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Laplace Transform of Heat Equation

  • #1

Homework Statement


[itex]T(x,t)[/itex] is the temperature distribution for [itex]t > 0[/itex] in a semi-infinite slab occupying [itex]x > 0[/itex]
[itex]T(x, 0) = T_0 e^{-ax}[/itex] for [itex]x > 0[/itex] (with [itex]a[/itex] positive constant)
[itex]T(0, t) = T_1[/itex] for [itex]t > 0[/itex]
[itex]\tau(x, s)[/itex] is the Laplace transform of [itex]T(x, t)[/itex]
show that [itex]\tau(x, s) = \frac{T_0}{s - Ka^2}e^{-ax} + [\frac{T_1}{s} - \frac{T_0}{s - Ka^2}] e^{-x\sqrt{\frac{s}{K}}}[/itex]

Homework Equations





The Attempt at a Solution


the non-homogeneous solution to the Laplace transform of the heat equation is [itex]\tau(x, s) = A(s) e^{-x\sqrt{\frac{s}{K}}}[/itex]
so i assumed that because [itex]T(x, 0) = T_0 e^{-ax}[/itex] then our general solution is
[itex]\tau(x, s) = A(s) e^{-x\sqrt{\frac{s}{K}}} + \frac{T_0 e^{-ax}}{s}[/itex]
is this correct???

we have [itex]T(0, t) = T_1[/itex], so [itex]\tau(0, s) = T_1[/itex]

thus [itex]T_1 = A(s) + \frac{T_0}{s}[/itex]

so [itex]A(s) = T_1 - \frac{T_0}{s}[/itex]

but this gives me a final solution of
[itex]\tau(x, s) = (T_1 - \frac{T_0}{s}) e^{-x\sqrt{\frac{s}{K}}} + \frac{T_0 e^{-ax}}{s}[/itex]
WHICH IS NOT RIGHT!!!
please somebody help me!!! i'm confused as to where the [itex]s - Ka^2[/itex] comes from...
 

Answers and Replies

  • #2
307
3

The Attempt at a Solution


the non-homogeneous solution to the Laplace transform of the heat equation is [itex]\tau(x, s) = A(s) e^{-x\sqrt{\frac{s}{K}}}[/itex]
so i assumed that because [itex]T(x, 0) = T_0 e^{-ax}[/itex] then our general solution is
[itex]\tau(x, s) = A(s) e^{-x\sqrt{\frac{s}{K}}} + \frac{T_0 e^{-ax}}{s}[/itex]
is this correct???
Hi wik_chick,

To start with, this is actually incorrect. Because it is important that you understand how to apply a Laplace transform to a differential equation, it's best if you start from the original heat equation to derive an answer. To start you off, applying a LT to [itex] u_{t} = k u_{xx}[/itex], gives you:
[tex](s\tau(x,s) - T(x,0)) - k\frac{d^2\tau(x,s)}{dx^2} = 0 [/tex]
or,
[tex]\frac{d^2\tau}{dx^2} - s\tau = -T_0e^{-ax}[/tex]

What is the solution to this ODE? It is in fact different than the solution you propose, and if you solve it and plug in your boundary condition, you'll arrive at the correct solution.
 
  • #3
assuming you just forgot to put the K before the [itex]\frac{d^2 \tau}{dx^2}[/itex] this is what i got:

[itex]K \frac{d^2 \tau}{dx^2} - s \tau = -T_0 e^{-ax}[/itex]
try [itex]\tau = Me^{-ax}, so \frac{d^2 \tau}{dx^2} = Ma^2 e^{-ax}[/itex]
so [itex]KMa^2 e^{-ax} - sMe^{-ax} = -T_0 e^{-ax}[/itex]
[itex]e^{-ax} (KMa^2 - sM) = -T_0 e^{-ax}[/itex]
[itex]M(Ka^2 - s) = -T_0[/itex]
[itex]M = -\frac{T_0}{Ka^2 - s} = \frac{T_0}{s-Ka^2}[/itex]
thus [itex]\tau(x, s) = \frac{T_0}{s - Ka^2} e^{-ax}[/itex]

am i correct so far? and what do i do next?
 
Last edited:
  • #4
307
3
(Yes, I did forget the k).

You've found the particular solution to the ODE, but you still need to write down the complimentary solution. That is, your solution should be of the form [itex]\tau(x,s) = \tau_C + \tau_P[/itex], where [itex]\tau_C[/itex] is the solution of the homogeneous ODE, and [itex]\tau_P[/itex] is the solution you've determined above.

In other words, you've only written down half of the solution to that ODE, and to find the other half, you have to solve the ODE:
[tex]k\frac{d^2\tau}{dx^2} - s\tau = 0.[/tex]
The solution is the combination of [itex]\frac{T_0}{s - Ka^2} e^{-ax}[/itex] (what you've already found), and the result you get from solving the above ODE.
 
