heat equation - New

[logistic_guy]

Solve the heat equation $\displaystyle k\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, \ \ \ 0 < x < L, \ \ \ t > 0 \ \$ subject to the given conditions. Assume a rod of length $\displaystyle L$.


$\displaystyle u(0,t) = 0, \ \ \ u(L,t) = 0, \ \ \ t > 0$

$\displaystyle u(x,0) =\begin{cases}1 & \ \ \ 0 <x<L/2\\0 & \ \ \ L/2 < x < L\end{cases}$

 

[logistic_guy]

Let us try to solve this partial differential equation by separation of variables.

Let $\displaystyle u(x,t) = X(x) \ T(t)$

Then,

$\displaystyle kT\frac{\partial^2 X}{\partial x^2} = X\frac{\partial T}{\partial t}$


$\displaystyle \frac{1}{X}\frac{\partial^2 X}{\partial x^2} = \frac{1}{kT}\frac{\partial T}{\partial t}$


$\displaystyle \frac{1}{X}\frac{\partial^2 X}{\partial x^2} = \frac{1}{kT}\frac{\partial T}{\partial t} = -\lambda$

This will give us two ordinary differential equations:

$\displaystyle \frac{\partial^2 X}{\partial x^2} +\lambda X = 0$

$\displaystyle \frac{\partial T}{\partial t} + \lambda kT = 0$

Or

$\displaystyle \frac{d^2 X}{dx^2} +\lambda X = 0$

$\displaystyle \frac{d T}{dt} + \lambda kT = 0$

We will continue in the next post!

 

[logistic_guy]

Experience tells us that when we have Dirichlet Boundary Conditions, we will get a non trivial solution only when $\displaystyle \lambda > 0$.

Let $\displaystyle \lambda = \mu^2$.

Then,

$\displaystyle \frac{d^2 X}{dx^2} +\mu^2 X = 0$

This is an elementary differential equation and its solution is:

$\displaystyle X(x) = A\cos \mu x + B\sin \mu x$

Apply the first boundary condition.

$\displaystyle 0 = A + 0$

So, the solution becomes:

$\displaystyle X(x) = B\sin \mu x$

Apply the second boundary condition.

$\displaystyle 0 = B\sin \mu L$

It's either $\displaystyle B = 0$ or $\displaystyle \sin \mu L = 0$. If $\displaystyle B = 0$, we will get the trivial solution $\displaystyle X(x) = 0$, so it must be $\displaystyle \sin \mu L = 0$.

Then,

$\displaystyle \mu L = \pi, 2\pi, 3\pi, \cdots$

Or

$\displaystyle \mu L = n\pi$, where $\displaystyle n = 1,2,3,\cdots$

Or

$\displaystyle \mu = \frac{n\pi}{L}$

And the solution becomes:

$\displaystyle X_n(x) = B\sin \frac{n\pi}{L} x$

 

[logistic_guy]

Back to the second equation.

$\displaystyle \frac{d T}{dt} + \lambda kT = 0$

This is also an elementary differential equation and it has the solution:

$\displaystyle T(t) = Ce^{-\lambda k t}$

 

[logistic_guy]

$\displaystyle u_n(x,t) = X_n(x) \ T_n(t) = D\sin \frac{n\pi}{L} x \ e^{-\frac{n^2\pi^2}{L^2} k t}$

where $\displaystyle D = BC$ and $\displaystyle \lambda = \mu^2 = \frac{n^2\pi^2}{L^2}$

 

[logistic_guy]

The solution can be written in a different form.

$\displaystyle u(x,t) = \sum_{n = 1}^{\infty}D_n\sin \frac{n\pi}{L} x \ e^{-\frac{n^2\pi^2}{L^2} k t}$

 

[logistic_guy]

It's time to find the coefficient $\displaystyle D_n$.

We will apply the initial condition.

$\displaystyle u(x,0) = \sum_{n = 1}^{\infty}D_n\sin \frac{n\pi}{L} x$

$\displaystyle u(x,0)\sin \frac{n\pi}{L} x = \sum_{n = 1}^{\infty}D_n\sin^2 \frac{n\pi}{L} x$

$\displaystyle \int_{0}^{L} u(x,0)\sin \frac{n\pi}{L} x \ dx = \sum_{n = 1}^{\infty}D_n \int_{0}^{L}\sin^2 \frac{n\pi}{L} x \ dx$

$\displaystyle \int_{0}^{L/2} u(x,0)\sin \frac{n\pi}{L} x \ dx + \int_{L/2}^{L} u(x,0)\sin \frac{n\pi}{L} x \ dx = \sum_{n = 1}^{\infty}D_n \int_{0}^{L}\sin^2 \frac{n\pi}{L} x \ dx$

$\displaystyle \int_{0}^{L/2} \sin \frac{n\pi}{L} x \ dx + 0 = \sum_{n = 1}^{\infty}D_n \ \frac{L}{2}$

$\displaystyle \frac{L}{\pi n}\left(1 - \cos \frac{n\pi}{2}\right) = \sum_{n = 1}^{\infty}D_n \ \frac{L}{2}$

This gives:

$\displaystyle D_n = \frac{2}{\pi n}\left(1 - \cos \frac{n\pi}{2}\right)$

Then, the final solution is:

$\displaystyle u(x,t) = \frac{2}{\pi}\sum_{n = 1}^{\infty}\frac{1}{n}\left(1 - \cos \frac{n\pi}{2}\right)\sin \frac{n\pi}{L} x \ e^{-\frac{n^2\pi^2}{L^2} k t}$