Differential equation has exactly one solutionLet n,d∈N and τ&gt;0.f:[0,nτ]×Rd×Rd→Rd,(t,x,y)→f(t,x,y) continuous with the condition: There exists a L&gt;0 so that |f(t,x1,y)−f(t,x2,y)|≤L|x1−x2|∀t∈[0,nτ],x1,x2,y∈Rd.Furthermore, let ϕ∈C1([−τ,0],Rd).{y′(t)=f(t,y(t),y(t−τ)), t∈[0,nτ],y(t)=ϕ(t), t∈[−τ,0]with the condition ϕ′(0)=f(0,ϕ(0),ϕ(−τ)). Show, that this problem has exactly one solution y∈C1([−τ,nτ],Rd).

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Differential equation has exactly one solutionLet n,d∈N and τ&amp;gt;0.f:[0,nτ]×Rd×Rd→Rd,(t,x,y)→f(t,x,y) continuous with the condition: There exists a L&amp;gt;0 so that |f(t,x1,y)−f(t,x2,y)|≤L|x1−x2|∀t∈[0,nτ],x1,x2,y∈Rd.Furthermore, let ϕ∈C1([−τ,0],Rd).{y′(t)=f(t,y(t),y(t−τ)), t∈[0,nτ],y(t)=ϕ(t), t∈[−τ,0]with the condition ϕ′(0)=f(0,ϕ(0),ϕ(−τ)). Show, that this problem has exactly one solution y∈C1([−τ,nτ],Rd).

etsahalen5tt · Accepted Answer

Answer: Iterating on the intervals of length τ, on each sub-interval [kτ,(k+1)τ], the second argument y(t)=x(t−τ)  is a completely known function by the iteration assumption, so the DDE x′=f(t,x(t),x(t−τ)) can be considered as ODE x′(t)=F(t,x(t)),F(t,x)=f(t,x,y(t)), and the existence-and-uniqueness theorem for ODE can be applied.

Differential equation has exactly one solution Let \(\displaystyle{n},{d}\in{\mathbb{{N}}}\)

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