Arapura diffforms

Introduction to dierential forms Donu Arapura March 9, 2015 The calculus of dierential forms give an alternative to vector calculus which is ultimately simpler and more exible. Unfortunately it is rarely encountered at the undergraduate level. However, the last few times I taught undergraduate advanced calculus I decided I would do it this way. So I wrote up this brief supplement which explains how to work with them, and what they are good for. By the time I got to this topic, I had covered a certain amount of standard material, which is brie y summarized at the end of these notes. My thanks to Jo~ ao Carvalho, John Crow, Mat u s Goljer and Josh Hill for catching some typos. Contents 1 1-forms 2 2 Exactness in R23 3 Parametric curves 3 4 Line integrals 4 5 Work 7 6 Green's theorem for a rectangle 8 72-forms 8 8 Exactness in R3and conservation of energy 11 9 \d" of a 2-form and divergence 12 10 Parameterized Surfaces 13 11 Surface Integrals 16 12 Surface Integrals (continued) 17 1 13 Length and Area 18 14 Green's and Stokes' Theorems 20 15 Cauchy's theorem 22 16 Triple integrals and the divergence theorem 24 17 Gravitational Flux 25 18 Laplace's equation 27 19 Beyond 3dimensions 29 20 Maxwell's equations in R432 21 Further reading 33 A Essentials of multivariable calculus 34 A.1 Dierential Calculus . . . . . . . . . . . . . . . . . . . . . . . . . 34 A.6 Integral Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 1 1-forms A dierential 1-form (or simply a dierential or a 1-form) on an open subset of R2is an expression F(x;y)dx+G(x;y)dywhereF;G areR-valued functions on the open set. A very important example of a dierential is given as follows: If f(x;y) isC1R-valued function on an open set U, then its total dierential (or exterior derivative) is df=@f @xdx+@f @ydy It is a dierential on U. In a similar fashion, a dierential 1-form on an open subset of R3is an expressionF(x;y;z )dx+G(x;y;z )dy+H(x;y;z )dzwhereF;G;H areR-valued functions on the open set. If f(x;y;z ) is aC1function on this set, then its total dierential is df=@f @xdx+@f @ydy+@f @zdz At this stage, it is worth pointing out that a dierential form is very similar to a vector eld. In fact, we can set up a correspondence: Fi+Gj+Hk$Fdx +Gdy+Hdz where i;j;kare the standard unit vectors along the x;y;z axes. Under this set up, the gradientrfcorresponds to df. Thus it might seem that all we are doing is writing the previous concepts in a funny notation. However, the notation is 2 very suggestive and ultimately quite powerful. Suppose that that x;y;z depend on some parameter t, andfdepends on x;y;z , then the chain rule says df dt=@f @xdx dt+@f @ydy dt+@f @zdz dt Thus the formula for dfcan be obtained by canceling dt. 2 Exactness in R2 Suppose that Fdx +Gdy is a dierential on R2withC1coecients. We will say that it is exact if one can nd a C2functionf(x;y) withdf=Fdx +Gdy Most dierential forms are not exact. To see why, note that the above equation is equivalent to F=@f @x; G=@f @y: Therefore if fexists then @F @y=@2f @y@x=@2f @x@y=@G @x But this equation would fail for most examples such as ydx. We will call a dierential closed if@F @yand@G @xare equal. So we have just shown that if a dierential is to be exact, then it had better be closed. Exactness is a very important concept. You've probably already encountered it in the context of dierential equations. Given an equation dy dx=F(x;y) G(x;y) we can rewrite it as FdxGdy = 0 IfFdxGdy is exact and equal to say, df, then the curves f(x;y) =cgive solutions to this equation. These concepts arise in physics. For example given a vector eld F= F1i+F2jrepresenting a force, one would like nd a function P(x;y) called the potential energy, such that F=rP. The force is called conservative (see section 8) if it has a potential energy function. In terms of dierential forms, F is conservative precisely when F1dx+F2dyis exact. 3 Parametric curves Before discussing line integrals, we have to say a few words about parametric curves. A parametric curve in the plane is vector valued function C: [a;b]!R2. In other words, we let xandydepend on some parameter trunning from ato b. It is not just a set of points, but the trajectory of particle travelling along the 3 curve. To begin with, we will assume that CisC1. Then we can dene the the velocity or tangent vector v= (dx dt;dy dt). We want to assume that the particle travels without stopping, v6= 0. Then vgives a direction to C, which we also refer to as its orientation . IfCis given by x=f(t); y=g(t);atb then x=f(u); y=g(u);bua will be calledC. This represents the same set of points, but traveled in the opposite direction. Suppose that Cis given depending on some parameter t, x=f(t); y=g(t) and thattdepends in turn on a new parameter t=h(u) such thatdt du6= 0. Then we can get a new parametric curve C0 x=f(h(u)); y=g(h(u)) It the derivativedt duis everywhere positive, we want to view the oriented curves CandC0as the equivalent. If this derivative is everywhere negative, then C andC0are equivalent. For example, the curves C:x= cos; y= sin;02 C0:x= sint; y= cost;0t2 represent going once around the unit circle counterclockwise and clockwise re- spectively. So C0should be equivalent to C. We can see this rigorously by making a change of variable ==2t. It will be convient to allow piecewise C1curves. We can treat these as unions ofC1curves, where one starts where the previous one ends. We can talk about parametrized curves in R3in pretty much the same way. 4 Line integrals Now comes the real question. Given a dierential Fdx+Gdy, when is it exact? Or equivalently, how can we tell whether a force is conservative or not? Checking that it's closed is easy, and as we've seen, if a dierential is not closed, then it can't be exact. The amazing thing is that the converse statement is often (although not always) true: THEOREM 4.1 IfF(x;y)dx+G(x;y)dyis a closed form on all of R2with C1coecients, then it is exact. 