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 $[\kappa]=\mathrm{length}^2/\mathrm{time}$.

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

\[XG^{\prime}=\kappa X^{\prime\prime}G.\]

After division by $\kappa XG$,

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

Thus

\[X^{\prime\prime}+k^2X=0, \qquad G^{\prime}+\kappa k^2G=0.\]

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.

If the endpoints are held at unequal constants $T_a$ and $T_b$, first remove the steady profile

\[S(x)=T_a+\frac{T_b-T_a}{L}x.\]

Because $S^{\prime\prime}=0$, the difference $\theta=T-S$ satisfies the same heat equation with $\theta(0,t)=\theta(L,t)=0$. Its initial profile is $f(x)-S(x)$, so

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

where

\[C_n=\frac2L\int_0^L[f(x)-S(x)] \sin\frac{n\pi x}{L}\,dx.\]

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.

Solved Problems

1. A rod initially at uniform excess temperature

A rod has both ends held at zero excess temperature and

\[T(x,0)=T_0, \qquad 0<x<L.\]

The sine coefficients are

\[\begin{aligned} B_n &=\frac{2T_0}{L}\int_0^L\sin\frac{n\pi x}{L}\,dx\\ &=\frac{2T_0}{n\pi}\left[1-(-1)^n\right]. \end{aligned}\]

Thus only odd modes occur:

\[\boxed{ T(x,t)=\frac{4T_0}{\pi} \sum_{\substack{n=1\\ n\ \mathrm{odd}}}^{\infty} \frac1n\sin\frac{n\pi x}{L} e^{-\kappa n^2\pi^2t/L^2}}.\]

Each term satisfies the fixed-end conditions and the heat equation. For every $t>0$ the exponential factors make the series convergent, and $T\to0$ as $t\to\infty$. The initial value is understood on the open interval; at the endpoints the imposed temperature changes at $t=0$.

2. Unequal endpoint temperatures with one transient mode

A $1.00\,\mathrm{m}$ rod has $T(0,t)=20^\circ\mathrm{C}$, $T(1,t)=80^\circ\mathrm{C}$ and diffusivity $\kappa=0.010\,\mathrm{m^2\,h^{-1}}$. Initially,

\[T(x,0)=20+60x+10\sin(\pi x) \quad (^\circ\mathrm{C}).\]

The steady profile is $S=20+60x$. Hence

\[\theta(x,0)=T(x,0)-S(x)=10\sin(\pi x),\]

so only the first transient mode is present:

\[\boxed{ T(x,t)=20+60x +10\sin(\pi x)e^{-0.010\pi^2t}}.\]

At the midpoint after $2.00\,\mathrm{h}$,

\[\boxed{ T(0.5,2)=50+10e^{-0.02\pi^2} =58.2087^\circ\mathrm{C}}.\]

The transient is positive and decays toward the steady midpoint temperature $50^\circ\mathrm{C}$. The exponent is dimensionless because $\kappa t/L^2$ is dimensionless.

Descriptive Questions

1. Starting from the given one-dimensional heat equation, derive the separated fixed-end temperature modes.
2. Derive the Fourier sine coefficients from an arbitrary initial temperature profile.
3. Explain why the decay time is proportional to $L^2$ and inversely proportional to $\kappa n^2$.
4. Show how subtracting a steady linear profile converts unequal fixed-end temperatures into homogeneous boundary conditions.

Numerical Problems

1. A rod has $L=0.50\,\mathrm{m}$ and $\kappa=1.2\times10^{-5}\,\mathrm{m^2\,s^{-1}}$. Find the half-life of its first mode. Final answer: $t_{1/2}=L^2\ln2/(\kappa\pi^2)=1463.14\,\mathrm{s}=24.3856\,\mathrm{min}$.
2. Two initial modes $n=1$ and $n=3$ have equal amplitudes. For $L=1.00\,\mathrm{m}$, $\kappa=0.010\,\mathrm{m^2\,h^{-1}}$, find their amplitude ratio $A_3/A_1$ after $2.00\,\mathrm{h}$. Final answer: $A_3/A_1=e^{-8\kappa\pi^2t/L^2}=0.206153$.
3. For $f(x)=T_0x(L-x)/L^2$ with $T_0=100\,\mathrm{K}$, find the first sine coefficient. Final answer: $B_1=8T_0/\pi^3=25.8012\,\mathrm{K}$.
4. For $L=0.40\,\mathrm{m}$ and $\kappa=2.0\times10^{-5}\,\mathrm{m^2\,s^{-1}}$, find the time for the $n=2$ amplitude to fall to $1\%$ of its initial value. Final answer: $t=L^2\ln100/(4\kappa\pi^2)=933.203\,\mathrm{s}=15.5534\,\mathrm{min}$.

The Fourier coefficients, transient solution, decay times, and numerical values are verified in the Unit II problem-check worksheet.

References

1. Heat equation — Wikipedia
2. Richard Haberman, Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th ed., Chapters 1–2, “Heat Equation” and “Method of Separation of Variables,” Pearson.
3. Erwin Kreyszig, Advanced Engineering Mathematics, 10th ed., Chapter 12, “Partial Differential Equations,” Wiley.
4. Mary L. Boas, Mathematical Methods in the Physical Sciences, 3rd ed., Chapter 13, “Partial Differential Equations,” Wiley.