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Solving Two-Variable Gradient Systems

Solve the two partial-derivative equations systematically, and verify that each proposed stationary point satisfies both equations.

Intermediate 10 min read

What You'll Learn

  • Write the stationary-point condition as a system of two partial-derivative equations
  • Use substitution to reduce a system to one equation in one variable
  • Factor carefully while retaining every possible solution branch
  • Use elimination to create simpler equations without treating them as the whole system
  • Check each candidate in both original gradient equations
  • Distinguish finding stationary points from classifying them as extrema

Introduction

A stationary point of a differentiable function of two variables is found by solving two equations at once: both first partial derivatives must be zero. The calculus gives the equations; algebra determines their simultaneous solutions.

This step can be more subtle than setting each derivative equal to zero separately. The variables are linked across the equations, and a solution of only one equation is not enough. Factoring, substitution, and elimination help organize the work, but every proposed point must ultimately satisfy both original equations.

The Stationary-Point System

Definition: Let \(f(x,y)\) be differentiable at an interior point \((a,b)\) of its domain. The point \((a,b)\) is a stationary point of \(f\) if \(f_x(a,b)=0\) and \(f_y(a,b)=0\). Equivalently, \(\nabla f(a,b)=\mathbf{0}\).

Thus, to find stationary points, calculate the first partial derivatives and solve the system

$$ f_x(x,y)=0, \qquad f_y(x,y)=0. $$

For a differentiable function on an open domain, Fermat’s necessary condition says that any interior local maximum or minimum must satisfy these equations. The condition is necessary, not sufficient: solving the system finds stationary points, but does not by itself tell us whether a point is a maximum, a minimum, or neither. That classification requires additional analysis.

The equations must be solved over the part of the plane where \(f\) is defined. Polynomial equations are defined everywhere in \(\mathbb{R}^2\), so their stationary-point systems are normally solved over real \(x\) and \(y\). Algebraic solutions involving nonreal coordinates are not points in the real plane.

Worked Example: Substitution and a Shared Variable

Worked Example: Substitution and a Shared Variable

Find the stationary points of \(f(x,y)=x^3+y^3-3xy\).

First compute the partial derivatives:

$$ f_x=3x^2-3y, \qquad f_y=3y^2-3x. $$

Set each equal to zero and divide both equations by \(3\):

$$ x^2-y=0, \qquad y^2-x=0. $$

The first equation gives \(y=x^2\). Substitute this into the second equation:

$$ (x^2)^2-x=0, \qquad x^4-x=0, \qquad x(x^3-1)=0. $$

Therefore \(x=0\) or \(x^3=1\). Over the real numbers, the latter gives \(x=1\), since \(x^3-1=(x-1)(x^2+x+1)\) and \(x^2+x+1=(x+\tfrac12)^2+\tfrac34>0\). For \(x=0\), the equation \(y=x^2\) gives \(y=0\). For \(x=1\), it gives \(y=1\).

Now check both candidates in the original derivative equations. At \((0,0)\), \(f_x=3(0)^2-3(0)=0\) and \(f_y=3(0)^2-3(0)=0\). At \((1,1)\), \(f_x=3(1)^2-3(1)=0\) and \(f_y=3(1)^2-3(1)=0\). Both are stationary points.

Why Substitution Can Be Complete

Substitution works when one equation can be rearranged to express one variable in terms of the other. In the example, \(x^2-y=0\) is equivalent to \(y=x^2\). Replacing \(y\) by \(x^2\) in the second equation produces a one-variable equation whose real solutions include every solution of the original system.

The word equivalent matters. The equation \(y=x^2\) describes exactly the same pairs as \(x^2-y=0\), so the substitution neither discards nor adds solutions at that stage. Once the resulting \(x\)-values are found, the corresponding \(y\)-values must still be recovered and checked in the original system.

Key takeaway: A substitution reduces the number of variables, but it does not remove the need to verify the resulting pairs in both original equations.

Factoring Without Losing Solutions

Factoring often reveals several possible values or branches. Each factor must be considered. In particular, dividing by an expression that might be zero can discard solutions, so factor first and examine every factor before dividing.

Worked Example: Factoring and Keeping Every Root

Find the stationary points of \(g(x,y)=x^4+y^4-4xy\).

The partial derivatives are

$$ g_x=4x^3-4y, \qquad g_y=4y^3-4x. $$

Setting them equal to zero and dividing by \(4\) gives

$$ x^3-y=0, \qquad y^3-x=0. $$

The first equation gives \(y=x^3\). Substituting into the second gives

$$ (x^3)^3-x=0, \qquad x^9-x=0, \qquad x(x^8-1)=0. $$

The factor \(x=0\) gives \(y=0\). For the other factor, use

$$ x^8-1=(x^4-1)(x^4+1) =(x^2-1)(x^2+1)(x^4+1). $$

For real \(x\), both \(x^2+1\) and \(x^4+1\) are positive, so they cannot equal zero. Thus \(x^2-1=0\), giving \(x=1\) or \(x=-1\). Since \(y=x^3\), these give \((1,1)\) and \((-1,-1)\).

