Lecture 14: 3/12/2004, CRITICAL POINTS Math21a, O. KnillHOMEWORK: Section 11.7: 8,18,20,22CRITICAL POINTS. A point (x0, y0) in a region G is called a critical point of f(x, y) if ∇f (x0, y0) = (0, 0).Remarks. Critical points are also called stationary points. Critical points are candidates for extrema becauseat critical points, the directional derivative is zero. It is usually assumed that f is differentiable.EXAMPLE 1. f(x, y) = x4+ y4− 4xy + 2. The gradient is ∇f(x, y) = (4(x3− y), 4(y3− x)) with critical points(0, 0), (1, 1), (−1, −1).EXAMPLE 2. f(x, y) = sin(x2+ y) + y. The gradient is ∇f(x, y) = (2x cos(x2+ y), cos(x2+ y) + 1). Fora critical points, we must have x = 0 and cos(y) + 1 = 0 which means π + k2π. The critical points are at(0, π), (0, 3π), ....EXAMPLE 3. (”volcano”) f(x, y) = (x2+ y2)e−x2−y2. The gradient ∇F = (2x − 2x(x2+ y2), 2y − 2y(x2+y2))e−x2−y2vanishes at (0, 0) and on the circle x2+ y2= 1. There are ∞ many critical points.EXAMPLE 4 (”pendulum”) f (x, y) = −g cos(x) + y2/2 is the energy of the pendulum. The gradient ∇F =(y, −g sin(x)) is (0, 0) for x = 0, π, 2π, ..., y = 0. These points are equilibrium points, where the pendulum is atrest.EXAMPLE 5 (”Volterra Lodka”) f(x, y) = a log(y) − by + c log(x) − dx. (This function is left invariant by theflow of the Volterra Lodka differential equation ˙x = ax − bxy, ˙y = −cy + dxy which you might have seen inMath1b.) The point (c/d, a/b) is a critical point.TYPICAL EXAMPLES.-2-1012-2-101201020-2-1012f(x, y) = x2+ y2-1-0.500.51-1-0.500.51-2-1.5-1-0.50-1-0.500.51f(x, y) = −x2− y2-1-0.500.51-1-0.500.51-1-0.500.51-1-0.500.51f(x, y) = x2− y2EXAMPLES WITH DISCRIMINANT D = det(H) = 0.-1-0.500.51-1-0.500.51-1-0.75-0.5-0.250-1-0.500.51f(x, y) = x2-1-0.500.51-1-0.500.5100.250.50.751-1-0.500.51f(x, y) = −x2-1-0.500.51-1-0.500.5100.511.52-1-0.500.51f(x, y) = x4+ y4CLASSIFICATION OF CRITICAL POINTS IN 1 DIMENSION.f0(x) = 0, f00(x) > 0, local minimum, f00(x) < 0 local maximum, f00= 0 undetermined.-1 -0.5 0.5 10.20.40.60.81-1 -0.5 0.5 1-1-0.8-0.6-0.4-0.2-1 -0.5 0.5 1-1-0.75-0.5-0.250.250.50.751-1 -0.5 0.5 1-1-0.75-0.5-0.250.250.50.751-1 -0.5 0.5 1-1-0.75-0.5-0.250.250.50.751CLASSIFICATION OF CRITICAL POINTS: SECOND DERIVATIVE TEST. Let f(x, y) be a function oftwo variables with a critical point (x0, y0). Define D = fxxfyy− f2xy, called the discriminant or Hessian.(Remark: With the Hessian matrix H =fxxfxyfyxfyywe can write D = det(H) as a determinant.If D > 0 and H11> 0 ⇒ local minimum (bottom of valley)If D > 0 and H11< 0 ⇒ local maximum (top of mountain).If D < 0 ⇒ saddle point (mountain pass).In the case D = 0, we would needhigher derivatives to determine the na-ture of the the critical point.EXAMPLE. (A ”napkin”).The function f (x, y) = x3/3 − x − (y3/3 − y) has the gradient ∇f(x, y) = (x2− 1, −y2+ 1). It is the zero vectorat the 4 critical points (1, 1),(−1, 1),(1, −1) and (−1, −1). The Hessian matrix is H = f00(x, y) =2x 00 −2y.H(1, 1) =2 00 −2H(−1, 1) =−2 00 −2H(1, −1) =2 00 2H(−1, −1) =−2 00 2D = −4Saddle pointD = 4, fxx= −2Local maximumD = 4, fxx= 2Local minimumD = −4Saddle pointGLOBAL MAXIMA AND MIN-IMA. To determine the maximumor minimum of f (x, y) on a domain,determine all critical points inthe interior the domain, andcompare their values with maximaor minima at the boundary.(A point is in the interior of G, if there is a small disc around (x0, y0) contained in G. A point is at the boundary of G,if any disc aound (x0, y0) contains both points in G and in the complement).EXAMPLE 5. Find the critical points of f(x, y) = 2x2− x3− y2. With ∇f(x, y) = 4x − 3x2, −2y), the criticalpoints are (4/3, 0) and (0, 0). The Hessian is H(x, y) =4 − 6x 00 −2. At (0, 0), the discriminant is −8 sothat this is a saddle point. At (4/3, 0), the discriminant is 8 and H11= 4/3, so that (4/3, 0) is a local maximum.WHY DO WE CARE ABOUT CRITICAL POINTS?• Critical points are candidates for extrema like maxima or minima.• Knowing all the critical points and their nature tells alot about the function.• Critical points are physically relevant. Examples are configurations with lowest energy).A CURIOUS OBSERVATION: (The island theorem) Let f(x, y)be the height on an island. Assume there are only finitely manycritical points on the island and all of them have nonzero de-terminant. Label each critical point with a +1 ”charge” if it isa maximum or minium, and with −1 ”charge” if it is a saddlepoint. Sum up all the charges and you will get 1, independent ofthe function. This property is an example of an ”index theorem”,a prototype for important theorems in physics and mathematics.CRITICAL POINTS IN PHYSICS. (informal) Most physical laws are based on the principle that the equationsare critical points of a functional (in general in infinite dimensions).• Newton equations. (Classical mechanics) A particle of mass m moving in a field V along a path γ : t 7→ r(t)extremizes the integral S(γ) =Rbamr0(t)2/2 − V (r(t)) dt. Critical points γ satisfy the Newton equationsmr00(t)/2 − ∇V (r(t)) = 0.• Maxwell equations. (Electromagnetism) The electromagnetic field (E, B) extremizes the Integral S(E, B) =18πR(E2− B2) dV over space time. Critical points are described by the the Maxwell equations in vacuum.• Einstein equations (General relativity) If g is a dot product which depends on space and time, and R isthe ”curvature” of the corresponding curved space time, then S(g) =R RRdV (g) is a function of g for whichcritical points g satisfy the Einstein equations in general relativity.OTHER WAYS TO FIND CRITICAL POINTS. Some ideas: walk in the direction of the gradient until youreach a local maximum or walk backwards to reach a local minima. To find saddle points, consider the shortestpath connecting two local minima and take the maximum along this
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