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MATH4604 Finite Element Method HELP（EXAM HELP， ONLINE TUTOR）
12-4. Show: Let $V, W \in \mathrm{TS}, q, u, v \in V, f: V \rightarrow W, \phi:=f \circ \alpha_q^{u v}$.
Then $\left(\nabla_u \nabla_v f\right)q=\left(\partial_1 \partial_2 \phi\right){(0,0)}$.
We have:
\begin{align*} \left(\nabla_u \nabla_v f\right)q &= \nabla_u \left(\nabla_v f\right)q \ &= \nabla_u \left(\left(\nabla f\right)q(v)\right) \ &= \nabla_u \left(\lim{t \rightarrow 0} \frac{f(q+tv)-f(q)}{t}\right) \ &= \lim{t \rightarrow 0} \frac{\nabla_u\left(f(q+tv)-f(q)\right)-\nabla_u f(q)}{t} \ &= \lim{t \rightarrow 0} \frac{\left(\nabla f\right){q+tv}(u) \cdot v – \left(\nabla f\right)q(u) \cdot v}{t} \ &= \lim{t \rightarrow 0} \frac{\left(\nabla f\right){q+tv}(u) – \left(\nabla f\right)q(u)}{t} \cdot v \ &= \left(\nabla_v \left(\nabla f\right)q(u)\right) \ &= \left(\nabla_v \left(\lim{s \rightarrow 0} \frac{f(q+su)-f(q)}{s}\right)q(u)\right) \ &= \left(\nabla_v \left(\lim{s \rightarrow 0} \frac{\phi(s,0)-\phi(0,0)}{s}\right)\right){(0,0)} \ &= \left(\partial_2 \left(\lim_{s \rightarrow 0} \frac{\phi(s,0)-\phi(0,0)}{s}\right)\right){(0,0)} \ &= \left(\partial_2 \partial_1 \phi\right){(0,0)} \ &= \left(\partial_1 \partial_2 \phi\right)_{(0,0)} \end{align*}
Therefore, $\left(\nabla_u \nabla_v f\right)q=\left(\partial_1 \partial_2 \phi\right){(0,0)}$, as desired.
12-5. Show: Let $V, W \in \mathrm{TS}, f: V \rightarrow W, q \in \mathbb{D}_f^{\prime \prime}, u, v \in V$.
Then: $\quad\left(\nabla_u \nabla_v f\right)_q=\left(\nabla_v \nabla_u f\right)_q$.
To prove the equality $\left(\nabla_u \nabla_v f\right)_q=\left(\nabla_v \nabla_u f\right)_q$, we use the definition of the second derivative of a smooth map $f: V \rightarrow W$ between two tangent spaces $V$ and $W$. Let $\gamma: (-1,1) \rightarrow V$ be a smooth curve with $\gamma(0) = q$ and $\gamma'(0) = u$, and let $\delta: (-1,1) \rightarrow V$ be a smooth curve with $\delta(0) = q$ and $\delta'(0) = v$. Then we define
$$
\left(\nabla_u \nabla_v f\right)q=\left(\left.\frac{d}{d t}\right|{t=0} \nabla_v\left(\gamma^{\prime}(t)\right) f\right)q $$ and $$ \left(\nabla_v \nabla_u f\right)_q=\left(\left.\frac{d}{d t}\right|{t=0} \nabla_u\left(\delta^{\prime}(t)\right) f\right)_q
$$
Our goal is to show that these two expressions are equal.
First, we evaluate $\nabla_v (\gamma'(t))f$: \begin{align*} \nabla_v (\gamma'(t))f &= \left(\frac{d}{ds}\Big|{s=0} (\gamma'(t)+sv)f\right)q \ &= \left(\frac{d}{ds}\Big|{s=0} \frac{d}{dt}\Big|{t=0} (\gamma(t)+st)f\right)q \ &= \left(\frac{d^2}{ds,dt}\Big|{s=t=0} (\gamma(t)+st)f\right)_q \ &= \left(\nabla_u \nabla_v f\right)_q + \left(R(u,v)f\right)_q, \end{align*} where $R$ is the Riemann curvature tensor. In the second line, we used the chain rule for taking derivatives of composite functions, and in the third line we used the definition of the first derivative of $f$ and the fact that $\gamma'(0) = u$.
Similarly, we evaluate $\nabla_u (\delta'(t))f$: \begin{align*} \nabla_u (\delta'(t))f &= \left(\frac{d}{ds}\Big|{s=0} (\delta'(t)+su)f\right)q \ &= \left(\frac{d}{ds}\Big|{s=0} \frac{d}{dt}\Big|{t=0} (\delta(t)+st)f\right)q \ &= \left(\frac{d^2}{ds,dt}\Big|{s=t=0} (\delta(t)+st)f\right)_q \ &= \left(\nabla_v \nabla_u f\right)_q + \left(R(v,u)f\right)_q. \end{align*}
Since $R$ is a symmetric tensor, we have $R(u,v)f = R(v,u)f$. Therefore, we can subtract $\left(R(u,v)f\right)_q$ from both sides of the first equation and $\left(R(v,u)f\right)_q$ from both sides of the second equation to obtain
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