10. Critical Points

Local Maximum

Let $f(x,y)$ be a function defined on a subset $A\subset {\mathbb{R}}^2$

Point $(x_0,y_0)\in A$ is local maximum of $f$ if there is

$$\text{open disc }{D}_r(x_0,y_0)\text{ of radius }r>0\text{ s.t. }$$ $$f(x,y)\le f(x_0,y_0)\forall (x,y)\in D_r(x_0,y_0)\cap A$$

Note that global maximum is for all $(x,y)\in A$

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Local Minimum

Let $f(x,y)$ be a function defined on a subset $A\subset {\mathbb{R}}^2$

Point $(x_0,y_0)\in A$ is local minimum of $f$ if there is

$$\text{open disc }{D}_r(x_0,y_0)\text{ of radius }r>0\text{ s.t. }$$ $$f(x,y)\ge f(x_0,y_0)\forall (x,y)\in D_r(x_0,y_0)\cap A$$

Note that global minimumis for all $(x,y)\in A$

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Gradient Vector on Critical Points

Suppose that $f(x,y)$ is defined on an open subset $U$ with

• continuous partial derivatives
• $(x_0,y_0)\in U$ being a local maximum or local minimum

Then

$$\frac{\partial f}{\partial x}(x_0,y_0)=\frac{\partial f}{\partial y}(x_0,y_0)=0$$

Hence the gradient vector $\nabla f(x_0,y_0)=0$

Proof

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Type of Critical Point from Partial Derivatives

Suppose that $f(x,y)$ is defined on an open subset $U$ with

• continuous partial derivatives

• $(x_0,y_0)\in U$ is a critical point ($\nabla f(x_0,y_0)=0$)

1. If

$$\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)\frac{{\partial }^{2}f}{\partial y^2}(x_0,y_0)-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}(x_0,y_0)\right)}^2>0,\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)<0$$

Then $(x_0,y_0)$ is a local maximum

2. If

$$\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)\frac{{\partial }^{2}f}{\partial y^2}(x_0,y_0)-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}(x_0,y_0)\right)}^2>0,\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)>0$$

Then $(x_0,y_0)$ is a local maximum

Classifying Stationary Points

All stationary points have $f_x=0$ and $f_y=0$ with

1. If $f_{xx}{f}_{yy}-f_{xy}^2>0$ and $f_{xx}>0$ (or $f_{yy}>0$) then

$$\text{local maximum}$$

2. If $f_{xx}{f}_{yy}-f_{xy}^2>0$ and $f_{xx}<0$ (or $f_{yy}<0$) then

$$\text{local maximum}$$

3. If $f_{xx}{f}_{yy}-f_{xy}^2<00$ then

$$\text{saddle point}$$

Proof

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Hessian Matrix $H$

For a function $f(x_1,\cdots ,x_n)$ of $n$ variables with continuous partial derivatives to second order then

It is a $n\times n$ symmetric matrix with entry $\frac{{\partial }^{2}y}{\partial x_i\partial x_j}$ at the $i$th row and $j$th column

Hence

$$\mathbf{H}(x_1,\cdot ,x_n)=\left(\begin{array}{ccc}\frac{{\partial }^{2}f}{\partial x_1^2}& \cdots & \frac{{\partial }^{2}f}{\partial x_1\partial x_n}\\ ⋮& \ddots & ⋮\\ \frac{{\partial }^{2}f}{\partial x_n\partial x_1}& \cdots & \frac{{\partial }^{2}f}{\partial x_n^2}\end{array}\right)$$

Hessian Matrix on Classifying Stationary Points

Suppose $f(\mathbf{x})$ is a function with $n$ variables so

$$\mathbf{x}=(x_1\cdots ,x_n)$$

which is defined on an open subset $U\subset {\mathbb{R}}^n$ which has

• continuous partial derivatives up to the second order

Let $\mathbf{a}=(a_1,\cdots ,a_n)$ be a critical point such that $\nabla f(\mathbf{a})=0$

1. If the Hessian Matrix $\mathbf{H}(\mathbf{a})$ is positive definite then

$$\mathbf{a}\text{ is a local mininum of }f$$

2. If the Hessian Matrix $\mathbf{H}(\mathbf{a})$ is negative definite then

$$\mathbf{a}\text{ is a local maximum of }f$$

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