11/5/2002, CRITICAL POINTS Math 21a, O. KnillCRITICAL POINTS. A point (x0, y0) is called a critical point of F (x, y) if ∇F (x0, y0) = (0, 0). A point(x0, y0, z0) is called a critical point of F (x, y, z) if ∇F (x0, y0, z0) = (0, 0, 0).EXAMPLE 1. F (x, y) = sin(x2+ y) + y. The gradient is ∇F (x, y) = (2x cos(x2+ y), cos(x2+ y) + 1). Whereare the critical points? We must have x = 0 and cos(y) + 1 = 0 which means π + k2π. The critical points areat (0, π), (0, 3π).EXAMPLE 2. (”volcano”) F (x, y) = (x2+ y2)e−x2−y2. The gradient ∇F = (2x − 2(x2+ y2), 2y − 2(x2+y2))e−x2−y2vanishes at (0, 0) and on the circle x2+ y2= 1. There are ∞ many critical points.EXAMPLE 3 (”pendulum”) F (x, y) = −g cos(x) + y2/2 is the energy of the pendulum. The gradient∇F = (y, −g sin(x)) vanishes for x = 0, π, 2π, ..., y = 0. These points are equilibrium points, where thependulum is at rest.EXAMPLE 4 (”Volterra Lodka”) F (x, y) = a log(y) − by + c log(x) − dx. The maximum occurs at c/d, a/b.EXAMPLE 5 (”From a Advanced Multivariable Calculus Midterm somewhere in the US”) Find all the criticalpoints of f(x, y) = 2x2− x3− y2.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 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 FOR FUNCTIONS OF TWO VARIABLES.Let F (x, y) be a function of two variables with a critical point (x0, y0). Let H =FxxFxyFyxFyybe the Hessianat (x0, y0).det(H) > 0 H11> 0 local minimum (valley)det(H) > 0 H11< 0 local maximum (top).det(H) < 0 saddle point (pass).det(H) = 0, tr(H) < 0 mountain linedet(H) = 0, tr(H) > 0 long valley.det(H) = 0, tr(H) = 0 need higher derivatives.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 vanishes at the4 critical points (1, 1),(−1, 1),(1, −1) and (−1, −1). The Hessian matrix is H = F00(x, y) =2x 00 −2y.F00(1, 1) =2 00 −2F00(−1, 1) =−2 00 −2F00(1, −1) =2 00 2F00(−1, −1) =−2 00 2det(H) = −4Saddle pointdet(H) = 4, H11= −2Local maximumdet(H) = 4, H11= 2Local minimumdet(H) = −4Saddle pointMAXIMA-MINIMA. To determinethe maximum or minimum of f(x, y)on a domain domain, determine allcritical points inside the domain,and compare their values with max-ima or minima at the boundary.EXAMPLE 5 Solution. Find all the critical points of f(x, y) = 2x2−x3−y2. We have ∇f(x, y) = 4x−3x2, −2y).The critical points are at (4/3, 0) and (0, 0). The Hessian is H(x, y) =4 − 6x 00 −2. At (0, 0), the Hessiandeterminant is −8 so that this is a saddle point. At (4/3, 0), the Hessian determinant is 8 and H11= 4/3, sothat (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 determine a lot about the function.• Critical points are physically relevant (i.e. configurations with lowest energy).A CURIOUS OBSERVATION: (The island theorem) Let F (x, y) be theheight on an island. Assume there are only finitely many critical pointson the island and all of them have nonzero determinant. Label eachcritical point with a +1 ”charge” if it is a maximum or minium, andwith −1 ”charge” if it is a saddle point. Sum up all the charges and youwill get 1, independent of the function. This property is an example ofan ”index theorem”, a prototype for important theorems in physics andmathematics.CRITICAL POINTS IN PHYSICS. (informal) Most physical laws are based on a simple principle: the equa-tions are critical points of a functional (in general in infinite dimensions). Examples:• 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 electomagnetic 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 vaccuum.• Einstein equations (General relativity) If g is a dot product which depends on space and time, and R is the”curvature” of the corresponding curved space time, then S(g) =RRdV (g) is a function of g for which criticalpoints g satisfy the Einstein equations in general relativity.OTHER WAYS TO FIND CRITICAL POINTS. Some ideas: walk in the direction of the gradient until reachinga local maximum or walkd backwards to reach local minima. To find saddle points, consider the shortest pathconnecting two local minima and take the maximum along this
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