Exercises — Solving wave equation — D'Alembert's solution

The wave equation is $u_{tt}=c^2{u}_{xx}$. Compare $u_{tt}=9u_{xx}$: so $c^2=9\Rightarrow ==c=3==$ (we take the positive root because $c$ is a speed).

• $f(x)=x^2$ (the shape at $t=0$),
• $g(x)=\sin x$ (the velocity at $t=0$).

What we did: matched symbols to slots. Why: every later step needs these three named correctly.

"At rest" means $g(x)=0$ everywhere. The velocity term is $\frac{1}{2c}{\int }_{x-ct}^{x+ct}gds$; with $g=0$ the integrand is $0$, so the integral is $0$ and the entire second term vanishes. What remains: $u(x,t)=\frac{1}{2}[f(x-ct)+f(x+ct)].$ Picture: the starting bump is cut into two half-height copies gliding apart at speed $c$.

A profile $\Phi (x-ct)$ keeps a fixed value where its argument is constant. For $x-ct=\text{const}$, as $t$ increases $x$ must increase → it moves in $+x$. The rule: minus sign ⇒ moves $+x$.

• (a) $x-4t$: minus ⇒ moves in $+x$ at speed $4$.
• (b) $x+2t$: plus ⇒ moves in $-x$ at speed $2$.
• (c) $3t-x=-(x-3t)$: rewrite as $-(x-3t)$; the argument $x-3t$ has a minus ⇒ moves $+x$ at speed $3$. (The outer minus just flips the profile upside-down; it does not change direction.)

Here $c=1$ and $g=0$, so only the shape term survives: $u(x,t)=\frac{1}{2}\left[\frac{1}{1+(x-t{)}^2}+\frac{1}{1+(x+t{)}^2}\right].$ Check at $t=0$: both fractions become $\frac{1}{1+x^2}$, sum is $\frac{2}{1+x^2}$, halved gives $\frac{1}{1+x^2}=f$. ✓

$c^2=4\Rightarrow c=2$. Only the integral survives:

$u=\frac{1}{2}[\cos (x-t)+\cos (x+t)]+\frac{1}{2}{\int }_{x-t}^{x+t}\sin sds.$ First term (sum-to-product $\cos A+\cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}$): $\cos x\cos t$. Integral: ${\int }_{x-t}^{x+t}\sin sds=[-\cos s{]}_{x-t}^{x+t}=-\cos (x+t)+\cos (x-t)=2\sin x\sin t$. Times $\frac{1}{2}$: $\sin x\sin t$. $u(x,t)=\cos x\cos t+\sin x\sin t=\cos (x-t).$ Check: $u(x,0)=\cos x=f$ ✓; $u_t=-\cos x\sin t\cdot (\dots )$ → at $t=0$, $u_t=\sin x=g$ ✓. Neat: the answer is a single right-moving wave $\cos (x-t)$.

$u(x_0,t_0)$ depends on $f$ at $x_0\pm c{t}_0$ and on $g$ over $[x_0-c{t}_0,x_0+c{t}_0]$. With $x_0=5,t_0=3,c=2$: $x_0-c{t}_0=5-6=-1,x_0+c{t}_0=5+6=11.$ So the domain of dependence is $==[-1,11]==$. Now test each point:

• $x=-2$: outside $[-1,11]$ → no influence.
• $x=0$: inside → yes (via $g$).
• $x=10$: inside → yes (via $g$).
• $x=11$: an endpoint → yes (it feeds both $f(x_0+c{t}_0)$ and the integral edge).

See Domain of Dependence and Influence for the geometry of these light-cone edges.

With $g=0$: $u(x,t)=\frac{1}{2}[f(x-t)+f(x+t)]$. Both $f$-values are $\ge 0$ (given $f\ge 0$), and $\frac{1}{2}$ is positive, so a sum of non-negatives times a positive number is $\ge 0$. Therefore $u(x,t)\ge 0$ always — never negative. What this teaches: a pure-displacement, non-negative pluck can only produce non-negative displacement; you need a velocity term (or a signed $f$) to dip below zero.

At $x=0$: $u(0,t)=\frac{1}{2}{e}^{-t^2}+\frac{1}{2}{e}^{-t^2}=e^{-t^2}$.

• At $t=0$: $u(0,0)=1$ — the full original height (the halves overlap).
• As $t\to \infty$: $e^{-t^2}\to 0$. The two bumps have marched apart, leaving the origin flat.

Picture: height at the centre decays like a Gaussian in $t$ — the energy has travelled away, not vanished. This is finite propagation speed in action.

