Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V resit / Question 5

Assume $\rho$ is mass per unit equilibrium length, the tension has constant magnitude $T$, the string has negligible bending stiffness, and there is no damping or external transverse force. Assume small slopes $|y_x|\ll 1$ and purely transverse motion. Then a short segment has mass $\rho \Delta x$ to first order. If $\alpha$ is the angle of its tangent, the small-slope approximation gives $\sin \alpha \simeq \tan \alpha =y_x$.

The vertical components of tension at the two ends give net transverse force $T[y_x(x+\Delta x,t)-y_x(x,t)]$. Newton's second law therefore givesto first order. Dividing by $\Delta x$ and taking its limit to zero yieldsThus $c$ is the propagation speed of small transverse waves.

Write $y(x,t)=Y(\xi ,\eta )$ with $\xi =x-ct$, $\eta =x+ct$. The chain rule gives the differential operatorsTheir squared difference isConsequently the wave equation becomes $Y_{\xi \eta }=0$. Integrating first in $\eta$ gives $Y_{\xi }=F^{\prime }(\xi )$; integrating in $\xi$ gives the general solutionThe arbitrary twice differentiable functions describe right-moving and left-moving waves, respectively. Conversely the chain rule shows that each such sum satisfies the wave equation.

Enforce the fixed-end Dirichlet boundary condition by taking the odd extension of the initial velocity to the whole line:This agrees with the specified initial condition for $s\ge 0$. To derive the relevant form of D'Alembert's formula, the zero initial displacement in b gives $G=-F$, up to a cancelling constant. The initial velocity then gives $-2c{F}^{\prime }=v_o$. HenceAn antiderivative, valid across $s=0$, isEvaluation therefore gives the solution throughout $x>0,t>0$:For $ct\le x$, replace $|x-ct|$ by $x-ct$; for $ct\ge x$, replace it by $ct-x$. Thus the same expression includes the reflected wave after it reaches the fixed end.

At $x=0$ the two terms agree, so $y(0,t)=0$. At $t=0$ they again agree, so $y(x,0)=0$. Differentiating the integral with respect to $t$ giveswhich recovers $y_t(x,0)=\kappa x{e}^{-x/L}$. Since $v_o$ is a continuously differentiable function, the formula is twice continuously differentiable even across $x=ct$, and satisfies the wave equation there as well.