Past exam of the geometry course of the University of Oxford / 2008 / A / 3 / ii / Solution

Since $E{u}^{\prime }+F{v}^{\prime }={\gamma }^{\prime }\cdot r_u$, differentiation givesThe corresponding calculation with ${\gamma }^{\prime }\cdot r_v=F{u}^{\prime }+G{v}^{\prime }$ givesThus the two requested equations hold exactly when ${\gamma }^{\prime \prime }$ is perpendicular to both tangent basis vectors, which is precisely the geodesic condition.

On expanding the left sides, the coefficient matrix of $(u^{\prime \prime },v^{\prime \prime })$ is the positive-definite metric matrix. It is invertible, so the equations form a smooth second-order ordinary differential equation in local coordinates. The local existence and uniqueness theorem gives a unique solution for each initial point and coordinate velocity. For an arbitrary initial tangent vector $w$ this is an affine parameter of constant speed $|w|$; it is arclength when $|w|=1$. For $w=0$ the solution is the constant curve. This specifies how the arbitrary-velocity statement relates to the preceding unit-speed convention.