Buck Meta Chapter 6 Applications in Geometry and Analysis

Buck Chapter 6 Meta Notes PhL 5.16.15 Chapter 6: Applications to Geometry and Analysis [ pp 296-365 ] 1 6.1 Transformations of Multiple Integrals [296] 1 6.2 Curves and arc length [313] 7 6.3 Surfaces and surface area [330] 14 6.4 Extremal Properties of functions of several variables [349] 23 6.1 Transformations of Multiple Integrals [296] This section is the Buck's version of deriving the Jacobian volume element for integration rule. Their approach is very different from my approach in tensor doc. What they do in this section: Claim (no proof) the J = detS finite volume ratio rule for linear transformation L, Theorem 1. Example: verify this rule for a finite tetrahedron with one corner at the origin. Show a complicated area transformation rule for linear transformation L: E2 → En. Claim the J volume rule for a general transformation and prove this in Theorems 2,3,4. Compare the J rule with change of variables in 1D Restate their pet box notation for integration volume on page 306 Examples 1,2,3 showing how to compute certain 2D integrals two ways. One way is the brute force way doing integral as stated. Second way is to use some kind of simplifying transformation and then do the integral in the new variables with J added. Example 4 shows how you might deal with an integration whose area includes a place where J = 0. In their proof of Theorem 3, they get to make use of their fancy set-derivative idea developed in an earlier chapter. They are adamant in their approach, just as I am with my N-piped approach. The Bucks are much more interested than I am in fine details and rigor in this business. They even claim the subject was one of current research in 1956 when the first Buck was written. Phrase algebraic topology is mentioned, as is "measure theory". I think the Bucks know all about measure theory and the fancy methods, just as they know about complex variables, but they are trying to explain things in this chapter in their dumbed-down world for the student reader. They wanted to say at least something about the Jacobian Integration Rule and their section title is "transformations of multiple integrals". I recall when I first encountered the J integration volume rule. I was very uncomfortable about it, teachers just used it without explaining why it worked in general. Students understood the idea in polars from simple geometry, but student had no notion of what I have presented in tensor doc on this subject. The basic idea is very simple: (1) you could first show that finite volumes transform with J under a linear transformation, like their tetrahedron example. (2) If J≠0, then this would be true for a differential volume since the transformation is linear locally; (3) then you just make J(x) be a function of x for the general case. Now they never prove item (1) and neither have I, but I think I could. My method fills a finite region with abutting differential N-pipeds (think polars) and for each of these we have a little J(x) rule. 6.2 Curves and arc length [313] You should think of a curve as a mapping like γ : E1 → En . Different mappings can give the same trace, so do not think of the trace itself as the curve. The curve is γ(t) [ bolded since vector in En]. The curve can cross itself with different tangents γ'(t) at the same point in En, for example, a simple crossing is a double point. Curve can be closed on itself so start and end are at same point. A curve with no self-crossings is a simple curve = a Jordan curve. Think of a curve as r(t) maybe in better notation. The tangent at domain value t is dr/dt = v(t). Sometimes if the curve has a sharp corner, you find v = 0 at the corner, that is a zero-speed point. I made an example with a corner which is not a zero speed point, but they don't mention that possibility. They have some nice examples. Example 1 = page 315 trace V like has a zero speed point Example 2 = a straight line Example 3 = page 316 trace in 2D has a double point, curve open Example 4 = page 317 trace in 3D has a double point, curve closed Curvature. I rolled my own on this topic. For any smooth non-planar curve, I show that local to a point on the curve the curve is planar and can be fitted to a circle and thus has a curvature radius R, but the curvature itself is called k = 1/R. In (6.17) Bucks produce an expression for k in terms of the velocity and acceleration of the curve, that is, in terms of γ'(t) and γ''(t) at point t. They expend 3 pages proving this, but I did my own derivation with some Maple help and my result duplicates theirs. Arc Length. Some curves have infinite arc length and are thus not rectifiable. The formula