx(1-x) y(1-y) + \alpha y(1-y) + \beta x(1-x) + \alpha + \beta$

[split flare]

Rion

[rustic bone]

Then:
$\frac{\partial^2 u}{\partial x^2} (x, y) = -2Ky(1-y) - \beta$

[split flare]

Rion

[rustic bone]

And:
$\frac{\partial^2 u}{\partial y^2} (x, y) = -2Kx(1-x) - \alpha$

[split flare]

Rion

[rustic bone]

So:

$\frac{\partial^2 u}{\partial x^2} (x, y) + \frac{\partial^2 u}{\partial y^2} (x, y) + k^2 u(x, y) = k^2 Kx(1-x)y(1-y) + (\alpha - 2K)y(1-y) + (\beta-2K)x(1-x)$

[split flare]

Rion

[rustic bone]

Pick $\alpha = 2K$ and $\beta = 2K$ and then you've got yourself:

$\frac{\partial^2 u}{\partial x^2} (x, y) + \frac{\partial^2 u}{\partial y^2} (x, y) + k^2 u(x, y) = k^2 Kx(1-x)y(1-y)$

[split flare]

Rion

[rustic bone]

With k = 100² = 10^4, you just gotta choose K = -10² to get the result @sullen drum

That will get you a particular solution

But I haven't checked at all whether it fits the boundary conditions

$u(x, y) = K \left( x(1-x) y(1-y) + 2y(1-y) + 2x(1-x) + 4 \right)$

[split flare]

Rion

[rustic bone]

doesn't look like it though, maybe you need some additional term

or i guess adjust the homogeneous solution

[sullen drum]

This doesnt seem quite right

thank you tho

[rustic bone]

Why

[sullen drum]

@rustic bone

[rustic bone]

I just fail to understand why we need to go through such a complicated method to find a particular solution

Or even how my particular solution is invalid

but I mean if that's what's expected of you then sure

[sullen drum]

Thank you @rustic bone +close