4 To prove this, we would need solve the equation df=Fdx +Gdy. In other words, we need to undo the eect of dand this should clearly involve some kind of integration process. To dene this, we rst have to choose a parametric C1 curveC. Then we dene: DEFINITION 4.2 Z CFdx +Gdy =Zb a F(x(t);y(t))dx dt+G(x(t);y(t))dy dt dt IfCis piecewise C1, then we simply add up the integrals over the C1pieces. Although we've done everything at once, it is often easier, in practice, to do this in steps. First change the variables from xandyto expresions in t, then replacedxbydx dtdtetc. Then integrate with respect to t. For example, if we parameterize the unit circle cbyx= cos;y= sin, 02, we see y x2+y2dx+x x2+y2dy=sin(cos)0d+ cos(sin)0d=d and therefore Z Cy x2+y2dx+x x2+y2dy=Z2 0d= 2 From the chain rule, we get LEMMA 4.3 Z CFdx +Gdy =Z CFdx +Gdy IfCandC0are equivalent, then Z CFdx +Gdy =Z C0Fdx +Gdy While we're at it, we can also dene a line integral in R3. Suppose that Fdx +Gdy+Hdz is a dierential form with C1coeents. Let C: [a;b]!R3 be a piecewise C1parametric curve, then DEFINITION 4.4 Z CFdx +Gdy+Hdz = Zb a F(x(t);y(t);z(t))dx dt+G(x(t);y(t);z(t))dy dt+H(x(t);y(t);z(t))dz dt dt The notion of exactness extends to R3automatically: a form is exact if it equalsdffor aC2function. One of the most important properties of exactness is its path independence: 5 PROPOSITION 4.5 If!is exact and C1andC2are two parametrized curves with the same endpoints (or more acurately the same starting point and ending point), thenZ C1!=Z C2! It's quite easy to see why this works. If !=dfandC1: [a;b]!R3then Z C1df=Zb adf dtdt by the chain rule. Now the fundamental theorem of calculus shows that the last integral equals f(C1(b))f(C1(a)), which is to say the value of fat the endpoint minus its value at the starting point. A similar calculation shows that the integral over C2gives same answer. If the Cis closed, which means that the starting point is the endpoint, then this argument gives COROLLARY 4.6 If!is exact and Cis closed, thenR C!= 0. Now we can prove theorem 4.1. If Fdx +Gdy is a closed form on R2, set f(x;y) =Z CFdx +Gdy where the curve is indicated below: (0,0)(x,0)(x,y) We parameterize both line segments seperately by x=t; y = 0 andx= x(constant ); y=t, and sum to get f(x;y) =Zx 0F(t;0)dt+Zy 0G(x;t)dt Then we claim that df=Fdx +Gdy. To see this, we dierentiate using the fundamental theorem of calculus. The easy calculation is @f @y=@ @yZy 0G(x;t)dt =G(x;y) 6 Slightly trickier is @f @x=@ @xZx 0F(x;0)dt+@ @xZy 0G(x;t)dt =F(x;0) +Zy 0@G(x;t) @xdt =F(x;0) +Zy 0@F(x;t) @tdt =F(x;0) +F(x;y)F(x;0) =F(x;y) The same proof works if if we replace R2by an open rectangle. However, it will fail for more general open sets. For example, y x2+y2dx+x x2+y2dy isC11-form on the open set f(x;y)j(x;y)6= (0;0)gwhich is closed. But it is not exact, since its integral along the unit circle is not 0. In more advanced treatments, this failure of closed forms to be exact can be measured by something called the de Rham cohomology of the set. 5 Work Line integrals have many important uses. One very direct application in physics comes from the idea of work. If you pick up a rock o the ground, or perhaps roll it up a ramp, it takes energy. The energy expended is called work. If you're moving the rock in straight line for a short distance, then the displacement can be represented by a vector d= (
x;
y;
z) and the force of gravity by a vector F= (F1;F2;F3). Then the work done is simply F
d=(F1
x+F2
y+F3
z): On the other hand, if you decide to shoot a rocket up into space, then you would have to take into account that the trajectory cmay not be straight nor can the force Fbe assumed to be constant (it's a vector eld). However as the notation suggests, for the work we would now need to calculate the integral Z cF1dx+F2dy+F3dz One often writes this as Z cF
ds (think ofdsas the \vector" ( dx;dy;dz ).) 7 6 Green's theorem for a rectangle LetDbe the rectangle in the xy-plane with vertices (0 ;0);(a;0);(a;b);(0;b). LetCbe the boundary curve of the rectangle oriented counter clockwise. Given C1functionsP(x;y);Q(x;y) onD, the fundamental theorem of calculus yields ZZ D@Q @xdxdy =Zb 0[Q(a;y))Q(0;y)]dy=Z CQ(x;y)dy ZZ D@P @ydydx =Za 0[P(x;b)P(x;0)]dx=Z CP(x;y)dx Subtracting yields Green's theorem for R THEOREM 6.1 Z CPdx +Qdy =ZZ D@Q @x@P @y dxdy Our goal is to understand, and generalize to 3 dimensions, the operation which takes the one form Pdx+Qdy to the integrand on the right. In traditional vector calculus this is handled using the curl(r) which a vector eld dened so that r(Pi+Qj+Rk)
k=@Q @x@P @y is the integrand of the right in Green's theorem. In general, one can discover the formula for the other components of r(Pi+Qj+Rk) by expressing the integrals of Pi+Qj+Rkaround the boundaries of rectangles in the xzandyz planes and rewriting them as double integrals. To make a long story short, r(Pi+Qj+Rk) = (RyQz)i+ (QxPy)k+ (PzRx)j (In practice, this is often written as a determinant r(Pi+Qj+Rk) =i j k @ @x@ @y@ @z P Q R: But this should really be treated as a memory aid and nothing more.) 72-forms Our goal in this section is to understand the operation Pdx +Qdy7!@Q @x@P @y dxdy in a more direct way than was done above. But rst we need to understand how to work with expressions of the form Fdxdy . In fact, for reasons that will 8 be clear later, we wish to introduce symbols dx^dy;::: which carries slightly more information; namely a sense of direction. The cross product of vectors uvis a very useful operation in 3 dimensional geometry. Its length gives the area of the parallelogram spanned by u;v, and it determines the plane containing this parallelogram. There is no direct analogue of the cross product in higher dimensions. However, there are certain features which do generalize. The essential idea is to dene new objects called 2-vectors which are to planes what ordinary vectors are to lines. A vector in the usual sense is a directed line segment. It is determined by its length, the line it lies on, and a choice of one of two possible directions along the line. The basic building block of a 2-vector is an oriented parallelogram (we usually omit the modier \oriented"). It has three attributes: its area, the plane on which it lies, and the orientation which is a choice of direction for walking around its perimeter. Two oriented parallelograms are considered equal if the areas are equal, the planes are parallel, and the orientations match. A parallelogram is equal to zero precisely when its area is, in which case the other attributes can be arbitrary. Given a parallelogram Pand a number c, we dene cPto be a parallelogram with the same plane, and area given by jcj
