# How to find the Gateaux differential of this functional?

• Math100
In summary: Yes, that's correct. Now, to evaluate it you would use the fundamental theorem of calculus and the chain rule repeatedly. For example, I'll do one bit in detail and then leave the rest for you to work out (I'm assuming you know how to differentiate ##(y + \tau \psi)^2## with respect to \tau).\frac{d}{d\tau} \int_a^b (y(x) + \tau \psi(x))^2\,dx = \int_a^b \frac{\partial}{\partial \tau} (y(x) + \tau \psi(x))^2\,dx = \int_a^b 2(y(x) + \tau \psi(x

#### 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 ##.

Math100 said:
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 ##.

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##.

Math100 said:
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 $x$.

SammyS
pasmith said:
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?

Math100 said:
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 $f'(t_0)$.

Last edited:
pasmith said:
$$\begin{split} S[y + \tau \psi] &= \int_a^b (y(x)' + \tau \psi(x)')^2 + \omega^2(y(x) + \tau\psi(x))^2 + 2(y(x) + \tau \psi(x))x^4\,dx \\ \left.\frac{d}{d\tau} S[y + \tau \psi]\right|_{\tau=0} &= \int_a^b \left.\frac{\partial}{\partial \tau}\left(((y'(x) + \tau\psi'(x))^2 + \omega^2(y(x) + \tau\psi(x))^2 + 2(y(x) + \tau\psi(x))x^4\right)\right|_{\tau=0}\,dx. \end{split}$$ 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?

Math100 said:
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.

SammyS

## 1. What is the Gateaux differential of a functional?

The Gateaux differential of a functional is a linear map that describes the rate of change of the functional with respect to a given direction in the function space. It is used to analyze the behavior of functionals in calculus of variations and optimization problems.

## 2. How do I find the Gateaux differential of a functional?

To find the Gateaux differential of a functional, you will need to take the directional derivative of the functional in the given direction. This can be done by first expressing the functional in terms of a parameter, then differentiating with respect to that parameter, and finally evaluating the derivative at a specific point.

## 3. What is the difference between Gateaux differential and Fréchet derivative?

The Gateaux differential and Fréchet derivative are both used to describe the rate of change of a functional. However, the Gateaux differential is defined for a specific direction in the function space, while the Fréchet derivative is defined for all possible directions. Additionally, the Gateaux differential is a linear map, while the Fréchet derivative is a functional.

## 4. Can the Gateaux differential of a functional be negative?

Yes, the Gateaux differential of a functional can be negative. It represents the direction of steepest descent for the functional, so it can have a negative value if the functional is decreasing in that direction.

## 5. What are the applications of the Gateaux differential in science?

The Gateaux differential is used in various fields of science, such as physics, engineering, and economics. It is particularly useful in problems involving optimization, where it helps to determine the optimal direction for a given functional. It is also used in calculus of variations to analyze the behavior of functionals and find solutions to minimization problems.