Coordinate Systems and Examples of the Chain Rule

Transcription

1 Coordinate Sstems and Eamples of the Chain Rule Ale Nita Abstract One of the reasons the chain rule is so important is that we often want to change coordinates in order to make difficult problems easier b eploiting internal smmetries or other nice properties that are hidden in the Cartesian coordinate sstem. We will see how this works for eample when tring to solve first order linear partial differential equations or when working with differential equations on circles, spheres and clinders. It is one of the achievements of vector calculus that it has a develped language for describing all of these different aspects in a unified wa, in particular using coordinate changes and differentiating functions under these coordinate changes to provide simplified epressions. The ke idea is the following: if ϕ : R n R n is a change-of-coordinates function and f : R n R is a differentiable function, then instead of working with Df, which ma be difficult in practice, we ma work with Df ϕ DfϕDϕ. Different Coordinate Sstems in R 2 and R 3 Eample. Let s start with something simple, like rotating our coordinates through an angle of θ. In a previous eample, in the section on differentiabilit, we showed that a rotation through θ is a linear function R : R 2 R 2 which operates on vectors,, written here as column vectors, b left multipling them b the rotation matri R, u R v cos θ sin θ sin θ cos θ cos θ sin θ sin θ + cos θ. v u, u, v θ In terms of the component functions of R, we have u R, cos θ sin θ v R 2, sin θ + cos θ We have thus introduced new rotated coordinates u, v in R 2.

2 Eample.2 There are other coordinate sstems in R 2 besides rotated Cartesian coordinates, but which are still linear in a sense. Consider for eample two vectors u and v in R 2 which are not scalar multiples of each other i.e. the are not collinear. Then I claim that an vector in R 2 can be described as a scalar multiple of u plus a scalar multiple of v hence the describe a coordinate sstem, telling ou how to find b going some wa in the direction of u then some wa in the direction of v, much like in regular Cartesian coordinates we go some wa in the direction of i and some wa in the direction of j. Let s see how this works. Let u a, b and v c, d and suppose the are not scalar multiples of each other i.e. c, d ka, b ka, kb for an k. Then consider an,. To sa that k u + k 2 v is to sa, k a, b + k 2 c, d k a + k 2 c, k b + k 2 d which, if we want to discover what k and k 2 are, means we have to solve a sstem of two equations in two unknowns, ak + bk 2 bk + dk 2 Recall that here a, b, c, d and, are all known, onl k and k 2 are unkown! OK, well, we know how to solve such a sstem, namel add c/a times the first equation to the second equation, { ak + bk 2 ck + dk 2 and then solve the second equation for k 2, k 2 a c a ak + bk 2 c a + 0k + a a c ad bc ad bc bc a + d k 2 Finall, substitute k 2 back into either equation and solve the result for k. For eample, substitute it into the first equation: a c ak + bk 2 ak + b k b a c ad bc ad bc a d b ad bc Thus, k u + k 2 v d b u + ad bc a c v.2 ad bc If we want to define a function ϕ : R 2 R 2 changing ij-coordinates to uv-coordinates, it is the function sending, to k, k 2, namel ϕ : R 2 R 2 k, k 2 ϕ, d b a c, ad bc ad bc.3 We can epress this in matri notation and this eplains the linearit of the uv-coordinates: k d b ϕ.4 k 2 ad bc c a 2

3 Eample.3 Let s tr this out with some specific vectors. Let u, and v, 2 and let 3, 5. v k 2 k u Let s find k and k 2 such that k u + k 2 v, and check that the above formula works. Well, and indeed d b k ad bc a c k 2 ad bc k u + k 2 v, + 2, 2, + 2, 4 + 2, + 4 3, 5 And we can do this for an in R 2. There are of course other coordinate sstems, and the most common are polar, clindrical and spherical. Let us discuss these in turn. Eample.4 Polar coordinates are used in R 2, and specif an point other than the origin, given in Cartesian coordinates b,, b giving the length r of and the angle which it makes with the -ais, r and θ arctan.5 Notice that the domain of arctan is naturall π 2, π 2, so this set of coordinates onl works on the right half-plane {, > 0}. We can adjoin the -ais and the left half of the plane, but there we have to define θ differentl. We leave the tedious details to the reader. We then get the composite function ϕ : R 2 R 2 2 r, θ ϕ, + 2, arctan.6 The thing to note is that we can go back and forth between Cartesian and polar coordinates. The reverse direction, from r, θ to, is ψ ϕ : R 2 R 2, ψr, θ r cos θ, r sin θ.7 3