area(P) and the same orientation if c >0 and the opposite orientation if c <0. Given two vectors u;vwe dene the parallelogram u^vto be given as follows ( ^is pronounced \wedge"). uv Evidently v^u=u^v u^u= 0 and c(u^v) = (cu)^v=u^(cv) The sum of vectors can be dened geometrically using the parallelogram law. There is a limited version of this for 2-vectors: if two parallelograms share one side, then the remaining sides can be added using the parallelogram law. In terms of algebra, this is just the distributive law: u^v+u^w=u^(v+w) Unfortunately there is no simple geometric rule for adding more complicated pairs of parallelograms, so we just add them formally without worrying about 9 the meaning. A 2-vector is a nite sum of oriented parallelograms. The set of 2-vectors with these operations form a vector space . In other words, all the expected rules for + and
apply. We want to emphasize that the formalism of 2-vectors works in any dimension. In R3, we can identify 2-vectors with vectors viau^v!uv. However, we will usually refrain from doing this. A 2-form is like a 2-vector but built using forms. On R3, this would be an expression: F(x;y;z )dx^dy+G(x;y;z )dy^dz+H(x;y;z )dz^dx whereF;G andHare functions dened on an open subset of R3. Any wedge product of two 1-forms can be put in this format. For example, using the above rules, we can see that (3dx+dy)^(dx+ 2dy) = 6dx^dy+dy^dx= 5dx^dy The real signicance of 2-forms will come later when we do surface integrals. A 2-form will be an expression that can be integrated over a surface in the same way that a 1-form can be integrated over a curve. In order to make the comparison with traditional vector calculus, we note that we can convert vector elds to 2-forms and back F1i+F2j+F3k$F1dy^dz+F2dz^dx+F3dx^dy; Earler, we learned how to convert a vector eld to a 1-form: F1i+F2j+F3k$F1dx+F2dy+F3dz To complete the triangle, we can interchange 1-forms and 2-forms using the so called Hodge star operator. (F1dx+F2dy+F3dz) =F1dy^dz+F2dz^dx+F3dx^dy (F1dy^dz+F2dz^dx+F3dx^dy) =F1dx+F2dy+F3dz Given a 1-form F(x;y;z )dx+G(x;y;z )dy+H(x;y;z )dz. We want to dene its derivative d!which will be a 2-form. The rules we use to evaluate it are: d(+) =d+d d(f) = (df)^+fd d(dx) =d(dy) =d(dz) = 0 whereandare 1-forms and fis a function. Recall that df=fxdx+fydy+fzdz wherefx=@f @xand so on. Putting these together yields a formula d(Fdx+Gdy+Hdz) = (GxFy)dx^dy+(HyGz)dy^dz+(FzHx)dz^dx 10 A 2-form can be converted to a vector eld by replacing dx^dybyk=ij, dy^dzbyi=jkanddz^dxbyj=ki. If we start with a vector eld V=Fi+Gj+Hk, replace it by a 1-form Fdx +Gdy +Hdz, applyd, then convert it back to a vector eld, we end up with the curl of V rV= (HyGz)i+ (GxFy)k+ (FzHx)j 8 Exactness in R3and conservation of energy AC11-form!=Fdx +Gdy +Hdz is called exact if there is a C2function (called a potential) such that !=df. A 1-form !is called closed if d!= 0, or equivalently if Fy=Gx; Fz=Hx; Gz=Hy These equations must hold when F=fx; G=fy; H=fz Therefore: THEOREM 8.1 Exact 1-forms are closed. We have a converse statement which is sometimes called \Poincar e's lemma". THEOREM 8.2 If!=Fdx +Gdy +Hdz is a closed form on R3withC1 coecients, then !is exact. In fact if f(x0;y0;z0) =R C!, whereCis any piecewiseC1curve connecting (0;0;0)to(x0;y0;z0), thendf=!. This can be rephrased in the language of vector elds. If F=Fi+Gj+Hk isC1vector eld representing a force, then it is called conservative if there is a C2real valued function P, called potential energy, such that F=rP. The theorem implies that a force F, which isC1on all of R3, is conservative if and only ifrF= 0.P(x;y;z ) is just the work done by moving a particle of unit mass along a path connecting (0 ;0;0) to (x;y;z ). To appreciate the importance of this concept, recall from physics that the kinetic energy of a particle of constant mass mand velocity v=dx dt;dy dt;dz dt is K=1 2mjjvjj2=1 2mv
v: Also one of Newton's laws says mdv dt=F 11 IfFis conservative, then we can replace it by rPabove, move it to the other side, and then dot both sides by vto obtain mv
dv dt+v
rP= 0 which simplies1to d dt(K+P) = 0: This implies that the total energy K+Pis constant. 9 \d" of a 2-form and divergence Earlier we introduced 2-vectors which correspond to sums of oriented parallel- ograms. We also have 3-vectors which correspond to oriented parallelopipeds. Given three vectors u;v;w2R3, we think of the 3-vector u^v^was the oriented parallelopiped with u;v;was the rst, second and third sides. The only attribute that will distinguish one from another is the oriented volume, which is the usual volume if u;v;wis right-handed, otherwise it is minus the usual volume. With these rules, we see that u^v^w=v^u^w=v^w^u=::: A 3-form is simply an expression f(x;y;z )dx^dy^dz=f(x;y;z )dy^dx^dz=f(x;y;z )dy^dz^dx=::: These are things that will eventually get integrated over solid regions. The important thing for the present is an operation which takes 2-forms to 3-forms once again denoted by \d". d(Fdy^dz+Gdz^dx+Hdx^dy) = (Fx+Gy+Hz)dx^dy^dz It's probably easier to understand the pattern after converting the above 2- form to the vector eld V=Fi+Gj+Hk. Then the coeecient of dx^dy^dz is the divergence r
V=Fx+Gy+Hz So far we've applied dto functions to obtain 1-forms, and then to 1-forms to get 2-forms, and nally to 2-forms. The real power of this notation is contained in the following simple-looking formula PROPOSITION 9.1 d2= 0 1This takes a bit of work that I'm leaving as an exercise. It's probably easier to work backwards. You'll need the product rule for dot products and the chain rule. 12 What this means is that given a C2real valued function dened on an open subset of R3, thend(df) = 0, and given a 1-form !=Fdx +Gdy+Hdz with C2coecents dened on an open subset of R3,d(d!) = 0. Both of these are quite easy to check: d(df) = (fyxfxy)dx^dy+ (fzyfyz)dy^dz+ (fxzfzx)dz^dx= 0 d(d!) = [GxzFyz+HyxGzx+FzyHxy]dx^dy^dz= 0 In terms of standard vector notation this is equivalent to r(rf) = 0 r