Check the original derivatives. At \((0,0)\), \(g_x=4(0)^3-4(0)=0\) and \(g_y=4(0)^3-4(0)=0\). At \((1,1)\), both derivatives equal \(4-4=0\). At \((-1,-1)\), \(g_x=4(-1)^3-4(-1)=-4+4=0\) and \(g_y=4(-1)^3-4(-1)=-4+4=0\). The stationary points are exactly \((0,0)\), \((1,1)\), and \((-1,-1)\).

Elimination and the Need to Return to the System

Elimination combines equations to obtain a condition that may be easier to use. For example, subtracting one equation from another can produce a factorized expression. Such a consequence is useful, but it need not be equivalent to the original pair by itself. It can identify branches to investigate without proving that every point on those branches solves both equations.

Here is the logic: if \(P(x,y)=0\) and \(Q(x,y)=0\), then \(P(x,y)-Q(x,y)=0\). The reverse implication does not generally hold. A point where \(P-Q=0\) might have \(P=Q\ne 0\). Therefore, after elimination suggests candidates, use the original equations to decide which candidates actually work.

Worked Example: Elimination into Two Branches

Find the stationary points of \(h(x,y)=\frac{x^3}{3}+xy+\frac{y^3}{3}\).

Its partial derivatives are

$$ h_x=x^2+y, \qquad h_y=x+y^2. $$

The stationary-point system is \(x^2+y=0\) and \(x+y^2=0\). Subtract the second left-hand side from the first:

$$ (x^2+y)-(x+y^2)=0, \qquad x^2-y^2-x+y=0, \qquad (x-y)(x+y-1)=0. $$

Every solution of the original system must therefore lie on at least one of two branches: \(x=y\) or \(x+y=1\).

On the branch \(x=y\), substitute into the first original equation:

$$ x^2+x=0, \qquad x(x+1)=0. $$

This gives \(x=0\) or \(x=-1\), so the candidates on this branch are \((0,0)\) and \((-1,-1)\).

On the branch \(x+y=1\), write \(y=1-x\) and substitute into the first original equation:

$$ x^2+(1-x)=0, \qquad x^2-x+1=0. $$

The discriminant is \((-1)^2-4(1)(1)=-3\), so this equation has no real solutions. This branch contributes no stationary points in \(\mathbb{R}^2\).

Finally, check the candidates in both original equations. At \((0,0)\), \(x^2+y=0\) and \(x+y^2=0\). At \((-1,-1)\), \(x^2+y=1-1=0\) and \(x+y^2=-1+1=0\). Thus the stationary points are \((0,0)\) and \((-1,-1)\).

A Reliable Solving Procedure

For a polynomial function, a consistent sequence of steps keeps the algebra complete and makes missed branches easier to detect.

1
Compute both partial derivatives.
Keep track of which derivative corresponds to \(x\) and which to \(y\).
2
Set up the simultaneous equations.
Write \(f_x=0\) and \(f_y=0\); neither equation can be ignored.
3
Choose an algebraic method.
Use substitution when one equation isolates a variable, or elimination when combining equations creates a useful condition.
4
Keep every factor and branch.
Do not divide by a variable or factor unless its zero case has already been checked.
5
Recover and verify the pairs.
Substitute each candidate into both original derivative equations and reject any pair that fails either one.

Common Mistake: Dividing Away a Solution

In the first example, substitution led to \(x(x^3-1)=0\). A tempting but incorrect next step is to divide by \(x\), leaving \(x^3-1=0\). That step assumes \(x\ne 0\), even though \(x=0\) is a possible solution of the factored equation.

The correction is to use the zero-product property: \(x(x^3-1)=0\) implies \(x=0\) or \(x^3-1=0\). The first case gives \((0,0)\), which checks in both original equations. The second gives \((1,1)\). Dividing by \(x\) without first handling \(x=0\) would incorrectly omit a stationary point.

A related error is to stop after finding a point that satisfies an eliminated equation. Elimination provides a necessary condition, not always a complete replacement for the original system. Verification in both original equations is the final safeguard.

Why Solving the System Matters

For an interior local extremum of a differentiable function, both partial derivatives must vanish. Consequently, solving the paired equations is an essential step in locating possible extrema. The algebra may produce no real solutions, one solution, several solutions, or—depending on the system—infinitely many. There is no general promise that a polynomial gradient system has just one stationary point.

Finding the points is not the same as classifying them. A stationary point might be a local extremum or might not be one. The system \(f_x=f_y=0\) identifies where to investigate; it does not determine the local shape of the surface. Keep these tasks separate: first solve and verify the system, then apply an appropriate classification method.

Takeaway: A stationary point is a real solution of both first-partial-derivative equations. Algebraic methods find candidates; substitution into the original system confirms which candidates qualify.

Check Your Understanding

For each question, focus on whether the proposed algebra preserves every solution of the system.

  1. What two equations must a stationary point of a differentiable function \(f(x,y)\) satisfy?
  2. After substituting \(y=x^2\) into a second equation, why should the resulting pairs still be checked in the original system?
  3. If factoring gives \(x(x-2)=0\), what values of \(x\) must be considered, and why is dividing by \(x\) unsafe?
  4. Why does subtracting two equations give a necessary condition for their common solutions but not necessarily an equivalent system?
  5. Does finding a stationary point establish that it is a local maximum or minimum? Explain briefly.

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