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

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

Y(0) = C + D = 100

D = 100 - C

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

Ce^{3n\pi} = -De^{-3n\pi}

C = -De^{-6n\pi}

C = -(100 - C)e^{-6n\pi}

C = (C - 100)e^{-6n\pi}

C = Ce^{-6n\pi} - 100e^{-6n\pi}

C - Ce^{-6n\pi} = -100e^{-6n\pi}

C(1 - e^{-6n\pi}) = -100e^{-6n\pi}

C = \dfrac{-100e^{-6n\pi}}{1-e^{-6n\pi}}

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

D = 100 - \dfrac{100e^{-6n\pi}}{e^{-6n\pi} - 1}

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

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

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

but then...

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})

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 C = -\dfrac{1}{2}e^{-30k} and when D = \dfrac{1}{2}e^{30k}.

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:
Y'' - \dfrac{n^2 k^2}{10^2}Y = 0

Then we find from it:

Y = Ce^{\dfrac{n \pi y}{10}} + De^{-\dfrac{n \pi y}{10}}

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

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

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

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.