and likewise fdy = and we have fdx = f((x, g(x))) 1 dx. (0.1)

Transcription

1 On line integrals in R 2 and Green s formulae. How Cauchy s formulae follows by Green s. Suppose we have some curve in R 2 which can be parametrized by t (ζ 1 (t), ζ 2 (t)), where t is in some interval and ζ : R 2 is a C 1 -function. We denote the independent variable in R 2 by (x, y). f f is some function on then we define fdx = f(ζ(t))ζ 1(t)dt, and likewise fdy = f(ζ(t))ζ 2(t)dt. Exercise: Prove that these definitions are independent of the parametrization, except for that the sign depends on the orientation, that is, which end of that is considered as starting point and which is endpoint. To avoid technicalities, we assume that each curve comes with an orientation which is implicit in the notation. The choice of notation dx and dy on the left is natural. To see this, suppose that is the graph of a function g on some interval on R. A parametrization is then given by ζ(x) = (x, g(x)) and we have (.1) fdx = f((x, g(x))) 1 dx. n other words, if we make a partition (x k ) K k= of and write this integral as the limit of Riemann sums, then the infinitesimal weight at a point x k associated with the contribution f((x k, g(x k ))) is precisely x k+1 x k, i.e. the length of the projection onto the x-axis of the piece of above [x k, x k+1 ]). Now imagine a toy domain and let be the boundary (with counterclockwise orientation). Let u C 1 () be given and integrate e.g. y u over, using area-measure (which is the same as Lebesgue-measure which is the same as dx dymeasure ). We then obtain the following, which we leave as an exercise for you to verify on your toy domain: y u dxdy = u dx. Hint: This is just an application of the fundamental theorem of analysis and Fubini s theorem, i.e. the fact that f darea = f dxdy = f dydx, which holds under mild conditions on f. (But not always!! Read integration theory one day!!) 1

2 2 Now, the same game as above can obviously be played with x and y interchanged, yielding x v dxdy = v dy, where v is some other function. Green s theorem is nothing but the two above formulas combined into one by writing (.2) u dx + v dy = y u + x v darea. All this can be formalized to higher dimensions, and is then called Stokes formula. Conceptually it is the same thing, but the necessary mathematical formalism that comes along is terrible. We leave this aside and focus on the connection with line integrals in C. Note that all the above can be transported to C by identifying it with R 2. Also, we can use the above definitions on functions with values in C, by simply treating the real and imaginary parts separately. Thus (.2) holds in this case as well. We will now focus on the connection between the above integrals and the integrals fdz, which we have recently learned in complex analysis. Note that the definition of fdz can be applied to any (continuous) function f : C C, holomorphic or not. Since z = x + iy, it would be nice if f dz = f dx + i f dy. We leave it as an exercise to verify the above identity. However, we stress that the above identity needs to be proven, since the left and right hand sides have been defined independently of each other. Green s formula can now be reformulated as (.3) f dz = y f + i x f darea (another exercise...) Now, we write f = u + iv. f f is holomorphic, then we know that it satisfies the Cauchy-Riemann equations: x u = y v and y u = x v. Exercise: Show that these two precisely amount to the fact that the right-hand side in (.3) is. We thus conclude that f dz = for holomorphic functions. But this is Cauchy s theorem!!!! This theorem is thus nothing but a consequence of Green s theorem! Why does the book prove it via Goursat s theorem? suppose there are several reasons: the students may not know Green s theorem, and even if they do the intuition is clearly lost. One wants to give a self-contained holomorphic proof where the integrals fdz appear as natural objects when studying holomorphic functions. But pedagogics aside, Goursat s theorem is stronger than what one gets from applying Green s theorem, since the latter assumes that f is continuously differentiable.

3 (Goursat only needs that the partial derivatives exist at all points, but there is no demand that they be continuous...) But why is our definition of f dz natural. First of all, it mimics that of Riemann s definition for integrals on R. To see this, let (z k ) K k= be an enumeration of consecutive nearby points on, (starting at the starting point and ending at the ending point). Then K f dz f(z k )(z k+1 z k ), k= and the limit of the right-hand side as the enumeration gets finer becomes the integral. (Exercise: Show this). This is the only possible definition of a line integral if we want the fundamental theorem of analysis to extend to C. That is, if f is holomorphic and is a small straight segment between nearby points z and w, then (.4) f(z)dz f(z )(w z ), and the derivative of this (as a function of w) then becomes the limit of f(z)dz w z as w z. We see that we need (.4) in order for this limit to yield f(z ), in analogy with the fundamental theorem of analysis. The connection between line integrals in R 2 and C. So far we have defined integration over a curve against dz, dx and dy, and we have also seen that the first can be expressed as a linear combination of the two latter. However, whereas we argued above that integration of analytic functions against dz is natural, the line integrals against dx and dy are not natural. For example, they change under rotations, which is not natural for many physical applications. However, there is one type of integration along curves in R 2 which is natural, namely that of arc-length integration, which is completely independent of e.g. how we choose to put the coordinate axes. As well you may have encountered that can integrate vector fields, and clearly a complex function f : C C can be seen as a vector field over R 2, (by identifying f = u + iv with (u, v)). Let us see how these concepts are connected. n the following, is a curve in C, and we fix a parametrization ζ : [, 1] C. First of all, integrating a function f over against arc-length is defined as (.5) 1 f(ζ(t)) dt. We now need to do 3 things; verify that this definition is independent of ζ, verify that the infinitesimal weight corresponds to the length of the corresponding curve segment (in analogy with the argument following (.1)), and finally we need to find a notation for this new integral. The latter is the hardest, and we leave the other two as exercises. n real applications, people often denote the integral (.5) by f ds, 3

