gradient descent for non convex function like $-x^2$

I know how to calculate gradient descent for a convex function where there is only one global minima. Also, I know methods to handle cases where the function is a non-convex function. What is really bothering me that for a non-convex function like $$y = -x^2$$, how gradient descent is actually calculated as here descent would go to minus infinity instead of directly converging to global maxima. Thus, it contradicts the point of getting stuck in a saddle point for a function like $$(x^2 - y^2)$$.

The calculation of the gradient is the same no matter the optimization algorithm you use. Gradient descent is a method to find (local) minima of a function. If no such minimum exists, any algorithm to find it will render wrong results or never converge. $$-x^2$$ only has a maximum, so looking for a minimum does not make much sense and to find the maximum you would need to run gradient ascent, i.e. going uphill.
Getting stuck in a saddle point is a different issue. For $$f(x,y) = x^2 - y^2$$ the gradient is $$\nabla f(x,y) = 2[x, -y]^T$$ and thus equals zero at $$x=y=0$$, so gradient ascent/descent would just sit there and not move. But $$f(0,0) = 0$$ is neither a minimum nor a maximum ($$f$$ actually doesn't have either of these).
• if I use ∇f(x,y) (where f(x,y)=x2$−y2$ is my loss function) to calculate gradient descent with respect to each dimension, which is x, y in this case, I will reach local minima with respect to x, but what about y? Do we calculate gradient ascent wrt y? But that doesn't make any sense. Oct 27 '19 at 14:07