Objective and Feasible Family
Each position (a,b) represents an entire function y(x), not a point on its graph. Surface height is the objective J[y]. The amber path contains the functions that satisfy the constraint; a constrained minimum is a lowest point along this path, not necessarily the bottom of the whole surface.
Two-Coefficient Slice
Blue contours have equal J; the amber constraint fixes K. Green marks an allowed infinitesimal direction. At stationarity, the objective gradient has no tangent component. Unless that gradient is zero, its contour is tangent to the constraint and their normals lie on the same line. The multiplier supplies the sign and length needed for the blue and amber arrows to coincide.
The current curve
Curve and Admissible Variation
Black is the actual function y. Dashed green is a variation η, describing the infinitesimal change y + εη. It vanishes at the endpoints and satisfies δK[η] = 0. It is a direction in function space, not a second candidate curve.
First-Variation Densities
The first variation is a pairing: δJ[η] = ∫gJη dx, and similarly for K. If these two density curves agree at every x, then δJ + λδK vanishes for every endpoint-preserving variation, not just the selected green one.
J Along the Constraint
Only exactly feasible functions appear here. The dot marks the current curve. A horizontal tangent means stationarity within this family; a low point suggests a minimum. Classification in the full function space requires a separate argument.
From vectors to constrained variations
Why the tangent matters
A functional assigns a number to a whole function: J measures its objective value, while K measures the constrained quantity. An allowed first-order variation changes y without changing K, so δK[η] = 0. In the coefficient picture, this is ∇K · t = 0: the tangent is perpendicular to the constraint gradient.
At stationarity, J must also have zero first-order change along every allowed tangent. Its gradient therefore has no tangent component. For a normal one-constraint problem, it must be a scalar multiple of the constraint gradient.
What the multiplier can cancel
With the article’s convention J + λ(K − k₀), the balance is ∇J + λ∇K = 0. Thus the amber arrow is −λ∇K. In the convention ∇J = μ∇K, the equivalent multiplier is μ = −λ; a negative λ is not a failed solution.
The normal-fit value cancels the part of ∇J parallel to ∇K. If a tangent component remains, no choice of this single scalar can remove it. That is why matching a multiplier away from a stationary curve does not solve the constrained problem.
Vectors and density functions
The upper arrows are derivatives with respect to the two coefficients a and b. The lower gJ and gK are functions of x representing the first variations after integration by parts. They describe the same stationarity condition in different representations; the coefficient arrows are not point samples of those density curves.
The L² residual is the square root of ∫(gJ + λgK)2 dx. Zero means the full density balance holds. The polynomial basis is not orthonormal, so this norm need not equal the coefficient-vector residual shown above.