Extrems values on armonic function

[krakatoa]

Homework Statement

f[/B] is an armonic function and C²(R²)
a) Suposse a point P₀ / f_xx(P₀) > 0. Prove that P₀ is not a extreme value.
b) consider D = {(x,y) / x² + y² < 1} and suposse f_xx > 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.

Homework Equations

In armonic funcions laplacian = 0

The Attempt at a Solution

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

 

[StoneTemplePython]

You need to show a bit more of work here. This may help. http://math.mit.edu/~jorloff/suppnotes/suppnotes02/v7.pdf

How can you relate the laplacian operator to coordinates in the Hessian?

 

[krakatoa]

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

 

[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?)

 

[krakatoa]

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

 

[StoneTemplePython]

and why evaluate on zero?

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

if i can see that f_xx = - f_yy implies that is a saddle point I can do the second part... sory I am confused

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.)

 

[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.

 

[StoneTemplePython]

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.

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 $\mathbf H$?

 

[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?

 

[StoneTemplePython]

The trace IS the laplacian operator, and is equal to zero

Yes. This is a very important.

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?

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

 

[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 f_xx > 0 ?
thanks!

 

[StoneTemplePython]

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?
thanks!

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 $f_{xx} > 0$).