Laplace's Equation: Steady-State Temperature in a Rectangular Plate

[mliuzzolino]

Homework Statement

A long rectangular metal plate has its two long sides and top at 0°. The base is at 100°. The plate's width is 10cm and its height is 30cm. Find the stead-state temperature distribution inside the plate.

Homework Equations

∇²T = 0

T(x,y) = X(x)Y(y)

X(x) = Acos(kx) + Bsin(kx)
Y(y) = Ce^ky+De_-ky

The Attempt at a Solution

Using boundary conditions to obtain X(x): T(0,y) = T(10,y) = 0

[itex] X(x) = Bsin(\dfrac{n\pi x}{10}) [/itex]
Using boundary conditions to obtain Y(y): T(x, 0) = 100; T(x, 30) = 0

Y(0) = C + D = 100

D = 100 - C

[itex] Y(30) = Ce^{3n\pi} + De^{-3n\pi} = 0 [/itex]

[itex] Ce^{3n\pi} = -De^{-3n\pi} [/itex]

[itex] C = -De^{-6n\pi} [/itex]

[itex] C = -(100 - C)e^{-6n\pi} [/itex]

[itex] C = (C - 100)e^{-6n\pi} [/itex]

[itex] C = Ce^{-6n\pi} - 100e^{-6n\pi} [/itex]

[itex] C - Ce^{-6n\pi} = -100e^{-6n\pi} [/itex]

[itex] C(1 - e^{-6n\pi}) = -100e^{-6n\pi} [/itex]

[itex] C = \dfrac{-100e^{-6n\pi}}{1-e^{-6n\pi}} [/itex]

[itex] C = \dfrac{100e^{-6n\pi}}{e^{-6n\pi} - 1} [/itex]
Then by D = 100 - C

[itex] D = 100 - \dfrac{100e^{-6n\pi}}{e^{-6n\pi} - 1} [/itex]

[itex] D = \dfrac{-100e^{-6n\pi}}{e^{-6n\pi} - 1} [/itex]
So
[itex] Y(y) = \dfrac{100e^{-6n\pi}}{e^{-6n\pi} - 1} (e^{\dfrac{n\pi y}{10}} - e^{\dfrac{-n\pi y}{10}})[/itex]

[itex] Y(y) = \dfrac{200e^{-6n\pi}}{e^{-6n\pi} - 1} sinh(\dfrac{n \pi y}{10}) [/itex]

So by T(x,y) = X(x)Y(y)[itex] T(x,y) = B \dfrac{200e^{-6n\pi}}{e^{-6n\pi} - 1} sinh(\dfrac{n \pi y}{10}) sin(\dfrac{n \pi y}{10}) [/itex]

but then...

[itex] T(x,y) = \sum_{n=1}^{\infty} B \dfrac{200e^{-6n\pi}}{e^{-6n\pi} - 1} sinh(\dfrac{n \pi y}{10}) sin(\dfrac{n \pi y}{10}) [/itex]

Up until here, aside from the nastiness of the problem, I decided to read the book and see if I was on track, but they went a completely different way and ended up with something else.

They didn't solve it analytically at all, and just said that we can notice a solution, namely when [itex]C = -\dfrac{1}{2}e^{-30k} [/itex] and when [itex]D = \dfrac{1}{2}e^{30k} [/itex].

This makes sense to me, and when we analyze the summation and check the condition when T(x,0) = 100, my solution shows that T --> 0 rather than 100.

What did I do wrong, and how do I analytically acquire the solution they achieved? I'm not well versed enough to just pick a solution out of a hat like that, so how could I achieve it along the methods I was using by applying boundary conditions to the Y(y) ODEs?

 

[LCKurtz]

After you have solved the ##X## equation you have eigenfunctions ##X_n(x)=\sin(\frac {n\pi x}{10}x)## and your next step is to solve ##Y''-\frac{n^2\pi^2}{10^2}Y=0## with ##Y(0)=100,\, Y(30)=0##. Frequently in this type of problem it is better to use the {sinh,cosh} family than the exponentials. In particular, I would suggest writing the solution as$$
Y(y) = C\sinh(\frac {n\pi}{10}y)+ D\sinh(\frac {n\pi}{10}(30-y))$$Then using ##Y(30)= 0## immediately tells you ##C=0## so you have ##Y## eigenfunctions$$
Y_n(y) = \sinh(\frac {n\pi}{10}(30-y))$$Now you have the solution$$
T(x,y) = \sum_{n=1}^\infty X_n(x)Y_n(y)=
\sum_{n=1}^\infty c_n \sinh(\frac {n\pi}{10}(30-y))\sin(\frac {n\pi x}{10})$$Now apply your last boundary condition ##T(x,0)=100##:$$
100 = \sum_{n=1}^\infty c_n \sinh(\frac {n\pi}{10}(30))\sin(\frac {n\pi }{10}x)$$Now set ##c_n \sinh(\frac {n\pi}{10}(30))##equal to the Fourier coefficient and it should all work to give you an analytical solution.

 

[mliuzzolino]

Thanks for the response!

I understand where the following comes from:
[itex] Y'' - \dfrac{n^2 k^2}{10^2}Y = 0 [/itex]

Then we find from it:

[itex] Y = Ce^{\dfrac{n \pi y}{10}} + De^{-\dfrac{n \pi y}{10}} [/itex]

I understand that [itex] sinh(x) = \dfrac{1}{2}(e^x - e^{-x}) [/itex].

However, I don't understand how we can arrives at

[itex] Y(y) = C [/itex] sinh [itex] (\dfrac{ n \pi}{10}y) + D [/itex] sinh [itex] ( \dfrac{n \pi}{10} (30 - y)) [/itex]

I think this it the step that I'm completely lost on.

 

[LCKurtz]

sinh(a-y) has an addition formula similar to the trig formula, so it is a linear combination of sinh(y) and cosh(y), which in turn are linear combinations of ##e^y## and ##e^{-y}##. As long as the pair is linearly independent you can use them for the general solution. The pair I have chosen just makes the arithmetic a bit easier.

 

[mliuzzolino]

sinh(a-y) has an addition formula similar to the trig formula, so it is a linear combination if sinh(y) and cosh(y), which in turn are linear combinations of ##e^y## and ##e^{-y}##. As long as the pair is linearly independent you can use them for the general solution. The pair I have chosen just makes the arithmetic a bit easier.

I apparently need to familiarize myself more with hyperbolic functions! I should have assumed there was some trick like this. Thanks a lot for your help! It all makes sense now. :)

 

[LCKurtz]

I apparently need to familiarize myself more with hyperbolic functions! I should have assumed there was some trick like this. Thanks a lot for your help! It all makes sense now. :)

You're welcome. There is one little thing I mis-wrote. I said your ##Y## boundary value problem had ##Y(30)=0## and ##Y(0)= 100##. I shouldn't have included ##Y(0)= 100## at that point because ##100=T(x,0)=X(x)Y(0)## does not imply ##Y(0)= 100##. The nonhomogeneous condition ##100=T(x,0)## was used after the series for ##T(x,y)## was developed.