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Heat Conduction Equation

A diffusion equation of the form

{\partial T\over\partial t} = \kappa\nabla^2T.
\end{displaymath} (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}.
\end{displaymath} (2)

This can be solved by Separation of Variables using
T(x,t) = X(x)T(t).
\end{displaymath} (3)

X {dT\over dt}=\kappa T{d^2X\over dx^2}.
\end{displaymath} (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},
\end{displaymath} (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},
\end{displaymath} (6)

and the $X$ solution is
X(x)=C\cos\left({x\over \lambda}\right)+D\sin\left({x\over \lambda}\right).
\end{displaymath} (7)

The general solution is then
$\displaystyle T(x,t)$ $\textstyle =$ $\displaystyle T(t)X(x)$  
  $\textstyle =$ $\displaystyle A e^{-\kappa t/\lambda^2}\left[{C\cos\left({x\over\lambda}\right)+D\sin\left({x\over\lambda}\right)}\right]$  
  $\textstyle =$ $\displaystyle e^{-\kappa t/\lambda^2}\left[{D\cos\left({x\over\lambda}\right)+E\sin\left({x\over\lambda}\right)}\right].$ (8)

If we are given the boundary conditions

T(0,t) = 0
\end{displaymath} (9)

T(L,t) = 0,
\end{displaymath} (10)

then applying (9) to (8) gives
D\cos\left({x\over \lambda}\right)=0 \Rightarrow D=0,
\end{displaymath} (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},
\end{displaymath} (12)

so (8) becomes
T_n(x,t)= E_ne^{-\kappa (n\pi/L)^2 t} \sin\left({n\pi x\over L}\right).
\end{displaymath} (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}.
\end{displaymath} (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).
\end{displaymath} (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.
\end{displaymath} (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)
c_n = {2\over \pi} \int_0^L \sin\left({m\pi x\over L}\right)T(x,0)\,dx.
\end{displaymath} (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
\end{displaymath} (19)

\left.{\partial T\over \partial x}\right\vert _{(L,t)} = 0.
\end{displaymath} (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},
\end{displaymath} (21)

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.
\end{displaymath} (22)

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© 1996-9 Eric W. Weisstein