4 We usuall see this in the form, r cos θ and r sin θ.8 r, r cos θ, r sin θ θ Eample.5 Clindrical coordinates are like polar coordinates, but we include a third dimension as well, which we call z we ma as well identif it with the z-coordinate in Cartesian coordinates, though we can just as well identif it with an line through the origin. Thus, our change-of-coordinates functions changing Cartesian into clindrical coordinates is ϕ : R 3 R 3 2 r, θ, z ϕ,, z + 2, arctan, z.9 and its inverse is the map ψ ϕ : R 3 R 3,, z ψr, θ, z r cos θ, r sin θ, z.0 z 4

5 Eample.6 Spherical coordinates are used when working with a sstem having inherent spherical smmetr, for eample the gravitational or the electric field surrounding a point particle. z z ϕ ϕ θ θ Figure.: Spherical Coordinates The basic idea behind spherical coordinates is that a point,, z can be entirel determined not onl b the coordinates, and z in the coordinate directions i, j and k, but b it s length ρ and two angles, one between and k the z-ais, k z ϕ arccos arccos k ρ and one between the projection of onto the, -plane and the -ais, i.e. between,, 0 and i, 0, 0,,, 0, 0, 0 θ arccos arccos,, 0, 0, This could also be described as the arctangent of /, if we draw a right triangle with legs and and hpotenuse θ arctan and there is a similar epression involving arcsine. This gives us a function from R 3 to R 3, converting Cartesian coordinates into spherical coordinates: C : R 3 R 3 z ρ, θ, ϕ C,, z, arccos, arctan. 5

6 In components this is ρ z θ arccos ϕ arctan arccos arcsin The reverse transformation, S C, converting spherical to Cartesian coordinates, is which, in components is S C : R 3 R 3,, z Sρ, θ, ϕ.3 ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ ρ sin ϕ cos θ ρ sin ϕ sin θ z ρ cos ϕ.4 Refer to Figure.6 for the derivation to follow: First, note that in that figure, the projection of,, z onto the -plane is,, 0, so θ is eas to compute using the triangle θ r,, 0 r cos θ, r sin θ, 0 Indeed, θ is easil seen to be the arctangent of /, or the arccosine of / 2 + 2, or the arcsine of / In fact, in the -plane we have the standard polar coordinates. Hence we also have r cos θ and r sin θ. Now let s look at the 3D picture. The vertical triangle has horizontal leg r 2 + 2, 6

7 vertical leg z, and hpotenuse ρ r 2 + z z 2, z ϕ ρ ϕ z r From this picture we can see that r ρ sin ϕ, so that r ρ sin ϕ. We can substitute this epression into our earlier epressions for and in polar coordinates: r cos θ ρ sin ϕ cos θ r sin θ ρ sin ϕ sin θ Finall, z ρ cos ϕ, so we also have z ρ cos ϕ 2 New Coordinates and the Chain Rule What are coordinates good for? The answer is the are basicall the most important thing in vector calculus, because the are precisel what simplifies complicated problems. The onl difficult in switching to more convenient coordinates is that ou have to then translate the whole problem ou re looking at into these new coordinates, and there ma well be need to use the chain rule. The reason is most interesting problems in phsics and engineering are equations involving partial derivatives, that is partial differential equations. These equations normall have phsical interpretations and are derived from observations and eperimentation. Once handed to us, however, we must treat the problems mathematicall, and our purpose here is to elucidate some of the basic methods of this mathematical treatment. Let us look at some eamples. 2. Application to Partial Differential Equations Eample 2. Suppose ou are asked to solve the following first order linear partial differential equation, 3u + 4u 0 or, in other notation, 3 u + 4 u 0 A solution of this equation is a function u : R 2 R, z u,, whose partial derivatives satisf this equation at all points, in the plane. How are we to solve such a problem? 7

