Exercises — Wave equation (hyperbolic) 1D — derivation

• $u_{tt}$ = vertical acceleration of a point on the string (second time-derivative).
• $u_{xx}$ = curvature of the string at that point (second space-derivative): how the slope changes as you move along $x$.
• $c$ = wave speed, and $c=\sqrt{T/\rho }$. The equation reads: acceleration of a point is proportional to how curved the string is there.

(b).

• (a) has one time-derivative → that is the heat equation (parabolic).
• (c) has no time at all → the Laplace equation (elliptic).
• (b) has two time-derivatives matched to two space-derivatives → the wave equation (hyperbolic). See Classification of second-order PDEs.

What are $A$, $B$, $C$? Every second-order PDE in two variables can be written in the standard form $A{u}_{tt}+B{u}_{tx}+C{u}_{xx}+(\text{lower-order terms})=0,$ where $A$ is the number multiplying the pure second time-derivative $u_{tt}$, $C$ the number multiplying the pure second space-derivative $u_{xx}$, and $B$ the number multiplying the mixed derivative $u_{tx}$. For the wave equation $u_{tt}-c^2{u}_{xx}=0$ we read off $A=1$, $B=0$, $C=-c^2$. The discriminant $B^2-4AC$ classifies the PDE just like the discriminant of a quadratic classifies its roots.

$c=\sqrt{\frac{T}{\rho }}=\sqrt{\frac{120}{0.006}}=\sqrt{20000}\approx 141.4\text{m/s}.$ Why this formula? From the derivation, $c^2=T/\rho$: tighter string (bigger $T$) → faster; heavier string (bigger $\rho$) → slower. Matches tuning a guitar (tightening raises pitch, i.e. speeds the wave).

Since $c=\sqrt{T/\rho }$, we have $c\propto \sqrt{T}$. To double $c$: $2c=\sqrt{\frac{T^{\prime }}{\rho }}\Rightarrow 4c^2=\frac{T^{\prime }}{\rho }=4\cdot \frac{T}{\rho }\Rightarrow T^{\prime }=4T.$ You must quadruple the tension. Speed grows only with the square root of tension, so doubling speed costs a factor of $4$ in pull.

Differentiate. Treat $2x-6t$ as the inside; the chain rule multiplies by the derivative of the inside each time. $u_t=6\sin (2x-6t),u_{tt}=-36\cos (2x-6t).$ $u_x=-2\sin (2x-6t),u_{xx}=-4\cos (2x-6t).$ For $u_{tt}=c^2{u}_{xx}$ we need $-36\cos (\cdot )=c^2(-4)\cos (\cdot )$, so $c^2=36/4=9$, giving $c=3\text{m/s}.$ Sanity check: the argument is $2x-6t=2(x-3t)$, a function of $x-3t$, so it travels right at speed $3$. Consistent.

d'Alembert's formula is $u(x,t)=\frac{1}{2}[f(x-ct)+f(x+ct)]+\frac{1}{2c}{\int }_{x-ct}^{x+ct}g(s)ds.$ With $g\equiv 0$ the integral vanishes. Put $c=2$: $u(x,t)=\frac{1}{2}\left[\frac{1}{1+(x-2t{)}^2}+\frac{1}{1+(x+2t{)}^2}\right].$ Physically: the single bump splits into two half-height copies, one running right, one left, each at speed $2$. See d'Alembert solution.

Use $u_{tt}=c^2{u}_{xx}$ with $c^2>0$, so sign of $u_{tt}$ = sign of curvature $u_{xx}$.

• $P$ sits in a $\cup$-shaped dip: concave up, $u_{xx}>0$, so $u_{tt}>0$ → accelerates upward. The dip fills in.
• $Q$ sits on a $\cap$-shaped hump: concave down, $u_{xx}<0$, so $u_{tt}<0$ → accelerates downward. The hump flattens. Every point is pulled toward straight; that perpetual chasing is what propagates the shape.

Acceleration is $u_{tt}=c^2{u}_{xx}$, which depends on curvature $u_{xx}$, not on slope $u_x$. At the steepest point of a smooth bump the graph is passing through an inflection-like region where curvature can be small or even zero, while the flat top of the hump (zero slope) can have large curvature. So steepness (slope) and acceleration (curvature) are different things. This is the classic slope/curvature confusion from the parent note.

A newton is $\text{kg}\cdot \text{m}/{\text{s}}^2$. So $\frac{T}{\rho }=\frac{\text{kg}\cdot \text{m}/{\text{s}}^2}{\text{kg}/\text{m}}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}\cdot \frac{\text{m}}{\text{kg}}=\frac{{\text{m}}^2}{{\text{s}}^2}.$ Taking the square root gives $\text{m}/\text{s}$. A genuine speed — the formula is dimensionally honest.

A pure travelling wave has the form (function of $x-ct$) or (function of $x+ct$).

