The Wave Equation — Question 1

Question 1 – Solution

Strategy. Transverse acceleration is driven by the difference of vertical tension components, not by a first time derivative.

Step 1: Derive the momentum equation. The vertical tension component is $\mathcal T\sin\theta\approx\mathcal T u_x$ under the small-slope approximation. For an interval $[a,d]$, $$\int_a^d\mu u_{tt}\,dx =\mathcal T u_x(d,t)-\mathcal T u_x(a,t)+\int_a^d(p-bu_t)\,dx.$$ Localization gives $$\boxed{\mu u_{tt}+b u_t=\mathcal T u_{xx}+p.}$$ The linearization neglects changes of length and tension of higher order in the slope.

Step 2: Check dimensions and speed. The units are $\mu:\mathrm{kg/m}$, $\mathcal T:\mathrm N$, $p:\mathrm{N/m}$ and $b:\mathrm{Ns/m^2}$. Since $u$ is a length, each term in the local equation has units $\mathrm{N/m}$. Without drag or forcing, $$\boxed{u_{tt}=c^2u_{xx},\qquad c=\sqrt{\mathcal T/\mu}.}$$ The speed is in $\mathrm{m/s}$. Quadrupling $\mu$ at fixed tension halves $c$.

Step 3: Explain the two initial data. Acceleration determines how velocity changes, so displacement alone cannot specify the motion. For example, on the whole line, both $u=0$ and $u=Vt$, with constant $V\ne 0$, solve the source-free undamped equation and have $u(x,0)=0$. Their initial velocities differ. A prescribed pair $u(x,0)=f(x)$, $u_t(x,0)=g(x)$ removes this particular ambiguity.

Step 4: Verify a moving profile. For $u=F(x-vt)$, $u_{tt}=v^2F''$ and $u_{xx}=F''$. Hence $$\boxed{(v^2-c^2)F''=0.}$$ If $F''$ is not identically zero, the admissible speeds are $v=\pm c$, corresponding to opposite directions. If $F$ is affine, both second derivatives vanish for every $v$; a translating straight line cannot by itself identify a propagation speed. A non-affine smooth profile is needed to draw that inference from the PDE.