they give is fine by me: L(r(t)) = !Syntax Error, I | dr/dt| dt = !Syntax Error, Iv(t)dt = !Syntax Error, I ds = !Syntax Error, I|dr| (6-20) They give an elaborate derivation of this formula for arc length as Theorem 5. [321] Example 4: [322] just shows the computation of a simple arc length integral, a form of helix in 3D. Equivalent Curves. Have the same trace, one is just a respeeding of the other. One curve will have unit speed at every point. Theorem 6; [323] Two equivalent curves have the same direction at the same point p in space. Theorem 7; [323] Smoothly equivalent curves have the same arc length. The comment that there is ongoing theory work on classes of curves. Curve mapped through a transformation. Closed goes to closed, but map pick up extra loops. Curves placed on surfaces to make things clearer. Examples given. Conformal. If the R matrix has the form R = then the relation between dx' and dx says dx' = a dx + b dy => ∂x'/∂x = a ∂x'/∂y = b dy' = -bdx + ady => ∂y'/∂x = -b ∂y'/∂y = a One normally takes x' = u and y' = v so ∂u/∂x = a ∂u/∂y = b => ∂u/∂x = ∂v/∂y and ∂u/∂y = - ∂v/∂x ∂v/∂x = -b ∂v/∂y = a Cauchy Riemann equations When z = x+iy and w = u+iv, these are the Cauchy Riemann equations and w(z) is analytic. In this case, the mapping above preserves angles as long as J ≠ 0. Conformal means preserving angles. Concept cannot be applied to E3→E3 it turns out. Theorem 8. [326] If a mapping E2→ E2 is conformal, the linear matrix they call dT must have the form shown top page 327, which is R = . They want J > 0 as a condition. the elements of the R matrix can of course vary from point to point as can J. 6.3 Surfaces and surface area [330] Whereas a curve is γ : E1 → En, a surface is a mapping Σ : E2 → En. As with curves, the surface is the mapping, not the trace. Different surfaces (mappings) can have the same trace (semantic problems). New feature is ∂Σ which is the boundary of a surface. As with all mappings, there are "problem points" and for the surface map, those are where rank of R falls below 2. A simple mapping Σ: E2(u,v) → E3(x,y,z) is r = r0 + ua + vb which creates a plane in E3 which has normal n = axb and passes through point r0 . Can then write this plane as nr = d where d = nr0, so this is my familiar planar form. In this case, the normal components can be expressed in terms of little 2x2 Jacobians. For sample n1 = a2b3- a3b2 = = = ∂(y,z)/∂(u,v) But this is a 2x2 subdeterminant of the R matrix. If rank(R) = 2, then at least one of these Jacobians must not vanish, and so n does not vanish. If rank = 1, then you get n = 0 which does not sound healthy. Theorem 9 [336] : For any curve r(t) passing through a point r on a smooth surface, one must have r'(t)n = 0 , which is to say, v n = 0 . The proof of this seemingly obvious fact is a bit messy as shown on page 336. I write up this proof in the raw notes. They have a fancy chain rule proof. Page 337 B shows the equation of the tangent plane to a surface at some point on the surface. And then C shows you higher order correction terms as you might try to approximate the surface in a Taylor expansion manner. Theorem 10: [338] If surface is defined by g(x,y,z) = 0, then g is the normal. My ancient proof of this fact works just fine. Area Formula. [338] Area = ∫dudv | n(u,v)| where n is the non-unit vector normal given in (6-29). This is really a theorem that Buck's don't prove, but if I identify Bucks' n = a x b with n = e1 x e2 I verify the formula using tensor doc stuff. dA3 = J dA'3 e3 = J dA'3 E3 = det(S) dA'3 det(R) e1 x e2 = e1 x e2 dA'3 = n dA'3 = n(u,v) dudv dA3 = | n(u,v)| dudv But |n|2 = sum of three squared 2x2 matrices as shown (6-30). I show in the raw notes how this matrix sum form of tensor doc agrees with that Bucks give. The new feature is this: In my work, dA3 is the area of a face of a differential 3-piped, and you have to imagine a finite volume and its surface made up of lots of tiny such pipeds (think spherical surface), so this integral then gives you the area of a finite arbitrary surface. Bucks then do some Example 1 and Example 2 area calculations of real world areas, p 341. Note 1: Bucks comment that "area theory" is still in the process of development, their books 1956 and 1965 editions. Hard to believe such a claim but must be true. Again, the world of a mathematician. Bucks close this section with a rather obscure discussion of manifolds. They want you to think of a 2-manifold as a certain set or class of mappings (not just as a surface), but I don't grok that idea much. A certain fancier statement is that if you consider a certain combination of transformations from domain to range, if you find that a certain Jacobian