4 4 although it is not clear what s refers to. (m is another popular choice...) Many times, s is referred to as the arc-length measure, which sounds good but one needs to understand integration theory quite well to make sense out of it. n any case, it is just notation, since the definition (.5) is clear. n complex analysis people usually write f dz for (.5), which is quite natural when comparing with the definition of f dz; 1 f dz = f(ζ(t))ζ (t) dt. We will use both notations depending on the circumstance. As long as f is just a function, there is no way of relating f dz with f dx or f dy, i.e. the former can not be written as a function of the latter by any formula. However, if we consider vector fields, things change. Let (P (x, y), Q(x, y)) be a vector field and a curve in R 2 parameterized by ζ : [, 1] R 2. Typically (P, Q) is thought of as describing some static flow. Then 1 P dx + Qdy = P (ζ(t))ζ 1(t) + Q(ζ(t))ζ 2(t) dt = (.6) 1 (P, Q), (ζ 1(t), ζ 2(t)) dt = (P, Q), (ζ 1(t), ζ 2(t)) Above, the brackets refer to scalar product and the dependence of (P, Q) on ζ is left implicit for readability. Note that (ζ 1 (t),ζ 2 (t)) is a unit vector pointing in the direction of. Thus P dx + Qdy is the amount of flow along. f is a closed curve and we assume that there is no rotation, then P dx + Qdy =. As an exercise, use Green s theorem to show that this happens if (.7) y P = x Q. (t is also true that this happens for all closed curves only if (.7) holds, which requires some more work to show...) n a similar fashion as above, show that we have (.8) P dy Qdx = (P, Q), (ζ 2(t), ζ 1(t)) ds which, since (ζ 2 (t), ζ 1 (t)) is a unit vector perpendicular to the curve, can be interpreted as the amount of flow across. Again, if is closed it is natural to assume that this flow is, since otherwise we d have sources inside the domain. Via Green s formulae, the assumption that there are no sources thus boils down to (.9) x P = y Q, which we leave as another exercise. Now, (.7) and (.9) combined look like the Cauchy-Riemann equations. However, the attentive eye has spotted that the minus-sign sits on the wrong place. This has ds.

5 several consequences which are important for the conceptual understanding of what is going on: f we identify the vector field (P, Q) with the complex function f = P + iq, we do not get an analytic function! f we think of an analytic function f = u + iv as the vector field (u, v), then the two real integrals in f dz have no physical interpretation, although they very much resemble (.6) and (.8) How do we get out of this mess? The solution is to identify an analytic function f = u + iv with the vector field (P, Q) = (u, v), i.e. f = P iq! This way, f is analytic if and only if (P, Q) corresponds to an irrotational sourcefree vector field, and f dz = (P, Q), t dz + (P, Q), n dz where t denotes the unit tangent vector to and n the unit normal vector (we continue to ignore issues of orientation, i.e. ±). n other words, the real and imaginary parts correspond to the physically relevant integrals (.6) and (.8). This connection with physics and analytic functions is actually not very important for the theory of analytic functions, thinking of analytic functions as vector fields confuses more than it helps, but it is certainly something that one should have thought through at least once. A more elaborate explanation of these issues, with examples, exercises, connections to electric potential theory and the role of harmonic functions, is given in Chapter, Sections 6,7, of Gamelin s book. Finally, although physics certainly exists in (at least) 3 dimensions as well, it is important to know that there is no counterpart of complex numbers in higher dimensions. Hence, our lucky extension of the algebraic structure on R to R 2, as explained in my invitation to the course, was indeed lucky, since it will fail in all higher dimensions. All this was extensively investigated in the 19th century by physicist/mathematician W. R. Hamilton. He actually managed to cook up something decent in 4 dimensions, called quaternions, but they have never become very useful... ( just learned from wikipedia that a B. O. Rodrigues actually found them 3 years before Hamilton. This is common in mathematics, that the wrong guy gets his name on a new discovery.) 5