A diffusion equation
of the form
![\begin{displaymath}
{\partial T\over\partial t} = \kappa\nabla^2T.
\end{displaymath}](h_663.gif) |
(1) |
Physically, the equation commonly arises in situations where
is the
thermal diffusivity
and
the temperature.
The 1-D heat conduction equation is
![\begin{displaymath}
{\partial T\over\partial t} = \kappa{\partial^2T\over\partial x^2}.
\end{displaymath}](h_666.gif) |
(2) |
This can be solved by Separation of Variables using
![\begin{displaymath}
T(x,t) = X(x)T(t).
\end{displaymath}](h_667.gif) |
(3) |
Then
![\begin{displaymath}
X {dT\over dt}=\kappa T{d^2X\over dx^2}.
\end{displaymath}](h_668.gif) |
(4) |
Dividing both sides by
gives
![\begin{displaymath}
{1\over \kappa T} {dT\over dt}={1\over X}{d^2X\over dx^2} = -{1\over \lambda^2},
\end{displaymath}](h_670.gif) |
(5) |
where each side must be equal to a constant. Anticipating the exponential solution in
, we have picked a negative
separation constant so that the solution remains finite at all times and
has units of length. The
solution
is
![\begin{displaymath}
T(t)=A e^{-\kappa t/\lambda^2},
\end{displaymath}](h_672.gif) |
(6) |
and the
solution is
![\begin{displaymath}
X(x)=C\cos\left({x\over \lambda}\right)+D\sin\left({x\over \lambda}\right).
\end{displaymath}](h_673.gif) |
(7) |
The general solution is then
If we are given the boundary conditions
![\begin{displaymath}
T(0,t) = 0
\end{displaymath}](h_678.gif) |
(9) |
and
![\begin{displaymath}
T(L,t) = 0,
\end{displaymath}](h_679.gif) |
(10) |
then applying (9) to (8) gives
![\begin{displaymath}
D\cos\left({x\over \lambda}\right)=0 \Rightarrow D=0,
\end{displaymath}](h_680.gif) |
(11) |
and applying (10) to (8) gives
![\begin{displaymath}
E\sin\left({L\over \lambda}\right)=0 \Rightarrow {L\over \lambda}=n\pi\Rightarrow \lambda={L\over n\pi},
\end{displaymath}](h_681.gif) |
(12) |
so (8) becomes
![\begin{displaymath}
T_n(x,t)= E_ne^{-\kappa (n\pi/L)^2 t} \sin\left({n\pi x\over L}\right).
\end{displaymath}](h_682.gif) |
(13) |
Since the general solution can have any
,
![\begin{displaymath}
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}](h_683.gif) |
(14) |
Now, if we are given an initial condition
, we have
![\begin{displaymath}
T(x,0) = \sum_{n=1}^\infty c_n\sin\left({n\pi x\over L}\right).
\end{displaymath}](h_685.gif) |
(15) |
Multiplying both sides by
and integrating from 0 to
gives
![\begin{displaymath}
\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}](h_687.gif) |
(16) |
Using the Orthogonality of
and
,
|
|
|
(17) |
so
![\begin{displaymath}
c_n = {2\over \pi} \int_0^L \sin\left({m\pi x\over L}\right)T(x,0)\,dx.
\end{displaymath}](h_692.gif) |
(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
![\begin{displaymath}
\left.{\partial T\over \partial x}\right\vert _{(0,t)} = 0
\end{displaymath}](h_693.gif) |
(19) |
![\begin{displaymath}
\left.{\partial T\over \partial x}\right\vert _{(L,t)} = 0.
\end{displaymath}](h_694.gif) |
(20) |
Following the same procedure as before, a similar answer is found, but with sine replaced by cosine:
![\begin{displaymath}
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}](h_695.gif) |
(21) |
where
![\begin{displaymath}
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}](h_696.gif) |
(22) |
© 1996-9 Eric W. Weisstein
1999-05-25