30 Jun 2025

One-Dimensional Heat-Flow Equation

Separated temperature modes in a finite rod, Fourier coefficients, and exponential decay rates.

bsc semester-ii mathematical-physics heat-equation separation-of-variables

The syllabus asks for the solution, not the derivation, of the one-dimensional heat equation. Take

\[\boxed{\frac{\partial T}{\partial t} =\kappa\frac{\partial^2T}{\partial x^2}}\]

as given, where $[\kappa]={\rm length}^2/{\rm time}$.

Consider a rod $0<x<L$ whose ends are held at zero temperature:

\[T(0,t)=T(L,t)=0, \qquad T(x,0)=f(x).\]

Put $T(x,t)=X(x)G(t)$. Then

\[XG'=\kappa X''G.\]

After division by $\kappa XG$,

\[\frac{G'}{\kappa G}=\frac{X''}{X}=-k^2.\]

Thus

\[X''+k^2X=0, \qquad G'+\kappa k^2G=0.\]

The boundary conditions select

\[X_n=\sin\frac{n\pi x}{L}, \qquad k_n=\frac{n\pi}{L},\]

and the time equation gives

\[G_n(t)=\exp\!\left[-\kappa \left(\frac{n\pi}{L}\right)^2t\right].\]

Therefore

\[\boxed{ T(x,t)=\sum_{n=1}^{\infty}B_n \sin\frac{n\pi x}{L} e^{-\kappa(n\pi/L)^2t}}.\]

At $t=0$ this must equal $f(x)$. Multiply by $\sin(m\pi x/L)$ and integrate. Orthogonality gives

\[\boxed{B_n=\frac2L\int_0^L f(x) \sin\frac{n\pi x}{L}dx}.\]

As an exact two-mode example, take

\[f(x)=T_0\left(\sin\frac{\pi x}{L} +\frac12\sin\frac{3\pi x}{L}\right).\]

Only $B_1=T_0$ and $B_3=T_0/2$ are nonzero, so

\[\boxed{ \frac{T(x,t)}{T_0} =\sin\frac{\pi x}{L}e^{-\kappa\pi^2t/L^2} +\frac12\sin\frac{3\pi x}{L}e^{-9\kappa\pi^2t/L^2}}.\]

The plotted two-mode solution is shown at increasing \(\tau=\kappa\pi^2t/L^2\). The third spatial mode decays as \(e^{-9\tau}\), nine times faster in the exponent than the first.

For one mode, the decay time is

\[\tau_n=\frac{1}{\kappa k_n^2} =\frac{L^2}{\kappa n^2\pi^2}.\]

Its units are time, and the $n^{-2}$ dependence shows why fine spatial structure disappears first. The heat-equation and boundary residuals are zero in the Unit II Maxima worksheet.