0
$\begingroup$

Let $F : \mathbb{R}^2 \rightarrow \mathbb{R}^2, u : \mathbb{R}^2 \rightarrow \mathbb{R},$ both $C^{\infty}$ and $ g (x,y) = u(F(x,y))$. What is $\partial_x g$? And $\partial^2_{x,x} g$?

Let $ F(x,y) = (F_1(x,y), F_2(x,y)), u=u(a,b)$.
$ DF = \begin{pmatrix} \partial_x F_1 & \partial_y F_1\\ \partial_x F_2 & \partial_y F_2 \end{pmatrix}, \; Du= \begin{pmatrix} \partial_a u & \partial_b u \end{pmatrix} $ How should I apply $ ((Du) \circ F) \cdot (DF) $ ?

  • 0
    You presume you do know the multi-varate chain rule, i.e., $D(u \circ F) = ((Du) \circ F) \cdot (DF)$, where $D$ is the differential operator?2012-06-10

1 Answers 1

1

Try explicitly

$\partial_x g = \frac{ \partial u }{\partial F_1}\frac{\partial F_1}{\partial x} + \frac{ \partial u }{\partial F_2}\frac{\partial F_2}{\partial x} $ where $\frac{\partial u}{\partial F_1} = \frac{ \partial u}{\partial a} \quad \mbox{ and } \frac{\partial u}{\partial F_2} = \frac{ \partial u}{\partial b}$ where $u=u(a,b)$. Similarly for the $y$.