Visual walkthrough — Solving wave equation — D'Alembert's solution

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Step 1 — What is a "wave that keeps its shape"?

WHAT. Before any equation, look at a single bump on a rope. Call the height of the rope at position $x$ and time $t$ the letter $u(x,t)$. So $u$ is just "how high the rope is, here and now." At $t=0$ we have a shape; a little later the same shape sits somewhere else.

WHY. We need to know what "moving without changing shape" means as a formula before we can solve anything. That single idea is the seed of the whole solution.

PICTURE. In the figure the same bump is drawn at three times. Notice: the point that was at the top slides to the right. If the top was at $x=0$ at time $0$, and it moves at speed $c$ (distance over time), then at time $t$ the top is at $x=ct$. So the shape at time $t$ is the old shape shifted right by $ct$.

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Step 2 — The equation that only such travellers obey

WHAT. We want the differential equation whose solutions are exactly these shape-preserving travellers. Physics (Newton's law on a tiny string element) gives the 1D wave equation: $u_{tt}=c^2{u}_{xx}.$

WHY. We need one master equation. Every symbol:

• $u_{tt}$ ::: acceleration of a point on the rope (rate of rate of change of height in time) — the "force felt" side.
• $u_{xx}$ ::: curvature of the rope (how much it bends in space) — a bent rope pulls itself straight.
• $c^2$ ::: turns curvature into acceleration; $c$ carries units of speed.

The sentence is: "how fast a point accelerates = (speed²) × how curved the rope is there." A sharply curved dip snaps back hardest.

PICTURE. The figure shows a curved piece of rope. Where it curves upward (a valley, $u_{xx}>0$) the acceleration $u_{tt}$ is upward — it gets pushed up, filling the valley. That restoring push is what makes the disturbance travel.

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Step 3 — Factor the operator: two transports hiding inside

WHAT. Move everything to one side and read the derivatives as objects we can multiply: $({\partial }_{tt}-c^2{\partial }_{xx})u=0.$ Here ${\partial }_t$ means "take the time-derivative" and ${\partial }_x$ "take the space-derivative." Because $a^2-b^2=(a-b)(a+b)$, ${\partial }_{tt}-c^2{\partial }_{xx}=({\partial }_t-c{\partial }_x)({\partial }_t+c{\partial }_x).$

WHY this tool — difference of squares. We chose factoring (not the quadratic formula, not guessing) because the operator is literally a difference of two squares, and the two factors are simpler first-order operators. Each factor is a Transport Equation — the simplest PDE there is. Solving two easy problems beats one hard one.

• ${\partial }_t-c{\partial }_x$ ::: annihilates left-movers $G(x+ct)$.
• ${\partial }_t+c{\partial }_x$ ::: annihilates right-movers $F(x-ct)$.

Why we're allowed to factor: the operators commute — for a smooth $u$, ${\partial }_x{\partial }_{t}u={\partial }_t{\partial }_{x}u$ (order of differentiation doesn't matter), exactly as ordinary numbers commute.

PICTURE. Two slanted lines cross the $x$–$t$ plane: one for each factor. Values ride along these lines unchanged — that is what a transport equation says.

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Step 4 — Rotate onto the natural grid: characteristic coordinates

WHAT. Those two slanted line-families beg to be used as axes. Define $\xi =x-ct,\eta =x+ct.$ $\xi$ (chalk-blue lines) labels which right-going line you're on; $\eta$ (chalk-pink) which left-going line.

WHY. In the ordinary $x,t$ grid the wave equation is tangled. If we tilt our graph paper so its lines are the characteristics, the equation should simplify — because along each line nothing changes. This is the Method of Characteristics idea: choose coordinates that follow the physics.

PICTURE. The old square $x$–$t$ grid is redrawn as a diamond grid of $\xi =$const and $\eta =$const lines. A single point $(x,t)$ is now named by which two lines cross there.

To translate derivatives we invert: $x=\frac{\xi +\eta }{2}$, $t=\frac{\eta -\xi }{2c}$. The chain rule (each old variable feeds both new ones) gives ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta },{\partial }_t=-c{\partial }_{\xi }+c{\partial }_{\eta }.$

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Step 5 — The great cancellation: $u_{\xi \eta }=0$

WHAT. Square each translated operator and subtract: $u_{xx}=({\partial }_{\xi }+{\partial }_{\eta }{)}^{2}u=u_{\xi \xi }+2u_{\xi \eta }+u_{\eta \eta },$ $u_{tt}=c^2(-{\partial }_{\xi }+{\partial }_{\eta }{)}^{2}u=c^2(u_{\xi \xi }-2u_{\xi \eta }+u_{\eta \eta }).$ Now form $u_{tt}-c^2{u}_{xx}$: $c^2(u_{\xi \xi }-2u_{\xi \eta }+u_{\eta \eta })-c^2(u_{\xi \xi }+2u_{\xi \eta }+u_{\eta \eta })=-4c^2{u}_{\xi \eta }=0.$

WHY it's beautiful. Every "straight" second derivative ($u_{\xi \xi }$, $u_{\eta \eta }$) appears once with $+$ and once with $-$ — they annihilate. Only the cross term survives, and only its sign differs, leaving $\boxed{u_{\xi \eta }=0.}$ Term by term: $u_{\xi \eta }$ = "first slope along $\xi$, then along $\eta$." Setting it to zero says the $\xi$-slope doesn't change as you walk in $\eta$.

