Visual walkthrough — Wave equation (hyperbolic) 1D — derivation

Everything below is one long story: a tight string, a tiny piece of it, and the tug-of-war between neighbours that makes a bump travel.

---

Step 0 — The picture we are explaining

Before any maths, look at the thing itself.

• The horizontal line across the picture is the string at rest.
• The wavy magenta curve is the string right now, after we plucked it.
• We measure position along the string with the letter $x$ (how far to the right).
• We measure how far up or down the string has moved with the letter $u$.

Our whole job: find an equation that says how this height changes as time ticks forward.

---

Step 1 — Cut out one tiny bead of string

We cannot reason about the whole wavy curve at once. So we do what physicists always do: zoom in until the curve looks like a short, nearly-straight segment.

WHAT: We slice out the piece between position $x$ and position $x+\Delta x$.

WHY: Newton's law $F=ma$ (see Newton's second law) is about objects with mass. A single point has no mass. This little segment does have mass, so we can apply $F=ma$ to it.

PICTURE: In the figure the segment is the thick violet chunk; its two ends sit at $x$ (left) and $x+\Delta x$ (right).

---

Step 2 — The only force: tension pulling along the string

What pushes this bead around? Its two neighbours. A stretched string pulls on each piece along its own direction — like two people yanking a rope, one from each side.

WHAT: At the right end the string pulls the bead up-and-to-the-right; at the left end it pulls down-and-to-the-left. Both pulls have size $T$, but they point in different directions because the string is tilted differently at the two ends.

WHY the direction matters: A force pointing partly sideways and partly up must be split into two questions — how much of it is sideways? and how much is up? Those two pieces behave completely differently, so we handle them separately in Steps 3 and 4.

---

Step 3 — The tilt, and the two magic ratios

To split each pull into "sideways part" and "up part," we need to measure the tilt of the string.

Draw the right triangle whose slanted side (the hypotenuse) is the pulling force $T$. Its horizontal side and vertical side are exactly the sideways and upward parts we want.

WHY these tools and not others? We have a slanted arrow of known length ($T$) and a known tilt ($\theta$), and we want its horizontal and vertical shadows. That is precisely the question sine and cosine were invented to answer: they convert "length + angle" into "how much across, how much up."

PICTURE: In the figure, the orange arm is $T\cos \theta$ (sideways) and the magenta arm is $T\sin \theta$ (up). Together they rebuild the slanted tension arrow.

---

Step 4 — Sideways forces cancel (the string does not run away)

WHAT: Add up the two horizontal (sideways) pulls: the right end pulls right, the left end pulls left.

$$\underset{\text{sideways pull, right end}}{\underbrace{T\cos \theta (x+\Delta x)}}-\underset{\text{sideways pull, left end}}{\underbrace{T\cos \theta (x)}}=0.$$

WHY it must be zero: The string only moves up and down (assumption 4 — motion is purely transverse). If the sideways forces did not cancel, the bead would accelerate sideways, which we forbade. So they cancel.

---

Step 5 — Upward forces do NOT cancel — that is what moves the bead

Now the interesting direction: up. This is where the motion lives.

WHAT: Add the two upward pulls. The right end pulls up by $T\sin \theta (x+\Delta x)$; the left end pulls up by $T\sin \theta (x)$ (this can be negative = pulling down). The net upward force is the difference:

$$F_{\text{net}}=\underset{\text{up-pull, right end}}{\underbrace{T\sin \theta (x+\Delta x)}}-\underset{\text{up-pull, left end}}{\underbrace{T\sin \theta (x)}}.$$

WHY a difference? If both ends pull up by the same amount, they simply balance and the bead feels no net vertical force. It is the mismatch between the two ends that pushes the bead. Look at the figure: on a valley (concave-up piece) the right end tilts up more than the left, so the net pull is upward.

Now apply Newton, $F=ma$, in the vertical direction:

$$\underset{F=\text{net upward force}}{\underbrace{T[\sin \theta (x+\Delta x)-\sin \theta (x)]}}=\underset{m=\text{mass}}{\underbrace{\rho \Delta x}}\times \underset{a=\text{vertical acceleration}}{\underbrace{\frac{{\partial }^{2}u}{\partial t^2}}}.$$

---

Step 6 — Trade angles for slopes

Angles are awkward; slopes are calculus-friendly. Let us convert.

