# Determine the nature of a critical point (Lagrange multipliers)

Let $F \colon \mathbb R^2 \to \mathbb R$ be the function $$F(x,y):=xye^x + ye^y - e^x+1$$ and denote with $C$ the set of zeroes of $F$, i.e. $C:=\{(x,y) \in \mathbb R^2 : F(x,y)=0\}$. Let also $f \colon \mathbb R^2 \to \mathbb R$ be a function which is $C^2$ in a neighbourhood of $(0,0)$ and such that $$\nabla f(0,0) = (-2,2), \qquad \qquad H_f(0,0) = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}.$$ Question: is $(0,0)$ a minimum for $f$ on $C$?

Well, I show you what I have done so far. First of all, some routine calculations yield $$\nabla F(0,0) = (-1,1) \qquad \qquad H_F(0,0) = \begin{pmatrix} -1 & 1 \\ 1 & 2 \end{pmatrix}$$

Indeed, the fact that $$\nabla f(0,0) = 2\nabla F(0,0)$$ does agree with the theory of Lagrange multipliers: the gradients are parellel, so I think that $(0,0)$ is an extremum for $f$ on $C$. The problem is how to classify it without any information on $f$: we have only its hessian matrix, which is - I suppose - the key to solve this. Both $H_f(0,0)$ and $H_F(0,0)$ are indefinite.

How can we establish the nature of the critical point $(0,0)$? Thanks in advance for your help.

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You have to figure out how the Hessian looks on the space orthogonal to $\nabla f$. –  Fabian Aug 16 '12 at 21:12
@Fabian Thanks for your comment. Why do I have to look a the space orthogonal to $\nabla f$? It is of course the subspace of $\mathbb R^{2}$ generated by $(1,1)$ but I do not know how to use this information. Would you mind explaining me a little bit, please? Thanks. –  Romeo Aug 16 '12 at 21:21

## 3 Answers

Suppose you have a parametrization $x=X(t), y=Y(t)$ of the curve $C$ near $(0,0)$, with $X(0)=0$ and $Y(0)=0$. Thus if $V$ and $A$ are the corresponding velocity and acceleration vectors, $$\dfrac{d}{dt} f(X(t),Y(t)) = X' f_x(X,Y) + Y' f_y(X,Y) = (\nabla f)\cdot V$$

$$\dfrac{d^2}{dt^2} f(X(t),Y(t)) = X'' f_x(X,Y) + Y'' f_y(X,Y) + (X')^2 f_{xx}(X,Y) + 2 X' Y' f_{xy}(X,Y) + (Y')^2 f_{yy}(X,Y) = (\nabla f)\cdot A + V^T H(f) V$$

Similarly for $f$ replaced by $F$, but in this case $\dfrac{d}{dt} F(X(t),Y(t)) = \dfrac{d^2}{dt^2} F(X(t),Y(t)) = 0$.

Now since $\nabla f(0,0) = 2 \nabla F(0,0)$, $(\nabla F(0,0)) \cdot V(0) = 0$ implies $\dfrac{d}{dt} f(X(t),Y(t))|_{t=0} = (\nabla f(0,0)) \cdot V(0) = 0$. And $$\dfrac{d^2}{dt^2} f(X(t),Y(t))|_{t=0} = \dfrac{d^2}{dt^2} f(X(t),Y(t))|_{t=0} - 2 \dfrac{d^2}{dt^2} F(X(t),Y(t))|_{t=0} = V(0)^T (H(f) - 2 H(F)) V(0)$$

Now $V(0)$ is orthogonal to $\nabla F(0,0) = (-1,1)$, so it is a multiple of $(1,1)$, and $$(1,1) \left( \pmatrix{2 & 1\cr 1 & 0\cr} - 2 \pmatrix{-1 & 1\cr 1 & 2\cr}\right) \pmatrix{1\cr 1\cr} = -2$$ so the second derivative is negative, and you have a local maximum.

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In order to find the Hessian, we need to know the change in $f$ to second order in the deviation from $(0,0)$.

Expanding $F(x,y)$ around this point, we have up to second order $$F(x,y) = \nabla F(0,0)\begin{pmatrix} x \\y\end{pmatrix} + \tfrac12 \begin{pmatrix} x & y\end{pmatrix} H_F \begin{pmatrix} x \\y\end{pmatrix}.$$

Similarly, expanding $f(x,y)$ we obtain $$f(x,y) = \nabla f(0,0)\begin{pmatrix} x \\y\end{pmatrix} + \tfrac12 \begin{pmatrix} x & y\end{pmatrix} H_f \begin{pmatrix} x \\y\end{pmatrix}.$$

In the last step, we need to implement the constraint $F(x,y)=0$. To first order, we have $\nabla F(0,0) \begin{pmatrix} x \\y\end{pmatrix}=0$ and because $\nabla f \parallel \nabla F$, we have that $f(0,0)$ is an extremum. In particular, we have $y=x$ to first order.

To second order, we have $$\tfrac12 \begin{pmatrix} x & y\end{pmatrix} H_F \begin{pmatrix} x \\y\end{pmatrix} = \tfrac12 \begin{pmatrix} x & x\end{pmatrix} H_F \begin{pmatrix} x \\x\end{pmatrix} = \frac{3}{2} x^2.$$ In order that $F(x,y)=0$ to second order, we need to compensate this via the first term, so we have $y= x - \frac{3}{2} x^2$.

Plugging this into the expansion of $f(x,y)$, we obtain a term from the gradiant and a term from the Hessian (Tunococ just considered the term from the Hessian). In total, we have $$f(x,y) = -3 x^2 + 2 x^2 =-x^2.$$ So it is a local maximum.

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I don't understand how you came up with the correction factor $-\frac 32 x^2$. With your $y$, you get $y'(0) = -2$. That means you're going in the direction violating the constraint. –  Tunococ Aug 16 '12 at 23:07
@Tunococ: $y'(x) =1-3 x$, setting $x=0$, i have $y'(0)=1$. –  Fabian Aug 17 '12 at 5:27

Edit: I did not take into account the non-linearity of the constraint so it was wrong. This is the wrong version:

The complementary space of $\nabla F(0, 0)$ is spanned by $(1, 1)$. The (unscaled) reduced Hessian of $f$ is a $1$-by-$1$ matrix: $\begin{pmatrix}1 & 1\end{pmatrix}\begin{pmatrix}2 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix} = 4 > 0$, so it is a minimum.

For correct answer, see Robert's answer above.

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Thanks for your answer, but I haven't understood. Why are you looking at the quadratic form given by the hessian on the complementary space of $\nabla f(0,0)$? Are my calculations of $\nabla F$ and $H_F$ useless? Thanks a lot. –  Romeo Aug 16 '12 at 21:24
@Tunococ: I don't think that this answer is completely correct. To second order you should take into account that the space with $F(x,y)=0$ is curved. –  Fabian Aug 16 '12 at 21:41
Sorry I was wrong. I switched $f$ and $F$. I've fixed that. –  Tunococ Aug 16 '12 at 22:35