0
$\begingroup$

I consider an operator $A:H^1_0\to H^1_0$ defined by $$Au(t)=\int_0^1 G(t,s) f(s,u(s))ds$$ where $$ G(t,s)=\begin{cases} t(1-s), &t\leq s\\s(1-t), &s\leq t.\end{cases}$$

I want to know what can be the condition on $f$ to obtain that $A$ is continuously differentiable?

Can I do this: $$ \begin{align} A'(u)[v](t) &=\lim_{h\rightarrow0}\frac{A(u+hv)(t)-Au(t)}{h} \\ &=\lim_{h\rightarrow0}\int_0^1 G(t,s) \frac{f(s,(u+hv)(s))-f(s,u(s))}{h} ds\\ &=\int_0^1G(t,s) f_{u}(s,u(s)) v(s) ds ? \end{align} $$

So the condition on $f$ is to be continuously differentiable. Is it right ?

Thank you

$\endgroup$
  • $\begingroup$ I am not sure what you mean by continuously differentiable. Do you mean that the operator is differentiable or that its range consists of continuously differentiable functions? $\endgroup$ – Mark Joshi Jul 2 '15 at 22:12
  • $\begingroup$ We need that $A'u$ continuous, because i have a functional $J$ such that $J''(u)=Id-A'(u)$ so to obtain that $J''$ is continuous we need that $A'$ continuous $\endgroup$ – Vrouvrou Jul 2 '15 at 22:23
  • $\begingroup$ The derivative must be $f_u(s, u(s)) v(s)$ where $f_u$ is the partial derivative of $f$ with respect to the second argument $u$. $\endgroup$ – user251257 Jul 2 '15 at 22:40
0
$\begingroup$

Yes it is true.

As not specified clearly, from the context I infer and assume $$ H^1_0 = H^1_0[0,1] = \{ f\in L^2[0,1] \mid \text{the weak derivative} f'\in L^2[0,1], f(0)=f(1)=0 \}. $$ The key idea is, that $H^1_0$ is continuously embedded in $C[0,1]$, as for $u\in H^1_0$ we have $$ |u(x)| \le \int_0^x |u'(t)| \mathrm d t \le \sqrt{\int_0^1 |u'(t)|^2 \mathrm d t} \le \|u'\|_{L^2} \le \|u\|_{H^1_0}. $$

As $f$ is $C^1$, the limit of the differential quotient exists and takes that value.

Proof: Notice that $G$ is bounded by $1$. Further, for $0 < |h| \le 1$ we have $$ \begin{align} \left|\frac{f(s, (u+hv)(s)) - f(s, u(s))}{h}\right| &= \frac1{|h|}\left|\int_{u(s)}^{u(s)+hv(s)} f_u(s, \mu) \mathrm d \mu \right| \\ &= \frac1{|h|} \left|\int_{0}^{h} f_u(s, u(s)+\eta v(s)) v(s) \mathrm d \eta \right| \\ &\le \frac1{|h|} \int_{0}^{h} |f_u(s, u(s)+\eta v(s))| |v(s)| \mathrm d \eta \\ &\le M_v \|v\|_{\infty} \end{align} $$ with $$ M_v = \sup\{ |f_u(s, u(s)+\eta v(s))| : 0\le s, \eta\le 1 \}, $$ which is finite, as $f_u, u, v$ are continuous and $u,v$ are bounded. Thus, the dominated convergence theorem applies.

The operator $A'(u)$ is linearly bounded.

Proof: As $f_u(s, u(s))$ is uniformly bounded for $0\le s\le 1$ with the same argument before, it follows $$ \|A'(u)[v]\|_{L^2}^2 \le \int_0^1 \int_0^1 \left| G(t,s) f_u(s,u(s)) v(s) \right|^2 \mathrm d s \mathrm d t \le M_0^2 \int_0^1 |v(s)|^2 \mathrm d s \le M_0^2 \|v\|_{H^1_0}^2. $$ As $G$ is Lipschitz contiuous with $\|G_t\|_\infty \le 1$, we have $$ (A'(u)[v])'(t) = \int_0^1 G_t(t,s) f_u(s, u(s)) v(s) \mathrm d s $$ with $$ \|(A'(u)[v])'\|_{L^2}^2 \le \int_0^1 \int_0^1 \left| G_t(t,s) f_u(s,u(s)) v(s) \right|^2 \mathrm d s \mathrm d t \le M_0^2 \int_0^1 |v(s)|^2 \mathrm d s \le M_0^2 \|v\|_{H^1_0}^2. $$ That is $A'(u)$ is a linearly bounded operator from $H^1_0$ to $H^1_0$.

Finally, $A'$ is continuous at $u$.

Proof: Fix $u\in H^1_0$. Let $$ U = \overline{\{ \tilde u(s) : \| \tilde u - u \|_{H^1_0} \le 1 , 0\le s\le 1 \}}, $$ which is bounded and closed, thus compact. Then, $f'$ is uniformly continuous on $[0,1]\times U$.

Let $\epsilon > 0$. Then, there exists some $\eta > 0$ such that for every $(s_1,\mu_1),(s_2,\mu_2)\in[0,1]\times U$ with $\|(s_1,\mu_1)-(s_2,\mu_2)\| < \eta$ it follows $$ |f_u(s_1, \mu_1) - f_u(s_2, \mu_2)| < \epsilon.$$ Now, for this $\eta$ there exists some $\delta > 0$ such that for every $u,\tilde u\in H^1_0$ with $\| u - \tilde u\|_{H^1_0} < \delta$ it follows $\| u - \tilde u\|_\infty < \eta$. Thus, for this $\delta$ and every $\tilde u$ in the $\delta$-neighborhood of $u$, we have $$ |f_u(s, u(s)) - f_u(s, \tilde u(s))| < \epsilon $$ for every $0\le s\le 1$. Then, dominated convergence theorem yields the continuity of $A'$ at $u$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Please to introduce the limite under the integral we need to have that $G(t,s) \frac{f(s,(u+hv)(s))-f(s,u(s))}{h}$ is dominated by an $L^1$ function ? $\endgroup$ – Vrouvrou Jul 3 '15 at 11:49
  • $\begingroup$ @Vrouvrou I add the complete proof. It would also work for the codomain $L^2$. However, I am not sure about $H^1_0$. $\endgroup$ – user251257 Jul 3 '15 at 13:32
  • $\begingroup$ A is from $H^1_0$ to $H^1_0$ $\endgroup$ – Vrouvrou Jul 3 '15 at 13:34
  • $\begingroup$ @Vrouvrou, it will work, as $G_t$ is bounded. You only need to change the $A'(u)$ is bounded and $A'$ is continuous part, to include the norm of $(A'(u)[v])'$. $\endgroup$ – user251257 Jul 3 '15 at 13:36
  • $\begingroup$ I don't understand :"You only need to change the $A′(u)$ is bounded part" $\endgroup$ – Vrouvrou Jul 3 '15 at 13:43

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.