8 Well, the first thing to notice is that what we have here is in fact a matri or dot product on the left hand side, namel 3u + 4u 3 u u Du v 4 where v 3, 4 is thought of as a column vector here. But we also know that this product is precisel the directional derivative of u in the direction of v, D v u Du v 3u + 4u v Thus, our original equation boils down to solving the directional derivative equation D v u 0 Well, what does it sa about u that it s directional derivative in the direction of v 3, 4 is 0? It means u is not changing in that direction, at an point in R 2! Pick a line parallel to v, that is having slope 4/3, 4 + b or 4 3 3b 3 where b is arbitrar. Then, on this line, u will be constant. The onl change in u will occur if we move from one line to another, that is in a direction perpendicular to v for we note, of course, that the equation of the line can be epressed as a dot product, the vector 4, 3 with the vector,, and the vector 4, 3 is orthogonal to v!. Thus, an differentiable function f : R R taking η 4 3 as input will work for us, so a general solution is u f g, whereg : R 2 R, g, 4 3, and f : R R is arbitrar If we chose coordinates as in Eample.2 in such a wa that one of the coordinate vectors were v, then the fact that D v u 0 would mean that it s partial derivative with respect to that new coordinate would be zero, alwas! Consequentl, we would onl have to worr about the other partial of u, in the other coordinate direction. Let s call the other coordinate w, which we ma as well pick to be orthogonal to v. Thus, if sa ξ and η are our new coordinates, meaning that ξv + ηw, for all, in R 2, then our original partial differential equation would simplif to 3u + 4u 0 u η 0 Let us verif that this is indeed a solution: Let u, f g, f4 3. Then b the chain rule we have 3u + 4u 3 f f4 3 3f f f For eample, we could choose ft e t, or ft sin t, or ft t 3 + 5, or anthing we want as long as it s differentiable on R! In these cases, u, e 4 3, or u, sin4 3, or u, are all solutions. 8

9 There is another wa to view this problem which highlights the change of coordinates. This still involves the observation that u is constant along the lines 4 3 c. The idea, then, is to pick coordinates, one parallel to v, the other perpendicular to v, for then the partial in one direction will be identicall zero and the result will be a simpler partial differential equation. Let us eplain. Define new coordinates ξ η 4 3 Note that ξ is in the direction of v, and η is perpendicular to it. If we solve this sstem for and this will be clearer: and therefore, 3 25 ξ η 4 25 ξ 3 25 η 3 25 ξ η, 4 25 ξ 3 25 η ξ 3, 4 + η 4, so, is epressed as a multiple of v plus a multiple of w, the vector orthogonal to v. What happens when we consider uξ, η, that is the epression of u in these new coordinates? Well, let s find out b computing some partials: and consequentl, u u ξ ξ + u η η 3u ξ + 4u η u u ξ ξ + u η η 4u ξ 3u η 0 3u + 4u 33u ξ + 4u η + 44u ξ 3u η 25u ξ + 2 2u η 25u ξ Therefore, dividing b 25 gives u ξ 0 a much simpler partial differential equation than 3u + 4u 0! Indeed, we can solve this b integrating with respect to ξ. When we do, since u is a function of ξ and η, we will get an arbitrar differentiable function of η, uξ, η fη Converting back to, coordinates now we see that u, f4 3 Eercise 2.2 Show that an homogeneous linear partial differential equation with constant coefficients the technical term for these things, au + bu 0 has a general solution given b u, fb a for an twice differentiable function f : R R. 9

10 Eercise 2.3 Show that an inhomogeneous linear partial differential equation with constant coefficients the technical term for these things, au + bu c where c is an nonzero constant has a general solution given b u, fb ae c/a2 + 2 for an twice differentiable function f : R R. [Hint: An ordinar differential equation of the form a + b 0 for a function of which we hope to discover, for that would be solving the differential equation, and constants a and b can be solved b noticing that we can rewrite it as / b/a, then noting that / ln via the chain rule, so that in fact we have ln b/a. Integrating with respect to gives ln b/a + c for some arbitrar constant of integration c. Taking the eponential of both sides gives e b/a+c Ce b/a, where C e c. Thus, we have solved our differential equation, up to a constant C.] 2.2 Application to Quadratic Polnomials One nice application of the tpe of rotated coordinates considered in Eample. is that we can recognize common quadratic polnomials in two variables even when their epressions might not look it. Eample 2.4 The equation is the equation of an ellipse, , rotated through an angle of π/6 radians. Figure 2.: , the ellipse rotated through π/6. Eample is the parabola 2 rotated through an angle of π/6 radians. 0

