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Thread: What is the gradient of u(Ox) when O is orthogonal.

  1. #1
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    What is the gradient of u(Ox) when O is orthogonal.

    Suppose we know $\displaystyle \nabla{u(x)}$. Let O be an orthogonal matrix.

    What is $\displaystyle \nabla{u(Ox)}$?

    Thank you!
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  2. #2
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    See here.
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  3. #3
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    Yea, I don't understand exactly what that means though.

    Is it $\displaystyle (O^{-1}) \nabla u \cdot (Ox)$?
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  4. #4
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    I don't think there's a dot product there (except implicitly, in the definition of the gradient.) Looking at the equation applied to your case, we'd have

    $\displaystyle \nabla(u(O\mathbf{x}))=O^{T}(\nabla u)(O\mathbf{x}).$

    As I see it, the way to interpret the LHS of this equation is to take the vector $\displaystyle \mathbf{x}$, rotate by left-multiplying by $\displaystyle O$, then you stuff the result of the rotation into the function $\displaystyle u$, and finally, you take the gradient.

    I interpret the RHS of this equation as follows: first, you take the gradient of $\displaystyle u$, and then you evaluate that function at the point $\displaystyle Ox$, and finally you left-multiply by $\displaystyle O^{T}.$

    If I were you, I'd work out a simple example in, say, 2 dimensions. See how it works out.
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