never vanishes, then the manifold is differentiable. And if the Jacobian always has the same sign, then the manifold is orientable. The latter means that you cannot smoothly get from one side of the trace surface to the other. They show that a Mobius strip is not orientable because those Jacobians are not always the same sign, and of course you CAN smoothly get from one side to the other. A spherical surface is a closed manifold (no boundary!), but one with a hole in it is open. I guess a Mobius strip is then open. This is all a very cursory discussion and it is algebraic topology. They are just giving the student reader a little kick in the pants (5 pages), a little view of what is beyond this book. 6.4 Extremal Properties of functions of several variables [349] A critical point of the function f(x,y,,,) is a place where f1 = f2 ... = fn = 0. Example 1: [350] Take a simple 1D function f(x) = 4x3-15x2+18x, graph on the left: (1) There are visibly two critical points (zero slope places) in the interval, but the true max and min on the interval are at the endpoints! If you could not "see" this plot, you would check each of the critical points' f values and the two boundary point values before you concluded where the maximum of f was on this interval. (2) There are two interior places where f'(x) = 0, one is a local max, the other is a local min. Example 1A: f(x) ≈ 4x3-15x2+18.7x shown on the right above. This has a single place where f'(x) = 0, but that place is neither a local min nor a local max! It is just a pause. It is also an inflection point where the curvature changes. So this shows that f'(x) = 0 at interior point that point is a local min or max I don't think you can call this a 1D saddle point. Example 1A demonstrates: Theorem 12: [352] (1D Local Min or Max Theorem). Consider f which is C" on [a,b] and f'(c) = 0 for some interior c, so c is then a critical point. Then: f"(c) < 0 => c is a local maximum point f"(c) = 0 => c is not a local minimum Example 2: [351] 2D f(x,y) = 4xy - 2x2 - y4 . What are the critical points? Graph on the right above: f1 = 4y-4x So solve these equations: y-x = 0 y=x f2 = 4x - 4y3 plot on right below x - y3 = 0 x(1-x2) = 0 x = 0,1,-1 so there are three critical points which are (0,0), (1,1),(-1,-1) which lie in our ±2 domain square. Two of these critical points are local and global max's, the one at (0,0) is a saddle where curves up in one direction and down in the cross direction. The true min is somewhere on the boundary. Example 3: [352] f = xy Here is a plot of xy where (0,0) is a critical point At the origin you have f1 = f2 = f11 = f22 = 0. The origin definitely is a saddle point, but the up and down are at 45 degrees. This example demonstrates: saddle curvatures must be + on one axis and - on the other axis Theorem 13: [353] (2D Local Min or Max Theorem) Let Δ = f122 - f11f22 at some critical point. That is, we also know that f1 = 0 and f2 = 0. Then 1 Δ > 0 => saddle point 2 Δ < 0 => local min or max (max if f11< 0, min if f11> 0) // f11 = 0 or f22 = 0 Δ ≥ 0 not Case 2 3 Δ = 0 => no conclusion Apply this theorem to Example 2 above where we already know the three critical points: f1 = 4y-4x f11 = -4 f12 = -4 Δ = 16 - (-4)(-12y2) = 16 - 48y2 f2 = 4x - 4y3 f22 = -12y2 (0,0): Δ = 16, so saddle point (1,1): Δ = -32 so extremum, and f11 < 0 so its a max (-1-1): same as above case Pages 355 and 356 show topo maps of a peak, of a saddle point, and of a monkey saddle, and finally a saddle next to a peak which is Example 3 f(x,y) = x2 + y3 - 3xy. On page 356 Bucks consider f(a,b) as the cost function for a linear fit F(x) = ax + b to some data. What straight line fit causes f(a,b) to be a global minimum? They find the single critical point, and at that point find Δ < 0 and f11>0 so it really is a local min. It is also a global min and it solves this problem. They mention gradient ascent and gradient descent as iterative methods of finding such min and max. Theorem 14 [358] If 22D f(x,y) = 0 on bounded and open domain D and if f is C", then the min and max of f must occur on the boundary ∂D. That is to say, a harmonic function has its max and min on the boundary ∂D. Constraint Problem Example 1. You have u = f(x,y,z) and v = g(x,y,z). Constraint is g(x,y,z) = 0 and you want to find the max of f(x,y,z). What do you do? First, a thought experiment. Suppose you could solve g(x,y,z) = 0 in the form z = φ(x,y). Then your problem is to find a max of the function F(x,y) ≡ f(x,y,φ(x,y)) which we know how to do! The critical points are obtained by solving F1 = f1 + f3φ1 = 0 F2 = f2 + f3φ2 = 0 and then you could compute F11 F22 and F12 to get Δ and you have it. Example 1 on page 359 is