11 Figure 2.2: , the parabola 2 rotated through π/6. These eamples are a feature of a more general phenomenon: Theorem 2.6 The general quadratic equation A 2 + B + C 2 + D + E + F 0 2. can alwas be transformed into the simpler quadratic equation A 2 + C 2 + D + E + F thus eliminating the term b choosing an appropriate set of orthogonal coordinates or basis vectors, in the language of linear algebra. Completing the square will then reveal which of the well-known tpes of quadratics our original equation was, as we eplain below. The new - and -coordinates are obtained from the old - and -coordinates b a counterclockwise rotation of the standard orthogonal Cartesian coordinate aes through an angle θ, where θ is given b 0, if B 0 θ A C cot 2.3, if B 0 B The new coordinates are then related to the old b the equations or in matri notation cos θ sin θ 2.4 sin θ + cos θ 2.5 cos θ sin θ sin θ cos θ The quardratic 2., moreover, has the following invariants under rotation: As a consequence, F F 2.6 A + C A + C 2.7 B 2 4AC B 2 4A C 2.8

12 . Supposing B 2 4AC < 0: If F > D2 4A + E2 4C, then 2. is said to be degenerate and does not represent an points. If F D2 4A + E2 4C, then 2. represents a single point, D 2A, E 2C. If F < D2 4A + E2 4C, then 2. represents an ellipse or a circle. 2. Supposing B 2 4AC 0: If A C 0, then 2. represents a line, otherwise, if one of A or C is nonzero, 2. represents a parabola. 3. Supposing B 2 4AC > 0: If F D2 4A + E2 4C, then 2. represents a single point, D 2A, E 2C. If F < D2 4A + E2 4C or F > D2 4A + E2 4C, then 2. represents a hperbola. Proof: Let θ be given b 2.3 and appl Eample. to get the new rotated, coordinates: cos θ sin θ sin θ + cos θ These, along with the trig double angle identities, sin 2θ 2 sin θ cos θ 2.9 cos 2θ cos 2 θ sin 2 θ 2.0 allow us to evaluate 2. in terms of and, namel b plugging 2.4 and 2.5 into 2.: A 2 + B + C 2 + D + E + F A cos θ sin θ 2 +B cos θ sin θ sin θ + cos θ +C sin θ + cos θ 2 +D cos θ sin θ +E sin θ + cos θ +F A 2 cos 2 θ 2 sin θ cos θ + 2 sin 2 θ +B 2 sin θ cos θ + cos 2 θ sin 2 θ 2 sin θ cos θ +C 2 sin 2 θ + 2 sin θ cos θ + 2 cos 2 θ +D cos θ sin θ +E sin θ + cos θ +F 2 A cos 2 θ + B sin θ cos θ + C sin 2 θ + [ 2A C sin θ cos θ + Bcos 2 θ sin 2 θ ] + 2 A sin 2 θ B sin θ cos θ + C cos 2 θ + D cos θ + E sin θ + D sin θ + E cos θ +F 2 A cos 2 θ + B sin θ cos θ + C sin 2 θ + [ A C sin 2θ + B cos 2θ ] 2

