Let us first study the heat equation in 1 space (and, of course, 1 time) dimension. This is the standard example of a parabolic equation.
with boundary conditions
(5.2) |
and initial condition
(5.3) |
We shall attack this problem by separation of variables, a technique always worth trying when attempting to solve a PDE,
(5.4) |
This leads to the differential equation
(5.5) |
We find, by dividing both sides by , that
(5.6) |
Thus the left-hand side, a function of , equals a function of on the right-hand side. This is not possible unless both sides are independent of and , i.e. constant. Let us call this constant .
We obtain two differential equations
Question: What happens if is zero at some point ?
Answer: Nothing. We can still perform the same trick.
We now have to distinguish the three cases , , and .
Write
, so that the equation for
becomes
(5.9) |
This has as solution
(5.10) |
gives , or . Using we find that
(5.11) |
which has a nontrivial (i.e., one that is not zero) solution when , with a positive integer. This leads to .
We find that
. The boundary conditions give
, so there is only the trivial (zero) solution.
We write
, so that the equation for
becomes
(5.12) |
The solution is now in terms of exponential, or hyperbolic functions,
(5.13) |
The boundary condition at gives , and the one at gives . Again there is only a trivial solution.
We have thus only found a solution for a discrete set of “eigenvalues” . Solving the equation for we find an exponential solution, . Combining all this information together, we have
(5.14) |
The equation we started from was linear and homogeneous, so we can superimpose the solutions for different values of ,
(5.15) |
This is a Fourier sine series with time-dependent Fourier coefficients. The initial condition specifies the coefficients , which are the Fourier coefficients at time . Thus
The final solution to the PDE + BC’s + IC is
(5.17) |
This solution is transient: if time goes to infinity, it goes to zero.