  • #5
i found the solution to the homogenous ODE...i think!
[itex]K \tau^{''} - s \tau = 0[/itex]
try [itex]\tau_H = e^{rx}[/itex](where r is a constant)
thus [itex]\tau_H^{''} = r^2 e^{rx}[/itex]
sub into ODE:
[itex]Kr^2e^{rx} - se^{rx} = 0[/itex]
[itex]e^{rx}(Kr^2 - s) = 0[/itex]
[itex]Kr^2 - s = 0[/itex]
[itex]Kr^2 = s[/itex]
[itex]r^2 = \frac{s}{K}[/itex]
[itex]r = ±\sqrt{\frac{s}{K}}[/itex]
thus [itex]\tau_H(x, s) = A(s)e^{-x\sqrt{\frac{s}{K}}} + B(s)e^{x\sqrt{\frac{s}{K}}}[/itex]

as [itex]x → ∞, T(x, t) → 0[/itex]
thus as [itex]x → ∞, \tau(x, s) → 0[/itex]
thus [itex]B(s) = 0[/itex]

so [itex]\tau_H(x, s) = A(s)e^{-x\sqrt{\frac{s}{K}}}[/itex]

[itex]\tau_H(0, s) = A(s) = \int^\infty_0 T(0, t) e^{-st} dt[/itex]
[itex]= \int^\infty_0 T_1e^{-st} dt[/itex]
[itex]= T_1 [\frac{e^{-st}}{-s}]^\infty_0[/itex]
[itex]= T_1 (0 - \frac{1}{-s})[/itex]
[itex]= \frac{T_1}{s}[/itex]
therefore [itex]T_H(x, s) = \frac{T_1}{s} e^{-x\sqrt{\frac{s}{K}}}[/itex]

but this plus the solution [itex]\tau_N(x, s)[/itex] to the nonhomogenous eqn only gives:
[itex]\tau(x, s) = \tau_H + \tau_N = \frac{T_1}{s}e^{-x\sqrt{\frac{s}{K}}} + \frac{T_0}{s-Ka^2}e^{-ax}[/itex]

so where does the [itex]\frac{-T_0}{s-Ka^2}e^{-x\sqrt{\frac{s}{K}}}[/itex] come from??
 
  • #6
307
3
but this plus the solution [itex]\tau_N(x, s)[/itex] to the nonhomogenous eqn only gives:
[itex]\tau(x, s) = \tau_H + \tau_N = \frac{T_1}{s}e^{-x\sqrt{\frac{s}{K}}} + \frac{T_0}{s-Ka^2}e^{-ax}[/itex]

so where does the [itex]\frac{-T_0}{s-Ka^2}e^{-x\sqrt{\frac{s}{K}}}[/itex] come from??
Close, you've used your boundary condition to solve for A(s) before writing the full solution of the ODE (i.e. you applied it only to the homogenous part of your solution, when you need to apply it to the full solution).

In particular,
[tex]\tau(x,s) = \tau_H + \tau_N = A(s)e^{-x\sqrt{\frac{s}{K}}} + \frac{T_0}{s-Ka^2}e^{-ax}[/tex]
Now find A(s) by using your BC,
[tex] \tau(0,s) = \frac{T_1}{s} = \tau_H(0,s) + \tau_N(0,s) = A(s) + \frac{T_0}{s-Ka^2}[/tex]
 
  • #7
excellent! so that gave me
[itex]\frac{T_1}{s} = \tau_H (0, s) + \tau_N (0, s) = A(s) + \frac{T_0}{s - Ka^2}[/itex]
so [itex]A(s) = \frac{T_1}{s} - \frac{T_0}{s - Ka^2}[/itex]
so [itex]\tau (x, s) = \frac{T_0}{s - Ka^2} e^{-ax} + [\frac{T_1}{s} - \frac{T_0}{s - Ka^2}] e^{-x\sqrt{\frac{s}{K}}}[/itex]
which is the solution I am seeking!
THANKYOU!!

now i have to use the inverse Laplace transform to get the formula for [itex]T(x, t)[/itex]
so [itex]T(x, t) = \frac{1}{2 \pi i}\int^{γ+i\infty}_{γ-i\infty}(\frac{T_0}{s - Ka^2} e^{-ax + st} + [\frac{T-1}{s} - \frac{T_0}{s - Ka^2}]e^{-x\sqrt{\frac{s}{K}} + st}) ds[/itex]

where do i go from here?
 
  • #8
307
3
:) This is slightly more difficult. To solve an inverse Laplace transform requires some time and the ability to solve integrals in the complex plane (e.g. residue theory).

However, you don't need to solve that integral explicitly. Rather, you want to use the properties of the Laplace transform, as well as some already determined identities, to find the inverse Laplace transform.

See here:
http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems
http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms
 

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