Existence of Solutions of First Order Differential Equations

Subjects: Ordinary Differential Equations
Links: Picard–Lindelöf theorem, First Order Differential Equations, First Order Linear Differential Equations

Existence Theorem

Let f be a continuous real valued function on the rectangle R=[x0a,x0+a]×[y0b,y0+b], and let fM. Let f be Lipschitz continuous with a Lipschitz constant K in R. Then the succesive approximations

ϕ0=y0,ϕk+1(x)=y0+x0xf(t,ϕk(t))dt

converge on the interval I=[x0α,x+α], with α=min{a,b/M} to a solution ϕ of the initial value problem

y=f(x,y)y(x0)=y0

on I.

Th: The kth succesive approximation of the Picard iteratives (ϕk) to the solution x of the initial value problem satisfies

ϕϕkMK(Kα)k+1(k+1)!eKα

Non-local Existence of Solutions

Let S=[x0a,x0+a]×R, and f:SR be Lipschtiz continuous with constant K>0. The succesive approximations (ϕk) for the problem

y=f(x,y)y(x0)=y0

exists on the entire interval [x0a,x0+a], and converge there to solution ϕ of the initial value problem.

Cor: Suppose f:R2R be a continuous function on the plane which satisfies the Lipschitz condtion on the strip Sa=[a,a]×R, where a>0. Then every initial value problem

y=f(x,y)y(x0)=y0

has a solution which exists for all x

Approximations and uniqueness

Let f,g be continuous functions on R=[x0a,x0+a]×[y0b,y0+b], and suppose f satisfies the Lipschitz condition with a Lipschitz constant K. Let ϕ and ψ be solutions of

y=f(x,y)y(x0)=y1y=g(x,y)y(x0)=y2

respectively on an interval I containing x0, with graphs contained in R. Additionally fgε on R, and |y1y2|δ. Then

|ϕ(x)ψ(x)|δeK|xx0|+εK(eK|xx0|1)

for all x in I.

Cor (Uniqueness Theorem): Let f be continuous and satisfy a Lipschitz conditions on R. If ϕ and ψ are solutions of

y=f(x,y)y(x0)=y0

on an interval I containing x0, then ϕ=ψ on I.

Cor: : Let f be continuous and satisfy a Lipschitz conditions on R. Let (gk) be a sequence continuous functions on R, such that

fgkεk

and ψk be the solution to the initial value problem on the interval I containing x0

y=gk(x,y)y(x0)=yk

and ϕ be the solution to the initial value problem on the interval I containing x0

y=f(x,y)y(x0)=y0

and |yky0|δk. If εk,δk0 as k. Then ψϕ on I.