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Mathematics LibreTexts

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 \[X(x) T'(t) = kX"(x) T(t). \label{eq:5.5}\] 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 \[\begin{aligned} T'(t) &= & -\lambda k T(t) \\ X"(x) &=& -\lambda X(x)\end{aligned}\] 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.

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 ([eq:5.5])), substituting \(t=t_0\), that \(X''(x_0)=0\).

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

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 \(\lambda_n= \frac{n^2\pi^2}{L^2}\).

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

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.