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Consider the following equation which holds for all $w$ in some space, $$\langle v(t), w \rangle = \langle v(0), w \rangle - \int_0^t \langle F(s,v(s)), w \rangle$$ where $\langle F(s,v),w \rangle$ is locally Lipschitz continuous in $v$.

This integral equation has a solution by help of Banach's fixed point theorem in analogy to the existence theorem of Picard-Lindelöf.

I don't see how this is similar to the equation $y'(t) = f(t, y(t))$ or its integral, because we have the duality pairing $\langle v(t), w \rangle$ here. How to overcome this?

Source is Section 2.1 of http://link.springer.com/article/10.1007%2Fs10492-012-0004-0.

edit: Maybe I put the integral inside the duality pairing and then apply the ODE existence theory to the first component of the duality pairing??

EDIT: some text from the paper: http://imgur.com/1r2Ijil http://imgur.com/Frw3IsK http://imgur.com/drlwRgv

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  • $\begingroup$ You cannot do as you suggest as that would need Lipschitz continuity of $F$, while you only have Lipschitz continuity of $\langle F(\cdot, \cdot), w\rangle$. This is weaker. But the proof should push through almost unchanged. $\endgroup$ – Giuseppe Negro May 22 '14 at 16:33
  • $\begingroup$ With your last sentence I guess you mean that I can use the Banach fixed point theorem on my first displayed equation. I don't see what the operator should be defined.. $\endgroup$ – riem May 22 '14 at 16:36
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In the article it is written $w_j$ is a finite dimensional basis of a finite dimensional space $A := span(w_1...w_k)$ and that the equation you mention has a solution $[0, T]\to A$ (only, it is written just before (2.1))

The author showed that $\|\langle F(s,u),w\rangle - F(s,v),w\rangle\|\le C\|u-v\|$ for all $u,v$ in a bigger space than $A$. Then, as Alex has written, let $$ F_j(y) = \langle v(0),w_j\rangle + \int_0^t \langle F(s,y), w_j\rangle ds. $$ If you assume that $F_j\colon A\to C([0,T])$; let $y_n:\mathbb R\to A$ be given, and $(F_j(y_n))_{j\le k}$ is a family of $k$ elements, there exists then a unique $y_{n+1}(t)\in A$ s.t. $$ \forall j, t,\qquad F_j(y_n)(t) = \langle y_{n+1}(t), w_j\rangle $$ because the family $(w_j)_j$ is a basis!! This transition $y_n\to y_{n+1}$ is now clear, and with the relation $$ \forall j, \langle y_{n+1}, w_j\rangle = F_j(y_n) $$ you prove that there exists a fixed point for the sequence $(y_k)$...

Note: It seems for me that the finite basis hypothesis is the key for the proof... Moreover, i'm not totally sure about the spaces above, but you will correct me if needed...

EDIT: how to get the fixed point? As usually: Let $\|\cdot\|$ be the norm as explained in the article (like $L^{m'}$) norm and define the new norm in time $$ \|f\|_T = sup_{t\in[-T,T]}(e^{-Lt}\|f(t)\|) $$ Moreover, as the deimension is finite and all norm are equivalent, the $\|\cdot\|$ norm in $A$ is equivalent to the norm defined by $f\in A$, $\sum_j |\langle f, w_j\rangle|$. Therefore, exists $C'$ such that $$ \|y_{n+1}(t) - y_n(t)\| \le C' \int_0^t \|y_n(s)-y_{n-1}(s)\|ds $$ Then, $$ \begin{split} \|y_{n+1}-y_n\|_T &\le C'\sup_{[-T,T]} e^{-Lt}\int_0^t\|y_n(s)-y_{n-1}(s)\|e^{-Ls}e^{Ls}ds \\ &\le C'\|y_n-y_{n-1}\|_T \sup_{[-T,T]} \frac 1Le^{-Lt}[e^{Lt}-1] \\ &\le \frac{C'}{L}\|y_n - y_{n-1}\|_T \end{split} $$ Thus for $L<C'$ the sequence is contracting. We just have to check that the space $\mathcal C([0, T]; A)$ with $\|\cdot\|_T$ is a Banach space. $A$ with $\|\cdot\|$ is a Banach (it is $L^{m'}$ projected on a closed space), thus $C([0,T],A)$ with this sup norm is also a Banach: Everything is fine and you can apply Banach fixed point theorem!

