Exercises — Solving heat equation — separation of variables

These come straight from Steps 3–4 of the derivation.

• (a) ${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2$, for $n=1,2,3,\dots$
• (b) $X_n(x)=\sin \frac{n\pi x}{L}$.
• (c) $T_n(t)=e^{-{\alpha }^2{\lambda }_{n}t}=e^{-{\alpha }^2(n\pi /L{)}^{2}t}$.

Why these and nothing else? The zero boundary conditions kill the cosine part and force $\sin kL=0$, i.e. $kL=n\pi$. Only $\lambda >0$ survives (Step 3 tested all three signs).

Recognise: $f$ is already a single sine mode. Compare $f(x)={\sum }_n{B}_n\sin \frac{n\pi x}{L}$ term-by-term. Only $n=2$ appears, with $B_2=7$; every other $B_n=0$.

Attach that mode's decay (from 1.1 with $n=2$, so $n^2=4$): $u(x,t)=7\sin \frac{2\pi x}{L}{e}^{-4{\alpha }^2{\pi }^{2}t/L^2}.$

Cosines, $X_n(x)=\cos \frac{n\pi x}{L}$ for $n=0,1,2,\dots$ (including the flat $n=0$ mode).

Why? The condition $X^{\prime }(0)=0$ needs a function with zero slope at $x=0$; $\cos$ has a flat top there, $\sin$ does not. Matching the eigenfunction to the boundary condition is the whole game — see Boundary Conditions — Dirichlet vs Neumann.

Set (project onto sines): a constant is not a single sine, so we must integrate. $B_n=\frac{2}{\pi }{\int }_0^{\pi }{f}_0\sin (nx)dx=\frac{2f_0}{\pi }{\left[\frac{-\cos nx}{n}\right]}_0^{\pi }=\frac{2f_0}{\pi }\cdot \frac{1-\cos n\pi }{n}=\frac{2f_0}{\pi }\cdot \frac{1-(-1{)}^n}{n}.$ The factor $1-(-1{)}^n$ is $0$ for even $n$ and $2$ for odd $n$. So $B_n=\{\begin{array}{ll}\frac{4f_0}{\pi n},& n\text{ odd}\\ 0,& n\text{ even.}\end{array}$ Sum: $u(x,t)=\frac{4f_0}{\pi }{\sum }_{n\text{odd}}\frac{1}{n}\sin (nx)e^{-n^{2}t}.$ Why only odd? A flat profile is symmetric about the midpoint $x=\pi /2$; even sines are antisymmetric there, so they cannot help build a symmetric shape — their coefficient must vanish. (This is Fourier Series symmetry at work.)

$B_n=2{\int }_0^{1}x\sin (n\pi x)dx.$ Integrate by parts ($u=x$, $dv=\sin (n\pi x)dx$): ${\int }_0^{1}x\sin (n\pi x)dx={\left[-\frac{x\cos (n\pi x)}{n\pi }\right]}_0^1+\frac{1}{n\pi }{\int }_0^1\cos (n\pi x)dx.$ The last integral is ${\left[\frac{\sin (n\pi x)}{n\pi }\right]}_0^1=0$. So the integral is $-\frac{\cos n\pi }{n\pi }=-\frac{(-1{)}^n}{n\pi }=\frac{(-1{)}^{n+1}}{n\pi }$. $\boxed{B_n=\frac{2(-1{)}^{n+1}}{n\pi }}\Rightarrow u(x,t)={\sum }_{n=1}^{\infty }\frac{2(-1{)}^{n+1}}{n\pi }\sin (n\pi x)e^{-n^2{\pi }^{2}t}.$

