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Exercise :

Study the existence and the uniqueness of the solutions of the initial value problem :

$$y'' + p(x)y' + q(x)y = f(x)$$

$$y(x_0) = y_0, y'(x_0) = y_1$$

where $p(x),q(x),f(x)$ continuous functions in some domain $a<x<b$ with $x_0 \in (a,b)$.

Now, I do not know how to handle this. I know that for a simple case : $y' = f(x,y)$, you can study and determine the existence of a solution in a domain $D = \{(x,y) \in \mathbb R^2:|y-y_0|\leq ε,|x-x_0| = \delta\}$ if the function $f(x,y)$ is continuous over $D$, while for the uniqueness, you just check if the function is bounded/Lipschitz. But on this case, that it involves multiple function that we do not know if they're $\neq 0$ and also a second-derivative, how should I approach ? What exactly shall I do to answer the question thoroughly ?

Thanks in advance !

Rebellos
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1 Answers1

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First of all note, that the 'simple' case $y'=f(x,y)$ already encompasses your problem in the following way:

Set $z_1'(x)=z_2(x)$ and $z_2'(x)=f(x)-p(x)z_2(x)-q(x)z_1(x)$ as your new system. Here $z_1$ plays the role of $y$, while $z_2$ is for $y'$. Hence you can define an $g:\mathbb{R}\times\mathbb{R}^2\rightarrow\mathbb{R}^2$ by $$g_1(x,z)=z_2,\quad g_2(x,z)=f(x)-p(x)z_2-q(x)z_1.$$ Now note, that this $g$ plays the same role as your '$f(x,y)$'. Since $p,q,f$ are continuous, this $g$ is continuous in $x$ as well. Now you need to check a local Lipschitz condition for the $z$ variable, which is fullfilled if $g$ is differentiable in the $z$ variable. Please check this last fact for yourself.

humanStampedist
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  • I get how to fix a new differential system and show through continuous functions that a solution exists. But how would I find a Lipschitz constant or inequality since no information about bounds are listed for the functions ? – Rebellos Oct 23 '17 at 15:57
  • $p,q,f$ are continuous, so they are bounded on compact sets. For $x_0\in (a,b)$ choose an $\varepsilon$ such that $\overline{B_{\varepsilon}(x_0)}\subset(a,b)$ and use this $\overline{B_{\varepsilon}(x_0)}$ as your compact set. Remember you only need a local Lipschitz condition (see e.g. https://math.stackexchange.com/questions/518323/the-definition-of-locally-lipschitz) – humanStampedist Oct 24 '17 at 10:11