Verification of solution to Heat Equation

In summary, the equation u(x,y,t) = e-λtsin(αt)cos(βy) is satisfied if lambda = alpha**2 + beta**2.
  • #1
rexasaurus
14
0
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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  • #2
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:
  • #3
The original equation was posted incorrectly.

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

My apologies
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 
  • #7
the last substitution should read λ=α22
 
  • #8
Thx for your help.
 

FAQ: Verification of solution to Heat Equation

1. What is the Heat Equation?

The Heat Equation is a partial differential equation that describes the flow of heat in a given region over time. It is used to model and understand the behavior of heat in various physical systems, such as heat transfer in materials or the temperature distribution in a room.

2. How do you verify a solution to the Heat Equation?

To verify a solution to the Heat Equation, we can use the method of separation of variables to solve the equation and then substitute the solution back into the original equation to see if it satisfies the equation. Alternatively, we can also use numerical methods, such as finite difference or finite element methods, to simulate the behavior of the solution and compare it to the expected behavior.

3. What are some common boundary conditions for the Heat Equation?

Some common boundary conditions for the Heat Equation include specifying the temperature at certain boundaries, such as a fixed temperature at the boundaries of a metal rod, or specifying the flux of heat at the boundaries, such as the flow of heat through a wall. These boundary conditions are important in determining the behavior of the solution to the Heat Equation.

4. Can the Heat Equation be applied to non-steady state systems?

Yes, the Heat Equation can be applied to non-steady state systems, where the temperature is changing over time. In these cases, the equation becomes a time-dependent partial differential equation, and the solution will depend on both time and space variables.

5. What are some real-world applications of the Heat Equation?

The Heat Equation has many real-world applications, including predicting the temperature distribution in buildings, modeling heat transfer in materials and industrial processes, and understanding the behavior of heat in geothermal systems. It is also used in fields such as meteorology, oceanography, and astrophysics to study the flow of heat in natural systems.

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