(rV) = 0 The analogue of theorem 8.2 holds: THEOREM 9.2 If!is a2-form on R3such thatd!= 0, then there exists a 1-formsuch thatd=!. 10 Parameterized Surfaces Recall that a parameterized curve is a C1function from a interval [ a;b]R1to R3. If we replace the interval by subset of the plane R2, we get a parameterized surface. Let's look at a few of examples 1) The upper half sphere of radius 1 centered at the origin can be parame- terized using cartesian coordinates 8 >>< >>:x=u y=v z=p 1u2v2 u2+v21 2) The upper half sphere can be parameterized using spherical coordinates 8 >>< >>:x= sin() cos() y= sin() sin() z= cos() 0=2;0<2 3) The upper half sphere can be parameterized using cylindrical coordinates 8 >>< >>:x=rcos() y=rsin() z=p 1r2 0r1;0<2 An orientation on a curve is a choice of a direction for the curve. For a surface an orientation is a choice of \up" or \down". The easist way to make 13 this precise is to view an orientation as a choice of (an upward or outward pointing) unit normal vector eld nonS. A parameterized surface S 8 >>< >>:x=f(u;v) y=g(u;v) z=h(u;v) (u;v)2D is called smooth provided that f;g;h areC1, the function that they dene from D!R3is one to one, and the tangent vector elds Tu=@x @u;@y @u;@z @u Tv=@x @v;@y @v;@z @v are linearly independent. In this case, once we pick an ordering of the variables (sayurst,vsecond) an orientation is determined by the normal n=TuTv jjTuTvjj T uT vn FIGURE 1S v=constantu=constant 14 If we look at the examples given earlier. (1) is smooth. However there is a slight problem with our examples (2) and (3). Here T= 0, when = 0 in example (2) and when r= 0 in example (3). To deal with scenario, we will consider a surface smooth if there is at least one smooth parameterization for it. LetCbe a closed C1curve in R2andDbe the interior of C.Dis an example of a surface with a boundary C. In this case the surface lies at in the plane, but more general examples can be constructed by letting Sbe a parameterized surface 8 >>< >>:x=f(u;v) y=g(u;v) z=h(u;v) (u;v)2DR2 then the image of CinR3will be the boundary of S. For example, the boundary of the upper half sphere S 8 >>< >>:x= sin() cos() y= sin() sin() z= cos() 0=2;0<2 is the circle Cgiven by x= cos(); y= sin(); z= 0;02 In what follows, we will need to match up the orientation of Sand its boundary curve. This will be done by the right hand rule: if the ngers of the right hand point in the direction of C, then the direction of the thumb should be \up". n S C FIGURE 2 15 11 Surface Integrals LetSbe a smooth parameterized surface 8 >>< >>:x=f(u;v) y=g(u;v) z=h(u;v) (u;v)2D with orientation corresponding to the ordering u;v. The symbols dxetc. can be converted to the new coordinates as follows dx=@x @udu+@x @vdv dy=@y @udu+@y @vdv dx^dy=@x @udu+@x @vdv ^@y @udu+@y @vdv =@x @u@y @v@x @v@y @u du^dv =@(x;y) @(u;v)du^dv In this way, it is possible to convert any 2-form !touv-coordinates. DEFINITION 11.1 The integral of a 2-form onSis given by ZZ SFdx^dy+Gdy^dz+Hdz^dx=ZZ D F@(x;y) @(u;v)+G@(y;z) @(u;v)+H@(z;x) @(u;v) dudv In practice, the integral of a 2-form can be calculated by rst converting it to the form f(u;v)du^dv, and then evaluatingRR Df(u;v)dudv . LetSbe the upper half sphere of radius 1 oriented with the upward normal parameterized using spherical coordinates, we get dx^dy=@(x;y) @(;)d^d= cos() sin()d^d SoZZ Sdx^dy=Z2 0Z=2 0cos() sin()dd = On the other hand if use the same surface parameterized using cylindrical coordinates dx^dy=@(x;y) @(r;)dr^d=rdr^d ThenZZ Sdx^dy=Z2 0Z1 0rdrd = which leads to the same answer as one would hope. The general result is: 16 THEOREM 11.2 Suppose that an oriented surface Shas two dierent smooth C1parameterizations, then for any 2-form!, the expression for the integrals of !calculated with respect to both parameterizations agree. (This theorem needs to be applied to the half sphere with the point (0 ;0;1) removed in the above examples.) 12 Surface Integrals (continued) Complicated surfaces may be divided up into nonoverlapping patches which can be parameterized separately. The simplest scheme for doing this is to triangulate the surface, which means that we divide it up into triangular patches as depicted below. Each triangle on the surface can be parameterized by a triangle on the plane. boundary We will insist that if any two triangles touch, they either meet only at a vertex, or they share an entire edge. We dene the boundary of a surface to be the union of all edges which are not shared. The surface is called closed if the boundary is empty. Given a surface which has been divided up into patches, we can integrate a 2-form on it by summing up the integrals over each patch. However, we require that the orientations match up, which is possible if the surface has \two sides". Below is a picture of a one sided, or nonorientable, surface called the Mobius strip. 17 Once we have pick an orientation of S, we get one for the boundary using the right hand rule. In many situations arising in physics, one needs to integrate a vector eld F=F1i+F2j+F3kover a surface. The resulting quantity is often called a ux. We will simply dene this integral, which is usually written asRR SF
dS orRR SF
ndS, to mean ZZ SF1dy^dz+F2dz^dx+F3dx^dy It is probably easier to view this as a two step process, rst convert Fto a 2-form as follows: F1i+F2j+F3k$F1dy^dz+F2dz^dx+F3dx^dy; then integrate. As a typical example, consider a uid such as air or water. Associated to this, there is a scalar eld (x;y;z ) which measures the density, and a vector eld vwhich measures the velocity of the ow (e.g. the wind velocity). Then the rate at which the uid passes through a surface Sis given by the ux integralRR Sv
dS 13 Length and Area It is important to realize some line and surface integrals are not expressible as integrals of dierential forms in general. Two notable examples are the arclength and area integrals. DEFINITION 13.1 The arclength of C: [a;b]!R3is given by Z Cds=Zb as dx dt2 +dy dt2 +dz dt2 dt The symbol dsis not a 1-form in spite of the notation. For example Z Cds=Z Cds 18 whereas for 1-forms the integral would change sign. Nevertheless, ds(or more accurately its square) is a sort of generalization of a dierential form called a tensor . To get a sense what this means, let us calculate the arclength of a curve lying on a surface. Suppose that Sis a parameterized surface given by 8 >>< >>:x=f(u;v) y=g(u;v) z=h(u;v) (u;v)2D and suppose that Clies onS. This means that there are functions k;`: [a;b]! Rsuch thatx=f(k(t);`(t));:::determines C. We can calculate the arclength ofCby applying the chain to the above integral all at once. Instead, we want to break this down into a series of steps. dx=@x @udu+@x @vdv dy=@y @udu+@y @vdv dz=@z @udu+@z @vdv For the next step, we introduce a new product (indicated by juxtaposition) which is distributive and unlike the wedge product is commutative. We square the previous formulas and add them up. (The objects that we are getting are tensors.) dx2=@x @u2 du2+ 2@x @u@x @vdudv +@x @v2 dv2 dy2=@y @u2 du2+ 2@y @u@y @vdudv +@y @v2 dv2 dz2=@z @u2 du2+ 2@z @u@z @vdudv +@z @v2 dv2 dx2+dy2+dz2=Edu2+ 2Fdudv +Gdv2(1) where E=jjTujj2; F=Tu