For a small angle, the sine and the tangent are almost equal (see figure — the two curves hug each other near $0$):

$$\sin \theta \approx \tan \theta =\underset{\text{slope of the string}}{\underbrace{u_x}}.$$

WHY replace $\sin \theta$ with $u_x$ and not with $\theta$? Because the slope $u_x$ is the quantity we can differentiate again to build curvature. Chasing $\theta$ leads nowhere useful; chasing the slope leads straight to the wave equation. Substitute into Step 5:

$$T[\underset{\text{slope at right end}}{\underbrace{u_x(x+\Delta x,t)}}-\underset{\text{slope at left end}}{\underbrace{u_x(x,t)}}]=\rho \Delta x{u}_{tt}.$$

The physics is now pure calculus: the difference in slope across the bead drives its acceleration.

---

Step 7 — Shrink the bead to a point → curvature appears

We now let the width $\Delta x$ go to zero, turning a "difference of slopes" into a genuine derivative.

WHAT: Divide both sides by $\Delta x$:

$$T\underset{\text{change in slope, per unit width}}{\underbrace{\frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x}}}=\rho u_{tt}.$$

WHY divide? So the left side reads "how fast the slope changes as you move right." As $\Delta x\to 0$ that fraction is, by the very definition of a derivative, the derivative of the slope:

$$\underset{\Delta x\to 0}{\lim }\frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x}=\frac{{\partial }^{2}u}{\partial x^2}=u_{xx}.$$

So we arrive at:

$$T{u}_{xx}=\rho u_{tt}.$$

---

Step 8 — Name the constant, read the meaning

WHAT: Divide by $\rho$ and give the leftover ratio a name.

$$u_{tt}=\underset{\text{call this }{c}^2}{\underbrace{\frac{T}{\rho }}}{u}_{xx}⟹\boxed{u_{tt}=c^2{u}_{xx}},c=\sqrt{\frac{T}{\rho }}.$$

Edge & degenerate cases (never leave a gap)

Where the naive picture would break — large slopes — assumption 2 fails: then $\sin \theta \ne u_x$, the horizontal pull no longer cancels cleanly, and you get a nonlinear wave whose speed can depend on amplitude. Our whole derivation lives strictly in the small-slope world; outside it, this equation is only approximate.

---

The one-picture summary

The single diagram above compresses all eight steps: cut a bead → tension pulls tangentially at both ends → split into sideways (cancels) and upward (mismatched) → the mismatch is $T\Delta (\text{slope})$ → divide by $\Delta x$ and shrink → curvature $u_{xx}$ drives acceleration $u_{tt}$ → name $c^2=T/\rho$.

Recall Feynman retelling — the whole walkthrough in plain words

Grab a tight string and stare at one tiny bead of it. Two invisible hands hold that bead — the string on its left and the string on its right — and each pulls exactly along the string's slant. Split each pull into "sideways" and "up." The sideways pulls fight to a draw (the bead never runs off left or right), so ignore them. The up pulls are the whole show: if the right side of the bead tilts up more steeply than the left side, the bead gets a net yank upward. "Right tilt minus left tilt" is just how the slope is changing across the bead — and how the slope changes is exactly curvature. So: curved-up bead gets pulled up, curved-down bead gets pulled down, each bead forever chasing "straight," overshooting, and passing the nudge onward. Newton's $F=ma$ turns that sentence into $T\times (\text{curvature})=(\text{mass})\times (\text{acceleration})$, and after tidying the constants it reads $u_{tt}=c^2{u}_{xx}$ with $c=\sqrt{T/\rho }$. Tighten the string, waves fly; load it with mass, they crawl.

Recall

What single physical quantity, appearing in $u_{xx}$, drives a bead's vertical acceleration? ::: The curvature of the string (how the slope changes across the bead), not its steepness. Why do the horizontal tension components cancel? ::: Because the string moves only up and down (purely transverse), so there can be no net sideways force. Where in the derivation does the derivative actually appear? ::: When we divide the slope-difference by $\Delta x$ and let $\Delta x\to 0$, giving $u_{xx}$ by the definition of a derivative. Why replace $\sin \theta$ by $u_x$ rather than by $\theta$? ::: Because $u_x$ is the slope, which we can differentiate once more to get curvature $u_{xx}$.

Related paths onward: d'Alembert solution · Method of characteristics · Separation of variables for the wave equation.