# Confusion about wave equation on triangular membrane

I was reading the Wolfram MathWorld article Wave Equation--Triangle on the equation of motion for a membrane shaped as a right isosceles triangle.

My confusion is centred on the orientation of the triangle.
How was it determined that the superposition of two wave solutions for a square membrane with the indices reversed yields a triangular membrane with vertices at the points $$(0,0)$$, $$(0,c)$$ and $$(c,0)$$ (as opposed to $$(0,0)$$ , $$(0,c)$$ and $$(c,c)$$)?

Moreover, this doesn't appear to be the most general solution since this necessarily implies $$\psi=0$$ along the $$x=y$$ diagonal (which isn't necessarily true of all waves on a triangular membrane).

Any insight would be much appreciated.

Thanks

The derivation from the wave equation for a square membrane was understandable until the line:

"Since points on the diagonal which are equidistant from the center must have the same wave equation solution (by symmetry), this procedure gives a wavefunction which will vanish along the diagonal as long as p and q are both even or odd."

Why must p and q both be even or odd for the wave function to vanish along the diagonal (i.e. to satisfy the boundary condition there)?

The wave equation solution doesn't appear to imply that but I must be missing something.

Edit 2

I think I now understand why $$\psi=0$$ along the diagonal.
If that weren’t the case, the diagonal boundary could oscillate upwards and downwards.
So are we imposing the boundary condition that the diagonal vanishes?

I still don’t understand the general formula. In particular, why must $$p>q$$?

Thanks for any further insight.

• I deleted my answer because it became useless after you changed your question. Commented Nov 20, 2023 at 23:12
• @Gonçalo I included the original question in the edit. I thought your answer was helpful but I understand your decision. Thanks for your insight. Commented Nov 21, 2023 at 0:45