Find the solution of the heat conduction problem

100Uxx=Ut 0 < x < 1 t > 0

U(0,t) = 0 U(1,t) = 0 t > 0

U(x,0) = sin(2*pi*x) - sin(5*pi*x) 0 <= x <= 1

Not even sure where to begin with this problem :(

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- June 3rd 2007, 06:03 PMpakmanHeat conduction using partial differential equations
Find the solution of the heat conduction problem

100Uxx=Ut 0 < x < 1 t > 0

U(0,t) = 0 U(1,t) = 0 t > 0

U(x,0) = sin(2*pi*x) - sin(5*pi*x) 0 <= x <= 1

Not even sure where to begin with this problem :( - June 3rd 2007, 07:11 PMThePerfectHacker
The heat equation is the partial differencial equation:

The one you have the the simplest one (1 dimensional):

Given this PDE with the Boundary Value Problem (fixed-fixed):

The solution is given by:

Where,

That is the coefficients of the half-range Fourier sine expansion.

What I wrote above is the solution to this problem (which was derived by Seperation of Variables.)

In you equation we have,

(that is the length of the rod).

(the coefficient in front).

So the solution is given by,

Where,

Can you take if from there? - June 3rd 2007, 07:45 PMThePerfectHackerQuote:

Originally Posted by**Navier-Stokes**

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Use the Identity:

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Here is the catch, let, me just do the integrand to explain where division by zero appears.

This is true**expect**when because then we have:

Let us look at each of the four cosine term in the two integrals,

The first one we just did. We said the integral always vanishes (is zero) except when .

The second one always vanishes, because, for .

The third one vanished except for .

The four one always vanished, because for .

Let,

Then, by what explained above,

Thus,

Which means the infinite "series" solution,

Is actually a finite sum because almost everything vanishes. (Except ).

Thus,

Thus, the solution is:

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This was a really fun problem, I never expected the solution to be in closed form. Wow! - November 10th 2007, 01:17 PMAMIYY4UClarification
Thank you for this clear explanation of this heat conduction in a rod problem. I was just wondering one thing (at the moment), how did you get those 2 and 5 coefficients in front of your f(x) at the start of the problem. I thought the problem's f(x) would just be sin(2*Pi*x) - sin(5*Pi*x).

I just realized that I have the same textbook where this problem came from. I checked the answer in the back of the book with yours and it was slightly different.

The book's answer was the same thing as what you wrote except without the 2 and 5 coefficient. Did you just accidently add the 2 and 5 coefficients in?

Thanks - November 10th 2007, 02:00 PMThePerfectHacker
- November 11th 2007, 10:56 AMAMIYY4UHelp with another problem from the same textbook
Would you mind helping me out with another problem in this textbook... #23 from the same section. It says:

The heat conduction in two space dimensions may be expressed in terms of polar coordinates as

(alpha)^2 (Urr + (1/r)Ur + (1/r^2)U(theta)(theta)) = Ut

Assuming that u(r, theta, t) = R(r)(phi)(theta)T(t), find ordinary differential equations that are satisfied by R(r), (phi)(theta), and T(t).

If you could, I'm also a little confused by the insulated heat conduction problems. For example,

Find the steady-state equation fo the heat conduction equation (alpha)^2 * Uxx = Ut that satisfies the given set of boundary conditions.

U(0,t) = 10 , u(50,t)=40

Thanks - November 11th 2007, 11:15 AMThePerfectHacker
Post in a new thread. I might be able to help. See if you can use LaTeX also.