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Heat equation with Laplace transform

  1. Nov 23, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Problem 8-19 in Matthews and Walker's book on mathematical physics.
    A straight wire of radius a is immersed in an infinite volume of liquid. Initially the wire and the liquid have temperature T=0. At time t=0, the wire is suddenly raised to temperature ##T_0## and maintained at that temperature. Find F(r,s), the Laplace transform of the resulting temperature distribution T(r,t) in the liquid.


    2. Relevant equations
    Heat eq.: ##\frac{\partial T}{\partial t} - \kappa \triangle T =0##.
    Laplace transform: ##\mathcal{L}[f,s]=\int _0^\infty e^{-st} f(t)dt##.


    3. The attempt at a solution
    First, I assume that "r" stands for the direction of the straigth wire. Second, the Laplace transform will be with respect to t, since t goes from 0 to infinity just like the limits of the integral of the L. transform. Third, I know that the solution has symmetry from ##r=a/2##, that is when I place the middle point of the wire at ##r=0##.
    I tried to mathematically write down the boundary-initial condition. This gave me ##T(r,t)=T_0## for ##0 \leq r \leq a##. And ##T(r,0)=0## for ##r>a##. That division in space basically ruins my hopes to solve the problem.
    So ##F(r,s)=\int _0^\infty e^{-st}T(r,t)dt##.
    Therefore I must calculate ##T(r,t)##.
    I guess I have to Laplace transform the initial-boundary conditions and the heat equation itself.
    So ##\mathcal{L}\left [ \frac{\partial ^2 T}{\partial r^2} \right ]=s^2 \mathcal{L}[T(r,t)]-T_0s##, because I believe that ##T'(0,t)=0## due to the symmetry of the problem.
    Now, ##\mathcal{L} \left [ \frac{\partial T}{\partial t} \right ]=T_0+s\mathcal{L}[T(r,t)]## (I found that by integration by parts).
    So that the PDE reduces to ##\kappa s^2 \mathcal{L}[T(r,t)]-\kappa T_0 s -(T_0+s\mathcal{L}[T(r,t)])=0##. I've reached that ##\mathcal{L}[T(r,t)]=T_0 \frac{\left ( \frac{1}{s} +\frac{1}{\kappa s^2} \right )}{\left ( 1-\frac{1}{\kappa s} \right )}##.
    Would this be the answer to the problem? I'm quite skeptic since I never applied Laplace transform onto the boundary/initial conditions. I wonder what I did wrong.
     
  2. jcsd
  3. Nov 24, 2012 #2

    haruspex

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    I'm not going to be able to answer your question about L(initial conditions), but I'll make a few points.
    You mean, distance from the centre of the wire, yes?
    Symmetric about r=0, surely.
    That's not the right form for Δ in polar, is it? I thought it was ##\frac{1}{r}\frac{∂}{∂r}\left(r\frac{∂}{∂r}\right)##
     
  4. Nov 24, 2012 #3

    fluidistic

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    Thanks for the help haruspex.
    Good point, yes.

    Once again yes.

    Hmm I don't really understand why it should be in polar coordinates?! For me it was "r" just like "x" in Cartesian coordinates. And now that I think about it, why not "r" in spherical or cylindrical (ok this one would be the same as for polar)?
    But in spherical coordinates it would be ##\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right )##
     
  5. Nov 24, 2012 #4

    haruspex

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    Because it is a straight wire. (Doesn't say, but I presume it is infinite in length.) So this is a cylindrical case, which is the same as 2D polar.
    The form you assumed for Δ is only valid in Cartesian (and since you only had one independent variable, it would be 1D Cartesian, or an infinite wire in 2D, or an infinite plane in 3D...).
     
  6. Nov 24, 2012 #5

    fluidistic

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    Hmm I still don't see why this discard the spherical possibility.
    Anyway it has a radius "a" as stated in the problem statement, I assume Matthews and Walker mean "length". This is how I understood -and would love to solve- the problem.
     
  7. Nov 24, 2012 #6

    haruspex

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    No, I'm pretty sure they mean a long straight wire (maybe infinite) of radius a, and r is the distance from the centre of the wire (so will be measured perpendicularly to the wire axis).
     
  8. Nov 24, 2012 #7

    fluidistic

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    Oh... I get the picture now. Wow, totally different from what I understand. Indeed your understanding sounds way more plausible. Thanks. I'll rethink the problem.
     
