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This question pertains to Rabinowitz : Minimax methods in critical point theory. This is kind of a shot in the dark, since it's unlikely anyone actually has the book on hand and it's not on the web. The setup of defining the function is too lengthy to post (3 pages needed), but a similar setup can be found in Evans.

On p. 84, the author states : Since $g=0$ on $A_{c- \hat{\epsilon}}$, the orbit $\eta(t,u)$ cannot enter $A_{c-\hat{\epsilon}}$.

Why is this?

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1 Answer 1

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According to (A.7) on p. 83, the $\frac{d \eta}{dt} = W(\eta)$ , where $W(x) = 0$ on $A$ (since $g = 0$ on $A$ ). This means that $\frac{d \eta}{dt} = 0$ when $\eta$ enters $A$ and the evolution just stops.

Therefore, since $I(\eta(0,u)) = I(u) \geq c-\epsilon > c -\hat \epsilon$ and $\frac{d I(\eta(t,u))}{dt}\leq 0,$ and by the chain rule, $\frac{dI(\eta(t,u))}{dt} = 0$ as soon as $\eta(t,u)\in A$. Since $A_{c-\hat \epsilon}\subset A$, it follows that $I(\eta(t,u)) \geq c-\hat\epsilon,$ which is sufficient for the next inequality to work.

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I'm sorry, I still don't see it. Is $\frac{d\eta}{dt}=0$ a contradiction? I don't see why we can't have $\eta$ constant past some value of $t$. –  Euler....IS_ALIVE Jan 14 '13 at 22:26
    
Yes, $ t\mapsto \eta(t,u)$ is constant as soon as it enters $A$, which it may or may not do. The reason it is constant is because of the fact that $\frac{d\eta}{dt}$ is 0 on $A$, which follows just from the definition (look closely at how $W$ and $g$ are defined). This is not a contradiction, it is direct. –  fouryear Jan 15 '13 at 8:51
    
That's exactly what I mean. You just said it may or may not enter $A$. The book says precisely that it does NOT enter $A$. That's actually the whole point of my original question. I understand what the consequences are... but why can it not enter $A$? –  Euler....IS_ALIVE Jan 16 '13 at 0:48
    
It stops as soon as it hits the boundary of $A$ at energy $c-\hat{\epsilon}$. What is meant by "may or may not enter $A$" is that we do not know that $I$ even decreases all the way to $c - \hat{\epsilon}$. In the former case, we don't care. In the latter case, $I$ will not decrease beyond $c - \hat\epsilon$ since as soon as that happens $\eta\in\partial A\subset A$ and the evolution stops. Even though the boundary of $A$ is technically in $A$, the next inequality works. The trajectory $\eta(t,u)$ does not enter $A$ in the sense that its energy, is always greater than $c-\hat\epsilon$ –  fouryear Jan 16 '13 at 5:29
    
Ok well then the book should not say the orbit cannot enter $A$, but I agree with everything else. Thank you. –  Euler....IS_ALIVE Jan 16 '13 at 15:49

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