$f$ is a finite combination of sine modes. Match to $\sin \frac{n\pi x}{L}=\sin \frac{n\pi x}{2}$: the term $\sin \frac{\pi x}{2}$ is $n=1$, and $\sin \frac{5\pi x}{2}$ is $n=5$. So $B_1=3$, $B_5=-1$, rest zero. Decay rate is ${\alpha }^2(n\pi /L{)}^2=(n\pi /2{)}^2=\frac{n^2{\pi }^2}{4}$: $u(x,t)=3\sin \frac{\pi x}{2}{e}^{-{\pi }^{2}t/4}-\sin \frac{5\pi x}{2}{e}^{-25{\pi }^{2}t/4}.$

The slowest mode is $n=1$ (smallest $n^2$). Its term is $\frac{400}{\pi }\sin x{e}^{-t}$. At $x=\pi /2$: $\sin \frac{\pi }{2}=1$, so $u\left(\frac{\pi }{2},t\right)\approx \frac{400}{\pi }{e}^{-t}\approx 127.32e^{-t}.$ Decay rate $=1$ (units $1/\text{time}$): the amplitude drops by a factor $e$ each unit of time. The next mode ($n=3$) decays as $e^{-9t}$ — $9$ times faster — so it's already negligible once $t$ is even moderately large. See the figure below.

Mode $n$ has amplitude $e^{-(n\pi /L{)}^{2}t}$. We want the ratio $\frac{{\text{amp}}_4}{{\text{amp}}_1}=\frac{e^{-(4\pi /L{)}^{2}t}}{e^{-(\pi /L{)}^{2}t}}=e^{-(16-1)(\pi /L{)}^{2}t}=e^{-15(\pi /L{)}^{2}t}.$ Set this equal to $e^{-1}$: $15\frac{{\pi }^2}{L^2}{t}^{\star }=1\Rightarrow t^{\star }=\frac{L^2}{15{\pi }^2}.$ Interpretation: the gap between modes widens at rate proportional to $(16-1)=15$ times the base rate. Wiggly modes lose the race fast — this is exactly why the rod smooths.

By orthogonality, the cross terms integrate to $0$ and each ${\int }_0^L{\sin }^2\frac{n\pi x}{L}dx=\frac{L}{2}$: $E(t)={\int }_0^L{u}^{2}dx=\frac{L}{2}{\sum }_{n=1}^{\infty }{B}_n^2{e}^{-2{\alpha }^2(n\pi /L{)}^{2}t}.$ Every exponential $e^{-2{\alpha }^2(n\pi /L{)}^{2}t}$ is a decreasing function of $t$ (its exponent is negative for $t$ increasing). A sum of non-negative terms each decreasing is itself non-increasing, so $E(t)$ never grows and $E(t)\to 0$ as $t\to \infty$. Physically: with cold ends and no source, stored heat can only leak out. (This is the Sturm-Liouville Theory orthogonality paying off.)

Set: $B_n=\frac{2}{\pi }{\int }_0^{\pi }x(\pi -x)\sin (nx)dx.$ Split into $\pi {\int }_0^{\pi }x\sin (nx)dx-{\int }_0^{\pi }{x}^2\sin (nx)dx$.

Standard results (integration by parts): ${\int }_0^{\pi }x\sin (nx)dx=\frac{-\pi (-1{)}^n}{n}=\frac{\pi (-1{)}^{n+1}}{n},$ ${\int }_0^{\pi }{x}^2\sin (nx)dx=\frac{-{\pi }^2(-1{)}^n}{n}+\frac{2((-1{)}^n-1)}{n^3}.$ Combine: ${\int }_0^{\pi }x(\pi -x)\sin (nx)dx=\frac{{\pi }^2(-1{)}^{n+1}}{n}-\left(\frac{-{\pi }^2(-1{)}^n}{n}+\frac{2((-1{)}^n-1)}{n^3}\right)=-\frac{2((-1{)}^n-1)}{n^3}=\frac{2(1-(-1{)}^n)}{n^3}.$ So $B_n=\frac{2}{\pi }\cdot \frac{2(1-(-1{)}^n)}{n^3}=\frac{4(1-(-1{)}^n)}{\pi n^3}=\{\begin{array}{ll}\frac{8}{\pi n^3},& n\text{ odd}\\ 0,& n\text{ even.}\end{array}$ $\boxed{u(x,t)=\frac{8}{\pi }\sum _{n\text{odd}}\frac{1}{n^3}\sin (nx)e^{-n^{2}t}.}$ Sanity: $x(\pi -x)$ is symmetric about $x=\pi /2$ and vanishes at both ends — consistent with only odd sines and the zero BCs.

