Why Symmetry Can Simplify a Lagrange System
When an objective and its constraint treat two variables in the same way, their Lagrange equations often have a structure that can be exploited. Subtracting the equations may produce a factor such as \(x-y\). One branch then has \(x=y\), but the other factor gives additional possibilities that must also be solved.
The key point is that symmetry simplifies the system; it does not guarantee that every solution has equal coordinates. A point with \(x\ne y\) may be a valid stationary candidate, and the point obtained by swapping its coordinates may be another one.
The Lagrange System and Interchanging Variables
For a differentiable objective \(f(x,y)\) and a differentiable equality constraint \(g(x,y)=c\), the one-constraint Lagrange equations are
Here \(\lambda\) is an unknown multiplier. At a feasible point where \(\nabla g\ne\mathbf{0}\), a constrained local extremum must satisfy these equations. The equations identify stationary candidates; solving them alone does not classify a candidate as a maximum or minimum.
Interchanging \(x\) and \(y\) means replacing \((x,y)\) by \((y,x)\). The objective and constraint are symmetric if \(f(y,x)=f(x,y)\) and \(g(y,x)=g(x,y)\). In that case, their Lagrange equations have the same form after the variables are interchanged. Thus if \((x,y,\lambda)\) solves the system, then \((y,x,\lambda)\) does too. This gives paired solutions when \(x\ne y\), not a reason to discard either one.
For polynomial objectives and constraints, the subtraction method has a precise factorization behind it. Define the two stationarity residuals by
When \(f\) and \(g\) are symmetric, interchanging \(x\) and \(y\) interchanges these residuals. Consequently, \(P(x,y,\lambda)=R_x-R_y\) satisfies \(P(x,x,\lambda)=0\). As a polynomial in \(x\) and \(y\), it is divisible by \(x-y\): the factor theorem, applied with \(x\) as the variable and \(y,\lambda\) fixed, gives
for some polynomial \(H\). Therefore every solution belongs to at least one of two branches: \(x=y\), or \(H(x,y,\lambda)=0\). The second branch must be solved; it may contain points with unequal coordinates.
A Reliable Branch-by-Branch Method
Before simplifying, write down both stationarity equations and the constraint. Then subtract or factor the equations and record all resulting branches. In particular, check whether a coordinate can be zero before dividing by it. After solving each branch, substitute its candidates into the original equations and constraint.
Confirm that swapping the variables leaves the objective and constraint unchanged.
Subtract or rearrange them to expose factors such as \(x-y\), while keeping the full factored equation.
Handle equal-coordinate, unequal-coordinate, and zero-coordinate cases as applicable.
Check feasibility and both stationarity equations for each candidate.
Worked Example: A Linear Objective on the Unit Circle
Worked Example: A Linear Objective on the Unit Circle
Find the Lagrange stationary candidates for \(f(x,y)=x+y\) subject to \(x^2+y^2=1\).
Set \(g(x,y)=x^2+y^2\) and \(c=1\). The gradients are \(\nabla f=(1,1)\) and \(\nabla g=(2x,2y)\). The system is
Since \(1=2\lambda x\), \(\lambda\ne0\). Subtracting the first two equations gives \(2\lambda(x-y)=0\), so \(x=y\). The constraint then gives \(2x^2=1\), hence \(x=y=1/\sqrt{2}\) or \(x=y=-1/\sqrt{2}\).
For the first point, the first stationarity equation gives \(\lambda=1/\sqrt{2}\). For the second, it gives \(\lambda=-1/\sqrt{2}\). Check: at either point \(x^2+y^2=1/2+1/2=1\), and \(2\lambda x=2\lambda y=1\). Thus both points satisfy the original system. Their objective values are respectively \(\sqrt{2}\) and \(-\sqrt{2}\).
Worked Example: Keeping the Unequal-Coordinate Branches
Worked Example: Keeping the Unequal-Coordinate Branches
Find all Lagrange stationary candidates for \(f(x,y)=x^4+y^4\) subject to \(x^2+y^2=1\).
The stationarity equations and constraint are
First consider the case where both \(x\ne0\) and \(y\ne0\). Dividing the first equation by \(2x\) and the second by \(2y\) is valid in this case, and gives \(\lambda=2x^2\) and \(\lambda=2y^2\). Thus \(x^2=y^2\), so either \(y=x\) or \(y=-x\). In either case, the constraint gives \(2x^2=1\). The candidates are therefore \((1/\sqrt{2},1/\sqrt{2})\), \((-1/\sqrt{2},-1/\sqrt{2})\), \((1/\sqrt{2},-1/\sqrt{2})\), and \((-1/\sqrt{2},1/\sqrt{2})\). Each has \(\lambda=1\).
