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I need help. Consider $\int_S \nabla v(x) \cdot \nabla v(x)\;dx$. Here $x = (x_1, ..., x_n)$.

Use the substitution $x = \Phi(y)$, where $\Phi:T \to S$ is injective and $C^1$ and $y = (y_1, ..., y_n)$. So the integral becomes $$\int_T \nabla v(\Phi(y)) \cdot \nabla v(\Phi(y)) |\det D\Phi|\;dy\tag{1}$$ where $D\Phi$ is the matrix representing the derivative. How can I get this to the following form: $$\int_T \nabla v(D\Phi)^{-1}(D\Phi)^{-T}\nabla v|\det D\Phi|\;dy$$? I don't know how to get the inverse matrix there nor the transpose.. obviously I should apply the chain rule to the grad terms in (1) but not sure how. Thanks.

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1 Answer 1

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Yes, what you need here is the chain rule. Here is an appropriate form of the chain rule for your situation. Let there be some arbitrary vector $a$. Then,

$$a \cdot \nabla_y v(\Phi(y)) = [a \cdot \nabla_y \Phi(y)] \cdot \nabla_x v(x)$$

The term in square brackets is the definition of $D\Phi$. For brevity, though, I will call it $J$, the Jacobian matrix.

The chain rule is then rewritten as

$$a \cdot \nabla_y v(\Phi(y)) = J(a) \cdot \nabla_x v(x)$$

It's possible to switch things around so that the Jacobian acts on the gradient instead of on the arbitrary vector $a$. The cost to doing this is a transpose.

$$a \cdot \nabla_y v(\Phi(y)) = a \cdot J^T[\nabla_x v(x)]$$

Or, more simply,

$$\nabla_y v(\Phi(y)) = J^T[\nabla_x v(x)]$$

All you need to do from here is solve for $\nabla_x v(x)$ and substitute. The result will match what you wrote with a little manipulation.

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Thanks, but I am confused about your first equation.. on the LHS, you're taking the gradient wrt. $y$ of $v(x) = (v(\Phi(y))$, right? (i.e. with the substitution). And the rightmost gradient on the RHS is simply the evaluation of the gradient wrt. $x$ of $v$ at the point $Phi(y)$. Would be helpful if you clear this issue. –  michael kar Oct 29 '12 at 16:52
    
@michaelkar Yes, that is correct. I've edited the answer to clarify this. –  Muphrid Oct 29 '12 at 17:11
    
Thanks. I still can't show the final result though :| –  michael kar Oct 29 '12 at 18:43

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