Tv; G=jjTvjj2 in the notation of section 10. The expression in (1) is called the metric tensor of the surface. We can easily deduce a formula for arclength in terms of it: Z Cds=Zb as Edu dt2 + 2Fdu dtdv dt+Gdv dt2 dt The area of Scan also be expressed in terms of the metric tensor. First recall that 19 DEFINITION 13.2 The area of Sis given by ZZ SdS=ZZ DjjTuTvjjdudv THEOREM 13.3 The area is given by ZZ SdS=ZZ Dp EGF2dudv The proof is as follows jjTuTvjj2=jjTujj2jjTvjj2sin2 =jjTujj2jjTvjj2(1cos2) =jjTujj2jjTvjj2(Tu
Tv)2 =EGF2 IfSis sphere of radius 1 parameterized by spherical coordinates, a straight forward calculation gives the metric tensor as sin2d2+d2 which yields area(S) =Z2 0Z 0sindd = 4 as expected. 14 Green's and Stokes' Theorems Stokes' theorem is really the fundamental theorem of calculus of surface inte- grals. We assume that surfaces can be triangulated. THEOREM 14.1 (Stokes' theorem) LetSbe an oriented smooth surface with smooth boundary curve C. IfCis oriented using the right hand rule, then for anyC11-form!on an open set of R3containingS, ZZ Sd!=Z C! If the surface lies in the plane, it is possible make this very explicit: THEOREM 14.2 (Green's theorem) LetCbe a closedC1curve in R2ori- ented counterclockwise and Dbe the interior of C. IfP(x;y)andQ(x;y)are bothC1functions then Z CPdx +Qdy =ZZ D@Q @x@P @y dxdy 20 As an application, Green's theorem shows that the area of Dcan be com- puted as line integral on the boundary ZZ Ddxdy =Z Cydx IfSis a closed oriented surface in R3, such as the surface of a sphere, Stoke's theorem shows that any exact 2-form integrates to 0, where a 2-form is exact if it equalsd!for some 1-form !. To see this write Sas the union of two surfaces S1andS2with common boundary curve C. OrientCusing the right hand rule with respect to S1, then orientation coming from S2goes in the opposite direction. ThereforeZZ Sd!=ZZ S1d!+ZZ S2d!=ZZ C!ZZ C!= 0 In vector notation, Stokes' theorem is written as ZZ SrF
ndS=Z CF
ds where Fis aC1-vector eld. In physics, there a two fundamental vector elds, the electric eld Eand the magnetic eld B. They're governed by Maxwell's equations, one of which is rE=@B @t wheretis time. If we integrate both sides over S, apply Stokes' theorem and simplify, we obtain Faraday's law of induction: Z CE
ds=@ @tZZ SB
ndS To get a sense of what this says, imagine that Cis wire loop and that we are dragging a magnet through it. This action will induce an electric current; the left hand integral is precisely the induced voltage and the right side is related to the strength of the magnet and the rate at which it is being dragged through. Stokes' theorem works even if the boundary has several components. How- ever, the inner an outer components would have opposite directions. S C 1C 2 21 THEOREM 14.3 (Stokes' theorem II) LetSbe an oriented smooth sur- face with smooth boundary curves C1;C2:::. IfCiis oriented using the right hand rule, then for any C11-form!on an open set of R3containingS, ZZ Sd!=Z C1!+Z C2!+::: 15 Cauchy's theorem Recall that a complex number is an expression z=a+biwherea;b2Rand i=p1, so thati2=1. The components aandbare called the real and imaginary parts of z. We can identify the set of complex numbers Cwith the plane R2=f(a;b)ja;b2Rg. Addition and subtraction of complex numbers correspond to the usual vector operations: (a+bi)(c+di) = (ac) + (bd)i However, we can do more, such as multiplication and division: (a+bi)(c+di) = (acbd) + (ad+bc)i a+bi c+di=(a+bi)(cdi) (c+di)(cdi)=ac+bd c2+d2+bcad c2+d2i The next step is calculus. The power of complex numbers is evident in the beautiful formula of Euler ei= cos() +isin() which unies the basic functions of calculus. Given a function f:C!C, we can write it as f(z) =f(x+yi) =u(x;y) +iv(x;y) wherex;yare real and imaginary parts of z2C, andu;vare the real and imaginary parts of f.fis continuous at z=a+biifuandvare continuous at (a;b) in the usual sense. So far there are no surprises. However, things get more interesting when we dene the complex derivative f0(z) = lim h!0f(z+h)f(z) h Notice that his a complex number. For the limit to exist, we should get the same value no matter how it approaches 0. If h=
xapproaches along the x-axis, we get f0(z) = lim
y!0[u(x+
x;y)u(x;y)] +i[v(x+
x;y)v(x;y)]
x =@u @x+i@v @x 22 Ifh=
yiapproaches along the y-axis, then f0(z) = lim
y!0[u(x;y+
y)u(x;y)] +i[v(x;y+
y)v(x;y)] i
y =i@u @y+@v @y Setting these equal leads to the Cauchy-Riemann equations @u @x=@v @y;@v @x=@u @y These equations have to hold when the complex derivative f0(z) exists, and in factf0(z) exists when they do. fis called analytic atz=a+biwhen these hold at that point. For example, z2=x2+y2+ 2xyiand ez=excos(y) +iexsin(x) are analytic everywhere. But f(z) = z=xiyis not analytic anywhere. A complex dierential form is an expression +iwhere;are dierential forms in the usual sense. Complex 1-forms can be integrated by the rule Z C+i=Z C+iZ C Suppose that fis analytic. Then expanding f(z)dz= (u+iv)(dx+idy) = [udxvdy] +i[vdx+udy] Dierentiating and applying the Cauchy-Riemann equations shows d(f(z)dz) =@u @y+@v @x dx^dy+i@u @x@v @y dx^dy= 0 Therefore Stokes' theorem implies what may be thought of as the fundamental theorem of complex analysis: THEOREM 15.1 (Cauchy's theorem) Iff(z)is analytic on a region with boundaryCthen Z Cf(z)dz= 0 Suppose we replace f(z) byg(z) =f(z) z. This is analytic away from 0. Therefore the theorem applies to the boundary of any region not containing 0. IfCis a closed positively oriented curve whose interior Ucontains 0, then applying Cauchy's theorem to a UDr,whereDris a disk of small radius rin U, shows that Z Cg(z)dz=Z Crg(z)dz (2) 23 HereCris a circle of radius raround 0. We can parameterize this with the help of Euler's formula by z=rcos() +risin() =rei;02 Thendz=rieid, so that Z Crg(z)dz=riZ2 0f(rei) reieid=iZ2 0f(rei)d Asr!0,f(rei)!f(0), therefore the above integral approaches 2 r. Since (2) holds for all small r, it follows that this equality holds on the nose. Therefore: THEOREM 15.2 (Cauchy's Integral Formula) Iff(z)is analytic in the interior of positively oriented closed curve C, then f(0) =1 2iZ Cf(z) zdz Using a change of variable z!za, we get a more general formula: COROLLARY 15.3 (Cauchy's Integral Formula II) With the same as- sumptions, for any point ain the interior of C f(a) =1 2iZ Cf(z) zadz This formula has many uses. Among other things, it can be used to evaluate complicated denite integrals. This and more can be found in any book on complex analysis. 