Visual walkthrough — Solving heat equation — separation of variables

---

Step 0 — What the pictures are of

Before any algebra, let's fix the stage.

• The rod is a line from $x=0$ to $x=L$. The letter $x$ just means "how far along the rod you are."
• Temperature $u$ is a height: at each spot $x$ and each time $t$, $u(x,t)$ is how hot that spot is right now. So a temperature profile is a curve drawn above the rod — tall where hot, low where cold.
• The two ends are held at $0$ forever: $u(0,t)=0$ and $u(L,t)=0$. In the picture, the curve is pinned to the floor at both ends.

The rule of the game — the heat equation itself — is:

$\underset{\text{rate of warming}}{\underbrace{u_t}}=\underset{\begin{array}{c}\text{how fast heat}\\ \text{spreads}\end{array}}{\underbrace{{\alpha }^2}}\underset{\text{curvature}}{\underbrace{u_{xx}}}$

Read it in words: a spot warms fast exactly where the curve bends up (a dip), and cools where it bends down (a bump). That single sentence is the whole physics; everything below is solving it.

---

Step 1 — The guess: shape × dimmer-knob

WHAT. We guess the answer has a special shape: $u(x,t)=X(x)T(t).$ $X(x)$ is a fixed shape along the rod (depends only on where you are). $T(t)$ is a single dimmer knob (depends only on time) that scales that whole shape up or down.

WHY. If the shape never changes and only its overall brightness fades, then the hard PDE — which mixes space and time — might split into two separate easy problems. We try it because if it fails, the algebra tells us immediately, and we've lost nothing.

PICTURE. The frozen shape $X(x)$ is the same curve at every time; the knob $T(t)$ just squashes it toward the floor as time passes.

---

Step 2 — Force the equation to split

WHAT. Put the guess into $u_t={\alpha }^2{u}_{xx}$.

Since $X$ ignores time and $T$ ignores space: $u_t=X(x)T^{\prime }(t),u_{xx}=X^{\prime \prime }(x)T(t).$ Here $T^{\prime }$ means "how fast the knob changes," and $X^{\prime \prime }$ means "the curvature of the shape." Substituting: $X{T}^{\prime }={\alpha }^2{X}^{\prime \prime }T.$

Now divide both sides by ${\alpha }^{2}XT$: $\underset{\text{only }t\text{ lives here}}{\underbrace{\frac{T^{\prime }}{{\alpha }^{2}T}}}=\underset{\text{only }x\text{ lives here}}{\underbrace{\frac{X^{\prime \prime }}{X}}}.$

WHY. The left side changes only when time changes; the right side changes only when position changes. If I stand still ($x$ fixed) and let time run, the right side is frozen — so the left side can't move either. If I freeze time and walk the rod, the left is frozen — so the right can't move. The only way two such things stay equal is if both are the same constant.

PICTURE. Two dials wired to always read the same number: turning the "time" dial can't change the reading, and turning the "space" dial can't either — so the reading is locked.

We name that locked constant $-\lambda$: $\frac{X^{\prime \prime }}{X}=-\lambda ,\frac{T^{\prime }}{{\alpha }^{2}T}=-\lambda .$

---

Step 3 — The space problem, and why the sign of $\lambda$ decides everything

WHAT. From $\frac{X^{\prime \prime }}{X}=-\lambda$ we get the space ODE: $X^{\prime \prime }+\lambda X=0,X(0)=0,X(L)=0.$ The boundary conditions come from the pinned ends: $u(0,t)=X(0)T(t)=0$ for all $t$, and $T$ is not zero (that would be a dead rod), so $X(0)=0$. Same at $L$.

WHY test three signs? The shape of the solution to $X^{\prime \prime }+\lambda X=0$ is completely different depending on whether $\lambda$ is negative, zero, or positive. We must check all three — the reader should never wonder "but what if $\lambda <0$?"

PICTURE. Three candidate shapes, each trying to touch the floor at both ends:

• Case $\lambda <0$ (write $\lambda =-{\mu }^2$): solutions are $X=A{e}^{\mu x}+B{e}^{-\mu x}$ — pure growing/shrinking exponentials, no wiggle. A curve made only of exponentials can't dip down and come back to the floor twice. Pinning it at both ends forces $A=B=0$: the flat-zero line. Trivial — rejected.
• Case $\lambda =0$: solutions are straight lines $X=A+Bx$. A line through the floor at $x=0$ and again at $x=L$ is the floor. $A=B=0$. Trivial — rejected.
• Case $\lambda >0$ (write $\lambda =k^2$): solutions are $X=A\cos kx+B\sin kx$ — genuine waves. Waves can return to the floor. This is the only case with a life. Keep it.

Finishing the good case. $X(0)=0$ kills the cosine (since $\cos 0=1\ne 0$), leaving $X=B\sin kx$. Then $X(L)=0$ needs $\sin kL=0⟹kL=n\pi ,n=1,2,3,\dots$ Only these special $k$ let the wave land exactly on the floor at $x=L$.

So the surviving shapes — the eigenfunctions — and their eigenvalues are: $\boxed{X_n(x)=\sin \frac{n\pi x}{L},{\lambda }_n={\left(\frac{n\pi }{L}\right)}^2}$ $n=1$ is one hump, $n=2$ is two humps, and so on — each fits a whole number of half-waves into the rod. (This "which shapes survive the boundary" question is the heart of Sturm-Liouville Theory.)

