How do we find the minium of $f(z)=-|z|^2+|z|^4$?

Question

Consider a function $f(z)$ of a complex variable $z=re^{i\theta}$ given by $f(z)=-|z|^2+|z|^4$. This function, when plotted gives the famous mexican hat potential with the minima lying on a circle $r=1/\sqrt{2}$.

But the minimisation condition requires $f_{rr}>0$ and $D\equiv f_{rr}f_{\theta\theta}-f^2_{r\theta}>0$. But in this case, it turns out that $D=0$ for all values of $z$ since $f(z)=f(|z|)=f(r)$.

How does this clash with the minimization requirement?

Answer 1

$D$ is the determinant of the Hessian matrix. For the mexican hat, you don't have a point minimum, but a whole 1-dimensional subspace of minima. In this case, the Hessian will be zero along the tangent of this subspace. Thus, the whole 2x2 Hessian will be degenerate, having 0 among its eigenvalues, and its determinant is surely 0.

Answer 2

I'll try to incorporate both @lisyarus and @gimusi answers. Lets calculate the gradient and Hessian explicitly

$$f\left(z=re^{i\theta}\right)=-r^{2}+r^{4}$$

$$\boldsymbol{\nabla}f=\begin{pmatrix}-2r+4r^{3}\\0\end{pmatrix}$$

$$H\left(f\right)=\begin{pmatrix}-2+12r^{2}&0\\0&0\end{pmatrix}$$

As you can see, the gradient vanishes at $r=\frac{1}{\sqrt{2}}$ for every value of $\theta$ - this means, as @lisyarus said, that this extermum is not isolated. The Hessian is

$$H\left(f\right)_{r=\frac{1}{\sqrt{2}}}=\begin{pmatrix}4&0\\0&0\end{pmatrix}$$

The structure of this matrix tells you more than its determinant. The requirements that

$$\begin{cases}f_{rr}>0\\D\equiv f_{rr}f_{\theta\theta}-f^2_{r\theta}>0\end{cases}$$

are necessary and sufficient condition for $H\left(f\right)$ to be positive definite (which is only a sufficient for a point to be a minimum). This is equivalent to say that both eigenvalues of $H\left(f\right)$ are positive. But you can clearly see that this is not the case, since our eigenvalues are

$$\begin{cases}\lambda_{r}=4\\\lambda_{\theta}=0\end{cases}$$

The eigenvalue in the $r$ direction is $\lambda_{r}=4$, which tells you that as you walk on lines of $\theta=\rm const.$ you get a minimum at $r=\frac{1}{\sqrt{2}}$. That's what @gimusi showed in his\her answer. On the other hand, the zero eigenvalue in the $\theta$ direction, $\lambda_{\theta}=0$, tells you that the function is flat as you walk on the circle $r=\frac{1}{2}$.

Answer 3

Hint: $\;f(z)=\big(|z|^2\big)^2-|z|^2=\left(|z|^2-\frac{1}{2}\right)^2 - \frac{1}{4} \ge -\frac{1}{4}\,$ with equality iff $\,|z|^2=\frac{1}{2}\,$.