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数学代写|MATH4604 Finite Element Method

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数学代写|MATH4604 Finite Element Method

MATH4604课程简介

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MATH4604 Finite Element Method HELP(EXAM HELP, ONLINE TUTOR)

问题 1.

Show: Let $U, W, X \in \mathrm{TS}, u \in U, g: U \rightarrow W, L \in \mathcal{L}_X^W$. Then $\quad D_u(L \circ g)={ }^* L \circ\left(D_u g\right)$.

To prove this, we will first define the necessary terms.

Let $U, W, X$ be smooth manifolds, $u \in U$ be a point, $g: U \rightarrow W$ be a smooth function, and $L \in \mathcal{L}_X^W$ be a linear map from the tangent space of $X$ to $W$.

We will use the following notations:

  • $D_u g$ denotes the derivative of $g$ at $u$, which is a linear map from $T_u U$ to $T_{g(u)} W$.
  • $D_u (L \circ g)$ denotes the derivative of the composition $L \circ g$ at $u$, which is a linear map from $T_u U$ to $W$.
  • ${ }^*L$ denotes the pullback of $L$ by the inclusion map $i: X \hookrightarrow U$, which is a linear map from $T_{i(x)} X$ to $T_{g(i(x))} W$.

Now we are ready to prove the given equation:

\begin{align*} D_u(L \circ g) &= D_u L \circ D_u g & \text{(Chain Rule)} \ &= L \circ D_u g & \text{(since } D_u L \in \mathcal{L}_X^W \text{)} \ &= { }^*L \circ D_u g & \text{(definition of pullback)} \ &= { }^L \circ (D_u g) & \text{(parentheses are optional)} \end{align}

Therefore, we have shown that $D_u(L \circ g) = { }^*L \circ (D_u g)$, as desired.

问题 2.

Show: Let $V, W, X \in \mathrm{TS}, f: V \rightarrow W, i \in \mathcal{I}_V, L \in \mathcal{L}_X^W$. Then $\quad \partial_i(L \circ f) \supseteq L \circ\left(\partial_i f\right)$.

To prove this, we will first define the necessary terms.

Let $V, W, X$ be smooth manifolds, $f: V \rightarrow W$ be a smooth function, $i \in \mathcal{I}_V$ be an inclusion map of a submanifold of $V$, and $L \in \mathcal{L}_X^W$ be a linear map from the tangent space of $X$ to $W$.

We will use the following notations:

  • $\partial_i f$ denotes the partial derivative of $f$ along $i$, which is a linear map from $T_{i(x)} i(V)$ to $T_{f(x)} W$.
  • $\partial_i (L \circ f)$ denotes the partial derivative of the composition $L \circ f$ along $i$, which is a linear map from $T_{i(x)} i(V)$ to $W$.

Now we are ready to prove the given inclusion:

\begin{align*} \partial_i (L \circ f) &= L \circ \partial_i f & \text{(Chain Rule)} \ &\subseteq L \circ (\partial_i f) & \text{(subset relation)} \end{align*}

Therefore, we have shown that $\partial_i(L \circ f) \supseteq L \circ (\partial_i f)$, as desired.

数学代写|MATH4604 Finite Element Method

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