Mixing
Fréchet Derivative of functionals which depend on the derivative
$begingroup$
Let $Omega:=[x_a,x_b]times[t_a,t_b]subset mathbb{R}^2$ and consider the functional $D:C^2left(Omega,mathbb{R}right)rightarrow mathbb{R}$ such that
$$D[y]= iint_{Omega}d(x,t,y) , dx , dt$$
where
$$d(x,t,y)= (1+y(x,t))^pint_{x_a}^{x} dfrac{dy(z,t)}{dt},dz$$
where $pinmathbb{N}$.
Is such a functional Fréchet-differentiable at any point $y_0 in C^2left(Omega,mathbb{R}right)$? 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
$$D[y_0+eta]= iint_Omegaleft[(1+y_0(x,t)+eta(x,t))^pint_{x_a}^{x} left(dfrac{dy_0(z,t)}{dt}+dfrac{deta(z,t)}{dt}right)dzright], dx,dt quad?$$
integration functional-analysis frechet-derivative
asked Jan 21 at 16:37
William TomblinWilliam Tomblin
256111
$endgroup$
|
show 2 more comments
$begingroup$
Let $Omega:=[x_a,x_b]times[t_a,t_b]subset mathbb{R}^2$ and consider the functional $D:C^2left(Omega,mathbb{R}right)rightarrow mathbb{R}$ such that
$$D[y]= iint_{Omega}d(x,t,y) , dx , dt$$
where
$$d(x,t,y)= (1+y(x,t))^pint_{x_a}^{x} dfrac{dy(z,t)}{dt},dz$$
where $pinmathbb{N}$.
Is such a functional Fréchet-differentiable at any point $y_0 in C^2left(Omega,mathbb{R}right)$? 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
$$D[y_0+eta]= iint_Omegaleft[(1+y_0(x,t)+eta(x,t))^pint_{x_a}^{x} left(dfrac{dy_0(z,t)}{dt}+dfrac{deta(z,t)}{dt}right)dzright], dx,dt quad?$$
integration functional-analysis frechet-derivative
asked Jan 21 at 16:37
William TomblinWilliam Tomblin
256111
$endgroup$
$begingroup$
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
1
$begingroup$
Isnt the integeral just y(b)-y(a)?
$endgroup$
– lalala
Jan 21 at 17:00
1
$begingroup$
Isn’t $(1+y)^p$ a function, and not a scalar?
$endgroup$
– Mindlack
Jan 21 at 17:49
1
$begingroup$
Then yes, your expression for $D[y_0+eta]$ is correct.
$endgroup$
– Mindlack
Jan 21 at 22:42
|
show 2 more comments
$begingroup$
Let $Omega:=[x_a,x_b]times[t_a,t_b]subset mathbb{R}^2$ and consider the functional $D:C^2left(Omega,mathbb{R}right)rightarrow mathbb{R}$ such that
$$D[y]= iint_{Omega}d(x,t,y) , dx , dt$$
where
$$d(x,t,y)= (1+y(x,t))^pint_{x_a}^{x} dfrac{dy(z,t)}{dt},dz$$
where $pinmathbb{N}$.
Is such a functional Fréchet-differentiable at any point $y_0 in C^2left(Omega,mathbb{R}right)$? 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
$$D[y_0+eta]= iint_Omegaleft[(1+y_0(x,t)+eta(x,t))^pint_{x_a}^{x} left(dfrac{dy_0(z,t)}{dt}+dfrac{deta(z,t)}{dt}right)dzright], dx,dt quad?$$
integration functional-analysis frechet-derivative
asked Jan 21 at 16:37
William TomblinWilliam Tomblin
256111
$endgroup$
Let $Omega:=[x_a,x_b]times[t_a,t_b]subset mathbb{R}^2$ and consider the functional $D:C^2left(Omega,mathbb{R}right)rightarrow mathbb{R}$ such that
$$D[y]= iint_{Omega}d(x,t,y) , dx , dt$$
where
$$d(x,t,y)= (1+y(x,t))^pint_{x_a}^{x} dfrac{dy(z,t)}{dt},dz$$
where $pinmathbb{N}$.
Is such a functional Fréchet-differentiable at any point $y_0 in C^2left(Omega,mathbb{R}right)$? 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
$$D[y_0+eta]= iint_Omegaleft[(1+y_0(x,t)+eta(x,t))^pint_{x_a}^{x} left(dfrac{dy_0(z,t)}{dt}+dfrac{deta(z,t)}{dt}right)dzright], dx,dt quad?$$
integration functional-analysis frechet-derivative
integration functional-analysis frechet-derivative
asked Jan 21 at 16:37
William TomblinWilliam Tomblin
256111
asked Jan 21 at 16:37
William TomblinWilliam Tomblin
256111
edited Jan 21 at 18:15
William Tomblin
asked Jan 21 at 16:37
William TomblinWilliam Tomblin
256111
asked Jan 21 at 16:37
William TomblinWilliam Tomblin
256111
asked Jan 21 at 16:37
William TomblinWilliam Tomblin
256111
256111
$begingroup$
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
1
$begingroup$
Isnt the integeral just y(b)-y(a)?
$endgroup$
– lalala
Jan 21 at 17:00
1
$begingroup$
Isn’t $(1+y)^p$ a function, and not a scalar?
$endgroup$
– Mindlack
Jan 21 at 17:49
1
$begingroup$
Then yes, your expression for $D[y_0+eta]$ is correct.
$endgroup$
– Mindlack
Jan 21 at 22:42
|
show 2 more comments
$begingroup$
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
1
$begingroup$
Isnt the integeral just y(b)-y(a)?
