DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

§3. Differential Equations 113is the identity matrix. It follows from the Inverse Function Theorem that (locally) we can solve for(t, v) asaC k function of (x, y). Note that the level curves of v have tangent vector X, asdesired.Now we repeat this procedure with the vector field Y. Let y(s, y 0 )bethe solution of thedifferential equation dy/ds = Y and write(x, y) =y(s, (u, 0))for some unique s and u. We similarly obtain (s, u) locally as a C k function of (x, y). We claimthat (u, v) give the desired coordinates. We only need to check that on a suitable neighborhoodof the origin they are independent; but from our earlier discussion we have v x =0,v y =1attheorigin, and, analogously, u x =1and u y =0,aswell. Thus, the derivative matrix of (u, v) istheidentity at the origin and the functions therefore give a local parametrization. □EXERCISES A.31. Suppose M(s) isadifferentiable 3 × 3 matrix function of s, K(s) isaskew-symmetric 3 × 3matrix function of s, andM ′ (s) =M(s)K(s), M(0) = O.Show that M(s) =O for all s by showing that the trace of (M T M) ′ (s) isidentically 0.2. (Gronwall inequality and consequences)a. Suppose f :[a, b) → R is differentiable, nonnegative, and f(a) =c>0. Suppose g :[a, b) →R is continuous and f ′ (t) ≤ g(t)f(t) for all t. Prove that(∫ t)f(t) ≤ c exp g(u)du for all t.ab. Conclude that if f(a) =0,then f(t) =0for all t.c. Suppose now v :[a, b) → R n is a differentiable vector function, and M(t) isacontinuousn × n matrix function for t ∈ [a, b), and v ′ (t) =M(t)v(t). Apply the result of part bto conclude that if v(a) =0, then v(t) =0 for all t. Deduce uniqueness of solutions tolinear first order differential equations for vector functions. (Hint: Let f(t) =‖v(t)‖ 2 andg(t) =2n max{|m ij (t)|}.)d. Use part c to deduce uniqueness of solutions to linear n th order differential equations.(Hint: Introduce new variables corresponding to higher derivatives.)