Heat Conduction Equation

A diffusion equation of the form

$${\partial T\over\partial t} = \kappa\nabla^2T.$$ (1)

Physically, the equation commonly arises in situations where $\kappa$ is the thermal diffusivity and $T$ the temperature.

The 1-D heat conduction equation is

$${\partial T\over\partial t} = \kappa{\partial^2T\over\partial x^2}.$$ (2)

This can be solved by Separation of Variables using

$$T(x,t) = X(x)T(t).$$ (3)

Then

$$X {dT\over dt}=\kappa T{d^2X\over dx^2}.$$ (4)

Dividing both sides by $\kappa XT$ gives

$${1\over \kappa T} {dT\over dt}={1\over X}{d^2X\over dx^2} = -{1\over \lambda^2},$$ (5)

where each side must be equal to a constant. Anticipating the exponential solution in $T$, we have picked a negative separation constant so that the solution remains finite at all times and $\lambda$ has units of length. The $T$ solution is

$$T(t)=A e^{-\kappa t/\lambda^2},$$ (6)

and the $X$ solution is

$$X(x)=C\cos\left({x\over \lambda}\right)+D\sin\left({x\over \lambda}\right).$$ (7)

The general solution is then

$$T(x,t) = T(t)X(x)$$

$$= A e^{-\kappa t/\lambda^2}\left[{C\cos\left({x\over\lambda}\right)+D\sin\left({x\over\lambda}\right)}\right]$$

$$= e^{-\kappa t/\lambda^2}\left[{D\cos\left({x\over\lambda}\right)+E\sin\left({x\over\lambda}\right)}\right].$$ (8)

If we are given the boundary conditions

$$T(0,t) = 0$$ (9)

and

$$T(L,t) = 0,$$ (10)

then applying (9) to (8) gives

$$D\cos\left({x\over \lambda}\right)=0 \Rightarrow D=0,$$ (11)

and applying (10) to (8) gives

$$E\sin\left({L\over \lambda}\right)=0 \Rightarrow {L\over \lambda}=n\pi\Rightarrow \lambda={L\over n\pi},$$ (12)

so (8) becomes

$$T_n(x,t)= E_ne^{-\kappa (n\pi/L)^2 t} \sin\left({n\pi x\over L}\right).$$ (13)

Since the general solution can have any $n$,

$$T(x,t)= \sum_{n=1}^\infty c_n\sin\left({n\pi x\over L}\right)e^{-\kappa (n\pi/L)^2 t}.$$ (14)

Now, if we are given an initial condition $T(x,0)$, we have

$$T(x,0) = \sum_{n=1}^\infty c_n\sin\left({n\pi x\over L}\right).$$ (15)

Multiplying both sides by $\sin(m\pi x/L)$ and integrating from 0 to $L$ gives

$$\int_0^L \sin\left({m\pi x\over L}\right)T(x,0)\,dx = \int_0... ...ft({m\pi x\over L}\right)\sin\left({n\pi x\over L}\right)\,dx.$$ (16)

Using the Orthogonality of $\sin(nx)$ and $\sin(mx)$,

$\sum_{n=1}^\infty c_n \int_0^L \sin\left({n\pi x\over L}\right)\sin\left({m\pi x\over L}\right)\,dx = \sum_{n=1}^\infty {\textstyle{1\over 2}}\pi \delta_{mn} c_n$
$= {\textstyle{1\over 2}}\pi c_m = \int_0^L \sin\left({m\pi x\over L}\right)T(x,0)\,dx,\quad$	(17)

so

$$c_n = {2\over \pi} \int_0^L \sin\left({m\pi x\over L}\right)T(x,0)\,dx.$$ (18)

If the boundary conditions are replaced by the requirement that the derivative of the temperature be zero at the edges, then (9) and (10) are replaced by

$$\left.{\partial T\over \partial x}\right\vert _{(0,t)} = 0$$ (19)

$$\left.{\partial T\over \partial x}\right\vert _{(L,t)} = 0.$$ (20)

Following the same procedure as before, a similar answer is found, but with sine replaced by cosine:

$$T(x,t) = \sum_{n=1}^\infty c_n\cos\left({n\pi x\over L}\right)e^{-\kappa (n\pi/L)^2 t},$$ (21)

where

$$c_n = {2\over \pi} \int_0^L \cos\left({m\pi x\over L}\right)\left.{\partial T(x,0)\over \partial x}\right\vert _{t=0} \,dx.$$ (22)