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 :(

Printable View

- Jun 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 :( - Jun 3rd 2007, 07:11 PMThePerfectHacker
The heat equation is the partial differencial equation:

$\displaystyle \nabla^2 u = \frac{1}{\alpha ^2}\cdot \frac{\partial u}{\partial t}$

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

$\displaystyle \frac{\partial u^2}{\partial x^2} = \frac{1}{10^2} \cdot \frac{\partial u}{\partial t}$

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

$\displaystyle u(0,t) = u(L,t) = 0 \mbox{ for } t>0$

$\displaystyle u(x,0) = f(x)$

The solution is given by:

$\displaystyle u(x,t) = \sum_{n=1}^{\infty} c_n \exp (-n^2\pi ^2 \alpha^2 t/L^2) \sin \frac{n\pi x}{L}$

Where,

$\displaystyle c_n = \frac{2}{L}\int_0^L f(x) \sin \frac{n\pi x}{L} dx$

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,

$\displaystyle f(x) = 2\sin (2\pi x) - 5\sin (5\pi x)$

$\displaystyle L=1$ (that is the length of the rod).

$\displaystyle \alpha = 10$ (the coefficient in front).

So the solution is given by,

$\displaystyle u(x,t) = \sum_{n=1}^{\infty}c_n \exp (-100\pi^2 n^2 t) \sin n\pi x$

Where,

$\displaystyle c_n = \frac{2}{1}\int_0^1 [2\sin (2\pi x)-5\sin (5\pi x)] \sin n\pi x \, dx$

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

Originally Posted by**Navier-Stokes**

---

$\displaystyle 2\int_0^1 (2\sin 2\pi x - 5 \sin 5\pi x) \sin n \pi x \, dx =$$\displaystyle 2\int_0^1 2\sin 2\pi x \cdot \sin n\pi x \, dx - 5\int_0^1 2\sin 5\pi x \cdot \sin n\pi x \, dx$

---

Use the Identity:

$\displaystyle \boxed{2\sin A \sin B = \cos (A - B) - \cos (A+B)}$

---

$\displaystyle 2\int_0^1 \cos (2-n)\pi x - \cos (2+n)\pi x\, dx - 5\int_0^1 \cos (5-n)\pi x - \cos (5+n)\pi x \,dx$

---

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

$\displaystyle 2\int_0^1 \cos (2-n)\pi x dx = \frac{2}{(2-n)\pi x} \sin (2-n)\pi x \big|_0^1 = 0$

This is true**expect**when $\displaystyle n=2$ because then we have:

$\displaystyle 2\int_0^1 \cos 0\pi x dx = 2\int_0^1 dx = 2$

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

$\displaystyle \cos (2-n)\pi x \mbox{ and }\cos (2+n)\pi x \mbox{ and }\cos (5-n)\pi x \mbox{ and }\cos (5+n)\pi x$

The first one we just did. We said the integral always vanishes (is zero) except when $\displaystyle n=2$.

The second one always vanishes, because, $\displaystyle 2+n\not = 0$ for $\displaystyle n\geq 1$.

The third one vanished except for $\displaystyle n=5$.

The four one always vanished, because $\displaystyle 5+n\not =0$ for $\displaystyle n\geq 1$.

Let,

$\displaystyle I_n = 2\int_0^1 \cos (2-n)\pi x - \cos (2+n)\pi x\, dx - 5\int_0^1 \cos (5-n)\pi x - \cos (5+n)\pi x \,dx$

Then, by what explained above,

$\displaystyle I_n = \left\{ \begin{array}{ccc} 2 & \mbox{ for }& n=2 \\ -5 & \mbox{ for }& n =5 \\ 0 &\mbox{ for }& n\not = 2,5 \end{array} \right\}$

Thus,

$\displaystyle c_1 = 0, c_2 = 2, c_3 =0, c_4=0, c_5=-5, c_6=0,c_7=0,c_8=0,...$

Which means the infinite "series" solution,

$\displaystyle u(x,t) = \sum_{n=1}^{\infty} c_n \exp (-100 \pi^2 n^2 t) \sin n \pi x$

Is actually a finite sum because almost everything vanishes. (Except $\displaystyle n=2,5$).

Thus,

$\displaystyle u(x,t) = c_2 e^{-100 \pi^2 2^2 t} \sin 2\pi x+c_5 e^{-100 \pi^2 5^2 t}\sin 5\pi x$

Thus, the solution is:

$\displaystyle \boxed{ \boxed{ u(x,t) = 2e^{-400\pi^2 t}\sin 2\pi x - 5e^{-2500 \pi^2 t}\sin 5\pi x} } $

---

This was a really fun problem, I never expected the solution to be in closed form. Wow! - Nov 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 - Nov 10th 2007, 02:00 PMThePerfectHacker
- Nov 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 - Nov 11th 2007, 11:15 AMThePerfectHacker
Post in a new thread. I might be able to help. See if you can use LaTeX also.