Verification of solution to Heat Equation

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Homework Help Overview

The discussion revolves around verifying a solution to the 2-D Heat Equation, specifically examining the function u(t,x,y)=e-λtsin(αt)cos(βt) with λ=α²+β². Participants are analyzing the relationship between the function and the heat equation, which is expressed as ut=Δu.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the process of taking partial derivatives and simplifying the equations to check for equality. There is mention of the original equation being potentially transcribed incorrectly, leading to confusion about the variables involved. Some participants question whether a typo exists in the problem statement.

Discussion Status

There is ongoing exploration of the problem, with various interpretations of the original equation being considered. Some participants have provided insights into the implications of the variable substitutions and the conditions under which the equation holds true, but no consensus has been reached.

Contextual Notes

Participants note discrepancies in the transcription of the original equation, specifically regarding the variables used in the sine and cosine functions. This has led to discussions about the implications of these differences on the verification process.

rexasaurus
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1. verify that u(t,x,y)=e-λtsin(αt)cos(βt) (for arbitrary α, β and with λ=α22) satisfies the 2-D Heat Equation.



2. ut=Δu



3. I began with:
Δu=uxx+uyy.
note the equation does not contain variable "x"
so uxx=0 i.e. Δu=uyy

uy=e-λtsin(αt){-βsin(βt)}
uyy=e-λtsin(αt){-β2cos(βt)}

next I found ut
ut=cos(βy) {e-λtαcos(αt)+sin(αt)-λe-λt}

I have tried to reduce both equations but don't see how they are equal. I also have tried using the λ=α22 to re-write the eq. Any suggestions?

 
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rexasaurus said:
1. verify that (for arbitrary α, β and with λ=α22) satisfies the 2-D Heat Equation.
2. ut=Δu
3. I began with:
Δu=uxx+uyy.
note the equation does not contain variable "x"
so uxx=0 i.e. Δu=uyy

uy=e-λtsin(αt){-βsin(βt)}
uyy=e-λtsin(αt){-β2cos(βt)}

next I found ut
ut=cos(βy) {e-λtαcos(αt)+sin(αt)-λe-λt}

I have tried to reduce both equations but don't see how they are equal. I also have tried using the λ=α22 to re-write the eq. Any suggestions?

Homework Statement





Your equation is transcribed incorrectly. Your original equation should be:

u(t,x,y)=e-λtsin(αx)cos(βy)
 
Last edited:
The original equation was posted incorrectly.

The actual question states:
u(t,x,y)=e-λtsin(αt)cos(βy)

My apologies
 
The heat equation in 2D is the Fourier equation. I think the answer they want you to check out is:

u(x,y,t) = e**(-lambda*t) * sin(alpha*x) * cos(beta*y)

Take the partial derivatives and plug them back into the original equation and you'll see that the equation is satisfied if lambda = alpha**2 + beta**2. The original equation is the Laplacian of u equals the partial of u with respect to time.
 
I checked the problem statement

The question actually states:
u(t,x,y)=e-λtsin(αt)cos(βy)

Could the instructor possibly have made a typo??

When I set ut=uyy (from simplifying Δu) I was able to substitute: λ=22

My last line reads:
αcos(αt)+β2sin(αt)=(α22)sin(αt)

Any thoughts?
 
rexasaurus said:
I checked the problem statement

The question actually states:
u(t,x,y)=e-λtsin(αt)cos(βy)

Could the instructor possibly have made a typo??

Of course he could have. That t in the sine term should be x. Then it all works.
 
the last substitution should read λ=α22
 
Thx for your help.
 

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