# Homework Help: Extrems values on armonic function

Tags:
1. Dec 26, 2017

### krakatoa

1. The problem statement, all variables and given/known data

f
is an armonic function and C2(R2)
a) Suposse a point P0 / fxx(P0) > 0. Prove that P0 is not a extreme value.
b) consider D = {(x,y) / x2 + y2 < 1} and suposse fxx > 0 for all (x,y) in D. Prove that: if f(x,y) = 0 in x2 + y2 = 1, so f(x,y) = 0 for all (x,y) in D.

2. Relevant equations

In armonic funcions laplacian = 0

3. The attempt at a solution

I intuit that P0 is a saddle point, and I try to show this in the hessian matrix and I can´t

2. Dec 26, 2017

### StoneTemplePython

3. Dec 26, 2017

### krakatoa

With the hessian i can classify the critical points, and i know that fxx > 0 and fyy < 0 (because f is an armonic function)

4. Dec 26, 2017

### StoneTemplePython

so you know about the diagonal components of your 2 x 2 Hessian, $\mathbf H$. (and they cannot evaluate to zero, why?)

Claim:
1.) I can find some $\mathbf x_1$, where $\mathbf x_1^* \mathbf H \mathbf x_1 \gt 0$.
2.) I can find some $\mathbf x_2$, where $\mathbf x_2^* \mathbf H \mathbf x_2 \lt 0$.

Why? You should in fact be able to come up with these $\mathbf x$'s.

This quadratic form argument is enough to prove a saddle point. (Why?)

5. Dec 26, 2017

### krakatoa

and why evaluate on zero? if i can see that fxx = - fyy implies that is a saddle point I can do the second part... sory I am confused

6. Dec 26, 2017

### StoneTemplePython

I'm not sure what this means. Your posts in general are too light on details and a bit tough to figure out.

Hmmm, if you're not familiar with quadratic forms: what tests are you aware of to prove something is a saddle point? Please list them, preferably with blurb on the intuition for why it makes sense. There are a lot of recipes you can follow, but having some intuition for why... makes things a lot easier. (Are you aware of what a taylor polynomial would look like here if you truncate it to have a quadratic remainder? This is probably the most direct way of making sense of these different saddle point tests.)

7. Dec 26, 2017

### krakatoa

I usually use the sylvester´s criterion (or see the eigenvalues from the hessian matrix) so I try to use this automatically, I thought it would be more simple. The problem is from a curse that not uses linear algebra knowledge.
sorry my english is not good.

8. Dec 26, 2017

### StoneTemplePython

If you know about Sylvester's criterion, you're in good shape.

You don't actually directly need the eigenvalues, but let's go that route.

How does the $trace\big(\mathbf H\big)$ relate to the Laplacian operator? What does the trace tell you about the eigenvalues of $\mathbf H$?

9. Dec 26, 2017

### krakatoa

The trace IS the laplacian operator, and is equal to zero, and if the trace is equal to zero so H have not eigenvalues...
if H have not eigenvalues, what can i said about the saddle point?

10. Dec 26, 2017

### StoneTemplePython

Yes. This is a very important.

Keep in mind we're dealing with scalars in $\mathbb C$. In the case of finite dimensions there are always eigenvalues with complex numbers. Even better: with Hermitian (or in reals, symmetric) matrices the eigenvalues are always real. (Why?) This means if you have a real symmetric Hessian there are always eigenvalues. (Your Hessian should always have this kind of symmetry unless you have issues with Schwarz's / Clairaut's theorem.)

so what you have is

$trace\big(\mathbf H\big) = \lambda_1 + \lambda_2 = 0$

We know each eigenvalue is real. What does this tell you about the signs of the eigenvalues? If you know about the signs, what does that tell you about saddle points?

- - - -
I know you said this course doesn't use linear algebra here, but I think you would benefit from improving knowledge of spectral theory. In particular going through chapters 4-7 of Linear Algebra Done Wrong would be quite helpful. The book is outstanding and freely available from the author here: https://www.math.brown.edu/~treil/papers/LADW/LADW_2017-09-04.pdf

11. Dec 26, 2017

### krakatoa

Great, you really help me, the eigenvalues have diferent sign, and it means that is a undefinited cuadratic form, so, is a saddle point, its okey?
and about the b) part, if f(x,y)=0, how fxx > 0 ?
thanks!!

Last edited: Dec 26, 2017
12. Dec 26, 2017

### StoneTemplePython

I would say it's indefinite (which is what I think you said here albeit with a little language translation issues), i.e. we can find a case where $\mathbf x_1^* \mathbf H \mathbf x_1 \gt 0$ and $\mathbf x_2^* \mathbf H \mathbf x_2 \lt 0$.

So yes you have it right that it is a saddle point.

The only remaining nit: how do you know for certain that you have one positive eigenvalue and one negative eigenvalue? Put differently, how do you know it isn't the case where $0 = \lambda_1 = \lambda_2$?

(Hint: real symmetric matrices, and complex Hermitian ones are always diagonalizable... and we know that the top left corner of $\mathbf H$, contains $f_{xx} > 0$).

Last edited: Dec 27, 2017