Heat equation on a half line

The document I upload is the note I took. I have a problem. When I tried to verify the solution by checking the initial condition and boundary condition, I have some problem to see the solution can really give initial condition, i.e., to set t=0 in the equation 18 of the document. I know one should take the limit as t->0+, but I failed to reach that. Do you have any clue?There is a limit to how far you can take the limit as t->0+, but I'm not sure what that limit is. I would need to see the document to be able to help you more. In summary, the heat equation can be solved using either the sine or heat kernel transformation,f
  • #1
63
0
Heat equation on a half line!

Hi,

I am now dealing with the heat equation on a half line, i.e., the heat equation is subject to one time-dependent boundary condition only at x=0 (the other boundary condition is zero at the infinity) and an initial condition.

I searched online, it seems that for the half line problem, only the sine transformation can solve the heat equation, but in that case, the final result is always zero at x=0 since when doing sine transformation, one should assume that the to-be-transformed function is odd, so the function is zero at x=0.

My question is, do you know any other techniques to solve the heat equation on the half line without using sine transform?

Thanks.
 
  • #2
Sinusoidally varying temperatures don't really make sense. Try using the Laplace transform.
 
  • #3
What is the specific problem?
 
  • #4
I searched online, it seems that for the half line problem, only the sine transformation can solve the heat equation, but in that case, the final result is always zero at x=0 since when doing sine transformation, one should assume that the to-be-transformed function is odd, so the function is zero at x=0.

IF you tell us where you got that (wrong) idea, we might be able to explain what the website means, or confirm that it really is wrong.

Sinusoidally varying temperatures don't really make sense. Try using the Laplace transform.

Sinusoidal in time, or in space? For example if you were applying a heat flux at x = 0 which was a periodic function of time, it would make good sense to do a Fourier decomposition of it.

As Chestermiller said, posting the complete problem would help.
 
  • #5
Fourier, yes, decomposing it into sine waves, not so much. You need those exponential decays to make it die at infinity.
 
  • #6
Hi,

Thanks for all your replies!

I didn't say it's sinusoidal, it's just a time-dependent function, not periodic.

Sorry, I shouldn't say "only the sine transformation can solve the heat equation". Yesterday, I just found using heat kernel can also solve the problem on a half line.

The document I upload is the note I took. I have a problem. When I tried to verify the solution by checking the initial condition and boundary condition, I have some problem to see the solution can really give initial condition, i.e., to set t=0 in the equation 18 of the document. I know one should take the limit as t->0+, but I failed to reach that. Do you have any clue?

Thanks!
 

Attachments

  • Heat.pdf
    84 KB · Views: 274

Suggested for: Heat equation on a half line

Replies
3
Views
512
Replies
2
Views
493
Replies
3
Views
436
Replies
14
Views
521
Replies
1
Views
584
Replies
17
Views
848
Replies
28
Views
871
Back
Top