16 Triple integrals and the divergence theorem Recall that a 3-form is an expression f(x;y;z )dx^dy^dz. Given a solid region VR3, we dene ZZZ Vf(x;y;z )dx^dy^dz=ZZZ Vf(x;y;z )dxdydz THEOREM 16.1 (Divergence theorem) LetVbe the interior of a smooth closed surface Soriented with the outward pointing normal. If !is aC12-form on an open subset of R3containingV, then ZZZ Vd!=ZZ S! 24 In standard vector notation, this reads ZZZ Vr
FdV=ZZ SF
ndS where Fis aC1vector eld. As an application, consider a uid with density and velocity v. IfSis the boundary of a solid region Vwith outward pointing normal n, then the uxRR Sv
ndSis the rate at which matter ows out of V. In other words, it is minus the rate at which matter ows in, and this equals @=@tRRR VdV. On the other hand, by the divergence theorem, the above ux integral equalsRRR Sr
(v)dV. Therefore ZZZ Vr
(v)dV=@ @tZZZ VdV which yieldsZZZ V r
(v) +@ @t dV= 0: The only way this can hold for all possible regions Vis that the integrand r
(v) +@ @t= 0 (3) This is one of the basic laws of uid mechanics. We can extend the divergence theorem to solids with disconnected boundary. Suppose that S2is a smooth closed oriented surface contained inside another such surface S1. We use the outward pointing normal S1and the inner pointing normal onS2. LetVbe region in between S1andS2. Then, THEOREM 16.2 (Divergence theorem II) If!is aC12-form on an open subset of R3containingV, then ZZZ Vd!=ZZ S1!+ZZ S2! 17 Gravitational Flux Place a \point particle" of mass mat the origin of R3, then this generates a force on any particle of unit mass at r= (x;y;z ) given by F=mr r3 wherer=jjrjj=p x2+y2+z2. This has singularity at 0, so it is a vector eld onR3f0g. The corresponding 2-form is given by !=mxdy^dz+ydz^dx+zdx^dy r3 25 LetBRbe the ball of radius Raround 0, and let SRbe its boundary. Since the outward unit normal to SRis just n=r=r. One might expect that the ux ZZ SRF
ndS=m R2ZZ SRdS =m R2area(SR) =m R2(4R2) However, this is really just a proof by notation at this point. To justify it, we work in spherical coordinates. SRis given by 8 >>< >>:x=Rsin() cos() y=Rsin() sin() z=Rcos() 0;0<2 Rewriting!in these coordinates and simplifying: !=mR3sin()d^d Therefore, ZZ SRF
ndS=ZZ SR!=4m as hoped. We claim that if Sis any closed surface not containing 0 ZZ S!= 4m if 0 lies in the interior of S 0 otherwise By direct calculation, we see that d!= 0. Therefore, the divergence theorem yields ZZ S!=ZZZ Vd!= 0 if the interior Vdoes not contain 0. On the other hand, if Vcontains 0, let BR be a small ball contained in V, and letVBRdenote part of Vlying outside ofBR. We use the second form of the divergence theorem ZZ S!ZZ SR!=ZZZ VBRd!= 0 We are subtracting the second surface integral, since we are supposed to use the inner normal for SR. ThusZZ S!=4m Incidentally, this shows that !is not exact. Thus theorem 9.2 fails for R3f0g. From here, we can easily extract an expression for the ux for several par- ticles or even a continuous distribution of matter. If Fis the force of gravity 26 associated to some mass distribution, for any closed surface Soriented by the outer normal, then the ux ZZ SF
ndS is4times the mass inside S. For a continuous distribution with density , this is given byRRR dV. Applying the divergence theorem again, in this case, yields ZZZ V(r
F+ 4)dV= 0 for all regions V. Therefore r
F=4 (4) 18 Laplace's equation The Laplacian is a partial dierential operator dened by
f=@2f @x2+@2f @y2+@2f @z2 This can expressed using previous operators as
f=r
(rf). As an example of where this arises, suppose that Fis a gravitational force, this is known to be conservative so that F=rP. Substituting into (4) yields the Poisson equation
P= 4 In a vacuum, this reduces to Laplace's equation
P= 0 A solution to Laplace's equation is called a harmonic function. These are of fundamental importance both in pure and applied mathematics. If we write r=p x2+y2+z2, then P=m r+Const: is the potential energy associated to particle of mass mat0. This is harmonic away from the singularity 0. We express
in terms of forms as
fdx^dy^dz=ddf or simply
f=ddf once we dene(gdx^dy^dz) =g. This last formula also works in the plane provided we dene (fdx+gdy) =fdygdx (fdx^dy) =f 27 (The-operator in ndimensions always takes p-forms to (np)-forms.) As an exercise, let us work out the Laplace equation in polar coordinates, and use this to determine the radially symmetric harmonic functions on the plane. The key is the determination of the -operator: dr=@r @xdx+@r @ydy=x rdx+y rdy Similarly d=y r2dx+x r2dy So that dr=x rdyy rdx=rd d=y r2dyx r2dx=1 rdr (dr^d) =(1 rdx^dy) =1 r Thus
f=d@f @rdr+@f @d =d r@f @rd1 r@f @dr =1 r@f @r+@2f @r2+1 r2@2f @2 Iffis radially symmetric, then it depends only on rso we obtain 1 rdf dr+d2f dr2=1 rd dr rdf dr = 0 This dierential equation can be solved using standard techniques to get f(r) =C+Dlogr for constants C;D. By a similar, but more involved, calculation we nd that f(r) =C+D r are the only radially symmetric harmonic functions in R3, where as above we writerinstead offor the distance from the origin. These are precisely the physical solutions written at the beginning of this section. 