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  • $\begingroup$ Thanks. So you are just defining $y_{n+1}$ componentwise right? Its $j$th coefficient is equal to $F_j(y_n)$ . OK. What would $y_0$ be, the starting point? $\endgroup$ – riem May 22 '14 at 19:45
  • $\begingroup$ Also, can you expand on how to prove that there is a fixed point for the $y_n$? $\endgroup$ – riem May 22 '14 at 19:57
  • $\begingroup$ I edited the text for Banach fixed point theorem. Moreover, $u_0$ which is given in the statement of Lemma 2.1, $u_0$ is in $L^{m'}$, the starting point is rather its projection on $A$. Are you okay now? I also don't like the articles that give absolutely no details at all on how to proceed to prove "simple" things. However it gives good exercises, but only if we have time... $\endgroup$ – Vintarel May 23 '14 at 12:32
  • $\begingroup$ Thanks that's a great help. I think the space $A$ should be equipped with the norm $H^1 \cap L^{m'}$, because the Lipschitz condition in the proof of Lemma 2.1, at the end, involves the space $H^1$ on the right hand side, so your first inequality involving $C'$ would make sense with that norm. $\endgroup$ – riem May 23 '14 at 13:36
  • $\begingroup$ And I guess the map is $T(y_n) = y_{n+1}$. I've not seen a "discrete" mapping like this before. $\endgroup$ – riem May 23 '14 at 13:37
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I cannot read the linked paper, I would have to buy the paper :). If I understand right, the formulation you give is a weak formulation for the integral equation $$v(t)=v(0)+\int_{0}^{t}F(s,v(s))d s,$$ which is again a formulation for the ode you mentioned. In my opinion, it is essential that you name the test space for $w$ as well as what they (the authors) understand by a weak solution of this type of ode. Now let $w$ be fixed, then, we have to show that the mapping $$ C_{w}:A\rightarrow A,\ C_{w}(x)=\langle v_{0},w\rangle+\int_{0}^{t}\langle F(s,v(s)),w\rangle ds $$ is a contraction for $A$ a proper Banach space (you should know, which space to pick, or the paper should mention this fact). Let $\|\cdot\|_{A}$ be the norm of the space $A$, we have for $x,y\in A$ and $w$ arbitrary but fixed: $$\|C_{w}(x)-C_{w}(y)\|_{A}\leq \int_{0}^{t}L\|x-y\|_{A}\ d s \|w\|\leq Lt\|x-y\|_{A}\|w\|,$$ applying Cauchy Schwarz (if we have an inner product) and using the local Lipschitz continuity. So for sufficiently small $t$ this is a contraction also depending on $w$. Now one might be able to assume $\|w\|=1$ (depends on the definition of the weak solution) and we have the contraction, and thus a solution in the - still to be specified - space $A$.

Of course, first of all you have to show, that the $C_{w}$ is mapping $A$ to $A$, i.e. $C_{w}$ is a self-map. But I cannot show this, since I do not know $A$....

Maybe this is a bit helpful or gives some ideas for the solution...

Yours,

Alex

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  • $\begingroup$ Hi Alex thanks for answering. My biggest issue is: with your proof, we obtain a fixed point $y$ of $C_w$. But all we get is that $y = C_w(y) = \langle y_0, w \rangle + \int_0^t \langle F(s,y(s)),w$ which does not have the same left hand side since the duality product is missing. I hope I'm not being silly.. $\endgroup$ – riem May 22 '14 at 16:39
  • $\begingroup$ riem, you should get something like $\langle y,w\rangle=C_{w}(y)$, which should be fine?! $\endgroup$ – Alex May 22 '14 at 16:40
  • $\begingroup$ Now I understand, your problem is that $C_{w}:A\rightarrow \mathbb{R}$ and not as I claimed $C_{w}:A\rightarrow A$. Hm, strange..., right now I do not have an idea how to get the operator well defined. $\endgroup$ – Alex May 22 '14 at 17:05
  • $\begingroup$ Please see my edited post -- I added some images with the text which may be helpful! $\endgroup$ – riem May 22 '14 at 17:10

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