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I was reading a paper about the transformation from Lagrangian to Hamiltonian mechanics which is based on the Legendre transformation, I have quickly written out the part where I couldn't follow: (Note that i used $\mathcal{L}$ for Legendre transformation not Laplace,...)

Consider a function $F(x_1, \ldots, x_m; u_1, \ldots,u_n) \in C^2$and suppose that, \begin{equation} \label{req: implicit function theorem} \det \Big( \frac{\partial^2 F}{\partial x_k \partial x_i} \Big) \neq 0. \hspace{2.0cm}(\forall i\in\{1,\ldots,m\}) \end{equation} we define, \begin{align} y_k &= \frac{\partial F}{\partial x_k}(x_1, \ldots, x_m; u_1, \ldots,u_n) \hspace{0.5cm} &(k\in\{1,2,\ldots,m\}),\\ x_i &= \phi_i(y_1,\ldots,y_m;u_1,\ldots,u_n) &(i\in\{1,2,\ldots,m\}). \end{align} The Legendre transformation is now defined as, \begin{equation} G(y_1,\ldots,y_m;u_1,\ldots,u_n)= (\mathcal{L}F)(y_1,\ldots,y_m;u_1,\ldots,u_n) = \sum_k y_k \phi_k -F. \end{equation} This gives the following set of equations, \begin{equation} \label{eq: analog} \frac{\partial G}{\partial y_k} = \phi_k;\hspace{1.2cm} \frac{\partial G}{\partial u_i}=-\frac{\partial F}{\partial u_i}. \end{equation}

Now I couldn't follow why the derivative of $\sum_k y_k \phi_k$ to $u_i$ is zero, could someone give me a hint/answer? Is it correct that, $$ \frac{\partial \phi_k}{\partial u_i}=0 ,$$ since that $\phi_k$ is in fact $x_k$ and thus an independent parameter, but how does the other term then vanish? (Looked upon as derivation of a product)

Thanks in advance

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I believe you are on the wrong track. Nothing vanishes. There are always two terms cancelling each other.

In order to define $\phi_i$ uniquely, that is in order to invert $y_k =\partial F/\partial x_k$ one typically assumes that $F$ is convex. In any case, let us assume that everything you have written down is well defined. Then $G$ is defined via (always make sure to define the functions properly with all the arguments, as there are always chain rules needed for such problems) $$ G({\bf y}, {\bf u}) =\sum_j y_j \phi_j ({\bf y},{\bf u})- F({\boldsymbol \phi}({\bf y},{\bf u}), {\bf u}) .$$

Taking the derivative with respect to $y_k$, we obtain $$\frac{\partial G}{\partial y_k} ({\bf y}, {\bf u}) = \phi_{k} ({\bf y}, {\bf u}) + \sum_j y_j \frac{\partial \phi_j}{\partial y_k} ({\bf y}, {\bf u}) - \sum_j \underbrace{\frac{\partial F}{\partial x_j}({\boldsymbol \phi}({\bf y},{\bf u}),{\bf u})}_{=y_j} \frac{\partial \phi_j}{\partial y_k} ({\bf y}, {\bf u}) = \phi_{k} ({\bf y}, {\bf u}).$$

Similarly, we obtain $$\frac{\partial G}{\partial u_k} ({\bf y}, {\bf u}) = \sum_j y_j \frac{\partial \phi_j}{\partial u_k} ({\bf y}, {\bf u}) - \sum_j \underbrace{\frac{\partial F}{\partial x_j} ({\boldsymbol \phi}({\bf y},{\bf u}),{\bf u})}_{=y_j} \frac{\partial \phi_j}{\partial u_k} ({\bf y}, {\bf u}) - \frac{\partial F}{\partial u_k}({\boldsymbol \phi}({\bf y},{\bf u}),{\bf u}) = - \frac{\partial F}{\partial u_k}({\boldsymbol \phi}({\bf y},{\bf u}),{\bf u}).$$

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