# Prove existence and uniqueness for a Cauchy problem

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?

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.

• Oh, you're right... I found: $$\left|| f(y_1,y_1',x)-f(y_2,y_2',x) \right|| = \left|| \Bigl[ y_1'- y_2' , e^{x}(y_2 - y_1) \Bigr]^t \right\vert| = (y_1' - y_2')^2 + e^{2x}(y_2 - y_1)^2$$ Now I divide by $$(y_1' - y_2')^2 + (y_1 - y_2)^2$$, and obtain $$\frac{(y_1' - y_2')^2 + e^{2x}(y_2 - y_1)^2 }{(y_1' - y_2')^2 + (y_1 - y_2)^2}$$ Commented Aug 19, 2020 at 18:25
• continuing from the last expression, I add and subtract $(y_2-y_1)^2$ and I obtain $$1 + \frac{(y_2 - y_1)^2}{(y_1' - y_2')^2 + (y_1 - y_2)^2} (e^{2x} - 1)$$ Then, for $x$ in a neigbourhood of $x_0=0$, i.e. $|x|<a$, I can bound the last quantity with $$1 + \frac{(y_2 - y_1)^2}{(y_1' - y_2')^2 + (y_1 - y_2)^2} (e^{2a} - 1)$$ where the coefficient of $(e^{2a}-1)$ is less than $1$. Therefore it's locally Lipschitz w.r.t $(y,y')$. Is it right now? Commented Aug 19, 2020 at 19:07
• @andereBen Looks right to me. Commented Aug 19, 2020 at 19:58
• Therefore, it seems to me that it's not possible to prove in this way the global existence: there is no way to bound $(e^{2x} -1)$, right? Commented Aug 19, 2020 at 22:34
• could you check my last comment, please? Commented Aug 20, 2020 at 14:00

$$\frac{d^2y}{dx^2}+e^x y=0$$ 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 : $$y(x)=\frac{Y_1(2)J_0\big(2e^{x/2}\big)-J_1(2)Y_0\big(2e^{x/2}\big)}{Y_1(2)J_0(2)-J_1(2)Y_0(2)}$$