[PDF] Existence and uniqueness of solutions for first-order nonlinear

Why is y0 a good introduction to existence and uniqueness theorems?

One reason is it can be generalized to establish existenceand uniqueness results for higher-order ordinary di?erential equations and for systems ofdi?erential equations. Another is that it is a good introduction to the broad class of existenceand uniqueness theorems that are based on ?xed points. y0 =f(x, y), y(x0)=y0.

What is the argument for uniqueness theorem?

then the solution curves (t; y1(t)) and (t; y2(t))cannot intersectat any point in the(t; y) plane where the uniqueness theorem applies. The argument is simple (see left gure below): If the two curves intersected at a point(t0; a), then they would both satisfy the same ODE with the same initial condition (y1(t0) =y2(t0) =a).

Does the theorem apply if the Ode function fails?

t >0; y(0) =y02t1=2the ODE function is not continuous so the theorem does not apply. However, a solutionexists anyway. This shows that (C) is a sucient but not necessary condition, i.e. if itfails then it doesnot necessarily follow that no solution exists.

What is the extension theorem?

The extension theorem then tells us that the solution must exist inall of (a; b). Revisiting existence/uniqueness examples. Example 1: y0 =xy2withy(0) =y0 >0. clearly continuous, as is @f= 2xy. However, any rectangleRsuch thatj@f@yj L(condition (Ly)) must be boundediny; it cannot contain ally2R.