Given the application $u(r,\phi):=v(r\cos\phi, r\sin \phi)$ I need to find by the chain rule an expression for $\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2}$ in terms of second partial derivatives of $u.$ I found first the Jacobian matrix based on the transformation function $(r,\phi)\rightarrow (x,y)=(r\cos \phi, r \sin \phi).$ Since this application is not a function, I do not know if one has to go through the Hessian matrix to find the second partial derivatives. Can you provide me some support or a solution proposal ? Thanks.
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$\begingroup$ Where do $x,y$ feature in all this? $\endgroup$– copper.hatCommented Sep 20, 2022 at 23:53
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$\begingroup$ $(x,y)$ is represented by polar coordinates. I need to thus represent the two second order partial derivatives in terms of partial derivatives of $u$ and the polar coordinates. $\endgroup$– user996159Commented Sep 21, 2022 at 6:40
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2$\begingroup$ You will confuse yourself and everyone else unless you write something sensible like "$u(r,\phi)=v(x,y)$ where $x=r\cos\phi$, $y=r\sin\phi$". The result is in every book: what is the problem? $\endgroup$– ancient mathematicianCommented Sep 21, 2022 at 6:43
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$\begingroup$ @user996159 Have you made any progress? $\endgroup$– Graham KempCommented Oct 5, 2022 at 3:31
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1 Answer
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Hint: Since you want an expression of the second order partial derivatives of $u$, begin by finding what they are in terms of the p.d. of $v$ and the derivatives of $x$ and $y$.
Reason: the first and second order partial derivatives of $x, y$ with respect to $r, \phi$ are much easier to find than the inverse, and trigonometric identities may be useable.$$\dfrac{\partial x}{\partial r}=\cos\phi~, \dfrac{\partial x}{\partial \phi}=-r\sin\phi~, \dfrac{\partial^2 x}{\partial r~^2}=0~,\textit{et cetera}\ldots$$
answered Sep 27, 2022 at 11:59