Troubleshooting Heat Conductor Problem | Solution Attempt

[Jamin2112]

Homework Statement

Homework Equations

All explained in my solution attempt

The Attempt at a Solution

Let me know what's going wrong here. I don't feel like I should be getting something as small as t = .05 seconds for an answer.

 

[Jamin2112]

Or maybe that's not weird, given the massive amplitude of 100 in the initial condition?

 

[Jamin2112]

What do you geniuses think?

 

[Dick]

I think you made that more complicated than it needs to be. Your x dependence is already in the form of a sin. There's no need for the Fourier series part. Just do the separation of variables. And they gave you α^2=1.158, not α=1.158. You don't want to square it again.

 

[Jamin2112]

I think you made that more complicated than it needs to be. Your x dependence is already in the form of a sin. There's no need for the Fourier series part. Just do the separation of variables. And they gave you α^2=1.158, not α=1.158. You don't want to square it again.

My bad on the σ² thing.

So, from the fact that that G(t)=G(0)e^-σ2π2k2/L2sin(kπ/L) where k is any integer, can I finish this problem? I don't see how I can since I don't have a value for G(0) and since there are infinitely many solutions. Our course notes make the G(0) absorb into a constant in an infinite series (see below) in one example.

Thoughts?

 

[Jamin2112]

Additional problem:

(My answers are in bold.)

Consider the heat equation u_tt=u_xx, 0 < x < 1 with boundary conditions u(0,t)=u(1,t)=1 and initial condition u(x,0)=f(x).

(a) Using physical intuitions only, write down the steady state temperature profile u_ss(x). Explain your answer.

The "steady state temperature profile" is the one in which temperature does not depend on time. Thus it remains constant at each point in the conductor for all t. But since the ends remain a constant 1, it only makes physical sense that u(x,t)=1 for all t≥0 and x ε (0, 1). Thus u_ss(x)=1.


(b) Using the transformation v=u-u_ss, show that v satisfies the same heat equation with a different initial temperature distribution.

We have v(x,t)=u(x,t)-1; hence v(0,t)=v(1,t)=0, v(x,0)=0 and obviously v_t=v_xx.

(c) Solve for v(x,t) using separation of variables. What is the physical meaning of v?

Not sure how to do this one. Any ideas?

 

[Ray Vickson]

My bad on the σ² thing.

So, from the fact that that G(t)=G(0)e^-σ2π2k2/L2sin(kπ/L) where k is any integer, can I finish this problem? I don't see how I can since I don't have a value for G(0) and since there are infinitely many solutions. Our course notes make the G(0) absorb into a constant in an infinite series (see below) in one example.

Thoughts?

What is stopping you from setting t = 0 in the formula for u(x,t) and finding G(0) by knowing that u(x,0) = 100*sin(πx)?

RGV

 

[Jamin2112]

What is stopping you from setting t = 0 in the formula for u(x,t) and finding G(0) by knowing that u(x,0) = 100*sin(πx)?

RGV

I see. My general solution will still be an infinite sum, though. Right?

 

[Ray Vickson]

I see. My general solution will still be an infinite sum, though. Right?

What is the Fourier series of the function f(x) = sin(πx), 0 < x < 1?

RGV

 

[Jamin2112]

What is the Fourier series of the function f(x) = sin(πx), 0 < x < 1?

RGV

Let me know if my revised answer is correct.

 

[Dick]

Separation of variables is an 'ansatz'. Look it up, it's a handy term. You don't have to care one bit about the general solution if you can guess the form of the solution and verify it.

 

[Jamin2112]

Separation of variables is an 'ansatz'. Look it up, it's a handy term. You don't have to care one bit about the general solution if you can guess the form of the solution and verify it.

So my u(x,t)=100e^-σ2∏2tsin(∏x) checks out?

In physics and mathematics, an ansatz is an educated guess that is verified later by its results. (Wikipedia)

 

[Dick]

So my u(x,t)=100e^-σ2∏2tsin(∏x) checks out?

You can check that it satisfies the PDE and the boundary conditions, can't you? Just plug it in. That's the "verified later" part.

 

[Jamin2112]

You can check that it satisfies the PDE and the boundary conditions, can't you? Just plug it in. That's the "verified later" part.

Sweet. What about the additional question I posted in this thread? Any thoughts or concerns?

 

[Dick]

Additional problem:

(My answers are in bold.)

Consider the heat equation u_tt=u_xx, 0 < x < 1 with boundary conditions u(0,t)=u(1,t)=1 and initial condition u(x,0)=f(x).

(a) Using physical intuitions only, write down the steady state temperature profile u_ss(x). Explain your answer.

The "steady state temperature profile" is the one in which temperature does not depend on time. Thus it remains constant at each point in the conductor for all t. But since the ends remain a constant 1, it only makes physical sense that u(x,t)=1 for all t≥0 and x ε (0, 1). Thus u_ss(x)=1.


(b) Using the transformation v=u-u_ss, show that v satisfies the same heat equation with a different initial temperature distribution.

We have v(x,t)=u(x,t)-1; hence v(0,t)=v(1,t)=0, v(x,0)=0 and obviously v_t=v_xx.

(c) Solve for v(x,t) using separation of variables. What is the physical meaning of v?

Not sure how to do this one. Any ideas?

For c), now I think they want to do the Fourier series of f(x) and write down a general solution, like your were doing for the original problem.

 

[Jamin2112]

For c), now I think they want to do the Fourier series of f(x) and write down a general solution, like your were doing for the original problem.

Got it. Am I on the right track with parts a) and b)?

 

[Dick]

Got it. Am I on the right track with parts a) and b)?

Seems fine to me.

 

[Jamin2112]

Seems fine to me.

Bottom line: I get the trivial solution u(x,t)=0 for all 0≤x≤1 and t≥0. Right?

 

[Dick]

Bottom line: I get the trivial solution u(x,t)=0 for all 0≤x≤1 and t≥0. Right?

Well, no. You will only get the trivial solution if f(x)=1. Otherwise, you can expand f(x)-1 in a sine series and treat each term like you did with the first problem.