The strategy is to differentiate the circulation integral $C(t)$ under the integral sign, apply the product rule to the integrand $u \cdot \partial_s \Phi$, and evaluate the two resulting terms separately. The first term produces the material derivative $D_t u = \partial_t u + (u \cdot \nabla)u$, which the Euler equation replaces by $-\nabla p$; the line integral of $\nabla p$ over the closed curve then vanishes by the fundamental theorem of calculus. The second term is an exact $s$-derivative and vanishes by the periodicity $\gamma(0) = \gamma(1)$.

For brevity, write $\Phi_s := \Phi(t, \gamma(s))$ and suppress the argument $(t, \Phi_s)$ when the context is clear.

**Step 1: Differentiation under the integral.** Since $u \in C^1$, $\Phi \in C^1$ (by smooth dependence on initial data for the flow ODE), and $\gamma \in C^1$, the integrand $(t,s) \mapsto u(t, \Phi(t,\gamma(s))) \cdot \partial_s \Phi(t, \gamma(s))$ is continuous in $(t,s)$ and differentiable in $t$ with continuous derivative. Differentiation under the integral sign is therefore justified, and the product rule gives \begin{align*} \frac{d}{dt}C(t) &= \underbrace{\int_0^1 \frac{d}{dt}\bigl[u(t, \Phi_s)\bigr] \cdot \partial_s \Phi(t, \gamma(s))\,d\mathcal{L}^1(s)}_{=:\, I_1} + \underbrace{\int_0^1 u(t, \Phi_s) \cdot \frac{d}{dt}\bigl[\partial_s \Phi(t, \gamma(s))\bigr]\,d\mathcal{L}^1(s)}_{=:\, I_2}. \end{align*}

[claim:Material Derivative Along The Flow] $\frac{d}{dt}\bigl[u(t, \Phi(t,\gamma(s)))\bigr] = (\partial_t u + (u \cdot \nabla)u)(t, \Phi_s)$ for all $s \in [0,1]$. [/claim]

[proof] Define $h: [0,T] \to \mathbb{R}^n$ by $h(t) := u(t, \Phi(t,\gamma(s)))$ for fixed $s$. The $i$-th component is $h_i(t) = u_i(t, \Phi(t,\gamma(s)))$. By the chain rule, \begin{align*} \frac{d h_i}{dt}(t) &= \frac{\partial u_i}{\partial t}(t, \Phi_s) + \sum_{k=1}^{n} \frac{\partial u_i}{\partial y_k}(t, \Phi_s) \cdot \frac{d \Phi_k}{dt}(t, \gamma(s)). \end{align*} Since $\frac{d}{dt}\Phi(t,\gamma(s)) = u(t, \Phi_s)$ by the flow equation, the second term becomes $\sum_k \frac{\partial u_i}{\partial y_k}(t, \Phi_s) \cdot u_k(t, \Phi_s) = ((u \cdot \nabla)u)_i(t, \Phi_s)$. Hence $\frac{d}{dt}[u(t, \Phi_s)] = (\partial_t u + (u \cdot \nabla)u)(t, \Phi_s)$. [/proof]

**Step 2: Evaluation of $I_1$.** By the claim, $\frac{d}{dt}[u(t, \Phi_s)] = (\partial_t u + (u \cdot \nabla)u)(t, \Phi_s)$. The Euler momentum equation gives $(\partial_t u + (u \cdot \nabla)u)(t, \Phi_s) = -\nabla p(t, \Phi_s)$. Therefore \begin{align*} I_1 &= -\int_0^1 \nabla p(t, \Phi_s) \cdot \partial_s \Phi(t, \gamma(s))\,d\mathcal{L}^1(s). \end{align*} Define $q: [0,1] \to \mathbb{R}$ by $q(s) := p(t, \Phi(t,\gamma(s)))$. By the chain rule, \begin{align*} q'(s) &= \sum_{k=1}^{n} \frac{\partial p}{\partial y_k}(t, \Phi_s) \cdot \frac{\partial \Phi_k}{\partial s}(t, \gamma(s)) = \nabla p(t, \Phi_s) \cdot \partial_s \Phi(t, \gamma(s)). \end{align*} Hence \begin{align*} I_1 &= -\int_0^1 q'(s)\,d\mathcal{L}^1(s) = -\bigl[q(1) - q(0)\bigr] = 0, \end{align*} where the last equality holds because $\gamma(0) = \gamma(1)$ implies $\Phi(t,\gamma(0)) = \Phi(t,\gamma(1))$, so $q(0) = p(t, \Phi(t,\gamma(0))) = p(t, \Phi(t,\gamma(1))) = q(1)$.

[claim:Evaluation Of The Second Integral] $I_2 = 0$. [/claim]

[proof] Since $\Phi$ and $u$ are $C^1$ and $\gamma$ is $C^1$, the map $(t,s) \mapsto \Phi(t,\gamma(s))$ is $C^1$ in both variables. Its mixed partial derivatives $\partial_t \partial_s \Phi(t,\gamma(s))$ and $\partial_s \partial_t \Phi(t,\gamma(s))$ are therefore both continuous, so by the symmetry of mixed partials (Schwarz's theorem), \begin{align*} \frac{d}{dt}\bigl[\partial_s \Phi(t,\gamma(s))\bigr] &= \partial_s\bigl[\partial_t \Phi(t,\gamma(s))\bigr]. \end{align*} Since $\partial_t \Phi(t,\gamma(s)) = u(t, \Phi_s)$ by the flow equation, this gives \begin{align*} \frac{d}{dt}\bigl[\partial_s \Phi(t,\gamma(s))\bigr] &= \partial_s\bigl[u(t, \Phi(t,\gamma(s)))\bigr]. \end{align*} Substituting into $I_2$ and writing $v(s) := u(t, \Phi(t,\gamma(s)))$: \begin{align*} I_2 &= \int_0^1 v(s) \cdot v'(s)\,d\mathcal{L}^1(s) = \int_0^1 \frac{1}{2}\frac{d}{ds}|v(s)|^2\,d\mathcal{L}^1(s) = \frac{1}{2}\bigl[|v(1)|^2 - |v(0)|^2\bigr] = 0, \end{align*} where the second equality uses the product rule $v \cdot v' = \frac{1}{2}\frac{d}{ds}(v \cdot v) = \frac{1}{2}\frac{d}{ds}|v|^2$, and the last equality holds because $\gamma(0) = \gamma(1)$ implies $v(0) = u(t, \Phi(t,\gamma(0))) = u(t, \Phi(t,\gamma(1))) = v(1)$. [/proof]

**Step 3: Conclusion.** Combining Steps 1–2 with both claims, $\frac{d}{dt}C(t) = I_1 + I_2 = 0 + 0 = 0$.