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Classifying Coupled Polynomial Critical Points

Solve the gradient equations for coupled polynomials, classify the resulting points using the Hessian, and recognize when higher-order analysis is needed.

Intermediate 11 min read

What You'll Learn

  • Set up and solve a coupled three-variable critical-point system.
  • Check that the solution set includes every real critical point.
  • Use Hessian definiteness to distinguish local minima, local maxima, and saddle points.
  • Apply Sylvester’s criterion to test whether a symmetric Hessian is positive definite.
  • Recognize when a degenerate Hessian requires higher-order analysis.
  • Verify critical points and classifications by substitution and directional checks.

From a Polynomial to Its Critical Points

A polynomial in three variables can couple its variables through terms such as \(xy\), \(yz\), or \(x^2y\). Its critical points are found by solving all three first-partial-derivative equations together. Because those equations may share variables, solving them one at a time without preserving their relationships can miss solutions or introduce false ones.

The task has two parts: find every real solution of the gradient system, then classify each solution. The Hessian gives a decisive classification when it is positive definite, negative definite, or indefinite. If it is only semidefinite, the second derivative test may not decide the issue, and higher-order behavior must be examined.

Definition: For a differentiable function \(f:\mathbb{R}^3\to\mathbb{R}\), a critical point is a point \((a,b,c)\) where all three first partial derivatives vanish: $$ \nabla f(a,b,c) = \left\langle f_x(a,b,c),f_y(a,b,c),f_z(a,b,c)\right\rangle = \mathbf{0}. $$ The Hessian \(H_f(a,b,c)\) is the symmetric matrix of second partial derivatives evaluated at that point.

Polynomials have continuous derivatives of every order, so their Hessians are defined everywhere. A local minimum or maximum means that the function value is respectively no greater than, or no less than, nearby values. A saddle point has both larger and smaller function values arbitrarily close to it.

A Reliable Workflow

For a coupled polynomial, solve the gradient equations as a system and keep track of all branches created by factoring or substitution. Once the real solutions are found, evaluate the Hessian at each one. For a symmetric \(3\times3\) matrix, Sylvester’s criterion says it is positive definite exactly when its three leading principal minors are positive; it is negative definite exactly when their signs alternate, beginning with negative. An indefinite Hessian has both positive and negative directions.

1
Find the gradient.
Compute \(f_x\), \(f_y\), and \(f_z\), retaining every variable-dependent term.
2
Solve the full system.
Set all three partial derivatives equal to zero and determine every real solution.
3
Evaluate the Hessian.
Substitute each critical point into the second partial derivatives.
4
Classify or investigate further.
Use definiteness or indefiniteness when decisive; if the Hessian is semidefinite, examine higher-order terms or nearby values.

Why the Hessian Test Works

At a critical point \(\mathbf{p}\), the gradient term in the second-order Taylor expansion vanishes. For a twice continuously differentiable function, this gives

$$ f(\mathbf{p}+\mathbf{h})-f(\mathbf{p}) = \frac{1}{2}\mathbf{h}^{\top}H_f(\mathbf{p})\mathbf{h} + o(\|\mathbf{h}\|^2) \quad\text{as }\mathbf{h}\to\mathbf{0}. $$

Here the remainder divided by \(\|\mathbf{h}\|^2\) tends to zero. If the Hessian is positive definite, its quadratic form is bounded below by \(\lambda\|\mathbf{h}\|^2\) for some \(\lambda>0\). For sufficiently small nonzero \(\mathbf{h}\), the remainder has magnitude less than \(\lambda\|\mathbf{h}\|^2/4\). The positive quadratic term therefore dominates, and the point is a strict local minimum. Applying the same reasoning to a negative definite Hessian gives a strict local maximum.

If the Hessian has both a positive and a negative direction, take small displacements along those directions. The quadratic term has opposite signs, and in each direction it dominates the smaller-order remainder sufficiently near the point. Thus function values both above and below the critical value occur nearby: the point is a saddle. If the Hessian is semidefinite, this argument does not settle the classification, because the quadratic term may vanish in some directions.

Worked Example: A Coupled Quadratic with One Critical Point

Worked Example: A Coupled Quadratic with One Critical Point

Find and classify all critical points of \(f(x,y,z)=x^2+y^2+z^2+xy+yz\).

First compute the gradient and set it equal to zero:

$$ f_x=2x+y,\qquad f_y=x+2y+z,\qquad f_z=y+2z. $$

The equations give \(y=-2x\) and \(y=-2z\), so \(z=x\). Substituting \(y=-2x\) and \(z=x\) into the middle equation gives

$$ x+2(-2x)+x=-2x=0. $$

Thus \(x=0\), \(y=0\), and \(z=0\). These values satisfy all three gradient equations, so the only critical point is the origin.

The Hessian is constant:

$$ H_f= \begin{pmatrix} 2&1&0\\ 1&2&1\\ 0&1&2 \end{pmatrix}. $$

Its leading principal minors are \(2\), \(2\cdot2-1\cdot1=3\), and \(2(4-1)-1(2)=4\). They are all positive, so Sylvester’s criterion shows that the Hessian is positive definite. The origin is a strict local minimum. Direct substitution checks the critical point: \(\nabla f(0,0,0)=\mathbf{0}\), and \(f(0,0,0)=0\).

Worked Example: Three Critical Points with Different Classifications

Worked Example: Three Critical Points with Different Classifications

Find and classify all critical points of \(f(x,y,z)=(x^2-1)^2+(y-x)^2+(z-y)^2\).

