How Do You Solve Second Order Partial Derivatives at Critical Points?

[astenroo]

Homework Statement

Let f(x,y)= 8x$$^{4}$$ + y$$^{4}$$ -2xy$$^{2}$$, what is $$\partial^{2} f$$/$$\partial x^{2}$$ and $$\partial^{2} f$$/$$\partial y^{2}$$ for the points where $$\partial f$$/$$\partial x$$ = $$\partial f$$/$$\partial y$$ = 0?

Homework Equations

The Attempt at a Solution

The first partial derivative with respect to x: is 32x$$^{3}$$ - 2y$$^{2}$$
and the second with respect to x is: 96x$$^{2}$$

The first with respect to y is: 4y$$^{3}$$ - 4xy
The second with respect to y is: 12y$$^{2}$$ - 4x

And this is where I get stuck. I have no clue of how to continue from here, and I have tried to find rules for this kind of operation in Boas Mathematical methods in the physical sciences... Maybe I have the wrong literature...

 

[fzero]

You found

$$\frac{\partial f}{\partial x} = 32x^3-2y^2, ~\frac{\partial f}{\partial y} = 4y^3-4xy.$$

Can you find the points where they are both zero?

 

[astenroo]

You found

$$\frac{\partial f}{\partial x} = 32x^3-2y^2, ~\frac{\partial f}{\partial y} = 4y^3-4xy.$$

Can you find the points where they are both zero?

This is actually where I got stuck. I should put them both equal to zero, and i believe i would end up with an equation system with two unknowns?

 

[astenroo]

but solving for either seems to be a bit complex...

 

[astenroo]

So, no. I can't find the points where they are both zero. Not without help.

 

[micromass]

So we got the system

$$\left\{ \begin{array}{l}<br /> 16x^3=y^2\\<br /> y^3=xy<br /> \end{array}\right.$$

If y is zero, then from the first equation follows that also x is zero. So (0,0) is a solution.
If y is nonzero, then we got the equation

$$\left\{ \begin{array}{l}<br /> 16x^3=y^2\\<br /> y^2=x<br /> \end{array}\right.$$

So we have that $$16x^3=x$$. This gives us x=0 or x=1/4 or x=-1/4. You can now easily see what the corresponding y-values are.

 

[astenroo]

So we got the system

$$\left\{ \begin{array}{l}<br /> 16x^3=y^2\\<br /> y^3=xy<br /> \end{array}\right.$$

If y is zero, then from the first equation follows that also x is zero. So (0,0) is a solution.
If y is nonzero, then we got the equation

$$\left\{ \begin{array}{l}<br /> 16x^3=y^2\\<br /> y^2=x<br /> \end{array}\right.$$

So we have that $$16x^3=x$$. This gives us x=0 or x=1/4 or x=-1/4. You can now easily see what the corresponding y-values are.

Ouch...yep, needed to sleep on it. I recently came up with a similar solution. Thank you very much for the help.

 

[astenroo]

Ok so here goes, if x=0, then y=0 (0;0)
if x=.25, then y=.5 or -.5 (both satisfy the equations) (0.25;-0.5) and (0.25;0.5)
if x=-.25, then y=has only imaginary (or is it complex?) solutions.

For the y''=96x^2=0 for (0;0) and y''=6 if x=0.25

For the other partial:
12y^2-4x= 0 at (0;0)
12y^2-4x=2 at (0.25;0.5)
12y^2-4x=2 at (0.25;-0.5)

Can someone confirm this for me?

 

[micromass]

Seems alright...

 

[astenroo]

Seems alright...

Thank you very much for your help. This is one the best forums ever :)