Wave equation (hyperbolic) 1D — derivation

WHAT are we deriving?

WHY hyperbolic? Writing it as $u_{tt}-c^2u_{xx}=0$, the second-order coefficients are $A=1$ (for $u_{tt}$), $B=0$, $C=-c^2$. The discriminant $B^2-4AC = 0-4(1)(-c^2)=4c^2>0$, so it is hyperbolic — it has two real families of characteristics and supports travelling waves.

HOW: Derivation from first principles

We model the string with these assumptions (state them — they're where the physics lives):

1. Perfectly flexible string (no bending stiffness).
2. Small displacements & small slopes: $|u_x|\ll 1$.
3. Tension $T$ is constant in magnitude along the string.
4. Motion is purely transverse (vertical), no horizontal motion.
5. Linear density $\rho$ constant; gravity negligible vs tension.

Step 1 — Isolate a tiny element

Take a piece of string between $x$ and $x+\Delta x$. Tension pulls tangentially at each end. Let $\theta(x)$ be the angle the string makes with the horizontal.

Why this step? Forces act along the string, so we resolve them using the local slope.

Step 2 — Horizontal balance (no horizontal motion)

The horizontal components must cancel: $T\cos\theta(x+\Delta x) - T\cos\theta(x) = 0.$ For small slopes, $\cos\theta\approx 1$, so the horizontal tension is essentially $T$ everywhere — consistent with assumption 3.

Why this step? It confirms $T$ can be treated as constant horizontally, which we need next.

Step 3 — Vertical balance = Newton's 2nd law

The vertical force is the difference of the vertical components of tension: $F_{\text{net}} = T\sin\theta(x+\Delta x) - T\sin\theta(x).$ The mass of the element is $\rho\,\Delta x$ and its vertical acceleration is $u_{tt}$. So $F=ma$ gives $T\big[\sin\theta(x+\Delta x)-\sin\theta(x)\big] = \rho\,\Delta x\,\frac{\partial^2 u}{\partial t^2}.$

Why this step? This is literally $ma$: net upward pull equals mass times upward acceleration.

Step 4 — Small-angle: slope = derivative

For small angles, $\sin\theta\approx\tan\theta = \dfrac{\partial u}{\partial x}$ (the slope!). Substitute: $T\left[u_x(x+\Delta x,t) - u_x(x,t)\right] = \rho\,\Delta x\,u_{tt}.$

Why this step? We trade geometric angles for derivatives of $u$ — turning physics into calculus.

Step 5 — Divide by $\Delta x$ and shrink it

$T\,\frac{u_x(x+\Delta x,t)-u_x(x,t)}{\Delta x} = \rho\,u_{tt}.$ As $\Delta x\to 0$, the left fraction is the definition of $\dfrac{\partial}{\partial x}(u_x)=u_{xx}$: $T\,u_{xx} = \rho\,u_{tt}.$

Why this step? The difference-quotient of the slope is exactly the second derivative — curvature.

Step 6 — Name the constant

Divide by $\rho$ and set $c^2 = T/\rho$: $\boxed{u_{tt} = c^2\,u_{xx}}$

d'Alembert: the travelling-wave solution

WHY include this? It shows the equation truly produces waves moving at speed $c$.

Change variables to characteristics $\xi=x-ct$, $\eta=x+ct$. By the chain rule: $u_x=u_\xi+u_\eta,\quad u_{xx}=u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta},$ $u_t=-c\,u_\xi+c\,u_\eta,\quad u_{tt}=c^2(u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta}).$ Plug into $u_{tt}-c^2u_{xx}=0$: $c^2(u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta}) - c^2(u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta})=0 \;\Rightarrow\; -4c^2u_{\xi\eta}=0.$ So $u_{\xi\eta}=0$, integrate twice: $u=F(\xi)+G(\eta)=\underbrace{F(x-ct)}_{\text{moves right}}+\underbrace{G(x+ct)}_{\text{moves left}}.$

Worked examples

Common mistakes (Steel-manned)

Active-recall flashcards

Connections

• Heat equation (parabolic) 1D — derivation — same "neighbours influence" idea but with $u_t$ (first order in time) → diffusive, not oscillatory.
• Laplace equation (elliptic) — the steady state ($u_{tt}=0$) limit.
• Classification of second-order PDEs — where the discriminant test lives.
• Method of characteristics — generalises the $x\pm ct$ trick.
• Separation of variables for the wave equation — bounded-string Fourier-series solution.
• d'Alembert solution — full treatment of the travelling-wave form.
• Newton's second law — the physical seed of the whole derivation.

Concept Map

Hinglish (regional understanding)