  9. Nov 24, 2012 #8

    fluidistic

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    I reach that ##\frac{\partial ^2 }{\partial r^2}\mathcal{L}[T] +\frac{1}{r} \frac{\partial}{\partial r} \mathcal{L}[T]-\frac{s}{K}\mathcal{L}[T]=0## which to me does not seem a very simple ODE. I reached this because I calculated ##\mathcal{L} \left [ \frac{\partial T}{\partial t} \right ]=s\mathcal{L}[T]+Te^{-st} \big | _{t=0}^{t=\infty}=s\mathcal{L}[T]## because I'm interested in the region ##r>a##.
    And ##\mathcal{L} \left [ \frac{\partial T}{\partial r} \right ]=\frac{\partial}{\partial r} \mathcal{L}[T]##, furthermore ##\mathcal{L} \left [ \frac{\partial ^2 T}{\partial r^2} \right ]=\frac{\partial ^2 }{\partial r^2}\mathcal{L}[T]##. I made a mistake in my first post (I messed up the variable of integration).
    Now I'm a bit lost. I must solve for ##\mathcal{L}[T]## in the ODE. But that Laplace transform is a function of T and s. T turns out to be a function of r and t. So that ##\mathcal{L}[T]## does not depend explicitly on r and hence its partial derivative with respect to r should be 0? I guess not, but I don't see why not.
     
  10. Nov 25, 2012 #9

    haruspex

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    I think you dropped a 2:
    ##\frac{\partial ^2 }{\partial r^2}\mathcal{L}[T] +\frac{2}{r} \frac{\partial}{\partial r} \mathcal{L}[T]-\frac{s}{K}\mathcal{L}[T]=0##
    Try solutions like r-1e-λr, where λ is a function of s.
     
  11. Nov 25, 2012 #10

    fluidistic

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    Hmm I've rechecked the arithmetics, I don't think I've dropped a 2. The 2 would appear (I believe) if I had taken the Laplacian in spherical coordinates. I notice that the solution you propose is similar to the spherical Bessel function of the first kind and 0th order.
    But if there's no factor 2, maybe it should be a solution of the form of a Bessel function?
     
  12. Nov 25, 2012 #11

    TSny

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  13. Nov 25, 2012 #12

    haruspex

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    Sorry, you're right. I had gone back to look at the form of the equation I had suggested to you (2D), but by mistake read the 3D one you had proposed!
     
  14. Nov 25, 2012 #13

    Mute

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    Nevermind. (Don't seem to have the option to delete the post).
     
  15. Nov 25, 2012 #14

    fluidistic

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    Ah yes thanks a lot. First time I deal with it. So the solution to the problem is ##\mathcal{L}[T]=AI_0 \left ( \sqrt{\frac{\kappa}{s}}r \right )##.

    Ah, no problem. :smile:

    No problem, and I think you should have the option (at the bottom of the text part when you press Edit if I remember well).
     
  16. Nov 25, 2012 #15

    TSny

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    Do you want the "first kind" ##I_0## or the "second kind" ##K_0##? Which has the proper behavior at infinity?
     
  17. Nov 25, 2012 #16

    fluidistic

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    Hmm I see. But K_0 seems unbounded for r=0 (although I agree in this exercise I'm only interested in the region r>a.
    So here the answer would then be ##\mathcal{L}[T]=AK_0 \left ( \sqrt{\frac{\kappa}{s}}r \right )##.
     
  18. Nov 25, 2012 #17

    TSny

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    Right, you don't need to worry about r = 0. Choose A to satisfy boundary condition.
     
  19. Nov 25, 2012 #18

    fluidistic

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    Ok thanks a lot. I reach ##A=\frac{T_0}{s K_0 \left ( \sqrt{\frac{\kappa a}{s}} \right ) }##.
    So [tex]\mathcal{L}[T]= \frac{T_0 K_0 \left ( \sqrt{\frac{\kappa r }{s}} \right ) }{sK_0 \left ( \sqrt {\frac{\kappa a }{s}}\right ) }[/tex].
     
  20. Nov 25, 2012 #19

    TSny

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    You might check to see if you need to flip ##\frac{\kappa }{s}## to ##\frac{ s }{\kappa}## in your square roots and it looks like you made a typo by putting r and a inside the square roots.
     
  21. Nov 25, 2012 #20

    fluidistic

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    You are absolutely right. Fixed and fixed. :)
     
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