Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2025 / Paper 3 / 8

Since tension is a force and ${\lambda }_0$ is a linear mass density,Thus $c$ has units $\mathrm{m}{\mathrm{s}}^{-1}$. It is the transverse wave speed of the ideal uniform string. The wave equation on a string assumes small transverse slopes, constant positive tension, positive uniform linear mass density, and negligible damping and bending stiffness.

The wave equation is second order in time. Even one normal mode obeys a second-order ordinary differential equation, so its initial displacement and initial velocity are independent data. Physically they specify stored elastic deformation and the initial motion.

The heat equation is first order in time; its initial temperature already fixes the instantaneous time derivative through the spatial Laplacian. It needs one initial field. In either case spatial boundary conditions remain necessary and are not substitutes for initial data. Classical fixed-end string data must satisfy the appropriate endpoint compatibility, including $f(0)=f(L)=0$ and, when endpoint velocities exist, $g(0)=g(L)=0$.

Use separation of variables with a nontrivial product $\eta =XT$. Substitution into the wave equation on a string gives ${\tau }_0{X}^{\prime \prime }T={\lambda }_{0}X{T}^{\prime \prime }$. Wherever the factors are nonzero,The left side depends only on $x$ and the right only on $t$, so both are a constant. For nontrivial fixed-end solutions this constant is negative: multiplying $X^{\prime \prime }=\sigma X$ by $X$ and integrating, using the homogeneous Dirichlet boundary conditions, givesWrite it as $-k^2$. The separated equations are thereforeThese equations extend across the zeros of $X$ and $T$; the division is a derivation device, not a restriction excluding their nodes.

For $k>0$, the harmonic oscillator equation gives $X=C\sin (kx)+D\cos (kx)$. The condition at zero requires $D=0$, and the condition at $L$ requires $C\sin (kL)=0$. A nonzero normal mode therefore hasA multiplicative normalization has been absorbed into the temporal amplitude. The $k=0$ candidate is affine and the two zero endpoint values force it to vanish. A positive separation constant gives hyperbolic functions, whose two zero endpoint values likewise give only the trivial solution. Negative $n$ adds no new eigenfunctions because it only reverses their signs.

For each spatial normal mode, the temporal equation is $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$, with angular frequency ${\omega }_n=c{k}_n=cn\pi /L$. HenceThe ordinary frequency is ${\omega }_n/(2\pi )=cn/(2L)$; it should not be confused with the angular frequency used in the sine and cosine arguments.

The wave equation and its homogeneous Dirichlet boundary conditions are linear, so any finite linear combination of the separated normal modes is another solution. More strongly, the Fourier sine basis is complete in L2 space on $(0,L)$. Expanding the initial fields in this basis yieldsCompleteness of the Fourier sine basis is the reason the expansion covers general data; linearity alone would not prove that every solution has this form. The convergence and differentiation are understood in a function space appropriate to the data. For finite wave energy, one can take $f\in H_0^1(0,L)$ and $g\in L^2(0,L)$, obtaining an energy solution; stronger smoothness and endpoint compatibility give a classical solution.

At $t=0$, the displacement expansion is $f(x)={\sum }_n{A}_n\sin (n\pi x/L)$. Fourier orthogonality givesDifferentiating the modal solution at zero gives $g(x)={\sum }_n{\omega }_n{B}_n\sin (n\pi x/L)$, soThe factor $1/{\omega }_n$ is required because the derivative of $\sin ({\omega }_{n}t)$ at zero is ${\omega }_n$. The dimensions agree: $A_n$ and $B_n$ are displacements, while the sine coefficients of $g$ are velocities. These projections also establish uniqueness within the finite-energy solution class.