The contraction mapping principle becomes an existence theorem when a differential or integral equation is rewritten as a fixed point problem. The focus keyword Picard iteration names the successive approximation method for initial value problems. This lesson relates differential equations to integral equations and explains why contraction estimates yield existence and uniqueness. :::definition[Solution of an Initial Value Problem] Let $f$ be continuous on $R=\{(x,y):|x-x_0|\leq a, |y-y_0|\leq b\}$. A function $w$ is a solution of $y'=f(x,y)$, $y(x_0)=y_0$ on $I$ if $(x,w(x))\in R$, $w'(x)=f(x,w(x))$, and $w(x_0)=y_0$. ::: :::theorem[Equivalence of Differential and Integral Forms] A function $w$ solves $y'=f(x,y)$, $y(x_0)=y_0$ if and only if it solves $y(x)=y_0+\int_{x_0}^x f(t,y(t))\,dt$. ::: :::proof Given the initial value problem. To prove equivalence. If $w$ solves the differential equation, integrate $w'(t)=f(t,w(t))$ from $x_0$ to $x$ to obtain the integral equation. Conversely, if $w$ satisfies the integral equation, the fundamental theorem of calculus gives $w'(x)=f(x,w(x))$, and substituting $x=x_0$ gives $w(x_0)=y_0$. $\square$ ::: :::theorem[Complete Subspace of Bounded-Range Continuous Functions] Let $X$ be the set of continuous functions from $[a,b]$ into $[l,m]$ with the uniform metric. Then $X$ is complete. ::: :::proof Given a Cauchy sequence $\{f_n\}$ in $X$. To prove completeness. Since $C[a,b]$ is complete, $f_n\to f$ uniformly for some continuous $f$. For each $t$, $f_n(t)\in[l,m]$ and $[l,m]$ is closed, so $f(t)\in[l,m]$. Hence $f\in X$, and $X$ is complete. $\square$ ::: :::theorem[Picard's Theorem] Let $f$ be continuous on $R=\{(x,y):|x-x_0|\leq a, |y-y_0|\leq b\}$, let $K=\sup_R|f|$, and suppose $|f(x,y_1)-f(x,y_2)|\leq M|y_1-y_2|$. With $\delta=\min\{a,b/K\}$, the initial value problem has a unique solution on $[x_0-\delta,x_0+\delta]$. ::: :::proof Given Picard's hypotheses. To prove existence and uniqueness. Let $I=[x_0-\delta,x_0+\delta]$ and let $X$ be the complete space of continuous functions from $I$ into $[y_0-b,y_0+b]$. Define $(Tw)(x)=y_0+\int_{x_0}^x f(t,w(t))\,dt$. The bound by $K$ shows that $T$ maps $X$ into itself. The Lipschitz condition gives $|(Tw_1)(x)-(Tw_2)(x)|\leq M\int_{x_0}^x|w_1(t)-w_2(t)|\,dt$. Iterating gives $d(T^nw_1,T^nw_2)\leq M^n\delta^n d(w_1,w_2)/n!$. For large $n$, this constant is less than $1$, so $T^n$ is a contraction. Hence $T$ has a unique fixed point, which is the unique solution. $\square$ ::: :::definition[Picard Iteration] Starting with $w_0$, define $w_n(x)=y_0+\int_{x_0}^x f(t,w_{n-1}(t))\,dt$. Under Picard's hypotheses, $w_n$ converges uniformly to the unique solution. ::: :::example[Linear Differential Equation] For $dy/dx=x+y$, $y(0)=0$, Picard iteration gives $w_n(x)=x^2/2!+x^3/3!+\cdots+x^{n+1}/(n+1)!$. The solution is $w(x)=e^x-x-1$. ::: :::solved-problem Let $K$ be continuous on $[a,b]\times[a,b]$ and $g\in C[a,b]$. Show that $f(x)=\lambda\int_a^bK(x,y)f(y)\,dy+g(x)$ has a unique solution for sufficiently small $\lambda$. ::: :::solution Define $(Tw)(x)=\lambda\int_a^bK(x,y)w(y)\,dy+g(x)$ and let $M=\sup|K(x,y)|$. Then $d(Tw_1,Tw_2)\leq|\lambda|M(b-a)d(w_1,w_2)$. If $|\lambda|<1/(M(b-a))$, then $T$ is a contraction, so the equation has a unique solution. :::