13 + 2 A sin 2 θ B sin θ cos θ + C cos 2 θ + D cos θ + E sin θ + D sin θ + E cos θ +F A 2 + C 2 + D + E + F where A A cos 2 θ + B sin θ cos θ + C sin 2 θ 2. B A C sin 2θ + B cos 2θ C A sin 2 θ B sin θ cos θ + C cos 2 θ D D cos θ + E sin θ E D sin θ + E cos θ F F If B 0, then choosing θ 0 gives B 0, and we need not bother rotating anthing, for we alread have what we want. Otherwise which is the whole point of this theorem, we can make B 0 b choosing θ appropriatel as follows: 0 B A C sin 2θ + B cos 2θ A C sin 2θ B cos 2θ A C cos 2θ cot 2θ B sin 2θ A C θ cot 0, π/2 π/2, π B This is possible since A C B R and cot 2θ is a bijection between 0, π/2 and R, and likewise it is a bijection between π/2, π and R, whence there is a unique θ between 0 and π/2, and another unique θ between π/2 and π, such that cot 2θ A C B. We cannot have θ π/2 in the epression above, because cot π/2 is not defined. But we can do something else: we can note that according to 2. 0 B A C sin 2 π 2 + B cos 2π 2 A C sin π + B cos π A + C iff A C. But if A C, then 2. becomes A A cos 2 π 2 + B sin π 2 cos π 2 + A sin2 π 2 A B A A sin 2 π 2 + B cos 2π 2 B 0 C A sin 2 π 2 B sin π 2 cos π 2 + A cos2 π 2 A D D cos π 2 + E sin π 2 E E D sin π 2 + E cos π 2 D F F 3

14 so that 2. for this function is given b A 2 + A 2 + D + E + F A 2 + A 2 + E D + F 0 which shows that if 2.2 defines a conic section, then so does 2.. Thus we can have θ π/2 as well. Now, as we noted above, we choose our θ to lie in [0, π because for θ [π, 2π there is no difference among the terms in 2. if we replace this θ b θ π, ecept in the case of a parabola, which we can adjust b multipling the 2.2 b. This proves 2.3, and so we ma alwas convert 2. into 2.2 b this choice of θ and a change of coordinates 2.4 and 2.5, according to 2.. It remains to show, of course, that 2.2 does indeed represent a conic section when it isn t degenerate or when it doesn t represent a point. To do this we must first show that 2. has certain invariants of rotation. The easiest is F F, which is shown in 2.. Net, we show A + C A + C : b 2. we have A + C [ A cos 2 θ + B sin θ cos θ + C sin 2 θ ] + [ A sin 2 θ B sin θ cos θ + C cos 2 θ ] Acos 2 θ + sin 2 θ + Csin 2 θ + cos 2 θ A + C Finall, we show the trickiest, but most important invariant, the disriminant B 2 4AC B 2 4A C : b 2., b the double angle trig identities 2.9 and 2.0 given above, and b the trig identities cos 2 θ + cos 2θ sin 2 θ cos 2θ [ ] 2 cos 4 θ + cos 2θ cos 2θ + cos 2 2θ cos 2θ cos 4θ cos 2θ + cos 4θ 8 sin 4 θ cos 2θ + cos 4θ sin 2 θ cos 2 θ 2 + cos 2θ cos 2θ cos2 2θ 4 cos 4θ cos 4θ 8 sin θ cos 3 θ sin θ cos θ cos 2 θ 2.7 sin 2θ + cos 2θ sin 2θ + sin 2θ cos 2θ 4 4 sin 2θ + sin 4θ 8 4

15 sin 3 θ cos θ sin 2 θsin θ cos θ cos 2θ sin 2θ 2 4 sin 2θ sin 2θ cos 2θ 4 4 sin 2θ sin 4θ 8 we have B 2 4A C [ A C sin 2θ + B cos 2θ ] 2 4 [ A cos 2 θ + B sin θ cos θ + C sin 2 θ ][ A sin 2 θ B sin θ cos θ + C cos 2 θ ] cos 4θ +cos 4θ 2 sin 4θ 2 {}}{{}}{{}}{ C 2 2AC + A 2 sin 2 2θ +BC AB 2 sin 2θ cos 2θ +B 2 cos 2 2θ [ 4 A 2 sin 2 θ cos 2 θ AB sin θ cos 3 θ + AC cos 4 θ + AB sin 3 θ cos θ B 2 sin 2 θ cos 2 θ + BC sin θ cos 3 θ + AC sin 4 θ BC sin 3 θ cos θ ] +C 2 sin 2 θ cos 2 θ 2 A2 + 2 C2 AC 2 A2 cos 4θ 2 C2 cos 4θ + AC cos 4θ + BC sin 4θ AB sin 4θ + 2 B B2 cos 4θ 4 [A 2 8 cos 4θ AB 4 sin 2θ + 3 sin 4θ + AC cos 2θ + 8 cos 4θ +AB 4 sin 2θ 8 sin 4θ B cos 4θ +BC 4 sin 2θ + 3 sin 4θ + AC cos 2θ + 8 cos 4θ BC 4 sin 2θ 8 sin 4θ + C ] cos 4θ 2 A2 + 2 C2 AC 2 A2 cos 4θ 2 C2 cos 4θ + AC cos 4θ + BC sin 4θ AB sin 4θ + 2 B2 + 2 B2 cos 4θ 2 A2 + 2 A2 cos 4θ + AB 2 sin 4θ 3 2 AC 2 AC cos 4θ + 2 AB sin 4θ + 2 B2 2 B2 cos 4θ 2 BC sin 4θ 3 2 AC 2 AC cos 4θ 2 BC sin 4θ 2 C2 + 2 C2 cos 4θ AC 3 2 AC 3 2 AC + 2 B2 + 2 B2 B 2 4AC Thus we have proved the invariants of rotation Since we can choose θ according to 2.3, thereb setting B equal to 0, we have b the invariant 2.8 of the discriminant that 5