---

Step 4 — The time problem: the dimmer fades

WHAT. Now feed ${\lambda }_n$ into the time side $\frac{T^{\prime }}{{\alpha }^{2}T}=-{\lambda }_n$, i.e. $T^{\prime }=-{\lambda }_n{\alpha }^{2}T.$ This says "the knob's rate of change is a negative multiple of itself" — the signature of exponential decay. Its solution: $\boxed{T_n(t)=e^{-{\alpha }^2{\lambda }_{n}t}=e^{-{\alpha }^2(n\pi /L{)}^{2}t}}$

WHY. $T^{\prime }=(\text{negative})\times T$ is the one equation whose answer is a shrinking exponential. Each mode gets its own decay speed: the exponent carries ${\lambda }_n=(n\pi /L{)}^2$, so mode $n$ fades at a rate $\propto n^2$. Now you see the payoff of the minus sign in Step 2 — with $+\lambda$ this exponential would grow and burn the rod up, which is impossible with frozen ends.

PICTURE. The higher the mode number, the steeper the plunge to zero.

---

Step 5 — Superpose: add many fading humps

WHAT. Each single mode $u_n(x,t)=\sin \frac{n\pi x}{L}{e}^{-{\alpha }^2(n\pi /L{)}^{2}t}$ solves the PDE and the two pinned-end conditions. Because the heat equation is linear and homogeneous (no $u^2$, no stray constant), any sum of solutions is again a solution — this is the Superposition Principle. So we allow every mode at once, each with its own strength $B_n$: $\boxed{u(x,t)=\sum _{n=1}^{\infty }{B}_n\sin \frac{n\pi x}{L}{e}^{-{\alpha }^2(n\pi /L{)}^{2}t}}$

WHY. One pure sine almost never matches a real starting profile. But stacking sines can build any pinned-end shape — that is exactly what a Fourier Series does. The $B_n$ are the volume settings for each hump.

PICTURE. Stack a tall one-hump, a shorter two-hump, a smaller three-hump… and their sum is a lumpy profile — the initial temperature.

Each symbol, right where it sits: $u(x,t)={\sum }_{n\ge 1}\underset{\begin{array}{c}\text{how loud}\\ \text{mode }n\text{ is}\end{array}}{\underbrace{B_n}}\underset{\begin{array}{c}\text{its frozen}\\ \text{shape}\end{array}}{\underbrace{\sin \frac{n\pi x}{L}}}\underset{\begin{array}{c}\text{its personal}\\ \text{fade timer}\end{array}}{\underbrace{e^{-{\alpha }^2(n\pi /L{)}^{2}t}}}$

---

Step 6 — Fit the start: find every $B_n$

WHAT. At $t=0$ every fade timer equals $1$ ($e^0=1$), so the solution must become the initial profile $f(x)$: $f(x)={\sum }_{n=1}^{\infty }{B}_n\sin \frac{n\pi x}{L}.$ To pull out one coefficient $B_m$, multiply both sides by $\sin \frac{m\pi x}{L}$ and integrate across the rod. The magic is orthogonality: ${\int }_0^L\sin \frac{n\pi x}{L}\sin \frac{m\pi x}{L}dx=\{\begin{array}{ll}L/2,& n=m\\ 0,& n\ne m\end{array}$ Every mismatched pair integrates to zero, so only the $n=m$ term survives: ${\int }_0^{L}f(x)\sin \frac{m\pi x}{L}dx=B_m\cdot \frac{L}{2}⟹\boxed{B_m=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{m\pi x}{L}dx}$

WHY. Orthogonality is a sieve: multiplying by one mode and integrating filters out that mode's strength while erasing all the others. The $\frac{2}{L}$ is just dividing by the mode's own "squared length" $L/2$.

PICTURE. Matching-frequency sines overlap to a positive area (they survive the sieve); different-frequency sines cancel to zero area.

---

The one-picture summary

Read the pipeline top to bottom: guess → split → sine shapes → decay timers → sum → fit. The bottom panel shows the true payoff: a lumpy hot start, watched over time, always relaxes into a single half-sine hump (mode $n=1$ outlives everyone because it decays slowest), then that hump itself sinks to a cold rod.

Recall Feynman: the whole walkthrough in plain words

We had a hot rod with its ends frozen and wanted its future. First we guessed the temperature is a fixed shape times a fading dimmer-knob. Plugging that in, the equation split into a space part and a time part that were forced to equal the same locked number — because one part can only change with position and the other only with time, so neither can move. Testing that number's sign, only a positive choice gave real wave shapes that could touch the frozen floor at both ends; those shapes turned out to be sine humps that fit a whole number of half-waves in the rod. The time part became a fading exponential, and skinnier humps (more wiggles) fade far faster. Since the equation is linear, we added up all the humps, each fading on its own timer. Finally, to match the actual starting warmth, we found each hump's loudness with a Fourier sieve — multiply by one hump, integrate, and every other hump cancels out. Let time run, and the rod always melts down to a single fat half-sine bump before going cold. That's the whole story.

---

Related builds: Wave Equation (same separation, oscillating time part), Laplace Equation (steady state, no time at all), and Fourier Series for the coefficient machinery.