Past exam of the differential equations 1 course of the University of Oxford / 2017 / 2 / a

The contraction mapping theorem states that a contraction mapping $T:S\to S$ on a nonempty complete metric space has a unique fixed point. Explicitly, if $d(Tu,Tv)\le qd(u,v)$ for all $u,v\in S$ and one fixed $0\le q<1$, then the iterates $u_{n+1}=T{u}_n$ converge to that fixed point from every $u_0\in S$. Both self-mapping and completeness are needed; the set need not be a vector space.

For $0<\eta \le h$, work on the closed subsetIt is a nonempty complete metric space because it is closed in the Banach space of continuous functions with the supremum norm. Define the Picard iteration operatorChoose $\eta$ so that $M\eta \le k$ and $L\eta <1$, omitting a bound whose constant is zero. Such a positive $\eta$ exists. The first inequality makes $T$ a self-map, since $|(Ty)(x)-b|\le M|x-a|\le k$. The Lipschitz condition givesso the contraction mapping theorem yields a unique fixed point. The fundamental theorem of calculus makes it a differentiable solution of $y^{\prime }=f(x,y)$ with $y(a)=b$; conversely every solution staying in the rectangle satisfies the integral equation and is a fixed point. This is the local Picard-Lindelöf theorem construction. The interval-restricted space $C_{\eta }$ avoids treating functions defined on a shorter interval as elements of $C_h$.

Under $Mh\le k$, the same Picard iteration operator maps all of $C_h$ into itself. It need not be a contraction mapping in one step, since $Lh$ can exceed one. Instead repeated integration gives the factorial estimate for Picard iterationproved inductively by integrating the preceding bound between $a$ and $x$. ThusFor some $n$ the coefficient is less than one, so $T^n$ is a contraction. Its unique fixed point $y_{\ast }$ is also fixed by $T$: $T^n(T{y}_{\ast })=T(T^n{y}_{\ast })=T{y}_{\ast }$ and uniqueness forces $T{y}_{\ast }=y_{\ast }$. Conversely any fixed point of $T$ is one of $T^n$. This iterated contraction argument gives a unique solution on the full interval $[a-h,a+h]$, without the unjustified extra assumption $Lh<1$.

The uniqueness statement is for solutions in the specified rectangle, where $f$ is given and controlled. The bound $Mh\le k$ keeps the constructed solution inside that rectangle throughout the interval.