Verification of solution to Heat Equation

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SUMMARY

The discussion centers on verifying the solution to the 2-D Heat Equation using the function u(t,x,y) = e^(-λt)sin(αx)cos(βy), where λ = α² + β². Participants clarify that the original equation was incorrectly transcribed, leading to confusion. The correct form requires taking partial derivatives and substituting back into the equation to confirm that it satisfies the Heat Equation. The conclusion is that the equation holds true when the variables are correctly defined and λ is appropriately substituted.

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  • Familiarity with partial derivatives and the Laplacian operator.
  • Knowledge of trigonometric functions and their derivatives.
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  • Review the derivation of the 2-D Heat Equation and its solutions.
  • Study the properties of the Laplacian operator in multiple dimensions.
  • Learn about Fourier series and their application in solving partial differential equations.
  • Explore common mistakes in transcribing mathematical equations and how to avoid them.
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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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