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# 5.2: Parabolic Equation

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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{aligned} \dfrac{\partial}{\partial t} u = k \dfrac{\partial^2}{\partial x^2} u,\;\;0<x<L,\;t>0.\end{aligned}

with boundary conditions

$u(0,t) = 0,\;u(L,t)=0,\;\;t>0,$

and initial condition $u(x,0) = x,\;0<x<L.$ We shall attack this problem by separation of variables, a technique always worth trying when attempting to solve a PDE, $u(x,t) = X(x) T(t).$ This leads to the differential equation

$X(x) T'(t) = kX"(x) T(t). \label{eq5.5}$

We find, by dividing both sides by $$XT$$, that

$\frac{1}{k}\frac{T'(t)}{T(t)} = \frac{X"(k)}{X(k)}.$

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 $$-\lambda$$.

We obtain two differential equations

\begin{aligned} T'(t) &= -\lambda k T(t) \\ X''(x) &= -\lambda X(x)\end{aligned}

Exercise $$\PageIndex{1}$$

What happens if $$X(x)T(t)$$ is zero at some point $$(x=x_0, t=t_0)$$?

Nothing. We can still perform the same trick.

Note

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 $$T(t_0)\neq 0$$. In that case we find (from \ref{eq5.5})), substituting $$t=t_0$$, that $$X''(x_0)=0$$.

We now have to distinguish the three cases $$\lambda>0$$, $$\lambda=0$$, and $$\lambda<0$$.

 $$\lambda>0$$

Write $$\alpha^2=\lambda$$, so that the equation for $$X$$ becomes $X''(x)=-\alpha^2 X(x).$ This has as solution $X(x) = A\cos\alpha x +B\sin\alpha x.$ $$X(0) = 0$$ gives $$A\cdot 1 + B \cdot 0=0$$, or $$A=0$$. Using $$X(L)=0$$ we find that $B\sin\alpha L = 0$ which has a nontrivial (i.e., one that is not zero) solution when $$\alpha L = n\pi$$, with $$n$$ a positive integer. This leads to $$\lambda_n= \frac{n^2\pi^2}{L^2}$$.

 $$\lambda=0$$

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

 $$\lambda<0$$

We write $$\lambda=-\alpha^2$$, so that the equation for $$X$$ becomes $X''(x)=-\alpha^2 X(x).$ The solution is now in terms of exponential, or hyperbolic functions, $X(x) = A \cosh x + B \sinh x.$ 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” $$\lambda_n>0$$. Solving the equation for $$T$$ we find an exponential solution, $$T=\exp(-\lambda k T)$$. Combining all this information together, we have $u_n(x,t) = \exp\left(-k \frac{n^2\pi^2}{L^2}t\right)\sin\left(\frac{n\pi}{L}x \right).$ The equation we started from was linear and homogeneous, so we can superimpose the solutions for different values of $$n$$, $u(x,t) = \sum_{n=1}^\infty c_n \exp\left(-k \frac{n^2\pi^2}{L^2}t\right)\sin\left(\frac{n\pi}{L}x\right).$ 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{aligned} c_n &=& \frac{2}{L} \int_0^L x \sin\frac{n\pi x}{L} dx \nonumber\\ &=& - \frac{2L}{n\pi}(-1)^n = (-1)^{n+1} \frac{2L}{n\pi}.\end{aligned} The final solution to the PDE + BC’s + IC is $u(x,t) = \sum_{n=1}^\infty (-1)^{n+1} \frac{2L}{n\pi} \exp\left(-k \frac{n^2\pi^2}{L^2}t\right)\sin\frac{n\pi}{L}x.$

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