How to find the Gateaux differential of this functional?

[Math100]

Homework Statement

Let $a<b, A$ and $B$ be constants.
Find the Gateaux differential of the functional $S[y]=\int_{a}^{b}(y'^{2}+\omega^{2}y^{2}+2yx^{4})dx, y(a)=A, y(b)=B$, where $\omega$ is a positive constant.

Relevant Equations

Suppose $X$ and $Y$ are locally convex topological vector spaces, $U\subseteq X$ is open, and $F: X\rightarrow Y$. The Gateaux differential $dF(u; \psi)$ of $F$ at $u\in U$ in the direction $\psi\in X$ is defined as $dF(u; \psi)=\lim_{\tau\rightarrow 0}\frac{F(u+\tau\psi)-F(u)}{\tau}=\frac{d}{d\tau}F(u+\tau\psi)\biggr\rvert_{\tau=0}$. If the limit exists for all $\psi\in X$, then one says that $F$ is Gateaux differentiable at $u$.

I am not sure if this is correct, but here is my work by using the definition of the Gateaux differential:
\begin{align*}
&dS(y; \psi)=\lim_{\tau\rightarrow 0}\frac{S(y+\tau\psi)-S(y)}{\tau}=\frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}\\
&S(y+\tau\psi)=\int_{a}^{b}[(y+\tau\psi)'^{2}+\omega^{2}(y+\tau\psi)^{2}+2(y+\tau\psi)x^{4}]dx\\
&=[(y+\tau\psi)'^{2}x+\omega^{2}(y+\tau\psi)^{2}x+\frac{2}{5}(y+\tau\psi)x^{5}]_{a}^{b}\\
&=(y+\tau\psi)'^{2}(b-a)+\omega^{2}(y+\tau\psi)^{2}(b-a)+[\frac{2}{5}(y+\tau\psi)](b^{5}-a^{5})\\
\end{align*}
Thus $S(y+\tau\psi)\biggr\rvert_{\tau=0}=y'^{2}(b-a)+\omega^{2}y^{2}(b-a)+\frac{2}{5}y(b^{5}-a^{5})$.
So $\frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}=0$.

Can anyone please check/verify/confirm this?

 

[SammyS]

Homework Statement:: Let $a<b, A$ and $B$ be constants.
Find the Gateaux differential of the functional $S[y]=\int_{a}^{b}(y'^{2}+\omega^{2}y^{2}+2yx^{4})dx, y(a)=A, y(b)=B$, where $\omega$ is a positive constant.
Relevant Equations:: Suppose $X$ and $Y$ are locally convex topological vector spaces, $U\subseteq X$ is open, and $F: X\rightarrow Y$. The Gateaux differential $dF(u; \psi)$ of $F$ at $u\in U$ in the direction $\psi\in X$ is defined as $dF(u; \psi)=\lim_{\tau\rightarrow 0}\frac{F(u+\tau\psi)-F(u)}{\tau}=\frac{d}{d\tau}F(u+\tau\psi)\biggr\rvert_{\tau=0}$. If the limit exists for all $\psi\in X$, then one says that $F$ is Gateaux differentiable at $u$.

I am not sure if this is correct, but here is my work by using the definition of the Gateaux differential:
\begin{align*}
&dS(y; \psi)=\lim_{\tau\rightarrow 0}\frac{S(y+\tau\psi)-S(y)}{\tau}=\frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}\\
&S(y+\tau\psi)=\int_{a}^{b}[(y+\tau\psi)'^{2}+\omega^{2}(y+\tau\psi)^{2}+2(y+\tau\psi)x^{4}]dx\\
&=[(y+\tau\psi)'^{2}x+\omega^{2}(y+\tau\psi)^{2}x+\frac{2}{5}(y+\tau\psi)x^{5}]_{a}^{b}\\
&=(y+\tau\psi)'^{2}(b-a)+\omega^{2}(y+\tau\psi)^{2}(b-a)+[\frac{2}{5}(y+\tau\psi)](b^{5}-a^{5})\\
\end{align*}
Thus $S(y+\tau\psi)\biggr\rvert_{\tau=0}=y'^{2}(b-a)+\omega^{2}y^{2}(b-a)+\frac{2}{5}y(b^{5}-a^{5})$.
So $\frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}=0$.

Can anyone please check/verify/confirm this?

I'm not all that familiar with the Gateaux differential.

It seems to me that you need to first do the differentiation, $\dfrac{d}{d\tau} S(y+\tau\psi)$, then evaluate the result at $\tau =0$.