28 19 Beyond 3dimensions It is possible to do calculus in Rnwithn>3. Here the language of dierential forms comes into its own. While it would be impossible to talk about the curl of a vector eld in, say, R4, the derivative of a 1-form or 2-form presents no problems; we simply apply the rules we've already learned. For example, if x;y;z;t are the coordinates of R4, then a 1-form is a linear combination of the 4 basic 1-forms dx; dy; dz; dt a forms is a linear combination of the 6 basic 2-forms dx^dy=dy^dx dx^dz=dz^dx ::: dz^dt=dt^dz and a 3 form is a linear combination of dx^dy^dz=dy^dx^dz=dx^dz^dy=dy^dz^dx=::: ::: dy^dz^dt=dz^dy^dt=::: The higher dimensional analogue of a surface is a k-manifold . A parameter- izedk-manifoldMinRnis given by a collection of C1functions 8 >>>>< >>>>:x1=f1(u1;:::uk) x2=f2(u1;:::uk) ::: xn=fn(u1;:::uk) (u1;:::uk)2DRkopen such that the map D!Rnis one to one and the tangent vectors (@x1 @u1;:::@xn @u1);::: (@x1 @uk;:::@xn @uk) are linearly independent for all values of the coordinates ( u1;:::uk). Given ak-formonRn, we can express it as a linear combination of k-fold wedges ofdx1:::dxn, and then rewrite it as g(u1;:::uk)du1^:::^duk. The \surface" integral is dened as Z M=Z :::Z Dg(u1:::uk)du1:::duk (5) Notice that number of integrations is usually suppressed in the notation on the left, since it gets too cumbersome after a while. In practice, for computing inte- grals, it's convenient to relax the conditions a bit by allowing Dto be nonopen and allowing some degenerate points where the map D!Rnisn't one to one. 29 More generally, a k-manifoldMis obtained by gluing several parameterized manifolds as we did for surfaces. To be more precise, a closed set MRn is ak-manifold, if each point of Mlies in the image of a parameterized k- manifold called a chart. As with curves and surfaces, it is important to specify orientations. Things are a little trickier since we can no longer rely on our geometric intuition to tell us which way is \up" or \down". Instead we can think that an orientation is a rule for specifying whether a coordinate system on a chart is right or left handed. We'll spell this out in an example below. The integralR Mcan be dened by essentially summing up (5) over various non- overlapping right handed charts (we can use left handed charts provided we use the opposite sign). A k-manifold with boundary Mis a closed set which can be decomposed as a union of a ( k1)-manifold @M, called the boundary, and a k- manifoldM@M. We orient this by the rule that a coordinate system u2;:::uk of@Mis right handed if it can be completed to right handed coordinate system u1;u2;:::ukofMsuch that the tangent vector (@x1 @u1;:::@xn @u1) \points out". Then the ultimate form of Stokes' theorem is: THEOREM 19.1 (Generalized Stokes' theorem) IfMis an oriented k manifold with boundary @M and ifis a(k1)-form dened on (an open set containing) M, thenZ Md=Z @M In order to get a feeling for how this works, let's calculate the \volume" V of the 4-dimensional ball B=f(x;y;z;t )jx2+y2+z2+t2Rgof radius RinR4in two ways. This is a 4-manifold with boundary S=f(x;y;z;t )j x2+y2+z2+t2=Rg. Vcan be expressed as the integral V=ZZZZ Bdxdydzdt =Z Bdx^dy^dz^dt We use extended spherical coordinates ;;; , wheremeasures the distance of (x;y;z;t ) to the origin in R4, and the angle to the t-axis. So that t=cos and =sin is the distance from from the projection ( x;y;z ) to the origin. Then letting ; be the remaining spherical coordinates gives x=sincos=sin sincos y=sinsin=sin sinsin z=cos=sin cos 30 ρt ψ xyσ zIn these coordinate Bis described as 8 >>< >>:0  0 02 0R To simply computations, we note that form will get multiplied by the Jacobian when we change coordinates: dx^dy^dz^dt=@(x;y;z;t ) @(; ;; )d^d ^d^d =3sin2 sind^d ^d^d Note that the Jacobian is positive, and this what it means to say the coordinate system; ;; is right handed or positively oriented. The volume is now easily computed ZR 03dZ 0sin2 d Z2 0dZ 0sind=1 22R4 Alternatively, we can use Stokes' theorem, to see that V=Z Bdx^dy^dz^dt=Z Stdx^dy^dz The parameter x;y;z gives a left hand coordinate system on the upper hemi- sphereU=S\ft>0g. It is left handed because t;x;y;z is left handed on R4. 31 For similar reasons, x;y;z gives a right handed system on the lower hemisphere Lwheret<0. Therefore V=Z Utdx^dy^dzZ Ltdx^dy^dz = 2ZZZ x2+y2+z2Rp R2x2y2z2dxdydz = 2ZR 02p R22dZ 0sindZ2 0d = 8Z=2 0R4sin2cos2d =1 22R4 20 Maxwell's equations in R4 As exotic as higher dimensional calculus sounds, there are many applications of these ideas outside of mathematics. For example, in relativity theory one needs to treat the electric E=E1i+E2j+E3kand magnetic elds B=B1i+B2j+B3k as part of a single \eld" on space-time. In mathematical terms, we can take space-time to be R4- with the fourth coordinate as time t. The electromagnetic eld can be represented by a 2-form F=B3dx^dy+B1dy^dz+B2dz^dx+E1dx^dt+E2dy^dt+E3dz^dt If we compute dFusing the analogues of the rules we've learned: dF=@B3 @xdx+@B3 @ydy+@B3 @zdz+@B3 @tdt ^dx^dy+::: =@B1 @x+@B2 @y+@B3 @z dx^dy^dz+@E2 @x@E1 @y+@B3 @t dx^dy^dt+::: Two of Maxwell's equations for electromagnetism r
B= 0;rE=@B @t can be expressed very succintly in this language as dF= 0. The analogue of theorem 9.2 holds for Rn, and shows that F=d(A1dx+A2dy+A3dz+A4dt) for some 1-form called the potential. Thus we've reduced the 6 quantites to just 4. In terms of vector analysis this amounts to the more complicated looking equations B=r(A1i+A2j+A3k);E=rA4@A1 @ti+@A2 @tj+@A3 @tk 32 There are two remaining Maxwell equations r
E= 4;rB=@E @t+ 4J whereis the electric charge density, and Jis the electric current. The rst law is really an analog of (4) for the electric eld. After applying the divergence theorem, it implies that the electric ux through a closed surface equals ( 4) times the electric charge inside it. These last two Maxwell equations can also be replaced by the single equation dF= 4Jof 3-forms. Here F=E3dx^dy+E1dy^dz+E2dz^dxB1dx^dtB2dy^dtB3dz^dt and J=dx^dy^dzJ3dx^dy^dtJ1dy^dz^dtJ2dz^dx^dt (We have been relying on explicit formulas to avoid technicalities about the denition of the-operator. In principle however, it involves a metric, and in this case we use the so called Lorenz metric.) Let's see how the calculus of dierential forms can be used to extract a physically meaningful consequence of these laws. Proposition 9.1 (in extended form) implies that dJ=1 4d2F= 0. Expanding this out yields @ @tdt^dx^dy^dz@J3 @zdz^dx^dy^dt+:::= @ @t+r
J dx^dy^dz^dt= 0 Thus the expression in brackets is zero. This really an analog of the equation (3). To appreciate the meaning integrate@ @tover a solid region Vwith boundary S. Then this equals ZZZ Vr
JdV=ZZ SJ
