jkwiens.com: Answer: Heat Distribution in a Rod

Answer: Heat Distribution in a Rod

Wednesday, 26 of March , 2008 at 12:55 pm

Question: A 1D rod of length L has an initial heat distribution of

$u(x,0) = - cos(\frac{8 \pi x}{L})$

If the rod has insulated ends ( $\frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(L,t) = 0$ ) and obeys the heat equation

$\frac{\partial u(x,t)}{\partial t} = k \frac{\partial^2 u(x,t)}{\partial x^2}$ ,

What is the heat distribution of the rod as a function of time?

Answer:
First assume that the solution to the PDE

$\frac{\partial u(x,t)}{\partial t} = k \frac{\partial^2 u(x,t)}{\partial x^2}$ ,

has the form

$u(x,t) = \phi(x) G (t)$

Therefore, we can reduce the equation to the following:

$\frac{\partial u(x,t)}{\partial t} = k \frac{\partial^2 u(x,t)}{\partial x^2}$
$\phi(x) \frac{\partial G(t)}{\partial t} = k G(t) \frac{\partial^2 \phi(x)}{\partial x^2}$
$\frac{1}{k G(t)} \frac{\partial G(t)}{\partial t} =\frac{1}{\phi(x)} \frac{\partial^2 \phi(x)}{\partial x^2}$

Since this equation is true for all x and t, therefore both sides of the equation must equal a constant.

$\frac{1}{k G(t)} \frac{\partial G(t)}{\partial t} =\frac{1}{\phi(x)} \frac{\partial^2 \phi(x)}{\partial x^2} = - \lambda$

which can be written as two ODEs

$\frac{d G(t)}{dt} = - \lambda k G(t)$

$\frac{d^2 \phi(x)}{dx^2} = - \lambda \phi(x)$

Solving the first-order differential $\frac{d G(t)}{dt} = - \lambda k G(t)$ is trivial and can easily be shown to have the solution:

$G(t) = C e^{-\lambda k t}$

Next, we need to solve $\frac{d^2 \phi(x)}{dx^2} = - \lambda \phi(x)$

The second-order differential has 3 different cases ( $\lambda = 0$ , $\lambda > 0$ , $\lambda < 0$ ).

Case $\lambda = 0$
The ODE for this case would be

$\frac{d^2 \phi(x)}{dx^2} =0$

which has the solution

$\phi(x) = A_0 x + A_1$

If we apply the BC $\frac{d \phi}{dx}(0)=0$ , we get

$\frac{d \phi}{dx}(0) = A_0 = 0$

which implies

$\phi(x) = A_1$

Case $\lambda > 0$
The ODE for this case would be

$\frac{d^2 \phi(x)}{dx^2} = - \lambda \phi(x)$

which has the solution

$\phi(x) = C_1 cos(\sqrt{\lambda}x) + C_2 sin(\sqrt{\lambda}x)$

Next, we apply the BC $\frac{d \phi}{dx}(0)=0$ .

$\frac{d \phi}{dx}(0) = \sqrt{\lambda}(-C_1 sin(\sqrt{\lambda}0) + C_2 cos(\sqrt{\lambda}0))$

$\frac{d \phi}{dx}(0) = \sqrt{\lambda}C_2$

Since $\lambda > 0$ , this implies . Therefore,

$\phi(x) = C_1 cos(\sqrt{\lambda}x)$

If we apply the 2nd BC $\frac{d \phi}{dx}(L)=0$ , we find the following

$\frac{d \phi}{dx}(L) = -C_1 \sqrt{\lambda} sin(\sqrt{\lambda}L) = 0$

Therefore, for non-trivial solutions $\sqrt{\lambda}L = n \pi$ where $n=1,2,3 \ldots$ .

This means,

$\phi(x) = C_1 cos(\frac{n \pi x}{L})$

is a solution.

Case $\lambda < 0$
The ODE for this case would be

$\frac{d^2 \phi(x)}{dx^2} = s \phi(x)$ where $s = -\lambda$ and

which has the solution

$\phi(x) = B_1 e^{\sqrt{s}x} + B_2 e^{-\sqrt{s}x}$

If we apply the BC $\frac{d \phi}{dx}(0)=0$ ,we find

$\frac{d\phi}{dx}(0) = B_1 \sqrt{s} e^{\sqrt{s}0} - B_2 \sqrt{s} e^{-\sqrt{s}0} = 0$

$\sqrt{s}(B_1 - B_2) = 0$

Since , we know that which means

$\phi(x) = B_1 (e^{\sqrt{s}x} + e^{-\sqrt{s}x})$

is a solution.

If we apply the 2nd BC $\frac{d \phi}{dx}(L)=0$ , we find

$\frac{d\phi}{dx}(L) = B_1 \sqrt{s} (e^{\sqrt{s}L} - e^{-\sqrt{s}L}) = 0$

Since and $e^{\sqrt{s}L} \neq e^{-\sqrt{s}L}$ , we know that .

This means that there is no solution where $\lambda < 0$ .

Since the PDE will satify any linear combination of the above solutions, we find that

$u(x,t) = A_0 + \sum_{n=1}^{\infty} A_n cos(\frac{n \pi x}{L}) e^{-(\frac{n\pi}{L})^2 kt}$

If we apply the IC $u(x,0) = - cos(\frac{8 \pi x}{L})$ to the solution, we find that and every other is zero.

Therefore, the solution would be

$u(x,t) =-3 cos(\frac{8 \pi x}{L}) e^{-(\frac{8\pi}{L})^2 kt}$

Category: Answers, Differential Equations

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