Physics 307 Mathematical Physics Luis Anchordoqui 1
Bibliography L. A. Anchordoqui and T. C. Paul, ``Mathematical Models of Physics Problems (Nova Publishers, 2013) G. F. D. Duff and D. Naylor, ``Differential Equations of Applied Mathematics (John Wiley & Sons, 1966) G. B. Arfken, H. J. Weber, and F. E. Harris ``Mathematical Methods for Physicists (7th Edition) (Academic Press, 2012) 2
Provisional Course Outline (Please note this may be revised during the course to match coverage of material during lectures, etc.) 1st week - Analytic Functions 2rd week - Integration in the Complex Plane 3rd week - Isolated Singularities and Residues 4th week - Elements of Linear Algebra 5th week - Initial Value Problem (Picard s Theorem) 6th week - Initial Value Problem (Green Matrix) 7th week - Boundary Value Problem (Sturm-Liouville Operator) 8th week - Boundary Value Problem (Special Functions) 9th week - Fourier Series and Fourier Transform 10th week - Hyperbolic Partial Differential Equation (Wave equation) 11th week - Parabolic Partial Differential Equation (Diffusion equation) 12th week - Elliptic Partial Differential Equation (Laplace equation) Midterm-exams (October 6, November 7, December 12) Final-exam (December 19 -- 3:45 pm to 5:45 pm) 3
Complex Analysis I 1.1 Complex Algebra 1.2 Functions of a Complex Variable 1.3 Cauchy s Theorem and its Applications 1.4 Isolated Singularities and Residues 4
Complex Algebra Real number system is adequate for solving many mathematical and physical problems It is necessary to extend such a system to solve equation x 2 +1=0 we get a nonnegative number Definition 1.1.1. We define because when we square a real number i to be imaginary number equal to square root of 1 That is i = p 1 which implies i 2 = 1 Proposition 1.1.1. We can combine the set real numbers R with this new imaginary number to form set complex numbers C 5
Definition 1.1.2. A complex number z is an ordered pair (x, y) with x 2 R,y 2 R x is called real part of z x = <e y z y = =m z is called imaginary part of Geometric representation of z as a point in complex plane _4+2i Im i 2+3i _2-2i 0 _i 1 3-2i Re Herein C denotes set of all complex numbers C = {z : z = x + iy, x 2 R,y 2 R} 6
Proposition 1.1.2. Addition and subtraction is defined exactly as in R 2 for example if z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 then we define z 1 + z 2 =(x 1 + x 2 )+i(y 1 + y 2 ) We define Multiplication makes C different from R 2 z 1 z 2 =(x 1 + iy 1 )(x 2 + iy 2 )=(x 1 x 2 y 1 y 2 )+i(x 1 y 2 + x 2 y 1 ) We can define division of complex numbers if z 2 6=0 and therefore then we define 1 z 2 = x 2 iy 2 x 2 2 + y2 2 z 1 z 2 = x 1 + iy 1 x 2 + iy 2 = (x 1 + iy 1 )(x 2 iy 2 ) (x 2 + iy 2 )(x 2 iy 2 ) = (x 1x 2 + y 1 y 2 )+i(x 2 y 1 x 1 y 2 ) x 2 2 + y2 2 7
Definition 1.1.3. If z = x + iy is a complex number then its conjugate is defined by z = x iy Remark 1.1.1. Conjugation has following properties which follow directly from the definition: <e z =(z + z )/2 =m z = (z z )/2 (z 1 + z 2 ) = z! + z2;(z 1 z 2 ) = z1z 2 it follows from this last property that if 2 R ( z) = z Remark 1.1.2. Unlike real numbers complex numbers do not have a natural ordering so there is no analog of complex-valued inequalities 8