$endgroup$
– lalala
Jan 21 at 17:00
1
$begingroup$
Isn’t $(1+y)^p$ a function, and not a scalar?
$endgroup$
– Mindlack
Jan 21 at 17:49
1
$begingroup$
Then yes, your expression for $D[y_0+eta]$ is correct.
$endgroup$
– Mindlack
Jan 21 at 22:42
$begingroup$
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
$begingroup$
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
$begingroup$
@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
1
1
$begingroup$
Isnt the integeral just y(b)-y(a)?
$endgroup$
– lalala
Jan 21 at 17:00
$begingroup$
Isnt the integeral just y(b)-y(a)?
$endgroup$
– lalala
Jan 21 at 17:00
1
1
$begingroup$
Isn’t $(1+y)^p$ a function, and not a scalar?
$endgroup$
– Mindlack
Jan 21 at 17:49
$begingroup$
Isn’t $(1+y)^p$ a function, and not a scalar?
$endgroup$
– Mindlack
Jan 21 at 17:49
1
1
$begingroup$
Then yes, your expression for $D[y_0+eta]$ is correct.
$endgroup$
– Mindlack
Jan 21 at 22:42
$begingroup$
Then yes, your expression for $D[y_0+eta]$ is correct.
$endgroup$
– Mindlack
Jan 21 at 22:42
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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
$$varphi(f) = u_f, quad u_f(x) = (1 + f(x))^p intlimits_a^b mathbf{D}_2f(x, t) dt.$$
Now,
$$begin{align*}
u_{f + h}(x) &= (1 + f(x)+ h(x))^p intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= big[(1 + f(x))^p + p(1 + f(x))^{p-1} h(x) + o(1+f(x); h(x))big]\
× intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= u_f(x)+p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt \
&+ o(1+f(x); h(x))intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt, \\
end{align*}$$
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
$$xmapsto p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt$$
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.
answered Jan 21 at 17:38
Will M.Will M.
2,835315
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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
$$varphi(f) = u_f, quad u_f(x) = (1 + f(x))^p intlimits_a^b mathbf{D}_2f(x, t) dt.$$
Now,
$$begin{align*}
u_{f + h}(x) &= (1 + f(x)+ h(x))^p intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= big[(1 + f(x))^p + p(1 + f(x))^{p-1} h(x) + o(1+f(x); h(x))big]\
× intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= u_f(x)+p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt \
&+ o(1+f(x); h(x))intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt, \\
end{align*}$$
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
$$xmapsto p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt$$
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.
answered Jan 21 at 17:38
Will M.Will M.
2,835315
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$varphi(f) = u_f, quad u_f(x) = (1 + f(x))^p intlimits_a^b mathbf{D}_2f(x, t) dt.$$
Now,
$$begin{align*}
u_{f + h}(x) &= (1 + f(x)+ h(x))^p intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= big[(1 + f(x))^p + p(1 + f(x))^{p-1} h(x) + o(1+f(x); h(x))big]\
× intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= u_f(x)+p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt \
&+ o(1+f(x); h(x))intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt, \\
end{align*}$$
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
$$xmapsto p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt$$
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.
answered Jan 21 at 17:38
Will M.Will M.
2,835315
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$varphi(f) = u_f, quad u_f(x) = (1 + f(x))^p intlimits_a^b mathbf{D}_2f(x, t) dt.$$
Now,
$$begin{align*}
u_{f + h}(x) &= (1 + f(x)+ h(x))^p intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= big[(1 + f(x))^p + p(1 + f(x))^{p-1} h(x) + o(1+f(x); h(x))big]\
× intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= u_f(x)+p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt \
&+ o(1+f(x); h(x))intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt, \\
end{align*}$$
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
$$xmapsto p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt$$
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.
answered Jan 21 at 17:38
Will M.Will M.
2,835315
$endgroup$
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
$$varphi(f) = u_f, quad u_f(x) = (1 + f(x))^p intlimits_a^b mathbf{D}_2f(x, t) dt.$$
Now,
$$begin{align*}
u_{f + h}(x) &= (1 + f(x)+ h(x))^p intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= big[(1 + f(x))^p + p(1 + f(x))^{p-1} h(x) + o(1+f(x); h(x))big]\
× intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt \
&= u_f(x)+p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt \
&+ o(1+f(x); h(x))intlimits_a^b (mathbf{D}_2f(x, t) + mathbf{D}_2h(x, t)) dt, \\
end{align*}$$
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
$$xmapsto p(1+f(x))^{p-1} h(x)intlimits_a^b mathbf{D}_2f(x, t) dt + (1 + f(x))^p intlimits_a^b mathbf{D}_2h(x, t) dt$$
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.
answered Jan 21 at 17:38
Will M.Will M.
2,835315
answered Jan 21 at 17:38
Will M.Will M.
2,835315
answered Jan 21 at 17:38
Will M.Will M.
2,835315
answered Jan 21 at 17:38
Will M.Will M.
2,835315
2,835315
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Indeed, my notation was completely wrong, sorry! However you are now definitely solving another problem...
$endgroup$
– William Tomblin
Jan 21 at 18:13
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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
add a comment |
$begingroup$
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
$begingroup$
Indeed, my notation was completely wrong, sorry! However you are now definitely solving another problem...
$endgroup$
– William Tomblin
Jan 21 at 18:13
$begingroup$
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
$begingroup$
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
add a comment |
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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
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@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
1
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Isnt the integeral just y(b)-y(a)?
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– lalala
Jan 21 at 17:00
1
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Isn’t $(1+y)^p$ a function, and not a scalar?
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– Mindlack
Jan 21 at 17:49
1
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Then yes, your expression for $D[y_0+eta]$ is correct.
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– Mindlack
Jan 21 at 22:42