handled in exactly this manner. First aside: Think of these equations u = f(x,y,z) and v = g(x,y,z) as a transformation where perhaps we could add arbitrary w = 0 to get a tensor doc transformation. Then x' = (u,v,w). What does the R matrix look like? Then R11 = ∂u/∂x = ∂1u so R = ∂1u ∂2u ∂3u = f1 f2 f3 ∂1v ∂2v ∂3v g1 g2 g3 0 0 0 0 0 0 Second aside: On the constraint surface we know is we move dr along the surface, then dg = g1dx + g2dy + g3dz = g dr = 0 Now if we really can find z = φ(x,y) as above, then dz = φ1dx + φ2dy as rule for staying on the surface. So install this dz to get dg = g1dx + g2dy + g3[φ1dx + φ2dy] = 0 or dg = [g1 + g3φ1] dx + [g2 + g3φ2] dy = 0 I am not sure where this leads, so just stop here. Third aside: We can also define G(x,y) = g(x,y,φ(x,y)) = 0. This is valid for all x and y in the domain. Then G1 = g1 + g3φ1 = 0 G2 = g2 + g3φ2 = 0 and these equations are parallel to those for F shown above! Right now I am not sure any of these "asides" is relevant. Theorem 15: [360] . Imagine that there are some point r such that g(x,y,z) = 0 [r on surface S] and f(x,y,z) is a local max or min at this same r which is not on the boundary of S. The theorem claims that r must be a "critical point" of the combined transformation u = f(x,y,z) E3 → E2 v = g(x,y,z) "Critical point" here means a place where the R matrix rank drops down from maximum. As I just showed above, the non-critical rank of the R matrix is 2, so the theorem claims that if you find a point r which solves your problem of max f with constraint g = 0, then at that point the rank drops below 2. Proof #1. We have already shown above that (the G stuff from aside #3 and critical points from first step) F1 = f1 + f3φ1 = 0 G1 = g1 + g3φ1 = 0 F2 = f2 + f3φ2 = 0 G2 = g2 + g3φ2 = 0 Solve the top left for φ1 = -f1/f3 and then top right says g1 + g3[-f1/f3] = 0 or f3g1- f1g3 = 0. Solve the bot left for φ2 = -f2/f3 and then bot right says g2 + g3[-f2/f3] = 0 or f3g2- f2g3 = 0. Rewrite these as: f1g3 - f3g1 = 0 g1 = g3(f1/f3) f2g3 - f3g2 = 0 g2 = g3(f2/f3) Now consider f1g2 - f2g1 = f1 [g3(f2/f3)] - f2 [g3(f1/f3)] = 0 Thus we have shown that = 0 and = 0 and = 0 Recall that R = ∂1u ∂2u ∂3u = f1 f2 f3 ∂1v ∂2v ∂3v g1 g2 g3 so we have just shown all three three 2x2 determinants vanish, and therefore rank(R) = 1 or 0. We have thus shown that if you find a solution to your problem, then that solution is a point where rank drops off! We have not proven the converse however. But if you are hunting for a solution, you need only consider points where rank drops off!! The reduces the scaled of the search dramatically! Proof #2 of Theorem 15 is quite interesting, I summarize it in the raw notes and won't repeat that here. It is a proof based on contradiction. The Lagrange Multiplier Solution to this same class of problem. Define H(x,y,z,λ) = f(x,y,z) + λ g(x,y,z). Then consider these four equations H1 = f1+ λg1 = 0 H2 = f2+ λg2 = 0 H3 = f3+ λg3 = 0 H4 = g = 0 These four equations define a critical point for the new function H of its four variables. Recall again, R = ∂1u ∂2u ∂3u = f1 f2 f3 ∂1v ∂2v ∂3v g1 g2 g3 The first three H equations say that (f1,f2,f3) + λ(g1,g2,g3) = 0 or f + λg = 0 This says that the two rows of R are linearly dependent, and that means rank drops off from 2, and that in turn means that the r that makes these H equations be true is a candidate r for a solution point. Note that the last equation enforces g = 0, the constraint. Since we know f and g, we know all the derivatives so we can of course write out the four H equations shown above. We must then find a value of x,y,z,λ that solves these equations! That is, r,λ. Although the equations are linear in λ, they in general are non-linear on x,y,z so not a slam dunk to solve the equations. Example: Suppose f(x,y,z) = 4x2 + y2 + z2 g(x,y,z) = 2x+3y+z-12 constraint Ex 1 p 359. We first compute: (notice how simple this example is, in general that won't be the case) f1 = 8x g1 = 2 f2 = 2y g2 = 3 f3 = 2z g3 = 1 The four H equations are then: H1 = f1+ λg1 = 0 = 8x + 2λ H2 = f2+ λg2 = 0 = 2y + 3λ H3 = f3+ λg3 = 0 = 2z + λ H4 = g = 0 = 2x+3y+z-12 What next? We have 4 equations in 4 unknowns. Replace λ = -2z from the H3 and then we have H1 = f1+ λg1 = 0 = 8x - 4z z = 2x H2 = f2+ λg2 = 0 = 2y -6z y = 3z = 6x H4 = g = 0 = 2x+3y+z-12 2x + 18x + 2x - 12 = 0 So we then have to solve, 22x = 12 or 11x = 6 x = 6/11 Then z = 2x = 12/11 y = 6x = 36/11 λ = -2z = -2 ( 12/11) = -24/11 My solution is then x = 6/11 y = 36/11 z = 12/11 // x,y,z agree with p 360 top λ = -24/11 I will write up the whole Lagrange Multiplier Theory in a separate document, as if it were to go online. So Chapter 6 meta notes are concluded!