Proposition 1.1.3. Let z = x + iy x y be a complex number with and both nonzero exists r 2 (0, 1) and # 2 (, ] such that z = re i# with e i# = cos # + i sin # Coordinates of polar form of z are related to its Cartesian components according to # r = z = p x 2 + y 2 z and # = tan 1 (y/x) # = Arg z principal argument of usually written as Reason to restrict in is to get uniqueness of representation 2 # (, ] rotation does not change the point Argz = argz +2k Theorem 1.1.1. De Moivre's theorem If z = r (cos # + i sin #) and n is a positive integer z n =[r(cos # + i sin #)] n = r n (cos n# + i sin n#) n n This says that to take integer -th power of a complex number we take -th power of the modulus and multiply argument by n 9
Corollary 1.1.1. De Moivre s Theorem n -th root of complex number z is a complex number w such that w n = z writing these two numbers in polar form w = s(cos ' + i sin ') & z = r(cos # + i sin #) Using De Moivre s s n (cos n' + i sin n') =r(cos # + i sin #) equality of these two complex numbers shows that s = r 1/n, cos n' = cos # and sin n' =sin# sine and cosine have period 2 n' = # +2k complex number z has n distinct roots # +2k # +2k w k = r applecos 1/n + i sin n n (1.1.1.) with k =0, 1,...,n 1 10
Remark 1.1.3. n z w k = r 1/n Notice that each of -th roots of has modulus all n -th roots of z lie on circle of radius r 1/n since argument of each successive -th root exceeds argument of previous root by we see that n Example 1.1.1. -th roots of Six sixth roots of z = 8 z are shown here n 2 /n are equally spaced on this circle w Im œ 2i w w _œ 2 0 œ 2 Re w w _œ 2i w 11
Functions of a Complex Variable y w = f(z) v z domain of f x range of f w u Definition 1.2.1. f A A A function from a set to a set is a rule of correspondence that assigns to each element in A B one and only one element in When domain is a set of complex numbers we say that mapping is a function of a complex variable (or a complex function for short) which we denote w = f(z) =u(x, y)+iv(x, y) (1.2.2.) u v defined on subsets of R 2 u = <e f and v = =m f Functions and can be thought of as real valued functions B 12
Proposition 1.2.1. Given z 0 2 C and r>0 we denote ball of radius r around z 0 Definition 1.2.2. Let Then A C,B r (z 0 ) A f B r (z 0 )={z 2 C : z and f : A! C is differentiable at z 0 if limit z 0 <r} f(z 0 + z) f(z 0 ) lim z!0 z + z z = lim z!0 f(z) z = df dz = f 0 (z 0 ) (1.2.3.) is independent of direction of approach to point z 0 Recall that for a single real variable we require that right-hand limit (where one approaches x 0 from x>x 0 ) and left-hand limit (approaching x 0 from x<x 0 ) be equal for derivative df (x)/dx to exist at x = x 0 13
For z = z 0 some point in a plane approach must be generalized x y x y z = x + i y f = u + iv f = u + i v and so f z = Let & be increments of & writing u + i v x + i y Take limit (1.2.3.) using two different directions of approach y y y =0 x dx! 00 z 0 d y = 0 y! 0 x =0 d x = 0 d y 0 df dz Figure 1.3: Alternative approaches to. x x y =0 f lim z!0 z = lim x!0 x! 0 u x + i v x x =0 y! 0 f lim z!0 z = lim y!0 z 0 i u y + v y = @u @x + i @v @x = i @u @y + @v @y If derivative exists at these limits must be identical Equating real & imaginary parts u x = v y u y = v x Cauchy-Riemann conditions @u @x u x, @u @y u y, @v @x v x, @v @y v y 14