Why we can't separate directly: $u(0,t)=T_0\ne 0$ would force $X(0)T(t)=T_0$ for all $t$ — impossible unless $T$ is constant. The homogeneous (zero-BC) machinery needs zero ends.

Trick — subtract the steady state. Look for the eventual profile $v(x)$ that no longer changes in time: $v_t=0\Rightarrow {\alpha }^2{v}^{\prime \prime }=0\Rightarrow v^{\prime \prime }=0\Rightarrow v(x)=ax+b$, a straight line. Fit the ends: $v(0)=T_0=b,v(L)=T_1=aL+b\Rightarrow a=\frac{T_1-T_0}{L}.$ $v(x)=T_0+\frac{T_1-T_0}{L}x.$ Define the deviation $w(x,t)=u(x,t)-v(x)$. Then:

• $w_t=u_t$ and $w_{xx}=u_{xx}-v^{\prime \prime }=u_{xx}$, so $w$ solves the same heat equation.
• $w(0,t)=T_0-T_0=0$, $w(L,t)=T_1-T_1=0$ — zero BCs.
• $w(x,0)=g(x)-v(x)$.

So $w$ is the standard Dirichlet problem with initial data $g(x)-v(x)$: $w(x,t)={\sum }_{n=1}^{\infty }{B}_n\sin \frac{n\pi x}{L}{e}^{-{\alpha }^2(n\pi /L{)}^{2}t},B_n=\frac{2}{L}{\int }_0^L(g(x)-v(x))\sin \frac{n\pi x}{L}dx,$ and finally $u(x,t)=v(x)+w(x,t)$. As $t\to \infty$, $w\to 0$ and $u\to v(x)$: the rod settles to the straight-line temperature between its two fixed ends. This same steady-state-plus-decay split powers Laplace Equation.

(a) Sine + Sum + Set (by inspection). $f$ is already a sum of pure sine modes with $L=\pi$, so $\sin \frac{n\pi x}{L}=\sin (nx)$. Read off: $B_1=2$, $B_4=-5$, $B_7=1$; all others $0$. Decay rate $={\alpha }^2(n\pi /L{)}^2=n^2$. Hence $u(x,t)=2\sin x{e}^{-t}-5\sin 4x{e}^{-16t}+\sin 7x{e}^{-49t}.$

(b) The $n=4$ amplitude is $5e^{-16t}$; initial value $5$. Set $e^{-16t^{\star }}=0.01$: $-16t^{\star }=\ln (0.01)=-\ln 100\Rightarrow t^{\star }=\frac{\ln 100}{16}\approx \frac{4.60517}{16}\approx \mathrm{0.28782.}$

(c) Largest exponent (slowest decay) is $n=1$, rate $1$, vs $16$ and $49$ for the others — those are gone almost immediately. So for large $t$, $u(x,t)\approx 2\sin x{e}^{-t}.$ At $x=\pi /2$: $\sin \frac{\pi }{2}=1$, giving $u(\pi /2,t)\approx 2e^{-t}$. The profile is a single half-sine bump that fades to a cold rod — the universal late-time shape (uses Superposition Principle to isolate the surviving mode).

Split $u=XT$ → Solve two ODEs → Sine eigenfunctions from zero BCs → Sum with $e^{-{\alpha }^2(n\pi /L{)}^{2}t}$ → Set $B_n=\frac{2}{L}{\int }_0^{L}f\sin \frac{n\pi x}{L}dx$. Nonzero ends? Subtract the straight-line steady state first.