PICTURE. The figure lines up the two expansions and strikes out the matching terms in chalk, leaving one lonely surviving box.

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Step 6 — Integrate twice: two free travellers appear

WHAT. From $u_{\xi \eta }=0$: integrate in $\eta$ → $u_{\xi }$ depends on $\xi$ alone, call it $F^{\prime }(\xi )$. Integrate in $\xi$ → $u=F(\xi )+G(\eta )=F(x-ct)+G(x+ct).$

WHY two whole functions. Each integration of a PDE gives back a function of the leftover variable (not a mere constant). Two integrations → two arbitrary functions, exactly the freedom a second-order equation needs. $F$ = right-mover (Step 1), $G$ = left-mover.

PICTURE. A tall bump is drawn as the sum of a right-running half and a left-running half; add the heights and you recover the original.

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Step 7 — Fitting the initial data (where $f$, $g$ enter)

WHAT. Two unknown functions need two facts. The initial shape $f$ and initial velocity $g$ supply them.

At $t=0$ the two riders overlap: u(x,0)=F(x)+G(x)=f(x).\tag{1} Differentiate $u_t=-c{F}^{\prime }(x-ct)+c{G}^{\prime }(x+ct)$ and set $t=0$: -cF'(x)+cG'(x)=g(x)\ \Rightarrow\ -F'(x)+G'(x)=\tfrac{1}{c}\,g(x).\tag{2}

WHY the $\frac{1}{c}$. The chain rule pulls a factor $c$ out of every time-derivative of a traveller (because $\frac{d}{dt}F(x-ct)=-c{F}^{\prime }$). Dividing it back gives $\frac{1}{c}$ — this single $c$ is the ancestor of the $\frac{1}{2c}$ in the final formula.

Integrate (2) from $0$ to $x$: -F(x)+G(x)=\tfrac1c\int_0^x g(s)\,ds+K.\tag{3}

Adding/subtracting (1) and (3) and rebuilding $u=F(x-ct)+G(x+ct)$ (the $K$ cancels, the limits merge) yields:

PICTURE. The figure shows the two half-copies of $f$ splitting apart, and the velocity $g$ shaded over $[x-ct,x+ct]$ — the domain of dependence.

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Step 8 — Edge cases you must be able to draw

WHAT / WHY. A solution you can't stress-test isn't understood. Three degenerate inputs:

1. $g=0$ (pure shape). Integral vanishes → $u=\frac{1}{2}[f(x-ct)+f(x+ct)]$. The bump splits into two half-height twins.
2. $f=0$ (pure kick). First term gone → only the spreading integral. A flat, widening plateau grows between two edges at $x\pm ct$.
3. $t=0$. Both $x\pm ct=x$; the two $f$-halves stack to full height and the integral (zero-width) vanishes → $u=f(x)$, $u_t=g(x)$. The formula obeys its own initial conditions.
4. Outside the cone. A point on the initial line farther than $ct$ from $x$ cannot reach $(x,t)$: finite speed $c$ — contrast the Heat Equation, which spreads instantly.

PICTURE. Three mini-panels: split twins, spreading plateau, and the light-cone showing what can and cannot influence $(x,t)$.

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The one-picture summary

The single diagram threads it all: the $x$–$t$ plane, the two characteristic diagonals through $(x_0,t_0)$, their feet at $x_0\pm c{t}_0$ carrying the two halves of $f$, and the shaded base carrying $\int g$. Reading the picture is reading D'Alembert's formula.

Recall Feynman: the whole walkthrough in plain words

Flick a rope. Newton says every point accelerates in proportion to how curved the rope is right there — that's $u_{tt}=c^2{u}_{xx}$. Stare at it and you notice it factors into two "just carry the value along a slanted line" problems: one line runs right, one runs left. So we tilt our graph paper to use those lines as axes ($\xi ,\eta$). On that tilted paper the equation collapses to "the slope-along-one-line never changes along the other" — $u_{\xi \eta }=0$ — which means the answer is simply any right-runner plus any left-runner: $F(x-ct)+G(x+ct)$. Finally we ask the rope its two starting facts — its shape $f$ and its push $g$ — and solving two simple equations tells us: cut the shape in half and send the halves apart ($\frac{1}{2}$), and smear the push over the stretch the wave can reach, sliced by the speed ($\frac{1}{2c}$). Anything too far away at the start simply can't have arrived yet — the wave has a top speed $c$. That's the entire song.

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Connections

• Method of Characteristics — the $\xi ,\eta$ tilt is this method in action.
• Transport Equation — the two first-order factors from Step 3.
• Separation of Variables — Wave Equation — the Fourier route on bounded strings.
• Domain of Dependence and Influence — Step 8's light-cone.
• Classification of Second-Order PDEs — the wave equation as hyperbolic prototype.
• Heat Equation — the instant-spreading contrast.