We have seen that for f 0 (z 0 ) Conversely if CR conditions hold and partial derivatives of to exist CR conditions must be satisfied u(x, y) v(x, y) and are continuous derivative f 0 (z) =u x + iv x exists This can be seen by writing f =(u x + iv x ) x +(u y + iv y ) y (1.2.11.) dividing by z f z = (u x + iv x ) x +(u y + iv y ) y x + i y = (u x + iv x )+(u y + iv y ) y/ x 1+i y/ x (1.2.12.) 15
If f/ z is to have a unique value then dependence on y/ x has to be eliminated Applying Cauchy-Riemann conditions to y derivatives yields u y + iv y = v x + iu x (1.2.13.) Substituting (1.2.13.) into (1.2.12.) cancels out y/ x dependence and leaves us with f z = u x + iv x (1.2.14.) which shows that f/ z provided partial derivatives are continuous is independent of direction of approach Hence f 0 (z) exists 16
Definition 1.2.3. Iff(z) is differentiable at z = z 0 and in some small region around z 0 we say that f(z) is analytic or holomorphic at z = z 0 In addition if f 0 (z 0 ) 6= 0 we say that f(z) is conformal at z 0 If f(z) is analytic everywhere in (finite) complex plane we call it an entire function f 0 (z) If does not exist at z = z 0 z 0 then is labeled a singular point 17
Example 1.2.1. Let f(z) =z 2 Multiplying we identify real part (x + iy)(x + iy) =x 2 y 2 +2ixy u(x, y) =x 2 y 2 and imaginary part v(x, y) =2xy From (1.2.9.) u x =2x = v y and u y = 2y = v x (1.2.15.) We see that f(z) =z 2 satisfies Cauchy--Riemann conditions throughout complex plane Since partial derivatives are evidently continuous we conclude that f(z) =z 2 is analytic 18
Example 1.2.2. Let Now f(z) =z u = x and v = y Applying Cauchy--Riemann conditions u x =16= v y = 1 Cauchy--Riemann conditions are not satisfied and f(z) =z is not an analytic function of z However note that f(z) =z is continuous providing example of a function that is everywhere continuous but nowhere differentiable in complex plane 19
Definition 1.2.4. Let z = x + iy x, y 2 R the exponential function e z is defined for every e z = e x (cos y + i sin y) z 2 C (1.2.17.) if we write e z = u(x, y)+iv(x, y) u(x, y) =e x cos y and v(x, y) =e x sin y e z is an entire function Easy to check that Cauchy-Riemann conditions are satisfied Remark 1.2.3. Note that for every z 2 C d dz ez = u x + iv x = e x cos y + ie x sin y = e x (cos y + i sin y) =e z so that e z is its own derivative (1.2.18.) 20
Corollary 1.2.2. For every y 1,y 2 2 R we have e i(y 1+y 2 ) = cos(y 1 + y 2 )+isin(y 1 + y 2 ) = (cos y 1 + i sin y 1 )(cos y 2 + i sin y 2 ) = e iy 1 e iy 2 On the other hand if x 1,x 2 2 R (1.2.19.) e x 1+x 2 e i(y 1+y 2 ) =(e x 1 e x 2 )(e iy 1 e iy 2 )=(e x 1 e iy 1 )(e x 2 e iy 2 ) (1.2.20.) writing z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 e z 1+z 2 = e z 1 e z 2 we deduce Remark 1.2.4. addition formula Note that e z = e x (cos y + i sin y) = e x cos y + i sin y = e x Since e x is never zero it follows that e z is non-zero for every z 2 C i.e. exp z : C! C 0 21
Corollary 1.2.3. Exponential function e z is periodic in C It follows that! =2 ki k 2 Z is non-zero w = e x and arg w = y +2k Usually we make choice arg w = y with restriction that z lies on horizontal strip A? = {z 2 C : 1 <x<1, <yapple } (1.2.21.) This restriction means that Restriction <arg w apple w on complex Upper edge of cut corresponding to can also be indicated <yapple -plane by a cut along negative real axis is regarded arg w = as part of cut Lower edge of cut corresponding to arg w = is not regarded as part of cut w w -plane -plane 22
