1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    2T = 0

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

    X(x) = Acos(kx) + Bsin(kx)
    Y(y) = Ceky+De-ky

    3. 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?
     
  2. jcsd
  3. May 5, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. May 5, 2013 #3
    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.
     
  5. May 5, 2013 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
    Last edited: May 5, 2013
  6. May 5, 2013 #5
    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. :)
     
  7. May 5, 2013 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
    Last edited: May 5, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Laplace's Equation: Steady-State Temperature in a Rectangular Plate
Loading...