16 for this θ B 2 4AC 4A C 2.9 It is this equation that allows us to conclude that 2. represents a conic section. We proceed b cases, starting with. B 2 4AC 4A C < 0 In this case we must have A C > 0 so either A, C > 0 or A, C < 0. Choose A, C > 0, and note that since A C > 0 we can divide 2. b A C : A C A 2 + C 2 + D + E + F C 2 + D A + A 2 + E C + F A C C 2 + D A + D2 4A 2 + A 2 + E C + E2 4C 2 2 C + D 2A + A + E 2C 0 + D 2A 2 C E 2C 2 2 D2 4A 2 C E2 4A C 2 + F A C A 2 D2 4A 2 C E2 4A C 2 + F A C D2 4A 2 C E2 4A C 2 + F A C so that + D 2A 2 C E 2C 2 A 2 D 2 4A 2 C + E2 4A C 2 F A C 2.20 D 2 A C 4A + E2 4C F If F D2 4A + E2 4C, then this equation represents a point onl the single point D 2A, E 2C satisfies this equation. If F > D2 4A + E2 4C, then there are no points satisfing this equation, since the left side is positive and the right side is negative. If F < D2 4A + E2 4C, then letting K D2 4A 2 C + E2 4A C F 2 A C, we can rewrite 2.20 as + D 2A 2 KC E 2C 2 KA 2 which is the equation of an ellipse or circle when A C centered at B 2 4AC 4A C 0 D 2A, E 2C. 2. If then either A 0 or C 0. If A C 0, then the equation clearl represents a line. If, sa, A 0 and C 0, then 2.2 becomes C 2 + D + E + F 0 which is the equation of a parabola in. Likewise if A 0 and C 0, then 2.2 becomes A 2 + D + E + F 0 6

17 which is the equation of a parabola in. 3. If B 2 4AC 4A C > 0 then A C < 0, so A > 0 and C < 0 or vice-versa. Suppose A < 0 and C > 0 the other case is similar. Then, dividing 2.2 b A C completing the square as above as in, we have D 2A + E 2C C 2 D 2 A 2 4A 2 C + E2 4A C 2 F A C 2.2 A C D 2 4A + E2 4C F If F D2 4A + E2 4C, then as above this represents a point. If F < D2 4A + E2 4C, then letting K D2 4A 2 C + E2 4A C F 2 A C, we can rewrite 2.20 as + D 2A 2 KC 2 + E 2C 2 KA 2 which is the equation of a hperbola. If F > D2 4A + E2 4C, then K D2 a negative number, and we can rewrite 2.20 as D 2A + E KC + 2C 2 KA 2 4A 2 C + E2 4A C F 2 A C which is also the equation of a hperbola, though this one is perpendicular to the previous one it s ais of smmetr is perpendicular to the ais of smmetr of the previous one. We can repeat the above arguments for the case A > 0 and C < 0 as well. This completes the proof. is Remark 2.7 This theorem is the basis for the second derivative test. The second derivative test shows that a function f : R 2 R behaves, at critical points, a lot like quadratic polnomials not the equations, but the polnomials themselves. Essentiall, under appropriate rotated coordinates a nondegenerate second degree quadratic polnomial looks either like p, 2 + 2, p, 2 2 or p, 2 2, i.e. like a paraboloid opening either up or down, or a hperbolic paraboloid a saddle, or Pringle chip. These ma be stretched slightl b adding different coefficients, but the three classes are basicall canonical representatives of the tpes. 2.3 The Laplace Operator in Polar and Spherical Coordinates Eample 2.8 Consider the Laplace operator on R 2, This operator takes twice differentiable functions f : R 2 R to at least continuous functions from R 2 to R, f 2 f f 2 : R2 R 7