Effect of f(z) =e z on two domains y exp(z) v A? B x u Fundamental region of exponential function A? in a one-to-one fashion is mapped on a plane that is cut along Rectangle B = {0 <x<x 0, negative real axis is mapped in a one-to-one fashion <yapple } of height 2 on an annulus that is cut along negative real axis 23
Remark 1.2.5. Region A? is usually known as a fundamental region Easy to see that every set of type A k = {z 2 C : 1 <x<1, (2k 1) <yapple (2k + 1) } k 2 Z (1.2.22.) Corollary 1.2.4. has same representation than A? Since exp : A?! C 0 is one-to-one and onto there is an inverse function Definition 1.2.5. Function f : C 0!A? Log w = z with z 2A? and e z = w (1.2.23.) is called principal logarithmic function 24
Suppose that z = x + iy w = u + iv x, y, u, v 2 R and Suppose further that we impose restriction <yapple If w = e z from (1.2.17.) u = e x cos y and v = e x sin y w =(u 2 + v 2 ) 1/2 = e x and y = Arg w Arg w principal argument of w Therefore x =ln w and y = Arg w or equivalently (1.2.24.) Logw =ln w + i Arg w (1.2.25.) In many practical situations one can define log w =ln w + i arg w where argument is chosen in order to make logarithmic function continuous in its domain of definition (if this is at all possible) 25
Definition 1.2.6. Suppose that z 2 C Then trigonometric functions cos z and sin z are defined in terms of exponential the function cos z = eiz + e iz 2 and sin z = eiz e iz Since exponential function is an entire function cos z sin z 2i (1.2.26.) from (1.2.26) both and are entire functions Easy 2C from (1.2.26.) that d dz cos z = sin z and d dz We can define functions sin z = cos z tan z, cot z, sec z, csc z (1.2.27.) in terms of functions cos z and sin z as in real variables 26
Definition 1.2.7. Suppose that z 2 C Then hyperbolic functions cosh z = ez + e z cosh z and sinh z are defined in terms 2 and sinh z = ez e z of exponential function Since exponential function is an entire function 2 (1.2.28.) from (1.2.28.) both cosh z and sinh z are entire functions Easy 2C from (1.2.28.) that d dz cosh z =sinhz and d dz We can define functions in terms of functions cosh z Remark 1.2.6. sinh z = cosh z tanh z, coth z, sech z, csch z and sinh z (1.2.29.) as in real variables Comparing (1.2.26.) and (1.2.28.) we obtain cosh z = cos iz and sinh z = i sin iz (1.2.30.) 27
Definition 1.2.8. A series of form 1 X Theorem 1.2.2. If a power series n=0 a n (z z 0 ) n a n 2 C and z 0 2 C is called a power series around point z 0 1X n=0 a n z n converges for some z 0 2 C then it converges for all z 2 C such that z < z 0 (which is a disc without boundary around origin with radius z 0 ) Proof. It follows from hypothesis that there exist M 0 such that a n z n 0 applem for all n 2 N a n z n = a n z 0 n z z 0 n Remark 1.2.7. Note that if series 1X n=0 apple M z z 0 n (1.2.31.) a n z n 0 diverges then so does series for z > z 0 28
Definition 1.2.9. 1 X Radius of convergence of a power series defined as R = max{ z : 1X n=0 n=0 a n z n 0 a n z n, converges} (1.2.32.) R 0 BUT of course it is possible that R =0 z <R (or z >R ) then power series converges (or diverges) y z z 0 = R convergence z 0 R divergence x Circle of a convergence of a power series 29
