0
$\begingroup$

I'm investigating where the IVP $$y'=-y, \enspace y(0)=3$$

has an unique solution. The right hand side satisfies a Lipschitz-condition for the constant $K$ with respect to variable $y$, and clearly the function $-y$ is continuous. This implies there is an interval $I:=(-\delta, \delta$) where the solution is unique. According to the proof, we have that $\delta = \min\{a,\frac{b}{2M}\} $, where $M = \sup f(y)_{y\in R}$. Define $R$ as the rectangle $$R : \vert x \vert \leq a, \enspace \enspace \vert y-3\vert\leq b$$

which means $M=(b+3)$, the maximal value of $f(x,y)=-y$ in R. The interval $I$ is maximal, when $\frac{b}{2M}$ is maximal, for we can then set $a$ as its equal. However, as the function $$q(p)=\frac{b}{2(b+3)}$$

has no extrema and no maximal points, this implies that $I$ can be made as large as we wish. Is it now correct to say that the unique solution exists globally?

$\endgroup$
1
$\begingroup$

Your equation is linear, with continuous coefficients on $\mathbb{R}$. Therefore, your solutions can be extended to all of $\mathbb{R}$ and there is a unique solution through every point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.