Initial shape: $f(x)=u(x,0)=\sin x\cos 0=\sin x$. Initial velocity: $u_t=\sin x\cdot (-c\sin ct)$, so $g(x)=u_t(x,0)=\sin x\cdot (-c\sin 0)=0$. So $f=\sin x$, $g=0$. Verify the PDE: $u_{tt}=\sin x\cdot (-c^2\cos ct)=-c^2\sin x\cos ct$; $u_{xx}=(-\sin x)\cos ct$, so $c^2{u}_{xx}=-c^2\sin x\cos ct$. They match ⇒ $u_{tt}=c^2{u}_{xx}$. ✓ Cross-check with D'Alembert ($g=0$): $\frac{1}{2}[\sin (x-ct)+\sin (x+ct)]=\sin x\cos ct$ by sum-to-product. Same answer. ✓

What we do: use the general solution and the two data equations from the parent note: $F(x)+G(x)=f(x)=x,-F^{\prime }(x)+G^{\prime }(x)=\frac{g(x)}{c}=1.$ Integrate the second (from $0$ to $x$): $-F(x)+G(x)=x+K$. Add to the first: $2G(x)=2x+K\Rightarrow G(x)=x+\frac{K}{2}$. Subtract: $2F(x)=-K\Rightarrow F(x)=-\frac{K}{2}$... wait, recompute the subtraction: $[F+G]-[-F+G]=2F=x-(x+K)=-K$, so $F(x)=-\frac{K}{2}$ as a constant? That signals $F^{\prime }=0$; let's keep it honest by rebuilding $u$: $u=F(x-t)+G(x+t)=-\frac{K}{2}+[(x+t)+\frac{K}{2}]=x+t.$ The constants cancel, giving $u(x,t)=x+t$. Check: $u(x,0)=x=f$ ✓; $u_t=1=g$ ✓; $u_{tt}=0=u_{xx}$ ✓. (Same as plugging into D'Alembert: $\frac{1}{2}[(x-t)+(x+t)]+\frac{1}{2}{\int }_{x-t}^{x+t}1ds=x+\frac{1}{2}(2t)=x+t$.)

With $g=0$: $u(x,t)=\frac{1}{2}[f(x-t)+f(x+t)]$. Replace $x$ by $-x$: $u(-x,t)=\frac{1}{2}[f(-x-t)+f(-x+t)]=\frac{1}{2}[f(-(x+t))+f(-(x-t))].$ Since $f$ is even, $f(-(x+t))=f(x+t)$ and $f(-(x-t))=f(x-t)$. Hence $u(-x,t)=\frac{1}{2}[f(x+t)+f(x-t)]=u(x,t).$ So $u$ is even in $x$. Meaning: a symmetric pluck stays symmetric forever — the two travellers are mirror images and their sum is symmetric.

D'Alembert only evaluates $f$; it never differentiates it. So $u(x,t)=\frac{1}{2}[f(x-t)+f(x+t)]$ is perfectly well-defined even where $f$ has corners — this is a weak (generalised) solution. At $t=\frac{1}{2}$: two half-height tents, each of half-height $\frac{1}{2}$ at their peaks $\frac{1}{2}(1)=\frac{1}{2}$, centred at $x=+\frac{1}{2}$ (the $f(x-t)$ copy) and $x=-\frac{1}{2}$ (the $f(x+t)$ copy). Where they still overlap (near $x=0$) the values add. The corners just ride along; they do not blow up. Lesson: the wave equation propagates singularities along characteristics rather than smoothing them (unlike the Heat Equation, which instantly smooths everything).

As $c\to 0^{+}$: both $x-ct\to x$ and $x+ct\to x$, so the two characteristic lines collapse onto the single vertical line $x=\text{const}$. The shape term $\frac{1}{2}[f(x-ct)+f(x+ct)]\to f(x)$ — nothing travels. The velocity term is delicate: $\frac{1}{2c}{\int }_{x-ct}^{x+ct}gds\approx \frac{1}{2c}\cdot (2ct)g(x)=tg(x)$ for small interval (mean-value / narrow interval). So $u(x,t)\to f(x)+tg(x).$ Physical read: with zero wave speed there is no propagation; each point of the string just moves in a straight line — starting at $f(x)$ and drifting at constant velocity $g(x)$, exactly like free particles. The PDE degenerates to $u_{tt}=0$.

Let $w=u_1-u_2$. By linearity $w$ solves $w_{tt}=c^2{w}_{xx}$ with zero data: $w(x,0)=f-f=0$ and $w_t(x,0)=g-g=0$. Feed $f=0,g=0$ into D'Alembert: $w(x,t)=\frac{1}{2}[0+0]+\frac{1}{2c}{\int }_{x-ct}^{x+ct}0ds=0.$ So $w\equiv 0$, i.e. $u_1=u_2$ — the solution is unique. The clean fact enabling this is that D'Alembert gives an explicit formula depending only on the data; zero data forces zero solution. The wave equation is well-posed both forward and backward in time (it is time-reversible: replacing $t\to -t$ leaves $u_{tt}=c^2{u}_{xx}$ unchanged). The Heat Equation is not — its backward problem is ill-posed because heat irreversibly smooths, and tiny data changes explode backward.