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

  • $\begingroup$ 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

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 $u$.

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  • $\begingroup$ 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

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