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In the following let $g(x,r,u)$ be a continuous function $g:[a,b]\times \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$. Furthermore we assume that the partial derivative $\frac{\partial g}{\partial u}$ exists and is also continuous. Now I want to calculate the Frechet derivative $DG$ of the operator $G: \mathcal{C}[a,b] \rightarrow \mathcal{C}[a,b]$ defined by: $$G(u)(x) = u(x) - \int_{a}^{b} g(x,s,u(s)) ds$$

As a guess for $DG$ I use the Gateaux derivative of $G(u)(x)$. Which I calculate according to the definition as follows:

$$\frac{1}{t}\left(G(u+th)-G(u)\right) = h(x) - \frac{1}{t}\int_{a}^{b}g\left(x,s,u(s)+th(s)\right)-g\left(x,s,u(s)\right)$$

Now we can do a Taylor expansion of the integrand in the third argument: $$g(x,s, z(t)) = g(x,s,z(0)) + \frac{\partial g(x,s,u(s))}{\partial z}\dot{z}(0) \left(z(t)-z(0)\right) + \mathcal{O}(t^{2})$$

Where $z(t) = u(s) + th(s)$. Using this expansion in the above integral we end up with the following expression for the Gateaux derivative:

$$\lim\limits_{t\to 0}\frac{1}{t}\left(G(u+th)-G(u)\right) = h(x) -\int_{a}^{b}\frac{\partial g}{\partial u}h(s)^{2} ds$$

Now we have to show that:

$$\lim\limits_{\lVert h \rVert \rightarrow 0}\frac{\lVert G(u+h) - G(u) - DG(u) \rVert}{\lVert h \rVert} = 0$$

Using the Gateaux derivative from above for $DG$ and working out the numerator I get:

$$\lVert G(u+h) - G(u) - DG(u) \rVert = \lVert \int_{a}^{b}\frac{\partial g}{\partial u} h(s)^{2} ds ~ - ~ \int_{a}^{b} \left(g\left(x,s,u(s)+h(s)\right) - g(x,s,u(s))\right) ds \rVert$$

For the first term I can use:

$\lVert \int_{a}^{b}\frac{\partial g}{\partial u} h(s)^{2} ds \rVert \leq \lVert h\rVert^{2} \int_{a}^{b}\frac{\partial g}{\partial u} ds$

And the fact the the partial derivative is continuous, hence bounded on $[a,b]$ am I right? But I can't see how to proceed with the second term. Above $\lVert . \rVert$ is the sup-norm. Is this the right way to calculate $DG$ or am I completely wrong with my approach?

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    $\begingroup$ why $h(s)^2$ this is not linear in $h$. Using $h(s)$ instead of $h(s)^2$ should work. $\endgroup$
    – daw
    Oct 8, 2022 at 13:39
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    $\begingroup$ The $h(s)^{2}$ enters when I do a Taylor expansion in order to calculate the Gateaux derivative (I edited my question and filled in this part as well). $\endgroup$
    – user844389
    Oct 8, 2022 at 15:22

1 Answer 1

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The $h^2$ term is incorrect.

Note that $$\frac1t\int_a^b ds\ [g(x,s,u(s)+th(s)) -g(x,s,u(s))] =\int_a^b ds \frac1t\int_0^t dt\, \frac{\partial}{d\tau} g(x,s,u(s) +\tau h(s))$$ where $$\frac{\partial}{\partial \tau} g(x,s,u(s)+\tau h(s)) = g_u (x,s,u(s)+\tau h(s)) h(s) $$ Now note that by continuity of $g_u$ you have that $$\lim_{t\to0}\frac1t\int_0^tdt\ g_u(x,s,u(s)+\tau h(s)) = g_u(x,s,u(s))$$ and then the above expression converges to $$\int_a^b ds\ g_u(x,s,u(s))\,h(s)$$ and there is no square.

One final remark: This calculation shows that $\frac{G(u+th)-G(u)}{t}$ converges pointwise to $$x\mapsto h(x)- \int_a^b ds\ g_u(x,s,u(s))h(s).$$ But what you want is convergence in norm. I'm sure you can leverage the above calculation together with some continuity requirements on $g$ to get norm convergence.

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    $\begingroup$ Ah thank you! Now for the convergence in norm. We have $\left| G(u+h)(x) - G(u)(x) - DG(u)(x)\right| \leq \left| \int_{a}^{b} g_{u}(x,s,u(s)) h(s) \right| + \left| \int_{a}^{b}g(x,s,u(s)+h(s)) - g(x,s,u(s))\right| \leq M(b-1) \lVert h\rVert + \epsilon (b-a) \lVert h \rVert$. Where in the first integral we use that $g_{u}$ is cont. hence bounded on $[a,b]$ and for the second integral we use the boundedness of $g$. HOWEVER now I'm confused because I am supposed to show that $\lVert G(u+h) - G(u) -DG(u)h \rVert / \lVert h\rVert$ goes to zero, which is not deducible from this estimate right?? $\endgroup$
    – user844389
    Oct 9, 2022 at 15:13
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    $\begingroup$ So somehow in the above estimate I need to get $\lVert h \rVert^{2}$. What am I missing at this point? P.S. For the second integral above we use of course that g is cont. $\endgroup$
    – user844389
    Oct 9, 2022 at 15:17
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    $\begingroup$ As an analogy look at $$|(x+t)^2-x^2 -2x\cdot t| ≤ |2xt+t^2|+2|xt| ≤ 4(|x|+1) \,|t|$$ This bound is true, but it clearly cannot be used to imply that $\frac{(x+t)^2-x^2-2xt}{t}$ converges to $0$. Your bound is doing basically the same thing, you separate the $DG$ term from the $G-G$ term and bound both by $\|h\|$, this procedure cannot show convergence of the differential quotient unless $DG=0$. You need to look at all 3 terms in the integral. As a hint: the mean value theorem should be useful to get the bound you desire. $\endgroup$
    – s.harp
    Oct 9, 2022 at 16:35
  • $\begingroup$ I think I got it know :) We have $g(x,s,u(s)+h(s)) - g(x,s,u(s)) = g_{u}(x,s,\xi) h(s)$ with some $\xi \in \left(u(s),u(s)+h(s)\right)$. Then I can write the intgeral as $\int_{a}^{b} \left(g_{u}(x,s,u(s)) - g_{u}(x,s,\xi)\right)h(s) ds \leq \int_{a}^{b}h(s) C \left| u(s) - \xi\right| ds \leq C \int_{a}^{b} h(s) h(s) ds \leq C \lVert h \rVert^{2} (b-a) $. $\endgroup$
    – user844389
    Oct 9, 2022 at 18:14
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    $\begingroup$ The value of $C$ is kind of opaque here, but yes, thats morally correct. What I would do differently is use uniform continuity of $g_u$ to bound $$|g_u(x,s,u(s))- g_u(x,s,\xi(s))|<\epsilon$$ for an arbitrary $\epsilon$, provided you have made $t$ small enough. $\endgroup$
    – s.harp
    Oct 9, 2022 at 19:28

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