 

[pasmith]

Homework Statement:: Let $a<b, A$ and $B$ be constants.
Find the Gateaux differential of the functional $S[y]=\int_{a}^{b}(y'^{2}+\omega^{2}y^{2}+2yx^{4})dx, y(a)=A, y(b)=B$, where $\omega$ is a positive constant.
Relevant Equations:: Suppose $X$ and $Y$ are locally convex topological vector spaces, $U\subseteq X$ is open, and $F: X\rightarrow Y$. The Gateaux differential $dF(u; \psi)$ of $F$ at $u\in U$ in the direction $\psi\in X$ is defined as $dF(u; \psi)=\lim_{\tau\rightarrow 0}\frac{F(u+\tau\psi)-F(u)}{\tau}=\frac{d}{d\tau}F(u+\tau\psi)\biggr\rvert_{\tau=0}$. If the limit exists for all $\psi\in X$, then one says that $F$ is Gateaux differentiable at $u$.

I am not sure if this is correct, but here is my work by using the definition of the Gateaux differential:$$\begin{align*} &dS(y; \psi)=\lim_{\tau\rightarrow 0}\frac{S(y+\tau\psi)-S(y)}{\tau}=\frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}\\ &S(y+\tau\psi)=\int_{a}^{b}[(y+\tau\psi)'^{2}+\omega^{2}(y+\tau\psi)^{2}+2(y+\tau\psi)x^{4}]dx\\ &=[(y+\tau\psi)'^{2}x+\omega^{2}(y+\tau\psi)^{2}x+\frac{2}{5}(y+\tau\psi)x^{5}]_{a}^{b} \end{align*}$$

No. $y$ and $\psi$ are arbitrary functions of $x$ satisfying $y(a) = A$, $y(b) = B$ and $\psi(a) = \psi~(b) = 0$. You can't actually evaluate the integral because you don't know what $y$ and $\psi$ are. Instead you must bring the differentiation with respect to $\tau$ inside the integral, where it becomes a partial derivative at constant $x$.

 

[Math100]

No. $y$ and $\psi$ are arbitrary functions of $x$ satisfying $y(a) = A$, $y(b) = B$ and $\psi(a) = \psi~(b) = 0$. You can't actually evaluate the integral because you don't know what $y$ and $\psi$ are. Instead you must bring the differentiation with respect to $\tau$ inside the integral, where it becomes a partial derivative at constant $x$.

Do you mean $S(y+\tau\psi)=\int_{a}^{b}\frac{d}{d\tau}(y'^{2}+\omega^{2}y^{2}+2yx^{4})dx$? If so, then how to evaluate this?

 

[pasmith]

Do you mean $S(y+\tau\psi)=\int_{a}^{b}\frac{d}{d\tau}(y'^{2}+\omega^{2}y^{2}+2yx^{4})dx$? If so, then how to evaluate this?

$$\begin{split} S[y + \tau \psi] &= \int_a^b (y(x)' + \tau \psi(x)')^2 + \omega^2(y(x) + \tau\psi(x))^2 + 2(y(x) + \tau \psi(x))x^4\,dx \\ \left.\frac{d}{d\tau} S[y + \tau \psi]\right|_{\tau=0} &= \int_a^b \left.\frac{\partial}{\partial \tau}\left(((y'(x) + \tau\psi'(x))^2 + \omega^2(y(x) + \tau\psi(x))^2 + 2(y(x) + \tau\psi(x))x^4\right)\right|_{\tau=0}\,dx. \end{split}$$ Again, you need to calculate the (partial) derivative with respect to $\tau$ and then set $\tau = 0$. $\left.\dfrac{df}{dt}\right|_{t=t_0}$ is how we indicate where the derivative is to be evaluated using Leibnitz's notation; with Newton's notation we would just write $f'(t_0)$.

 

[Math100]

$$\begin{split} S[y + \tau \psi] &= \int_a^b (y(x)' + \tau \psi(x)')^2 + \omega^2(y(x) + \tau\psi(x))^2 + 2(y(x) + \tau \psi(x))x^4\,dx \\ \left.\frac{d}{d\tau} S[y + \tau \psi]\right|_{\tau=0} &= \int_a^b \left.\frac{\partial}{\partial \tau}\left(((y'(x) + \tau\psi'(x))^2 + \omega^2(y(x) + \tau\psi(x))^2 + 2(y(x) + \tau\psi(x))x^4\right)\right|_{\tau=0}\,dx. \end{split}$$ Again, you need to calculate the (partial) derivative with respect to $\tau$ and then set $\tau = 0$. $\left.\dfrac{df}{dt}\right|_{t=t_0}$ is how we indicate where the derivative is to be evaluated using Leibnitz's notation; with Newton's notation we would just write $f'(t_0)$.

