PDEs: Laplace's Equation over a Parallelogram

[Master1022]

TL;DR

This isn't a school problem, just a conceptual question.

Hi,

I have been learning about Laplace's equation recently, and have been wondering: how would we approach the problem if the region was a parallelogram (or some other shape that isn't a standard rectangle or circle)? Is this something that could feasibly be solved by hand, or would it require computational/numerical methods?

All the examples I have seen are either with rectangles or circles, but I was just wondering what the best approach would be if we were given some region (eg. 0 \leq y \leq 1 and y \leq x \leq y + 1)? I have only learned to solve Laplace's equation by separation of variables (we haven't used any computational methods, but we have learned about Laplace transforms for solving the heat and wave equations)

My thoughts on different approaches:
1. Try and fit a new coordinate system to align with the axes of our parallelogram and re-calculate the Laplacian. I didn't think that this would work as our generalised coordinate Laplacian is only defined for orthogonal coordinate systems.

2. Use a coordinate transformation to transform it to a square

3. Perhaps not feasible/ worth the time and better to just use a computer

This isn't a homework problem or anything, I was just wondering how to deal with a situation which I wouldn't initially think was a standard case.

Any help is greatly appreciated. Thanks in advance

 

[Chestermiller]

Who says that the Laplacian can only be expressed for orthogonal coordinate systems?

 

[Master1022]

Who says that the Laplacian can only be expressed for orthogonal coordinate systems?

The Laplacian can be expressed for non-orthogonal coordinate systems, but in our lectures we have only seen the derivations for the orthogonal systems (probably to keep the algebra simple). Would that be the best method - find the laplacian in those (non-orthogonal) coordinates and solve that PDE?

Thanks for your reply.

(NB. I only mention the content of my lectures to see whether it would be possible to do with my current knowledge)

 

[Antarres]

Well, your idea is fine, but why not do it in different order?

1. Find coordinates that span the parallelogram.
2. Find a coordinate transformation that transforms parallelogram into a rectangle.
3. Solve the equation
4. Transform back

This should work, it's just a sort of a substitution, a linear one even.

 

[Orodruin]

Well, your idea is fine, but why not do it in different order?

1. Find coordinates that span the parallelogram.
2. Find a coordinate transformation that transforms parallelogram into a rectangle.
3. Solve the equation
4. Transform back

This should work, it's just a sort of a substitution, a linear one even.

The Laplacian is not going to be invariant under this coordinate transformation.

 

[Antarres]

I know that it isn't, but I assumed that the following holds:

Let's say we have function $f(x,y)$ that satisfies differential equation $\Delta_{xy} f = 0$. Then let's say that there is a transformation of coordinates $u=u(x,y)$, $v=v(x,y)$.
Then function $g(u,v) = f(x(u,v),y(u,v))$ should satisfy $\Delta_{uv} g = 0$. That is, if we transform the whole equation, then we should get a transformed solution. And since transformation is linear, it should be easy to invert it, so I meant to form an equation which you know how to solve, and make a transformation like above to the new coordinates to acquire the transformed solution. Obviously boundary and initial conditions should also be transformed.

This was just intuitive idea, I guessed that there should be no preferent coordinate system in which a certain problem should be solved, just some coordinate systems make the problem easier. I checked that it works on a couple ordinary equations, but haven't checked them on PDEs really, so maybe I'm completely wrong. But I'd like to see the correction, in case I am.

 

[Orodruin]

I know that it isn't, but I assumed that the following holds:

Let's say we have function $f(x,y)$ that satisfies differential equation $\Delta_{xy} f = 0$. Then let's say that there is a transformation of coordinates $u=u(x,y)$, $v=v(x,y)$.
Then function $g(u,v) = f(x(u,v),y(u,v))$ should satisfy $\Delta_{uv} g = 0$. That is, if we transform the whole equation, then we should get a transformed solution. And since transformation is linear, it should be easy to invert it, so I meant to form an equation which you know how to solve, and make a transformation like above to the new coordinates to acquire the transformed solution. Obviously boundary and initial conditions should also be transformed.

This depends on what you mean with $\Delta_{uv}$. If you mean the Laplace operator expressed in the new coordinate system, yes. However, if you necessarily mean $\Delta_{uv} = \partial_u^2 + \partial_v^2$ then no. The Laplace operator takes this form only in Cartesian coordinates. For example, the Laplace operator expressed in polar coordinates $r$ and $\phi$ on $\mathbb R^2$ is given by
$$
\nabla^2 = \partial_r^2 + \frac 1 r \partial_r + \frac{1}{r^2} \partial_\phi^2.
$$

 

[Antarres]

Yes yes, I meant the transformed one, of course.

 

[Orodruin]

Yes yes, I meant the transformed one, of course.

The problem with going to an oblique coordinate system is that the Laplace operator will no longer be separable and therefore you are going to run into different problems by transforming to oblique coordinates.

 

[Antarres]

Well, what I meant was that we transform from the oblique coordinates into rectangular ones, which changes our function and our boundary conditions, but makes the equation maybe more easy to solve. And then when we get the solution, we transform back.

Of course, this may not really simplify the problem, since the function that is being acted on might take on a more complicated form, but in principle, it should be viable in some cases. So that's what I wanted to check if it's right.

Obviously Laplace operator is not invariant, so it will take on another form when we transform to paralellogram, but instead of trying to solve the equation in that form, we transform the coordinates so that Laplace operator becomes separable again. Isn't that possible?

 

[jasonRF]

Well, if you want to do a coordinate mapping, you preserve the Laplacian if the mapping is conformal. In principle you can use the Schwarz-Christoffel mapping to map the interior to the upper half-plane
https://en.m.wikipedia.org/wiki/Schwarz–Christoffel_mapping
it might be easier to solve there, but the transformation will not yield nice simple functions. You will get some kind of elliptic functions just for the coordinate transformation, I think. In theory this is an analytical technique, but as domains get more complicated this quickly requires a numerical method. Prof Trefethen (in UK, Oxford or Cambridge, I forget) developed software to do this. A free mat lab version is linked on that Wikipedia page.