$ u(x) = g(x) + \int_0^x f(x,t, u(t)) dt $

Question

$f: [0,k] \times [0,k] \times R \rightarrow R\ \ \ $ is continous in $ \{ (x,t,z) \in [0,k] \times [0,k] \times R , 0 \leq t \leq x \leq k \} $

$ g: [0,k] \rightarrow R \ \ $ is continous

And $\ \ \ | f(x,t,z) - f(x,t,y)| \leq C |z-y| \ \ \ \ $ for a $ 0 \leq C $

Now i have to show that the equation: $$ u(x) = g(x) + \int_0^x f(x,t, u(t)) dt $$ has exactly one continous solution in $[0,k]$.

I have tried using the Banach fixed-point theorem but i dont know how. is it the right approach at all?

Answer 1

Let $X = C[0, k]$. Let us define $\Phi:X \rightarrow X$ by \begin{align} \Phi(w)(x) = g(x) = \int^x_0 f(x, t, w(t))\ dt. \end{align} Indeed, $\Phi$ maps $X$ to $X$, since for any $0\le x \le y\le k$ \begin{align} |\Phi(w)(x)-\Phi(w)(y)| \le&\ |g(x)-g(y)| + \left|\int^x_0 f(x, t, w(t))\ dt-\int^y_0 f(y, t, w(t))\ dt \right|\\ \le&\ |g(x)-g(y)| + \int^y_0|f(x, t, w(t))-f(y, t, w(t))|\ dt + \int^x_y|f(x, t, w(t))|\ dt \\ \le&\ |g(x)-g(y)| + \int^y_0|f(x, t, w(t))-f(y, t, w(t))|\ dt + \|f\|_\infty|x-y|. \end{align} Hence, by continuity of $f$, we see that $\Phi(w)(x)$ is also continuous which means $\Phi$ maps $X$ to $X$.

Next, let us show that $\Phi: X\rightarrow X$ is a contraction on $X$. Notice that

\begin{align} \|\Phi(u)-\Phi(w)\|_\infty =&\ \sup_{x \in [0, k]}|\Phi(u)(x)-\Phi(w)(x)|\\ \le&\ \sup_{x \in [0, k]} \int^x_0 |f(x, t, u(t)) - f(x, t,w(t))| \ dt\\ \le&\ C\sup_{x, \in [0, k]} \int^x_0 |u(t) -w(t)|\ dt\\ \le&\ C k \|u -w\|_\infty. \end{align}

However, we can't conclude that $\Phi$ is a contraction since $Ck$ doesn't necessarily need to be less than $1$. So, instead of proving $\Phi$ is a contraction on $X$, we can prove that $\Phi$ is a contraction on $[0, k_1]$ for some $k_1<k$ and $Ck_1<1$. Then, by Banach fixed point theorem, we know that there exists a unique $u$ defined on $[0, k_1]$ such that \begin{align} u(x) = g(x) + \int^x_0 f(x, t, u(t))\ dt \end{align} for $x \in [0, k_1]$. Repeat the argument on the interval $[k_1, 2k_1]$. Since $[0, k]$ is compact, then we only need to repeat the argument a finite number of times. This shows that there is a unique $u \in C[0, k]$ that solves the above integral equation.