- #1
Dustinsfl
- 2,281
- 5
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.
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.
Last edited: