Laplace equation w/ dirichlet boundary conditions - Partial Diff Eq.

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SUMMARY

The discussion focuses on solving the Laplace equation for steady state temperature distribution T(x,y) in a flat metal sheet, represented by the equation ∂²T/∂x² + ∂²T/∂y² = 0. The boundary conditions specified are T(0,y) = T(S,y) = T(x,0) and T(x,S) = T₀. Participants noted the importance of hyperbolic functions, specifically sinh and cosh, in the Fourier series solution, correcting a sign mistake in the differential equation for Y(y). The solution involves applying boundary conditions to determine T everywhere on the square plate.

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Homework Statement



The steady state temperature distribution T(x,y) in a flat metal sheet obeys the partial differential equation:

\displaystyle \frac{\partial^2 T}{\partial x^2}+ \frac{\partial^2 T}{\partial y^2} = 0

separate the variables in this equation just like in the one-dimensional wave equation and find T everywhere on a square flat plate of sides S with the boundary conditions:

T(0,y)=T(S,y)=T(x,0) , T(x,S)=T_0

Homework Equations


The Attempt at a Solution



this is what I have so far..not sure what to do next?

[url=http://postimage.org/image/k0ujzisqz/][PLAIN]http://s17.postimage.org/k0ujzisqz/steady_state_heat_prob.jpg[/url][/PLAIN]
 
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I'm not bumping my post, sorry! I forgot to mention that classmates mentioned sinh shows up in the Fourier series and I don't know why also. Any help getting started would be greatly appreciated

many thanks
 
The differential equation for Y(y) has a sign mistake in it, though the solutions you wrote down are correct. Instead of exponentials, you can write down the solution in terms of the hyperbolic functions
\begin{align*}
\cosh ky &= \frac{e^{ky}+e^{-ky}}{2} \\
\sinh ky &= \frac{e^{ky}-e^{-ky}}{2}
\end{align*} It's analogous to how you can write down the solutions for X(x) in terms of sin kx and cos kx or in terms of eikx and e-ikx. You're just choosing a different basis.

Now you have to apply the boundary conditions.
 
Thanks vela !
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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