1
$\begingroup$

I have the IVP $$\ddot{x}+p(t)\dot x +q(t)x=r(t)$$ $$x(t_0)=x_0$$ $$\dot x(t_0)=x_1$$ where $p(t),r(t),q(t)$ are continuous functions defined for all t in the real numbers. $t_0,x_0,x_1$ are real numbers as well.

Given that there is a $\delta >0$ and a unique function $x(t)\in C^2([t_0,t_0+\delta])$ that satisfies the the IVP for $t_0\leq t \leq t_0+\delta$, (ie, we have a unique local solution) I want to show that a unique global solution exists.

My idea was that we can say that since the interval is arbitrary, that we have a unique solution in $t_0\leq t \leq t_0+\delta$ for all t. Then we can take the union of the local unique solutions and we have a unique global solution...

I am hesitant because this seems too easy, so I am sure I am missing something.

I have read that given a local solution, one has to show that there is a global lipshitz constant so that the solution doesn't blow up in order to get a global solution.

I guess I was just hoping someone could help me out with the question.

Thanks.

$\endgroup$
1
$\begingroup$

To see what you're missing, consider the d.e. $\dot{y} = y^2$. Again, you always have unique local solutions, but you can't just "put them together" to form a global solution on $\mathbb R$ because they blow up, going to $\pm\infty$ at some finite $t$: the general solution is $y = 1/(c - t)$.

EDIT: You may note that if $R = x^2 + \dot{x}^2$, $$ \dot{R} = \left | 2 x \dot{x} + 2 \dot{x}\ddot{x} \right| \le A(t) + B(t) R$$ for some continuous $A$ and $B$, and bounds on $A$ and $B$ in some interval can be used to give you bounds on $R$ there.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you Robert. Do you have a way I could show the global solution? I have had a tough time finding reading on how to do this. $\endgroup$ – MathIsHard May 25 '18 at 0:39

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.