I am reading some lecture notes about image science and active contours, and I come across the following:
$J_1(c)=\int_a^b\left|\frac{\partial c}{\partial q}(t,q)\right|^2dq + \int_a^b g^2(|\nabla I(c(t,q))|)dq$
$J_1'(t)=2\int^b\left<\frac{\partial c}{\partial q},\frac{\partial^2 c}{\partial t \partial q}\right>dq + 2\int^b\left< \frac{\partial c}{\partial t}, g\nabla g \right>dq$
where $C=\{c:[a,b]\rightarrow\Omega, c\ \rm{piecewise}\ C^1, c(a)=c(b)$, $c(q)=(c_1(q),c_2(q))$, $g(s)$ is the edge detection function.
I just want to know the meaning of the symbol $\left<a,b\right>$ inside the integral on the second line.
The symbol $\langle \cdot , \cdot \rangle$ is often used to mean "inner product." We have the integral $$ J_1(c)=\int_a^b\left|\frac{\partial c}{\partial q}(t,q)\right|^2dq + \int_a^b g^2(|\nabla I(c(t,q))|)dq $$ which, written this way, seems like it is meant to be a functional on $C$. I'll reinterpret it as a function of $t$, the argument of $c$, so that we can take a derivative with respect to $t$ in a meaningful way. We have $$ J_1(t)=\int_a^b\left|\frac{\partial c}{\partial q}(t,q)\right|^2dq + \int_a^b g^2(|\nabla I(c(t,q))|)dq $$ Then \begin{align} J_1'(t) &= \frac{d}{dt}\int_a^b\left|\frac{\partial c}{\partial q}(t,q)\right|^2dq + \frac{d}{dt}\int_a^b g^2(|\nabla I(c(t,q))|)dq \\&= \int_a^b \frac{\partial }{\partial t} \left( \left|\frac{\partial c}{\partial q}(t,q)\right|^2 \right) dq + \int_a^b \frac{\partial}{\partial t} \left( g^2(|\nabla I(c(t,q))|) \right) dq \end{align} Let the left term be denoted by $A(t)$ and let the right term be denoted by $B(t)$, i.e. $J_1 ' (t) = A(t) + B(t)$.
At this point, it will be useful to recall (or learn) that $|f(t)|^2 = \langle f(t) , f(t) \rangle$, where $f$ is a vector-valued function. If $f$ and $g$ are vector-valued functions, then $$ \frac{d}{dt} \langle f(t), g(t) \rangle = \langle f'(t) , g(t) \rangle + \langle f(t), g'(t) \rangle $$ This follows almost immediately from the product rule. In particular, we have \begin{align} \frac{d}{dt} |f(t)|^2 &= \frac{d}{dt} \langle f(t), f(t) \rangle = \langle f'(t) , f(t) \rangle + \langle f(t) , f'(t) \rangle \\&= \langle f(t), f'(t) \rangle + \langle f(t), f'(t) \rangle = 2 \langle f(t), f'(t) \rangle \end{align} Now, resuming our computation, we find \begin{align} A(t) &= \int_a^b \frac{\partial }{\partial t} \left( \left|\frac{\partial c}{\partial q}(t,q)\right|^2 \right) dq \\&= \int_a^b 2 \left\langle \frac{\partial c}{\partial q}(t,q), \frac{\partial }{\partial t} \left(\frac{\partial c}{\partial q}(t,q) \right) \right\rangle dq \\&= 2 \int_a^b \left\langle \frac{\partial c}{\partial q}, \frac{\partial c }{\partial t \partial q} \right\rangle dq \end{align} which is the desired form for $A(t)$.
I see no way to make sense of $B(t)$ as it is written in the OP. It seems like $g$ is a function from the reals to the reals since it is squared without being enclosed in magnitude symbols and takes a magnitude as its argument. This means that $g \nabla g = gg'$ is just a scalar, so you can't meaningfully take the inner product of it with $\frac{\partial c}{\partial t}$. If instead, we take $$ J_1(t)=\int_a^b\left|\frac{\partial c}{\partial q}(t,q)\right|^2dq + \int_a^b g^2(c(t,q))dq $$ then we find $$ B(t) = \int_a^b \frac{\partial}{\partial t} \left( g^2(c(t,q)) \right) dq $$ Making use of the chain rule, we compute \begin{align} B(t) &= \int_a^b 2g(c(t,q)) \left( Dg(c(t,q)) \cdot \left(\frac{\partial}{\partial t} c(t,q) \right)\right) dq \\&= \int_a^b 2g \left( Dg\cdot \frac{\partial c}{\partial t} \right) dq \end{align} When we work with the Euclidean inner product, we identify $Dg \cdot \frac{\partial c}{\partial t}$ with $\langle \nabla g , \frac{\partial c}{\partial t} \rangle$. This is justified since, with a Euclidean inner product, we have $\langle v, w \rangle = v^T w$ and $Dg^T = \nabla g$. Thus \begin{align} B(t) &= \int_a^b 2g \left\langle \nabla g , \frac{\partial c}{\partial t} \right\rangle dq \\&= 2 \int_a^b \left\langle g \nabla g , \frac{\partial c}{\partial t}\right\rangle dq \\&= 2 \int_a^b \left\langle \frac{\partial c}{\partial t}, g\nabla g \right\rangle dq \end{align} which is the desired form for $B(t)$.
Putting this all together, we find $$ J_1'(t)=2\int^b_a \left<\frac{\partial c}{\partial q},\frac{\partial^2 c}{\partial t \partial q}\right>dq + 2\int^b_a \left< \frac{\partial c}{\partial t}, g\nabla g \right>dq$$ as desired.