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Classifying the stationary points of $f(x, y) = 4xy-x^4-y^4 $

2

$begingroup$

$f(x, y) = 4xy-x^4-y^4 $

The gradient of this function is $0$ in $(-1, -1), (0, 0),(1, 1)$

I tried to compute the determinant of the Hessian Matrix, but it's 0 for every point, I always get a null eigenvalue for each point, so no matter what I can't find out what kind of points these are.

For $(0, 0)$, since $f(0, 0) = 0$, $f(x, 0) = -x^4$ and $f(0, y) = -y^4$. Since both will always be negative, isn't that supposed to be a maximum or a minimum point? According to Wolfram it's a saddle point and I can't understand why.

multivariable-calculus maxima-minima hessian-matrix

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3244937

asked Apr 6 at 9:59

Frost832Frost832

325

$endgroup$

add a comment | 

2

$begingroup$

$f(x, y) = 4xy-x^4-y^4 $

The gradient of this function is $0$ in $(-1, -1), (0, 0),(1, 1)$

I tried to compute the determinant of the Hessian Matrix, but it's 0 for every point, I always get a null eigenvalue for each point, so no matter what I can't find out what kind of points these are.

For $(0, 0)$, since $f(0, 0) = 0$, $f(x, 0) = -x^4$ and $f(0, y) = -y^4$. Since both will always be negative, isn't that supposed to be a maximum or a minimum point? According to Wolfram it's a saddle point and I can't understand why.

multivariable-calculus maxima-minima hessian-matrix

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3244937

asked Apr 6 at 9:59

Frost832Frost832

325

$endgroup$

add a comment | 

$begingroup$

$f(x, y) = 4xy-x^4-y^4 $

The gradient of this function is $0$ in $(-1, -1), (0, 0),(1, 1)$

I tried to compute the determinant of the Hessian Matrix, but it's 0 for every point, I always get a null eigenvalue for each point, so no matter what I can't find out what kind of points these are.

For $(0, 0)$, since $f(0, 0) = 0$, $f(x, 0) = -x^4$ and $f(0, y) = -y^4$. Since both will always be negative, isn't that supposed to be a maximum or a minimum point? According to Wolfram it's a saddle point and I can't understand why.

multivariable-calculus maxima-minima hessian-matrix

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3244937

asked Apr 6 at 9:59

Frost832Frost832

325

$endgroup$

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3244937

asked Apr 6 at 9:59

Frost832Frost832

325

share|cite|improve this question

edited Apr 6 at 12:19

YuiTo Cheng

2,3244937

asked Apr 6 at 9:59

Frost832Frost832

325

edited Apr 6 at 12:19

YuiTo Cheng

2,3244937

edited Apr 6 at 12:19

YuiTo Cheng

2,3244937

asked Apr 6 at 9:59

Frost832Frost832

325

asked Apr 6 at 9:59

Frost832Frost832

325

2 Answers
2

active

oldest

votes

2

$begingroup$

Note that:

• 
  $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $xinleft(-sqrt 2,sqrt2right)setminus0$;


• 
  $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $(1, 1)$, I computed the determinant $Delta f(x, y) = f(x, y) - f(1, 1) = 4xy-x^4-y^4-2)$, now to study its sign, can I consider $f(1, y) > 0$ and $f(x, 1) > 0$ ?
  $endgroup$
  – Frost832
  Apr 6 at 13:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At tht point, all eigenvalues of the Hessean are smaller than $0$. Therefore, $f$ has a local maximum there.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much.
  $endgroup$
  – Frost832
  Apr 6 at 13:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:57
  

  
  
  
  

add a comment | 

2

$begingroup$

Note that the gradient of $f$ is given by $nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=beginpmatrix-12x^2 &4\4 &-12y^2endpmatrix$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=beginpmatrix0 &4\4 &0endpmatrix, H_f(1,1)=beginpmatrix-12 &4\4 &-12endpmatrix,H_f(-1,-1)=beginpmatrix-12 &4\4 &-12endpmatrix$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

$endgroup$

add a comment | 

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2

$begingroup$

Note that:

