# Unique solution of IVP on I

I have been given the differential equation $dy/dx = f(x,y)$ with initial condition $y(x_0)=y_0$. Also $f$ is continuous in both $x$ and $y$ and Lipschitz continuous in $y$ with L-constant L. Let F be a mapping on the space $C^0(I)$ of continuous functions $u:I\rightarrow \mathbb{R}$. $I = [x_0,M]$. $$Fu(x) = y_0 + \int_{x_0}^x f(s,u(s))ds.$$ Let for $\alpha\geq 0$ the norm $||\cdot||_\alpha$ be given on $C^0(I)$ by $$||u||_\alpha = \max_{x\in I} |u(x)e^{-\alpha x}|,\quad u \in C^0(I).$$

I have found an $\alpha$ for which F is a Lipschitz contraction with respect to $||\cdot||_\alpha$. The next question is to show that the IVP has exactly 1 solution on I. But isn't this just the contraction theorem applied to $y = F(y)$ (which gives me a unique fixed point, hence a unique solution to this IVP).

If you have found $\alpha > 0$ for which the mapping $F$ is a contraction in the metric induced by the norm $\lVert \cdot \rVert_{\alpha}$ then yes, the contraction mapping theorem just gives you the existence and uniqueness of the fixed point of $F$.
Moreover, as there exist constants $0 < c_1 < c_2$ such that $$c_1 \lVert u \rVert_{\alpha} \le \lVert u \rVert \le c_2 \lVert u \rVert_{\alpha}, \quad \forall \,u \in C^0(I),$$ where $\lVert \cdot \rVert$ denotes the usual supremum norm, it follows that the sequence of successive approximations converges uniformly on $I$ to the (unique) solution of the IVP.