We assume the initial value problem : $y=f'(x,y),y(0)=0,0\leq x\leq a, |y|\le b$ where $a,b >0$. We assume that the function $f(x,y)$ is continuous and : $$|f(x,y)-f(x,z)|\leq h(x) |y-z|$$ where $h:(0,a]\to \mathbb R$ integratable function.
Prove that this problem has a unique solution.
Thoughts :
First of all, let's assume a number $0<a_1 <a $. Then, let's assume the set :
$$D=\{(x,y) \in \mathbb R^2: 0\leq x\leq a,|y|\leq b\}$$
Despite not having a relation like $|x| \leq a$, can we say that $f(x,y)$ is continuous over $D$ which means that it has a solution in it ?
If we also do a Mean Value Theorem for $f(x,y)$ with respect to $y$ handling it as a single variable function over a subset of its domain, $(z,y)$ :
$$f_y(x,y)|_{y=ξ} = \frac{f(x,y) - f(x,z)}{y-z}$$
$$\Leftrightarrow $$
$$f(x,y) - f(x,z) = f_y(x,y)|_{y=ξ} \cdot (y-z)$$
Taking absolute values :
$$|f(x,y) - f(x,z)| = |f_y(x,y)|_{y=ξ}| \cdot |y-z|$$
Now, let's commute :
$$f_y(x,y)|_{y=ξ} \equiv g(x)$$
So, we get :
$$|f(x,y) - f(x,z)| = |g(x)| \cdot |y-z|$$
which has the same form as our initial given condition, but without the $\le$ sign (which means we haven't bound it). I can't seem though how to proceed from here or if I need to do something else. I also don't get where I'll be using that the function $h(x)$ is integretable.
Please, provide me some help or a thorough solution, I'll be greatly appreciating it since I am stuck on this problem for a few days.
You can not assume that $f$ has partial derivatives. There may even not exist local Lipschitz constants, as there are unbounded integrable functions.
What you need to do is to take 2 solutions to the same initial value problem and consider the evolution of their distance. Using $$ y(x)=0+\int_0^xf(s,y(s))\,ds,\;and\;z(x)=0+\int_0^xf(s,z(s))\,ds, $$ you get $$ \|y(x)-z(x)\|\le\int_0^xh(s)\|y(s)-z(s)\|\,ds. $$ This you can now try to solve using the Grönwall lemma or similar considerations.
For instance, define $H(x)=\int_0^xh(s)\,ds$ and consider $$\phi(x)=\max_{s\in[0,x]}e^{-2H(s)}|y(s)-z(s)|.$$ Then $$ \|y(x)-z(x)\|\le\int_0^xh(s)e^{2H(s)}\phi(x)\,ds=\frac12(e^{2H(x)}-1)\phi(x) $$ which implies $$ ϕ(x)\le\frac12\phi(x) $$ which is only possible for $\phi(x)=0$.