Past exam of the year 1 of the computer science course of the University of Oxford / 2003 / Calculus, probability, linear algebra, discrete mathematics, logic and proof, and models of computation / 2

Assume all partial derivatives needed below are continuous. The multivariable chain rule iswhere the partial derivatives of $g$ are evaluated at $(u(x,y),v(x,y))$. Apply the product rule and the multivariable chain rule once more:Clairaut's theorem gives $g_{vu}=g_{uv}$, soDifferentiating $f_x$ with respect to $y$ instead gives

For this change of variables,and $u_{xx}+u_{yy}=0=v_{xx}+v_{yy}$. The second derivative chain rule therefore givesThus the transformation rule for the Laplacian isFor $f=x+y$, its Laplacian is zero. At every point with $(x,y)\ne (0,0)$, division by $4(x^2+y^2)$ provesThis conclusion applies to each local inverse branch of the change of variables: its Jacobian determinant is $4(x^2+y^2)$, so such a branch exists away from the origin by the inverse function theorem. A single globally defined $g$ cannot represent $x+y$ on the entire plane, since $(x,y)$ and $(-x,-y)$ give the same $(u,v)$ but generally opposite values of $x+y$. Nor can a differentiable $g$ represent this $f$ in a neighborhood of the origin: the multivariable chain rule there would give $f_x=f_y=0$, contradicting $f_x=f_y=1$. The requested harmonic function conclusion is therefore a local statement on nonsingular branches.^[ref]

Write $z(x,y)=Z(r,s)$ in the new coordinates. Sincethe partial differential equation becomesFor a twice continuously differentiable solution of a differential equation, $Z_r$ is independent of $s$, so integration first in $r$ and then in $s$ givesThis is the decomposition of the one-dimensional wave equation into waves traveling in opposite directions.

The first initial condition gives $\varphi (y)+\psi (y)=\cos y$. The second initial condition gives ${\varphi }^{\prime }(y)-{\psi }^{\prime }(y)=0$. Differentiating the first relation and combining the two relations yields ${\varphi }^{\prime }=ps{i}^{\prime }=-\frac{1}{2}\sin y$. Thus $\varphi (y)=\frac{1}{2}\cos y+C$ and $\psi (y)=\frac{1}{2}\cos y-C$. The constants cancel, and the cosine addition formula givesBoth initial conditions and the partial differential equation follow directly by differentiation.