ndS In other words, the rate of change of the electric charge in Vequals minus the ux of the current accross the surface. This is the law of conservation of electric charge. 21 Further reading For more information about dierential forms, see the books [Fl, S]. All the physics background can be found in [Fe]. A standard reference for complex analysis is [A]. The material of the appendix can be found in any book on advanced calculus. For a rigorous treatment, see [R, S]. 33 References [A] L. Ahlfors, Complex Analysis [Fe] R. Feynman et. al., Lectures on Physics [Fl] H. Flanders, Dierential forms and applications to the physical sciences [R] W. Rudin, Principles of mathematical analysis [S] M. Spivak, Calculus on manifolds A Essentials of multivariable calculus A.1 Dierential Calculus To simplify the review, we'll stick to two variables, but the corresponding state- ments hold more generally. Let f(x;y) be a real valued function dened on open subset of R2. Recall that the limit lim (x;y)!(a;b)f(x;y) =L means that f(x;y) is approximately Lwhenever (x;y) is close to ( a;b). The precise meaning is as follows. If we specied  > 0 (say= 0:0005), then we could pick a tolerance  >0 which would guarantee that jf(x;y)Lj<  (i.e.f(x;y) agrees with Lup to the rst 3 digits for = 0:0005) whenever the distance between ( x;y) and (a;b) is less than . A function f(x;y) iscontinuous at (a;b) if lim (x;y)!(a;b)f(x;y) exists and equals f(a;b). It is continuous if it is so at each point of its domain. We say that fisdierentiable , if near any point p= (x0;y0;f(x0;y0)) the graphz=f(x;y) can approximated by a plane passing through p. In other words, there exists quantities A=A(x0;y0);B=B(x0;y0) such that we may write f(x;y) =f(x0;y0) +A(xx0) +B(yy0) +remainder withremainder!0 as (x;y)!(x0;y0). We can see that the coecients are nothing but the partial derivatives A(x;y) =@f @x= lim h!0f(x+h;y)f(x;y) h B(x;y) =@f @y= lim h!0f(x;y+h)f(x;y) h 34 There is a stronger condition which is generally easier to check. fis called continuously dierentiable or C1if it and its partial derivatives exist and are continuous. Consider the following example f(x;y) =( x3 x2+y2if (x;y)6= (0;0) 0 if (x;y) = (0;0) This is continuous, however @f @x=3x2 x2+y22x4 (x2+y2)2 has no limit as ( x;y)!(0;0). To see this, note that along the x-axisy= 0, we have@f @x= 1. So the limit would have to be 1 if it existed. On the other hand, along they-axisx= 0,@f @x= 0, which shows that there is no limit. So f(x;y) is notC1. Partial derivatives can be used to determine maxima and minima. THEOREM A.2 If(a;b)is local maximum or minimum of a C1function f(x;y), then (a;b)is a critical point, i.e. @f @x(a;b) =@f @y(a;b) = 0 THEOREM A.3 (Chain Rule) Iff;g;h :R2!RareC1functions, then f(g(u;v);h(u;v))is alsoC1and ifz=f(x;y),x=g(u;v),y=h(u;v)then @z @u=@z @x@x @u+@z @y@y @u @z @v=@z @x@x @v+@z @y@y @v A function f(x;y) is twice continously dierentiable or C2if isC1and if its partial derivatives are also C1. We have the following basic fact: THEOREM A.4 Iff(x;y)isC2then the mixed partials @2f @y@x=@ @y@f @x @2f @x@y=@ @x@f @y are equal. IffisC2, then we have a Taylor approximation f(x;y)f(a;b) +@f @x(a;b) (xa) +@f @y(a;b) (yb) +@2f @x2(a;b) (xa)2 +2@2f @y@x(a;b) (xa)(yb) +@2f @y2(a;b) (yb)2 35 Since it is relatively easy to determine when quadratic polynomials have maxima or minima, this leads to the second derivative test. THEOREM A.5 A critical point (a;b)of aC2functionf(x;y)is a local minimum (respectively maximum) precisely when the matrix @2f @x2(a;b)@2f @y@x(a;b) @2f @y@x(a;b)@2f @y2(a;b)! is positive (respectively negative) denite. The above conditions are often formulated in a more elementary but ad hoc way in calculus books. Positive deniteness is equivalent to requiring @2f @x2(a;b)>0 @2f @x2(a;b)@2f @y2(a;b) @2f @y@x(a;b)2 >0 A.6 Integral Calculus Integrals can be dened using Riemann's method. This has some limitations but it's the easiest to explain. Given a rectangle R= [a;b][c;d]R2, choose integersm;n> 0 and let
x=ba m
y=dc n. Choose a set of sample points P=f(x1;y1);:::(xm;yn)gRwith (xi;yj)2Rij= [a+ (i1)
x;a+i
x][b+ (j1)
y;b+j
y] The Riemann sum S(m;n;P ) =X i;jf(xi;yj)
x
y Then the double integral is ZZ Rf(x;y)dxdy = lim m;n!1S(m;n;P ) This denition is not really that precise because we need to choose Pfor each pairm;n. For the integral to exist, we really have to require that the limit exists for any choice of P, and that any two choices lead to the same answer. The usual way to resolve the above issues is to make the two extreme choices. Dene upper and lower sums U(m;n) =X i;jMij
x
y L(m;n) =X i;jmij
x
y 36 where Mij= maxff(x;y)j(x;y)2Rijg mij= minff(x;y)j(x;y)2Rijg In the event that the maxima or minima don't exist, we should use the greatest lower bound and least upper bound instead. As m;n!1 the numbers L(m;n) tend to increase. So their limit can be understood as the least upper bound, i.e. the smallest number LL(m;n). Likewise we dene the limit Uas the largest numberUU(m;n). If these limits coincide, the common value is taken to be ZZ Rf(x;y)dxdy =L=U otherwise the (Riemann) integral is considered to not exist. THEOREM A.7RR Rf(x;y)dxdy exists iffis continuous. The integral of f(x;y) =( 1 if (x;y) has rational coordinates 0 otherwise would be undened from the present point of view, because L= 0 andU= 1. Although, in fact the integral can be dened using the more powerful Lebesgue theory [R]; in this example the Lebesgue integral is 0. For more complicated regions DR, set ZZ Df(x;y)dxdy =ZZ Rf(x;y)D(x;y)dxdy whereD= 1 insideDand 0 elsewhere. The key result is THEOREM A.8 (Fubini) IfD=f(x;y)jaxb; g(x)yh(x)gwith f;g;h continuous. Then the double integral exists and ZZ Df(x;y)dxdy =Zb a Zh(x) g(x)f(x;y)dy! dx A similar statement holds with the roles of xandyinterchanged. This allows one to compute these integrals in practice. The nal question to answer is how double integrals behave under change of variables. Let T:R2!R2be a transformation given by C1functions x=f(u;v); y=g(u;v) We think of the rst R2as theuv-plane and the second as the xy-plane. Given a regionDin theuv-plane, we can map it to the xy-plane by T(D) =f(f(u;v);g(u;v))j(u;v)2Dg 37 The Jacobian @(x;y) @(u;v)=@x @u@y @v@x @v@y @u =@x @u@x @v@y @u@y @v: THEOREM A.9 IfTis a one to one function, his continuous and Da region of the type occurring in Fubini's theorem, then ZZ T(D)h(x;y)dxdy =ZZ Dh(f(u;v);g(u;v))@(x;y) @(u;v)dudv 38