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Let $\Omega:=[x_a,x_b]\times[t_a,t_b]\subset \mathbb{R}^2$ and consider the functional $D:C^2\left(\Omega,\mathbb{R}\right)\rightarrow \mathbb{R}$ such that $$D[y]= \iint_{\Omega}d(x,t,y) \, dx \, dt$$ where $$d(x,t,y)= (1+y(x,t))^p\int_{x_a}^{x} \dfrac{dy(z,t)}{dt}\,dz$$ where $p\in\mathbb{N}$.

Is such a functional Fréchet-differentiable at any point $y_0 \in C^2\left(\Omega,\mathbb{R}\right)$? And how do we calculate its derivative?

In particular, I am not sure how the derivative in the integrand should be dealt with. When applying the limit definition, should I consider something like $$D[y_0+\eta]= \iint_\Omega\left[(1+y_0(x,t)+\eta(x,t))^p\int_{x_a}^{x} \left(\dfrac{dy_0(z,t)}{dt}+\dfrac{d\eta(z,t)}{dt}\right)dz\right]\, dx\,dt \quad?$$

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  • $\begingroup$ If $y$ were of class $\mathscr{C}^2,$ the standard product rules and differentiation under integral sign would apply. For a case like this, I believe you need to do the limit definition. $\endgroup$ – Will M. Jan 21 at 16:40
  • $\begingroup$ @WillM. I tried to edit my question, making my point a bit clearer. I actually thought that $y$ being in $C^1$ or $C^2$ didn't make any difference, am I wrong? $\endgroup$ – William Tomblin Jan 21 at 16:55
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    $\begingroup$ Isnt the integeral just y(b)-y(a)? $\endgroup$ – lalala Jan 21 at 17:00
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    $\begingroup$ Isn’t $(1+y)^p$ a function, and not a scalar? $\endgroup$ – Mindlack Jan 21 at 17:49
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    $\begingroup$ Then yes, your expression for $D[y_0+\eta]$ is correct. $\endgroup$ – Mindlack Jan 21 at 22:42
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All your notation seems wrong. So I am fixing it.

Let us consider the space $\mathrm{X} = \mathscr{B}^1_{\mathbf{R}}(\mathrm{U} \times \mathrm{I})$ of differentiable real valued bounded function with continuity defined on the product of the open set $\mathrm{U}$ of $\mathbf{R}^d$ and the compact interval $\mathrm{I} = [a, b]$ of $\mathbf{R}.$ Endow $\mathrm{X}$ with the structure of complete normed space by setting $\|f\|_\mathrm{X} = \|f\| + \left\| \mathbf{D} f \right\|$ (the sup-norm and the operator norm). (The fact that $\mathrm{X}$ is complete follows from the mean value theorem.)

Define $\varphi:\mathrm{X} \to \mathrm{Y} = \mathscr{C}^b_\mathbf{R}(\mathrm{U}),$ the right hand side is the space of continuous and bounded functions with the sup norm, by $$\varphi(f) = u_f, \quad u_f(x) = (1 + f(x))^p \int\limits_a^b \mathbf{D}_2f(x, t)\ dt.$$

Now, $$\begin{align*} u_{f + h}(x) &= (1 + f(x)+ h(x))^p \int\limits_a^b (\mathbf{D}_2f(x, t) + \mathbf{D}_2h(x, t))\ dt \\ &= \big[(1 + f(x))^p + p(1 + f(x))^{p-1} h(x) + o(1+f(x); h(x))\big]\\ &\times \int\limits_a^b (\mathbf{D}_2f(x, t) + \mathbf{D}_2h(x, t))\ dt \\ &= u_f(x)+p(1+f(x))^{p-1} h(x)\int\limits_a^b \mathbf{D}_2f(x, t)\ dt + (1 + f(x))^p \int\limits_a^b \mathbf{D}_2h(x, t)\ dt \\ &+ o(1+f(x); h(x))\int\limits_a^b (\mathbf{D}_2f(x, t) + \mathbf{D}_2h(x, t))\ dt, \\\\ \end{align*}$$ where $o(a; s)$ is a sum of powers of $s^\alpha,$ where $\alpha$ runs from 2 until p, and the coefficients are powers of $a$ as well. Because $f$ is bounded, there exists a constant $c = c(f)$ such that $$\|o(1 + f; h)\| \leq c(\|h\|^2+\ldots+\|h\|^p) = o(\|h\|).$$ It is easy to check that the function $L_f:h \mapsto L_f(h)$ given according to the rule $$x\mapsto p(1+f(x))^{p-1} h(x)\int\limits_a^b \mathbf{D}_2f(x, t)\ dt + (1 + f(x))^p \int\limits_a^b \mathbf{D}_2h(x, t)\ dt$$ belongs to $\mathrm{Y}$ (that is, $L_f:\mathrm{X} \to \mathrm{Y}$) and it is linear. The desired derivative of $\varphi$ is therefore $\mathbf{D}\varphi(f) = L_f.$ Q.E.D.

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  • $\begingroup$ Indeed, my notation was completely wrong, sorry! However you are now definitely solving another problem... $\endgroup$ – William Tomblin Jan 21 at 18:13
  • $\begingroup$ My problem contains yours, I believe. And I made a mistake too, instead of $(x)$ it should be $(x, t).$ $\endgroup$ – Will M. Jan 21 at 18:20

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