Theorem 1.2.3. 1X If a n z n has radius of convergence R>0 n=0 1X na n z n then series Proof. Fix z 0 Series n=1 1 has precisely same radius of convergence with z 0 <R and pick z such that z < z 0 <R X an z n 0 We can thus find a number converges and therefore lim n!1 a nz n 0 =0 M such that a n z n 0 applem, 8n We now write na n z n 1 = na n z n 1 z0 z 0 n = n z 0 a n z n 0 z z 0 n 1 (1.2.33.) Introducing = z/z 0 < 1 we have na n z n 1 = n a n z0 n n 1 apple M z 0 z 0 n n 1 (1.2.34.) 30
Now since series 1X n=0 lim n!1 na n n 1 z n 1 0 n +1 n = <1 converges (1.2.35.) Since z 0 was an arbitrary number with z 0 <R series converges uniformly for z <R of series of derivatives satisfies R 0 R This shows that radius of convergence R 0 Theorem 1.2.4. If 1X n=0 then F (z) = a n z n has radius of convergence R>0 1X n=0 a n z n is differentiable on S = {z 2 C : z <R} 1X and derivative is f(z) = na n z n 1 n=1 31
Proof. We will show that [F (z + h) F (z)]/h f(z)!0 as h! 0 whenever z <R nx Using binomial theorem (z + h) n = ( n k) h k z n k F (z + h) h F (z) f(z) = = = = 1X (z + h) n z n hnz n 1 a n h " 1X a n X n # n h k z n k h k k=2 " 1X X n # n a n h h k 2 z n k k k=2 2 3 1X nx 2 a n h 4 n h j z n 2 j 5 j +2 n=0 n=0 n=0 n=0 j=0 k=0 where in last line we have taken j = k 2 we get 32
Using F (z + h) h n j+2 apple n(n 1) n 2 j F (z) f(z) = h = h We already know that series we obtain 2 1X nx 2 n(n 1) a n 4 n=0 1X n(n 1) a n ( z + h ) n 2 1X n(n 1) a n z n 2 n=0 n=0 j=0 n 2 j 3 h j z n 2 j 5 (1.2.36.) converges for z <R Now for z <Rand h! 0 we have z + h <R eventually It thus follows Exercise 1.2.2. F (z + h) F (z) lim h!0 h Using theorems 2.2.3. and 2.2.4. show that e z = 1X n=0 z n /n! f(z) =0 whenever (1.2.37.) z <R 33
Definition 1.3.1. Suppose C is a curve parameterized by and A and B are points (f(a),g(a)) and (f(b),g(b)) We say that: x = f(t),y = g(t),aapple t apple b B C 2 B A = B C 1 C 3 A = B A A C C C C is a smooth curve if f 0 g 0 and and not simultaneously zero on (a, b) are continuous on is piecewise smooth if C = C 1 [ C 2 [ [C n C 1,C 2,...,C n is a closed curve if A = B is a simple closed curve if A = B [a, b] smooth curves and curve does not cross itself 34
Definition 1.3.2. Let f(z) =u(x, y)+iv(x, y) be defined at all points on a smooth curve C Divide C into n sub-arcs a = t 0 <t 1 < <t n = b of [a, b, ] z* n x = x(t), y = y(t), a apple t apple b z 0 z n 1 z* z 2 1 z z* 2 1 Form sum C z n lim kp k!0 Let z k = z k z k 1, with k =1, 2,...,n kp k norm of partition z k (maximum value of ) Choose sample point nx Z f(zk) z k = k=1 C f(z) dz on each sub-arc z k = x k + iy k nx nx f(zk) if lim f(z exists z k k) z k k=1 P!0 k=1 and is independent of details of choosing points z k (1.3.39) 35
Remark 1.3.1. As an alternative contour integral may be defined by Z z2 z 1 f(z) dz = = Z (x2,y 2 ) (x 1,y 1 ) Z (x2,y 2 ) (x 1,y 1 ) [u(x, y)+iv(x, y)][dx + idy] [u(x, y)dx v(x, y)dy]+i Z (x2,y 2 ) (x 1,y 1 ) [v(x, y)dx + u(x, y)dy] with path joining (x 1,y 1 ) and (x 2,y 2 ) specified This reduces complex integral to complex sum of real integrals Theorem 1.3.1. [ML- inequality] If f and if then is continuous on a smooth curve f(z) applem for all z on C Z f(z)dz appleml C C where L is length of C 36
Proof. using z 1 + z 2 + z 3 + + z n apple z 1 + z 2 + z 3 + z n nx f(z k) z k apple nx f(z k) z k applem nx z k (1.3.40.) k=1 k=1 k=1 z k is length of chord joining points z k and z k 1 Since sum of lengths of chords cannot be greater than C length (2.3.40.) becomes nx k=1 f(z k) z k apple ML kp k!0 If last inequality yields Z C f(z)dz appleml 37
38