After breaking down into smaller pieces, I got the following:
\begin{align*}
&(y'(x)+\tau\psi'(x))^{2}=y'^{2}(x)+2y'(x)\tau\psi'(x)+\tau^{2}\psi'^{2}(x)\\
&\omega^{2}(y(x)+\tau\psi(x))^{2}=\omega^{2}y^{2}(x)+2\tau\psi(x)\omega^{2}+\omega^{2}\tau^{2}\psi^{2}(x)\\
&2(y(x)+\tau\psi(x))x^{4}=2x^{4}y(x)+2x^{4}\tau\psi(x)\\
&\frac{d}{d\tau}[(y(x)'+\tau\psi(x)')^{2}]=2y'(x)\psi'(x)+2\tau\psi'^{2}(x)\\
&\frac{d}{d\tau}[\omega^{2}(y(x)+\tau\psi(x))^{2}]=2\psi(x)\omega^{2}+2\omega^{2}\tau\psi^{2}(x)\\
&\frac{d}{d\tau}[2(y(x)+\tau\psi(x))x^{4}]=2x^{4}\psi(x)\\
\end{align*}
So
\begin{align*}
&\frac{d}{d\tau}S[y+\tau\psi]_{\tau=0}\\
&=[2y'(x)\psi'(x)+2\tau\psi'^{2}(x)+2\psi(x)\omega^{2}+2\omega^{2}\tau\psi^{2}(x)+2x^{4}\psi(x)]_{\tau=0}\\
&=2y'(x)\psi'(x)+2\psi(x)\omega^{2}+2x^{4}\psi(x).\\
\end{align*}
Thus
\begin{align*}
&\int_{a}^{b}[2y'(x)\psi'(x)+2\psi(x)\omega^{2}+2x^{4}\psi(x)]dx\\
&=[2xy'(x)\psi'(x)+2x\psi(x)\omega^{2}+\frac{2}{5}x^{5}\psi(x)]_{a}^{b}\\
&=[2by'(x)\psi'(x)+2b\psi(x)\omega^{2}+\frac{2}{5}b^{5}\psi(x)]-[2ay'(x)\psi'(x)+2a\psi(x)\omega^{2}+\frac{2}{5}a^{5}\psi(x)]\\
&=(b-a)(2y'(x)\psi'(x)+2\psi(x)\omega^{2})+(b^{5}-a^{5})(\frac{2}{5}\psi(x)).\\
\end{align*}
Thus
\begin{align*}
&\frac{d}{d\tau}S[y+\tau\psi]_{\tau=0}=(b-a)(2y'(x)\psi'(x)+2\psi(x)\omega^{2})+(b^{5}-a^{5})(\frac{2}{5}\psi(x)).\\
\end{align*}
Is this correct?

 

[pasmith]

After breaking down into smaller pieces, I got the following:
\begin{align*}
&(y'(x)+\tau\psi'(x))^{2}=y'^{2}(x)+2y'(x)\tau\psi'(x)+\tau^{2}\psi'^{2}(x)\\
&\omega^{2}(y(x)+\tau\psi(x))^{2}=\omega^{2}y^{2}(x)+2\tau\psi(x)\omega^{2}+\omega^{2}\tau^{2}\psi^{2}(x)\end{align*}

You are missing a $y(x)$ from the second term on the right on the last line. But it is easier to use the chain rule rather than multiply everything out: $$\frac{\partial}{\partial \tau} (y(x) + \tau \psi(x)')^2 = 2(y'(x) + \tau \psi'(x))\psi'(x)$$ etc.

Thus
\begin{align*}

&\int_{a}^{b}[2y'(x)\psi'(x)+2\psi(x)\omega^{2}+2x^{4}\psi(x)]dx\\

&=[2xy'(x)\psi'(x)+2x\psi(x)\omega^{2}+\frac{2}{5}x^{5}\psi(x)]_{a}^{b}\\

&=[2by'(x)\psi'(x)+2b\psi(x)\omega^{2}+\frac{2}{5}b^{5}\psi(x)]-[2ay'(x)\psi'(x)+2a\psi(x)\omega^{2}+\frac{2}{5}a^{5}\psi(x)]\\

&=(b-a)(2y'(x)\psi'(x)+2\psi(x)\omega^{2})+(b^{5}-a^{5})(\frac{2}{5}\psi(x)).\\

\end{align*}

Aside from the factor of $y(x)$ missing from the second term, the first line is essentially correct. But the rest is wrong. You have a definite integral with respect to $x$; its value should therefore not depend on $x$, but should be a function of the limits $a$ and $b$. However, you can't evaluate the integral since you don't know how the integrand depends on $x$ (because you don't know what $y$ and $\psi$ are). It is not in general true that $\int_a^b xy(x)\,dx = by(b) - ay(a)$.

The most you can do is integrate the first term by parts to get $$\int_a^b 2y'(x)\psi'(x) + 2\omega^2y(x)\psi(x) + 2x^4 \psi(x)\,dx = 2 \int_a^b \left(-y''(x) + \omega^2y(x) + x^4\right)\psi(x)\,dx + 2\left[y'(x)\psi(x)\right]_a^b$$ but that is as far as you can go.