• (a) argument $x-5t$ → right-moving at speed $5$ (the shape at $x$ appears later at larger $x$).
• (b) argument $x+5t$ → left-moving at speed $5$.
• (c) $\cos x\cos 5t$ is a standing wave — neither purely left nor right. But it is still a solution: it is the sum of a left- and a right-mover, since $\cos x\cos 5t=\frac{1}{2}[\cos (x-5t)+\cos (x+5t)]$. This links to Separation of variables for the wave equation.

With $f\equiv 0$, only the integral term survives: $u(x,t)=\frac{1}{2c}{\int }_{x-ct}^{x+ct}\cos sds=\frac{1}{2c}[\sin s{]}_{x-ct}^{x+ct}=\frac{1}{2c}[\sin (x+ct)-\sin (x-ct)].$ Using $\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$ with $A=x+ct$, $B=x-ct$: $u(x,t)=\frac{1}{2c}\cdot 2\cos x\sin (ct)=\frac{\cos x\sin (ct)}{c}.$ Check $t=0$: $u=0$ ✓. And $u_t=\cos x\cos (ct)$, at $t=0$ gives $\cos x=g(x)$ ✓.

The idea. We rewrite $u$ as a function of the two new variables $\xi =x-ct$ (constant along right-movers) and $\eta =x+ct$ (constant along left-movers). To convert $x$- and $t$-derivatives into $\xi$- and $\eta$-derivatives we use the chain rule. Since both $\xi$ and $\eta$ depend on $x$ and on $t$, each derivative operator splits: $\frac{\partial }{\partial x}=\frac{\partial \xi }{\partial x}\frac{\partial }{\partial \xi }+\frac{\partial \eta }{\partial x}\frac{\partial }{\partial \eta }=\frac{\partial }{\partial \xi }+\frac{\partial }{\partial \eta },\frac{\partial \xi }{\partial x}=1,\frac{\partial \eta }{\partial x}=1.$ $\frac{\partial }{\partial t}=\frac{\partial \xi }{\partial t}\frac{\partial }{\partial \xi }+\frac{\partial \eta }{\partial t}\frac{\partial }{\partial \eta }=-c\frac{\partial }{\partial \xi }+c\frac{\partial }{\partial \eta },\frac{\partial \xi }{\partial t}=-c,\frac{\partial \eta }{\partial t}=+c.$ Where the signs come from: $\xi =x-ct$ so $\partial \xi /\partial t=-c$; $\eta =x+ct$ so $\partial \eta /\partial t=+c$. That opposite sign is the whole reason the two families of waves separate.

Second $x$-derivative. Apply ${\partial }_x=({\partial }_{\xi }+{\partial }_{\eta })$ twice: $u_{xx}=({\partial }_{\xi }+{\partial }_{\eta })({\partial }_{\xi }+{\partial }_{\eta })u=u_{\xi \xi }+u_{\xi \eta }+u_{\eta \xi }+u_{\eta \eta }.$ The two cross terms $u_{\xi \eta }$ and $u_{\eta \xi }$ are equal (mixed partials commute for smooth $u$), so they combine into $2u_{\xi \eta }$ — that is why the cross coefficient is $2$: $u_{xx}=u_{\xi \xi }+2u_{\xi \eta }+u_{\eta \eta }.$ Second $t$-derivative. Apply ${\partial }_t=(-c{\partial }_{\xi }+c{\partial }_{\eta })$ twice. Factor out $c$: ${\partial }_t=c(-{\partial }_{\xi }+{\partial }_{\eta })$, so $u_{tt}=c^2(-{\partial }_{\xi }+{\partial }_{\eta })(-{\partial }_{\xi }+{\partial }_{\eta })u=c^2(u_{\xi \xi }-u_{\xi \eta }-u_{\eta \xi }+u_{\eta \eta })=c^2(u_{\xi \xi }-2u_{\xi \eta }+u_{\eta \eta }).$ Here the cross terms carry a minus sign because each factor contributes one $-{\partial }_{\xi }$; multiplying $(-{\partial }_{\xi })({\partial }_{\eta })$ and $({\partial }_{\eta })(-{\partial }_{\xi })$ both give $-{\partial }_{\xi }{\partial }_{\eta }$.

Subtract. $u_{tt}-c^2{u}_{xx}=c^2(u_{\xi \xi }-2u_{\xi \eta }+u_{\eta \eta })-c^2(u_{\xi \xi }+2u_{\xi \eta }+u_{\eta \eta }).$ The $u_{\xi \xi }$ and $u_{\eta \eta }$ terms cancel; the cross terms give $c^2(-2-2)u_{\xi \eta }$: $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }.$ Setting this to zero forces $u_{\xi \eta }=0$. Integrating once in $\eta$ gives $u_{\xi }=(\text{function of }\xi \text{ only})$; integrating again in $\xi$ gives $u=F(\xi )+G(\eta ),$ where $F$ and $G$ are arbitrary twice-differentiable functions — "arbitrary" because integration introduces free functions of the other variable, and "twice-differentiable" so that $u_{tt}$ and $u_{xx}$ actually exist. These are exactly the right- and left-moving shapes. See Method of characteristics.

