Cauchy problem - unique solution

[evinda]

Hello! (Wave)

Theorem: Suppose that the function $f(x,y)$ is continuous as for $x$ in a space $\Omega \subset \mathbb{R}^2$ (not necessarily bounded) and $f$ satisfies the Lipschitz condition as for $y$ for any closed and bounded space $\Omega^1 \subset \Omega$. Then $\forall (x_0, y_0) \in \Omega$ there is an interval $[a,b] $ (such that $x_0 \in [a,b]$) where there is a unique solution of the Cauchy problem.Proof:$y'(\xi)=f(\xi, y(\xi))$

$\int_{x_0}^x y'(\xi) d{\xi}= \int_{x_0}^x f(\xi, y(\xi)) d{\xi} \Rightarrow y(x)=y_0+ \int_{x_0}^x f(\xi, y(\xi)) d{\xi} $

$\phi_0(x), \phi_0(x)=y_0, \phi_1(x)= y_0+ \int_{x_0}^x f(\xi, \phi_0(\xi))d{\xi}, \phi(x_0)=y_0$

$\phi_2(x)=y_0+ \int_{x_0}^x f(\xi, \phi_1(\xi)) d{\xi}$

$\dots$

$\phi_n(x)= y_0+ \int_{x_0}^x f(\xi, \phi_{n-1} (\xi)) d{\xi}$

$\phi_0(x), \phi_1(x), \dots, \phi_n(x) \to$ consecutive approximations

$\lim_{n \to +\infty} \phi_n(x)$ solutionHow do we deduce that $\lim_{n \to +\infty} \phi_n(x)$ is a solution?Also, why have we shown like that the Cauchy problem has a unique solution?

 

[gtoguy97]

By the definition of Lipschitz condition we have that:$|f(x,y_1)-f(x,y_2)|\leq K|y_1-y_2|$for some constant $K>0$ and for any $y_1,y_2 \in \Omega^1$This implies that the function $f(x,y)$ is uniformly continuous in $\Omega^1$. Therefore, by the mean value theorem, for each $x \in (a,b)$ there is some point $\xi \in [a,b]$ such that $f(x,\phi_{n-1}(x))=f(\xi,\phi_{n-1}(\xi))$This means that the integral $\int_{x_0}^x f(\xi, \phi_{n-1} (\xi)) d{\xi}$ can be evaluated using the same value of $f$ at all points $\xi \in [a,b]$, which implies that $\phi_n(x)$ converges to a unique solution. This proves that there is a unique solution of the Cauchy problem.