Now check cases where a coordinate is zero. If \(x=0\), the constraint gives \(y=1\) or \(y=-1\). The second stationarity equation then gives \(4y^3=2\lambda y\), so \(\lambda=2y^2=2\); the first equation is \(0=0\). This yields \((0,1,2)\) and \((0,-1,2)\), where the entries are \(x,y,\lambda\). If \(y=0\), the constraint gives \(x=1\) or \(x=-1\), and the first equation gives \(\lambda=2x^2=2\); the second equation is \(0=0\). This yields \((1,0,2)\) and \((-1,0,2)\).
For the four candidates with both coordinates nonzero, \(x^2=y^2=1/2\), so \(4x^3=2x=2\lambda x\) and \(4y^3=2y=2\lambda y\). For each axis candidate, the nonzero coordinate is \(1\) or \(-1\), making its stationarity equation \(4x^3=4x=2(2)x\), or \(4y^3=4y=2(2)y\); the equation for the zero coordinate holds automatically. All eight points satisfy the unit-circle constraint. The list includes unequal-coordinate candidates, including points on the axes.
Worked Example: A Symmetric Objective with Axis Candidates
Worked Example: A Symmetric Objective with Axis Candidates
Find all Lagrange stationary candidates for \(f(x,y)=(x^2-y^2)^2\) subject to \(x^2+y^2=1\). This objective is unchanged by interchanging \(x\) and \(y\), even though some of its stationary points will not have equal coordinates.
Let \(D=x^2-y^2\). The partial derivatives are \(f_x=4xD\) and \(f_y=-4yD\). With \(g=x^2+y^2\), the system becomes
Rearranging the first equation gives \(x(2D-\lambda)=0\); rearranging the second gives \(y(-2D-\lambda)=0\). The constraint rules out \(x=y=0\). If both coordinates are nonzero, then \(\lambda=2D\) and \(\lambda=-2D\), so \(D=0\) and \(\lambda=0\). Thus \(x^2=y^2\), and the constraint gives \(x^2=y^2=1/2\). The four candidates are \((1/\sqrt{2},1/\sqrt{2})\), \((-1/\sqrt{2},-1/\sqrt{2})\), \((1/\sqrt{2},-1/\sqrt{2})\), and \((-1/\sqrt{2},1/\sqrt{2})\), all with \(\lambda=0\).
If \(x=0\), then \(y=\pm1\), \(D=-1\), and the second stationarity equation becomes \(4y=2\lambda y\), giving \(\lambda=2\). If \(y=0\), then \(x=\pm1\), \(D=1\), and the first equation becomes \(4x=2\lambda x\), again giving \(\lambda=2\). These produce the four axis points \((0,1),(0,-1),(1,0),(-1,0)\), each with multiplier \(2\).
For the diagonal candidates, \(D=0\), so both stationarity equations equal zero, as required for \(\lambda=0\); their coordinates also give \(x^2+y^2=1\). At each axis candidate, the equation for the zero coordinate is \(0=0\), while the other equation is \(4x=4x\) or \(4y=4y\), and the constraint is satisfied. This verifies all eight solutions. In particular, symmetry does not force a stationary candidate to lie on \(x=y\).
A Common Mistake: Dividing Before Checking Zero Cases
In the quartic example, a tempting shortcut is to divide \(4x^3=2\lambda x\) by \(2x\), and divide \(4y^3=2\lambda y\) by \(2y\), then infer \(x^2=y^2\). That calculation assumes both \(x\ne0\) and \(y\ne0\). It therefore omits \((0,1),(0,-1),(1,0),(-1,0)\), all of which satisfy the original system.
Why the Method Matters
Symmetry can reduce algebra by exposing relationships between equations and by pairing solutions related by a variable swap. The reduction is useful only when it preserves every branch. The factor \(x-y\) identifies the equal-coordinate case; it does not say that the other factor has no solutions. Likewise, division can simplify a nonzero case, but it cannot replace checking zero-coordinate cases.
In practice, keep the original system visible while simplifying. Treat every factor as a branch, impose the constraint on each branch, and verify the resulting points. This gives a complete stationary-candidate list without mistaking a convenient symmetric family for the whole solution set.
Check Your Understanding
Use the branch-by-branch method to answer these questions.
- For a symmetric polynomial Lagrange system, what factor may appear when the two stationarity residuals are subtracted, and what branches must then be considered?
- Why does symmetry imply that a solution with \(x\ne y\) has a swapped partner, rather than imply that \(x=y\)?
- In the quartic circle example, why is it invalid to divide both stationarity equations by their respective coordinates before checking zero cases?
- For \(f(x,y)=(x^2-y^2)^2\) on \(x^2+y^2=1\), which candidates have zero multiplier?
- What two checks should be made after solving a branch of a Lagrange system?