2
$\begingroup$

I was recently studying Laplace's equation and set myself the problem to solve it on a isoceles, right triangular domain.

Two possible approaches came to my mind:

a) transform the coordinates into a nicer domain

b) exploit the symmetry or mean value property of harmonic functions

Regarding a): I am terribly bad at coordinate transformations, so I avoided this immediately. However it seems like a very useful approach for this and more complicated domains. I'd be very thankful if anyone could show how to transform the domain for this specific case and give me some guidance on how to approach these transformations.

For b), I defined the following BCs as to make life a little easy:

$$ u(x,y=0) = 1 - \frac{x}{2\pi}$$ $$ u(x=0,y) = 1 - \frac{y}{2\pi}$$ $$ u(x,2\pi-x) = 0$$

where the lengths of the cathetus are $2\pi$.

From the MVP for harmonic functions I know that the center of a circular domain is given by the average of the value of the boundary. This can obviously not be straightforwardly extended to a square domain, but my intuition was that it could be extended to the diagonal: the diagonal of a square domain is equal to the average of the squares boundary.

My whole attempt is based on this assumption. Since I am a mere engineer with next to no proficiency in proofs, I just went with it but I would be very thankful again if someone could point me to a proof or give a rough proof sketch.

Now using this idea, I decided to solve Laplace's equation on a square of length $2\pi$ and defined the following two additional BCs to exploit my intuitive notion of the MVP:

$$ u(x,y=0) = \frac{x}{2\pi}$$ $$ u(x=0,y) = \frac{y}{2\pi}$$

I then proceeded as usual for a square domain and split the problem into four subproblems, each with one of the specified BCs per side and the remaining sides set to zero, giving four infinite series of combinations of $sin$ and $sinh$ where the constants are given by the fourier sine expansion of the inhomogenous boundary.

The overall solution is the superposition of the four individual solutions, so I end up with a sum of four series. I can see that the satisfy the BCs on the cathetus, but I am unsure about the diagnonal because of the terms of the type

$$\frac{1}{\pi\mathrm{sinh}(2\pi n)}$$

they do tend to zero as n increases, so they would solve the equation if there was a negative sign in front of one of the two coordinate pairings. Have I just made a sign error or is my approach fundamentally flawed by my MVP assumption?

edit: realised its mean value property (MVP) not the MVT and changed accordingly

$\endgroup$
1
$\begingroup$

Consider the extended problem

\begin{cases} \nabla^2 v = 0 & \text{on } (x,y) \in [0,2\pi]^2 \\ v = u & \text{on } (x=0,y) \cup (x,y=0) \\ v = 0 & \text{on } (x=2\pi,y) \cup (x,y=2\pi) \end{cases}

Now, define $v^*(x^*,y^*)$ such that $(x^*,y^*)$ is the reflection of $(x,y)$ about the line $y=2\pi - x$ (third side of your triangle). Using analytic geometry, we find

\begin{align} x^* &= 2\pi - y \\ y^* &= 2\pi - x \end{align}

Consequently, $v^*(x^*,y^*) = v(2\pi-y,2\pi-x)$ is also harmonic and is zero on the 2 sides $(x=0,y)\cup (x,y=0)$. Therefore

$$ u(x,y) = v(x,y) - v(2\pi-y,2\pi-x) $$

is a solution of the original problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.