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5.2 parabolic equation

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.

\begin{eqnarray} {∂u\over ∂t} = k{{∂}^{2}u\over ∂{x}^{2}},\kern 2.77695pt \kern 2.77695pt 0 < x < L,\kern 2.77695pt t > 0.& & %&(5.1) \\ \end{eqnarray}

with boundary conditions

u(0,t) = 0,\kern 2.77695pt u(L,t) = 0,\kern 2.77695pt \kern 2.77695pt t > 0,
(5.2)

and initial condition

u(x,0) = x,\kern 2.77695pt 0 < x < L.
(5.3)

We shall attack this problem by separation of variables, a technique always worth trying when attempting to solve a PDE,

u(x,t) = X(x)T(t).
(5.4)

This leads to the differential equation

X(x)T'(t) = kX”(x)T(t).
(5.5)

We find, by dividing both sides by XT, that

{1\over k} {T'(t)\over T(t)} = {X”(k)\over X(k)} .
(5.6)

Thus the left-hand side, a function of t, equals a function of x on the right-hand side. This is not possible unless both sides are independent of x and t, i.e. constant. Let us call this constant − λ.

We obtain two differential equations

\begin{eqnarray} T'(t)& =& −λkT(t)%&(5.7) \\ X”(x)& =& −λX(x)%&(5.8) \\ \end{eqnarray}

Question: What happens if X(x)T(t) is zero at some point (x = {x}_{0},t = {t}_{0})?

Answer: Nothing. We can still perform the same trick.

This is not so trivial as I suggest. We either have X({x}_{0}) = 0 or T({t}_{0}) = 0. Let me just consider the first case, and assume T({t}_{0})\mathrel{≠}0. In that case we find (from (5.5)), substituting t = {t}_{0}, that X''({x}_{0}) = 0.

We now have to distinguish the three cases λ > 0, λ = 0, and λ < 0.

λ > 0
Write {α}^{2} = λ, so that the equation for X becomes

X''(x) = −{α}^{2}X(x).
(5.9)

This has as solution

X(x) = A\mathop{cos}\nolimits αx + B\mathop{sin}\nolimits αx.
(5.10)

X(0) = 0 gives A ⋅ 1 + B ⋅ 0 = 0, or A = 0. Using X(L) = 0 we find that

B\mathop{sin}\nolimits αL = 0
(5.11)

which has a nontrivial (i.e., one that is not zero) solution when αL = nπ, with n a positive integer. This leads to {λ}_{n} = {{n}^{2}{π}^{2}\over {L}^{2}} .

λ = 0
We find that X = A + Bx. The boundary conditions give A = B = 0, so there is only the trivial (zero) solution.

λ < 0
We write λ = −{α}^{2}, so that the equation for X becomes

X''(x) = −{α}^{2}X(x).
(5.12)

The solution is now in terms of exponential, or hyperbolic functions,

X(x) = A\mathop{cosh}\nolimits x + B\mathop{sinh}\nolimits x.
(5.13)

The boundary condition at x = 0 gives A = 0, and the one at x = L gives B = 0. Again there is only a trivial solution.

We have thus only found a solution for a discrete set of “eigenvalues” {λ}_{n} > 0. Solving the equation for T we find an exponential solution, T =\mathop{ exp}\nolimits (−λkT). Combining all this information together, we have

{u}_{n}(x,t) =\mathop{ exp}\nolimits \left (−k{{n}^{2}{π}^{2}\over {L}^{2}} t\right )\mathop{sin}\nolimits \left ({nπ\over L} x\right ).
(5.14)

The equation we started from was linear and homogeneous, so we can superimpose the solutions for different values of n,

u(x,t) ={ \mathop{∑ }}_{n=1}^{∞}{c}_{ n}\mathop{ exp}\nolimits \left (−k{{n}^{2}{π}^{2}\over {L}^{2}} t\right )\mathop{sin}\nolimits \left ({nπ\over L} x\right ).
(5.15)

This is a Fourier sine series with time-dependent Fourier coefficients. The initial condition specifies the coefficients {c}_{n}, which are the Fourier coefficients at time t = 0. Thus

\begin{eqnarray}{ c}_{n}& =& {2\over L}{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{L}x\mathop{sin}\nolimits {nπx\over L} dx %& \\ & =& −{2L\over nπ}{(−1)}^{n} = {(−1)}^{n+1}{2L\over nπ}.%&(5.16) \\ \end{eqnarray}

The final solution to the PDE + BC’s + IC is

u(x,t) ={ \mathop{∑ }}_{n=1}^{∞}{(−1)}^{n+1}{2L\over nπ}\mathop{exp}\nolimits \left (−k{{n}^{2}{π}^{2}\over {L}^{2}} t\right )\mathop{sin}\nolimits {nπ\over L} x.
(5.17)

This solution is transient: if time goes to infinity, it goes to zero.

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