Prove existence and uniqueness for a Cauchy problem

Question

I need a help in proving the existence and uniqueness for the following Cauchy problem:

\begin{cases}y''+e^{x}y=0 \\ y(0)=1 \\ y'(0)=0\end{cases}

This can be recasted as a first order system where $f$ is defined as $$f(x,y)=[-e^x y , y]^T$$

In order to prove (local) existence and uniqueness, I need to show that $f$ is locally Lipschitz w.r.t $y$, (it is the RHS of an ODE)

I compute:

$$\left|| f(x,y_1)-f(x,y_2) \right|| = \left|| [e^x (y_2 - y_1),y_1 - y_2]^T \right|| = (y_2 - y_1)^2 \Bigl( 1 + e^{2x} \Bigr) = \left|| y_1 - y_2\right|| \Bigl( 1 + e^{2x} \Bigr) $$

So, for $|x| < a$ (i.e. in a neigbourhood of $x_0=0$ I have $$\Bigl( 1 + e^{2x} \Bigr)\leq1+e^{2a}$$, so it's locally Lipschitz (but no globally)

Is everything correct?

Answer 1

You got the function $f(x,y)$ wrong. What you need to do is define a third variable to serve as the first derivative of $y$. The function you want is $$f([y,y']^T,x) = [y',-e^xy]^T$$. This is the function you want to show is Lipschitz.

Answer 2

$$\frac{d^2y}{dx^2}+e^x y=0$$ Change of variable :$\quad e^x=t\quad\implies\quad \frac{dt}{dx}=t$

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=t\frac{dy}{dt}$

$\frac{d^2y}{dx^2}=\frac{d\frac{dy}{dx}}{dt}\frac{dt}{dx}=(\frac{dy}{dt}+t\frac{d^2y}{dt^2})t=t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}$ $$t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}+ty=0$$ $$\frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+\frac{1}{t}y=0$$ This is a Bessel equation which solution is well known. See Eq.(6) and (7) in : https://mathworld.wolfram.com/BesselDifferentialEquation.html $$y(t)=c_1J_0\big(2\sqrt{t}\big)+c_2Y_0\big(2\sqrt{t}\big)$$ $J_0$ and $Y_0$ are the Bessel functions of the first and second kind respectively. The general solution of the ODE is : $$y(x)=c_1J_0\big(2e^{x/2}\big)+c_2Y_0\big(2e^{x/2}\big)$$ The coefficients $c_1$ and $c_2$ are determined according to the conditions $y(0)=1$ and $y'(0)=0$ which leads to the unique solution : $$y(x)=\frac{Y_1(2)J_0\big(2e^{x/2}\big)-J_1(2)Y_0\big(2e^{x/2}\big)}{Y_1(2)J_0(2)-J_1(2)Y_0(2)}$$