Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 0 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.