18 This is the epression of in standard Cartesian coordinates, but what if we wanted to eploit some internal smmetr of a problem. Suppose that we were tring to solve a second order differential equation, called the wave equation which obviousl models wave propagation through some medium, 2 u t 2 u Here u : R 3 R is given also as a function of time, ut,,, and the z-value of u at a point, in R 2 at time t represents the height of the wave. Suppose further that we know the wave has some circular smmetr, that is the wave propagates outward smmetricall about some point. Then it would obviousl be more convenient to treat this problem, that is to look for solutions to u tt u, if we convert to polar coordinates. M claim is that the Laplace operator in polar coordinates takes the epression r r + 2 r 2 + r 2 2 θ 2 Let s prove this b starting with the right-hand side and then showing it s equal to the left-hand side. Well, this means using the polar coordinates of Eample.4, r cos θ and r sin θ and the chain rule on an twice-differentiable function f : R 2 R, since what we want to compute is the partials of fr, θ, which means the partials of the composition f ψ : R 2 R where ψ ϕ : R 2 R 2, ψr, θ r cos θ, r sin θ is the change-of-coordinate function from polar to Cartesian. Well, the chain rule in components is f r r, θ f ψr, θ r r, θ + f ψr, θ r r, θ f θ r, θ f ψr, θ θ r, θ + f ψr, θ θ r, θ which we ll abbreviate, in the interest of space, to f r f r + f r f θ f θ + f θ thus avoiding mention of where we evaluate each but as ou have noted in our WebWork and homework, where ou evaluate e.g. f is different from where ou evaluate r!. Then, since r cos θ, r sin θ, θ r sin θ, θ r cos θ we have f r f cos θ + f sin θ f θ f r sin θ + f r cos θ Let s now compute the second partials. This will involve using the chain rule again on f and f, for eample f r f r + f r f r + f r f cos θ + f sin θ. So, f rr f r r f cos θ + f sin θ r f r cos θ + f r sin θ f r + f r cos θ + f r + f r sin θ f cos 2 θ + 2f sin θ cos θ + f sin 2 θ 8

19 and f θθ f θ θ f r sin θ + f r cos θ θ r f sin θ θ + r f cos θ θ f r θ sin θ + f f sin θ + r θ θ cos θ + f cos θ θ f f r θ + f θ sin θ + f cos θ + r θ + f θ cos θ f sin θ f r r sin θ + f r cos θ sin θ + f cos θ f +r r sin θ + f r cos θ cos θ f sin θ f r 2 sin 2 θ 2f r 2 sin θ cos θ + f r 2 cos 2 θ rf cos θ rf sin θ Therefore, the punchline is r f r + f rr + r 2 f θθ f cos θ + f sin θ + f cos 2 θ + 2f sin θ cos θ + f sin 2 θ r + f r 2 r 2 sin 2 θ 2f r 2 sin θ cos θ + f r 2 cos 2 θ rf cos θ rf sin θ r f cos θ + r f sin θ + f cos 2 θ + 2f sin θ cos θ + f sin 2 θ +f sin 2 θ 2f sin θ cos θ + f cos 2 θ r f cos θ r f sin θ f sin 2 θ + cos 2 θ + f sin 2 θ + cos 2 θ f + f This completes the proof. Eample 2.9 Show that the Laplace operator on R 3, z 2 has the following epression in spherical coordinates: 2 r 2 + r 2 2 θ 2 + r 2 sin 2 θ or, in even more compact notation, r 2 r 2 + r r r 2 sin θ 2 φ r sin θ + θ θ r + cos θ r 2 sin θ r 2 sin 2 θ You can check out the Plane Math page for a derivation of this, but I would recommend that onl as a wa to check our work. θ 2 φ 2 9