Fixed Point Analysis and Partial Differential Equations

Suppose $f : ℝ^2→ℝ$ is continous and globally Lipschitz in its second coordinate, i.e. there exists some $m∈ℝ$ such that for all $x,y_1,y_2∈ℝ$ $$|f(x,y_1)-f(x,y_2)|≤m·|y_1-y_2|$$ Further suppose $h > 0$ such that for the aforementioned $m$ we observe $mh < 1$. Then when $c[0,h]$ is the space of continuous functions $[0,h]→ℝ$ under the maximum absolute difference metric $d$, and $y_0∈ℝ$ is fixed, show that the operator $$T : c[0,h]→c[0,h]$$ $$T(φ)(x)=y_0+∫_0^x f(t,φ(t)) \mathrm{d}t$$ is well defined and has a unique fixed point.

Proof: Note that by the fundamental theorem of calculus, $T(φ)$ is continuous on $[0,h]$ so that $T$ is indeed well defined. Since $c[0,h]$ is a complete metric space, by Banach's theorem it suffices to show that $T$ is a contracting map to ensure existence and uniqueness of a fixed point. Observe the inequalities given $φ_1,φ_2∈c[0,h]$: $$d(T(φ_1),T(φ_2)) = \max_{x∈[0,h]} |(y_0+∫_0^x \! f(t,φ_1(t))~\mathrm{d}t) - (y_0+∫_0^x \! f(t,φ_2(t))~\mathrm{d}t)| =$$ $$\max_{x∈[0,h]} |∫_0^x \! f(t,φ_1(t)) - f(t,φ_2(t))~\mathrm{d}t| ≤ \max_{x∈[0,h]} ∫_0^x \! |f(t,φ_1(t)) - f(t,φ_2(t))|~\mathrm{d}t≤$$ $$\max_{x∈[0,h]} m·∫_0^x \! |φ_1(t)-φ_2(t)|~\mathrm{d}t ≤ mh·\max_{x∈[0,h]} |φ_1(x)-φ_2(x)| = mh·d(φ_1,φ_2)$$ so that $T$ is a contracting map of factor no larger than $mh < 1$.