Differentiate each term:

$$ f_x=4x(x^2-1)-2(y-x),\qquad f_y=2(y-x)-2(z-y),\qquad f_z=2(z-y). $$

From \(f_z=0\), we obtain \(z=y\). Then \(f_y=0\) gives \(y=x\). With \(y=x\), the remaining equation becomes \(4x(x^2-1)=0\), or \(4x(x-1)(x+1)=0\). Hence the complete list of critical points is \((0,0,0)\), \((1,1,1)\), and \((-1,-1,-1)\). Substitution into the gradient equations confirms that all three satisfy them.

The Hessian at a general point is

$$ H_f(x,y,z)= \begin{pmatrix} 12x^2-2&-2&0\\ -2&4&-2\\ 0&-2&2 \end{pmatrix}. $$

At either point with \(x=\pm1\), this becomes \(\begin{pmatrix}10&-2&0\\-2&4&-2\\0&-2&2\end{pmatrix}\). Its leading principal minors are \(10\), \(40-4=36\), and \(10(8-4)-(-2)(-4)=32\). The Hessian is positive definite, so both points are strict local minima. Indeed, the function is a sum of squares and equals zero at each, confirming the minimum value there.

At the origin, the Hessian is \(\begin{pmatrix}-2&-2&0\\-2&4&-2\\0&-2&2\end{pmatrix}\). Its quadratic form has both signs: along the \(x\)-axis, the function changes by

$$ f(t,0,0)-f(0,0,0) = (t^2-1)^2+t^2-1 = -t^2+t^4<0 \qquad (0<|t|<1), $$

while along the \(y\)-axis,

$$ f(0,t,0)-f(0,0,0) = 1+t^2+t^2-1 = 2t^2>0 \qquad (t\ne0). $$

Both changes occur arbitrarily close to the origin, so it is a saddle point. The critical value is \(f(0,0,0)=1\), and the calculations verify that nearby values lie on both sides of it.

Worked Example: A Degenerate Point That Is Still a Minimum

Worked Example: A Degenerate Point That Is Still a Minimum

Consider the coupled polynomial \(f(x,y,z)=(x-y)^4+(y-z)^4+(x+y+z)^4\). Find and classify every critical point.

Set \(u=x-y\), \(v=y-z\), and \(w=x+y+z\). By the chain rule,

$$ \nabla f = 4 \begin{pmatrix} 1&0&1\\ -1&1&1\\ 0&-1&1 \end{pmatrix} \begin{pmatrix} u^3\\v^3\\w^3 \end{pmatrix}. $$

The matrix has determinant \(3\), so it is invertible. Therefore \(\nabla f=\mathbf{0}\) if and only if \(u^3=v^3=w^3=0\), which is equivalent to \(u=v=w=0\). These equations imply \(x=y=z\) and \(x+y+z=0\), hence \(x=y=z=0\). Substitution confirms that the origin is a critical point, and there are no others.

At the origin, every second derivative vanishes: each term in the Hessian is a constant times the square of one of the linear forms \(u,v,w\), and those forms are zero there. Thus \(H_f(0,0,0)\) is the zero matrix, and the second derivative test is inconclusive.

Nevertheless, the origin is a strict global minimum. Each fourth power is nonnegative. If their sum is zero, then all three linear forms are zero; as just shown, that forces \(x=y=z=0\). Therefore \(f(x,y,z)>0=f(0,0,0)\) at every other point. This verifies the classification even though the Hessian provides no decisive information.

A Common Pitfall: Using Only the Hessian Determinant

The determinant alone does not determine whether a three-variable Hessian is positive definite. For instance, let \(g(x,y,z)=-x^2-y^2+z^2+xy\). At the origin, its gradient is zero and its Hessian is

$$ H_g(0,0,0)= \begin{pmatrix} -2&1&0\\ 1&-2&0\\ 0&0&2 \end{pmatrix}, \qquad \det H_g=(4-1)\cdot2=6>0. $$

It would be incorrect to conclude that the origin is a minimum merely because this determinant is positive. In fact, \(g(t,0,0)=-t^2<0\), while \(g(0,0,t)=t^2>0\) for every nonzero \(t\). The origin is a saddle. Check definiteness using the full Hessian, such as its principal minors or its directions of curvature, rather than relying on the determinant alone.

Completeness and the Role of Higher-Order Terms

A classification is complete only when the gradient system has been solved for all real critical points. In the second example, the equations \(z=y\) and \(y=x\) reduced the system to one factored equation, and every root of that equation was retained. A factor can represent a whole branch of solutions, so discarding a factor or dividing by an expression that might be zero can lose critical points.

After solving, verify each candidate in the original gradient equations. Then compute the Hessian at that point, not at a nearby or generic point. If it is positive definite, negative definite, or indefinite, the second derivative test gives the corresponding local classification. If it is semidefinite, do not treat that result as a classification: the fourth-degree example shows that a zero Hessian can occur at a strict minimum. Inspect the polynomial’s higher-order behavior or compare nearby values instead.

Takeaway: For a coupled polynomial, solve the entire gradient system, check every real candidate, and classify each point using its Hessian. A semidefinite Hessian leaves the classification unresolved; it does not mean the point is neither a minimum nor a maximum.

Check Your Understanding

For each question, distinguish solving for critical points from classifying them.

  1. What three equations define a critical point of a differentiable function of three variables?
  2. For a symmetric \(3\times3\) Hessian, what does Sylvester’s criterion require for positive definiteness?
  3. Why does an indefinite Hessian imply that a critical point is a saddle?
  4. Why does a semidefinite Hessian not by itself establish a local minimum or maximum?
  5. For \(g(x,y,z)=-x^2-y^2+z^2+xy\), what directional checks show that the origin is a saddle?

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