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Consider the following function:

$$f(x_1, x_2) = x_1^2x_2+x_1x_2^3$$

$f$ is a function of $(x_1, x_2)$. The conjugate variables $(u_1, u_2)$ to $(x_1, x_2)$ are $$u_1 = \partial f/ \partial x_1 = 2 x_1 x_2 + x_2^3$$ and $$u_2 = \partial f/ \partial x_2 = x_1^2 + 3x_1x_2^2.$$

One can construct $$g=f - u_1 x_1$$ which, to my understanding, replaces $x_1$ to its conjugate variable, $u_1$. The differential of $g$ is

$$dg = u_2 dx_2 - x_1 du_1$$ thus $g=g(u_1, x_2)$. In words, $g$ is a function of the old variable $x_2$ and the conjucate variable to $x_1$, $u_1$.

Now I fail to see this in the above example. I can compute

$$g = f - u_1x_1 = -x_1^2x_2$$ but $g$ is then still a function of $(x_1, x_2)$, not $(u_1, x_2)$. What am I missing here?

In theory, I could invert the $u_1=u_1(x_1, x_2)$ and $u_2=u_2(x_1, x_2)$ equations, then plug $x_1(u_1, u_2)$ and $x_2(u_1, u_2)$ into $f(x_1, x_2)$ to get $f(u_1, u_2)$. But that's just a simple change of variable. What's the point of the Legendre transformation concept then?

Also, in this particular example, inverting the $u(x)$ equations doesn't seem obvious. How would one get the expression for $g(u_1, x_2)$?

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    $\begingroup$ Not an answer, but the inverting isn't very hard in this case: just solve for $x_1 = \frac{u_1-x_2^3}{2x_2}$. Hamiltonians are usually quadratic so when you do this in that case all the inverting is just inverting linear functions as well. $\endgroup$
    – jacob1729
    Apr 27, 2019 at 10:54

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I can show you the steps that you need to calculate the Legendre transformation, you have to look at the theory behind .

for a given function of two variable $f(x,y)$ we are looking for the Legendre function $\tilde{f}(u,v)$

Step I:

The determinant of matrix $A$ must be unequal zero. where $A$ is:

$$A= \left[ \begin {array}{cc} {\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,y \right) & \left( {\frac {\partial }{\partial x}}f \left( x ,y \right) \right) {\frac {\partial }{\partial y}}f \left( x,y \right) \\ \left( {\frac {\partial }{\partial x}}f \left( x,y \right) \right) {\frac {\partial }{\partial y}}f \left( x ,y \right) &{\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,y \right) \end {array} \right] $$

Step II:

with the equations:

$$u=\frac{\partial f(x,y)}{\partial x}\tag 1$$ $$v=\frac{\partial f(x,y)}{\partial y}\tag 2$$

we get from equation (1) , $x=f_x(u,y)$ and from equation (2) $y=f_y(u,x)$ both solutions must be exist

Step III:

with this solutions $f_x$ and $f_y$ we obtain the Legendre function

$$\tilde{f}(u,v)=u\,f_x(u,v)+v\,f_y(u,v)-f(f_x(u,v),f_y(u,v))$$

Example:

$$f(x,y)={x}^{2}y+x{y}^{3}$$

$$f_x(u,y)=1/2\,{\frac {u-{y}^{3}}{y}}$$ $$f_y(u,x)=1/3\,{\frac {\sqrt {3}\sqrt { \left( x \right) \left( u-{x}^{2} \right) }}{x}} $$ $$\tilde{f}(u,v)=1/2\,{\frac {u \left( u-{v}^{3} \right) }{v}}+1/3\,\sqrt {3}\sqrt {v \left( u-{v}^{2} \right) }-1/12\,{\frac { \left( u-{v}^{3} \right) ^{ 2}\sqrt {3}\sqrt {v \left( u-{v}^{2} \right) }}{{v}^{3}}}-1/18\,{ \frac { \left( u-{v}^{3} \right) \sqrt {3} \left( v \left( u-{v}^{2} \right) \right) ^{3/2}}{{v}^{4}}} $$

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  • $\begingroup$ But you need to be able to get an analytic expression for $f_x(u, y)$ and $f_y(u, x)$, correct? What happens if you can't invert equations (1) and (2)? Can you still obtain your $\tilde{f}(u,v)$? $\endgroup$
    – Botond
    Apr 29, 2019 at 15:01
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    $\begingroup$ Correct . If you don’t have analytic solution for the invert functions , you don’t get close solution for the Legendre function, I don’t think that numerical solution make sense. $\endgroup$
    – Eli
    Apr 29, 2019 at 18:48

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