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!