# Partial Legendre transform: understanding a simple example

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 $$g(u_1, x_2)$$?

• 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. Apr 27, 2019 at 10:54

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:

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

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

• 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 $\tilde{f}(u,v)$? Apr 29, 2019 at 15:01
• 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.
– Eli
Apr 29, 2019 at 18:48