Differentiate: $u_{tt}=-{\omega }^2\sin (kx)\cos (\omega t),u_{xx}=-k^2\sin (kx)\cos (\omega t).$ Substituting into $u_{tt}=c^2{u}_{xx}$: $-{\omega }^2\sin (kx)\cos (\omega t)=c^2(-k^2)\sin (kx)\cos (\omega t)\Rightarrow {\omega }^2=c^2{k}^2\Rightarrow \boxed{\omega =ck}.$ This is the dispersion relation for the string. With $k=3$, $c=2$: $\omega =ck=6$. So $u=\sin (3x)\cos (6t)$ solves it. (Standing waves are the building blocks of Separation of variables for the wave equation.)

Differentiate under the integral (the interval is fixed): $\frac{dE}{dt}={\int }_0^L[\rho u_t{u}_{tt}+T{u}_x{u}_{xt}]dx.$ Replace $\rho u_{tt}=T{u}_{xx}$ in the first term: $={\int }_0^L[T{u}_t{u}_{xx}+T{u}_x{u}_{xt}]dx=T{\int }_0^L\frac{\partial }{\partial x}\left(u_t{u}_x\right)dx,$ because ${\partial }_x(u_t{u}_x)=u_{tx}{u}_x+u_t{u}_{xx}$ (product rule, and $u_{tx}=u_{xt}$). Integrate: $\frac{dE}{dt}=T[u_t{u}_x{]}_0^L=T(u_t(L,t)u_x(L,t)-u_t(0,t)u_x(0,t)).$ Fixed ends mean $u(0,t)=u(L,t)=0$ for all $t$, so $u_t=0$ at both ends. Both boundary terms vanish: $\frac{dE}{dt}=0.$ The kinetic + potential energy of the vibrating string is constant — the equation neither creates nor destroys energy. That $\rho u_{tt}=T{u}_{xx}$ is exactly Newton's second law applied to the element is what makes this work.

Let $w=u_1-u_2$. Because the wave equation is linear, $w$ also satisfies $w_{tt}=c^2{w}_{xx}$, with fixed ends $w(0,t)=w(L,t)=0$, and zero initial data: $w(x,0)=f-f=0$ and $w_t(x,0)=g-g=0$. Apply the energy result L5.1 to $w$: $E_w(t)$ is constant. At $t=0$, $w_t=0$ everywhere so the kinetic part is $0$; and $w(x,0)=0$ everywhere means $w_x(x,0)=0$ so the potential part is $0$. Hence $E_w(0)=0$, so $E_w(t)=0$ for all $t$. But $E_w={\int }_0^L[\frac{1}{2}\rho w_t^2+\frac{1}{2}T{w}_x^2]dx$ is a sum of squares — it can only be zero if $w_t\equiv 0$ and $w_x\equiv 0$ everywhere. Then $w$ is constant in both $x$ and $t$; since $w=0$ at the boundary, $w\equiv 0$. Therefore $u_1=u_2$: the solution is unique.

(i) Find $c$. A right-mover keeps its shape and slides right without changing form. Its peak occurs wherever the argument $x-ct$ equals the fixed number $s^{\star }$ that maximises $F$. So the peak position obeys $x_{\text{peak}}-ct=s^{\star }\Rightarrow x_{\text{peak}}=s^{\star }+ct,$ i.e. the peak moves along a straight line at speed $c$. Using the two snapshots: $c=\frac{\Delta x_{\text{peak}}}{\Delta t}=\frac{3-1}{0.4-0}=\frac{2}{0.4}=5\text{m/s}.$

(ii) Any smooth $F$ works. Let $s=x-ct$ be the inside. By the chain rule (derivative of $F$ times derivative of the inside): $u_t=\frac{\partial s}{\partial t}{F}^{\prime }(s)=-c{F}^{\prime }(s),u_{tt}=(-c{)}^2{F}^{\prime \prime }(s)=c^2{F}^{\prime \prime }(s),$ $u_x=\frac{\partial s}{\partial x}{F}^{\prime }(s)=F^{\prime }(s),u_{xx}=F^{\prime \prime }(s).$ Therefore $u_{tt}=c^2{F}^{\prime \prime }(s)=c^2{u}_{xx}.✓$ This holds for every twice-differentiable $F$ — not just sines. That is exactly why the equation supports arbitrary travelling shapes (a plucked corner, a Gaussian bump, anything), which is the content of the d'Alembert solution.

Recognise the equation → plug into $c=\sqrt{T/\rho }$ → read curvature to predict motion → build d'Alembert / standing-wave solutions → prove conservation and uniqueness from energy.