[SOLVED]Laplace on a plate with corners excluded

dwsmith

Well-known member
We have a square plate with length $$100$$. Three side are $$0$$ degrees and one side is kept at $$100^{\circ}$$. I left $$T(100, y) = 100$$ and the other than were zero. Small areas near the two corners must be consider excluded. So I took this as the corners at $$(100,100)$$ and $$(100,0)$$. However, I don't know what or how I am supposed to use this information. I solved the problem normally--maybe this is not how to do it with the corner stipulation or it could be correct.

So the highlights of the solution are:

$$T(x,y)=\varphi(x)\psi(y)$$ so $$\frac{\varphi''}{\varphi} = -\frac{\psi''}{\psi} = \lambda^2$$.
\begin{align}
\varphi(x) &\sim \{\cosh(\lambda x), \sinh(\lambda x)\}\\
\psi(y) &\sim \{\cos(\lambda y), \sin(\lambda)\}
\end{align}
Then $$\lambda = \frac{\pi n}{100}$$ and
$T(x,y) = \sum_{n=1}^{\infty}A_n\sin\left(\frac{\pi n}{100}y\right)\sinh\left(\frac{\pi n}{100}x\right).$
Using the last condition, we have
$A_n = \begin{cases} 0, & \text{if $$n$$ is even}\\ \frac{400}{\pi n\sinh(\pi n)}, & \text{if $$n$$ is odd} \end{cases}$
so
$T(x,y) = \frac{400}{\pi}\sum_{n=1}^{\infty}\frac{1}{(2n-1)\pi\sinh[(2n-1)\pi]}\sin\left(\frac{\pi (2n-1)}{100}y\right)\sinh\left(\frac{\pi (2n-1)}{100}x\right).$
Since we are exlcuding the corners, does this work or is there a tweak I needed to execute some where prior?
If the above is correct, would calculation the approximate value near the corners simply be evaluating at $$(100,100)$$ and $$(100,0)$$?

Or do we discuss the Gibbs phenomena? That is usually a 9% distortion at the corners.

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