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.

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

with boundary conditions

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

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

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

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

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

Exercise $\PageIndex{1}$

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.

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 $$X''(x)=-\alpha^2 X(x). \nonumber$$ This has as solution $$X(x) = A\cos\alpha x +B\sin\alpha x. \nonumber$$ $X(0) = 0$ gives $A\cdot 1 + B \cdot 0=0$, or $A=0$. Using $X(L)=0$ we find that $$B\sin\alpha L = 0 \nonumber$$ 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 $$X''(x)=-\alpha^2 X(x). \nonumber$$ The solution is now in terms of exponential, or hyperbolic functions, $$X(x) = A \cosh x + B \sinh x. \nonumber$$ 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 $$u_n(x,t) = \exp\left(-k \frac{n^2\pi^2}{L^2}t\right)\sin\left(\frac{n\pi}{L}x \right). \nonumber$$ The equation we started from was linear and homogeneous, so we can superimpose the solutions for different values of $n$, $$u(x,t) = \sum_{n=1}^\infty c_n \exp\left(-k \frac{n^2\pi^2}{L^2}t\right)\sin\left(\frac{n\pi}{L}x\right). \nonumber$$ 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

$$\begin{aligned} c_n &=& \frac{2}{L} \int_0^L x \sin\frac{n\pi x}{L} dx \nonumber\\ &=& - \frac{2L}{n\pi}(-1)^n = (-1)^{n+1} \frac{2L}{n\pi}.\end{aligned} \nonumber$$ The final solution to the PDE + BC’s + IC is $$u(x,t) = \sum_{n=1}^\infty (-1)^{n+1} \frac{2L}{n\pi} \exp\left(-k \frac{n^2\pi^2}{L^2}t\right)\sin\frac{n\pi}{L}x. \nonumber$$

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