• 
  $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $xinleft(-sqrt 2,sqrt2right)setminus0$;


• 
  $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $(1, 1)$, I computed the determinant $Delta f(x, y) = f(x, y) - f(1, 1) = 4xy-x^4-y^4-2)$, now to study its sign, can I consider $f(1, y) > 0$ and $f(x, 1) > 0$ ?
  $endgroup$
  – Frost832
  Apr 6 at 13:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At tht point, all eigenvalues of the Hessean are smaller than $0$. Therefore, $f$ has a local maximum there.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much.
  $endgroup$
  – Frost832
  Apr 6 at 13:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:57
  

  
  
  
  

add a comment | 

2

$begingroup$

Note that:

• 
  $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $xinleft(-sqrt 2,sqrt2right)setminus0$;


• 
  $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $(1, 1)$, I computed the determinant $Delta f(x, y) = f(x, y) - f(1, 1) = 4xy-x^4-y^4-2)$, now to study its sign, can I consider $f(1, y) > 0$ and $f(x, 1) > 0$ ?
  $endgroup$
  – Frost832
  Apr 6 at 13:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  At tht point, all eigenvalues of the Hessean are smaller than $0$. Therefore, $f$ has a local maximum there.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much.
  $endgroup$
  – Frost832
  Apr 6 at 13:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Apr 6 at 13:57
  

  
  
  
  

add a comment | 

$begingroup$

Note that:

• 
  $f(x,x)=4x^2-2x^4=2x^2(2-x^2)$ and therefore $f(x,x)>0$ of $xinleft(-sqrt 2,sqrt2right)setminus0$;


• 
  $f(x,0)=-x^4<0$ (unless $x=0$).

Therefore, yes, $(0,0)$ is a saddle point.

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

$endgroup$

share|cite|improve this answer

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

answered Apr 6 at 10:20

José Carlos SantosJosé Carlos Santos

173k23133242

2

$begingroup$

Note that the gradient of $f$ is given by $nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=beginpmatrix-12x^2 &4\4 &-12y^2endpmatrix$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=beginpmatrix0 &4\4 &0endpmatrix, H_f(1,1)=beginpmatrix-12 &4\4 &-12endpmatrix,H_f(-1,-1)=beginpmatrix-12 &4\4 &-12endpmatrix$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

$endgroup$

add a comment | 

2

$begingroup$

Note that the gradient of $f$ is given by $nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=beginpmatrix-12x^2 &4\4 &-12y^2endpmatrix$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=beginpmatrix0 &4\4 &0endpmatrix, H_f(1,1)=beginpmatrix-12 &4\4 &-12endpmatrix,H_f(-1,-1)=beginpmatrix-12 &4\4 &-12endpmatrix$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

$endgroup$

add a comment | 

$begingroup$

Note that the gradient of $f$ is given by $nabla f(x,y)=(4y-4x^3,4x-4y^3)$ and thus the Hessian of $f$ is given by

$$H_f(x,y)=beginpmatrix-12x^2 &4\4 &-12y^2endpmatrix$$

Now, we look at the Hessian for $(0,0),(1,1),(-1,-1)$:

$$H_f(0,0)=beginpmatrix0 &4\4 &0endpmatrix, H_f(1,1)=beginpmatrix-12 &4\4 &-12endpmatrix,H_f(-1,-1)=beginpmatrix-12 &4\4 &-12endpmatrix$$

Now, by checking the principal minors of the matrices, you can note that $H_f(1,1)$ and $H_f(-1,-1)$ are negative definite.

Thus, by the sufficient criteria for local extrema, we have that $(1,1)$ and $(-1,-1)$ are local maxima.

In a similar manner, $H_f(0,0)$ is indefinite, and thus $(0,0)$ is indeed a saddle point.

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

$endgroup$

share|cite|improve this answer

edited Apr 6 at 10:27

answered Apr 6 at 10:20

blubblub

3,299929

answered Apr 6 at 10:20

blubblub

3,299929

answered Apr 6 at 10:20

blubblub

3,299929