Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 4

For $r>0$, the chain rule gives $r_x=x/r$ and $r_y=y/r$. Applying it twice to the radial function $u=f(r)$ givesUsing $x^2+y^2=r^2$, Laplace's equation becomesTwo integrations now giveThese are the general solutions of the specified radial function form on an annulus or a punctured plane. If the solution must extend continuously to the origin, the logarithm term forces $A=0$, leaving only a constant function.

Use the product rule for the Laplace operator:Thus Laplace's equation, away from the origin, reduces to $g^{\prime \prime }+3g^{\prime }/r=0$. On the line $x=0$ the displayed expression vanishes automatically; requiring the equation on an open region gives the same radial equation by continuity from $x\ne 0$. Multiplying by $r^3$ gives $(r^3{g}^{\prime }{)}^{\prime }=0$, so $g^{\prime }=C{r}^{-3}$ and $g=A+B{r}^{-2}$. ThereforeIf extension through the origin is required, $B=0$: along the horizontal axis the second term is $B/x$. The remaining harmonic function is $Ax$.

For the wave equation with wave speed one, D'Alembert's formula with arbitrary twice continuously differentiable initial displacement $h(x)=u(x,0)$ isThe arbitrary function $h$ is essential: only the initial velocity is prescribed. For $g=0$, this givesBoth travelling terms solve the wave equation, and their time derivatives cancel at $t=0$. Conversely every classical solution has this form for its initial displacement $h$.

The velocity contribution in D'Alembert's formula is $\frac{1}{2}{\int }_{x-t}^{x+t}1ds=t$. Hence, with arbitrary twice continuously differentiable $h$,The added term has both second partial derivatives zero and has time derivative one, so it supplies precisely the required initial condition while preserving the wave equation.

Here the velocity contribution in D'Alembert's formula isThus, with arbitrary twice continuously differentiable $h$,For the added term, $u_{tt}=e^x\sinh t=u_{xx}$ and $u_t(x,0)=e^x\cosh 0=e^x$. Its displacement at $t=0$ is zero, leaving $h$ free as required.