How to check for local extrema or saddle point given an semidefinite matrix

Question

I've computed the Hessian of a given function $f(a,b,c) = y-a\sin(bx-c)$ and got the following result:

$\begin{pmatrix} 0 & -x\cdot\cos(bx - c) & \cos(bx - c) \\ -x\cdot\cos(bx - c) & ax^2\cdot\sin(bx - c) & -ax\sin(bx - c) \\ \cos(bx - c) & -ax\cdot\sin(bx - c) & a\cdot\sin(bx - c) \end{pmatrix}$

This matrix is positive semi-definite and thus one can not state for a given point $P=(a_i,b_i,c_i)$ if it is a local min, max or a saddle point. Is there any other way to explicitly determine if we have a loc. min, max or saddle point?

Answer 1

With reference to the given matrix, we have that

$$\det(a\sin(bx - c))=a\sin(bx - c)$$

$$\begin{vmatrix} 0 & -x\cos(bx - c) \\ -x\cos(bx - c) & ax^2\sin(bx - c) \\ \end{vmatrix}=-x^2\cos^2(bx - c)$$

$$\begin{vmatrix} 0 & -x\cdot\cos(bx - c) & \cos(bx - c) \\ -x\cdot\cos(bx - c) & ax^2\cdot\sin(bx - c) & -ax\sin(bx - c) \\ \cos(bx - c) & -ax\cdot\sin(bx - c) & a\cdot\sin(bx - c) \end{vmatrix}=$$ $$=x\cdot\cos(bx - c)(-ax\cos(bx - c)\sin(bx - c)+ax\cos(bx - c)\sin(bx - c))+\cos(bx - c)(ax^2\cos(bx - c)\sin(bx - c)-ax^2\cos(bx - c)\sin(bx - c))=0$$

therefore for $x^2\cos^2(bx - c)\neq 0$

• $a\sin(bx - c)\le 0$ the matrix is negative semidefinite

• $a\sin(bx - c)>0$ the matrix is indefinite

and for $x^2\cos^2(bx - c)= 0$

• $a\sin(bx - c)< 0$ the matrix is negative semidefinite

• $a\sin(bx - c)>0$ the matrix is positive semidefinite

Note that the condition for critical points implies

• $f_a=-\sin(bx - c)=0\implies \sin(bx - c)=0$

• $f_b=-ax\sin(bx - c)=0\implies \sin(bx - c)=0 \lor x=0$

• $f_c=a\sin(bx - c)=0\implies \sin(bx - c)=0$

therefore

• $\cos(bx - c)=\pm 1$

and the expression for the Hessian simplifies.