Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 2 pptx

Now we compute the derivative.The complex derivative in the coordinate directions is Since w is analytic, u and v satisfy the Cauchy-Riemann equations, ux = vy and uy = −vx... The comple

Now we compute the derivative.

The complex derivative in the coordinate directions is

Since w is analytic, u and v satisfy the Cauchy-Riemann equations,

ux = vy and uy = −vx

Using the chain rule we can write the derivatives with respect to x and y in terms of u and v.

φx− ıφy = dw

dz (Φu− ıΦv)Thus we see that,

∂Φ

∂u − ı∂Φ

∂v =

 dwdz

We write this in operator notation

∂

∂u − ı ∂

∂v =

 dwdz

The complex conjugate of this relation is

= dwdz

(w0)−1 is an analytic function Recall that for analytic functions f , f0 = fx = −ıfy So that fx+ ıfy = 0

dwdz

f (z) = log |z| + ı arg(z) = log r + ıθ

The Cauchy-Riemann equations in polar coordinates are

ur = 1

rvθ, uθ = −rvr.

We calculate the derivatives.

uθ = 0, −rvr = 0Since the Cauchy-Riemann equations are satisfied and the partial derivatives are continuous, f (z) is analytic in

|z| > 0, | arg(z)| < π The complex derivative in terms of polar coordinates is

2 Next we consider

f (z) =p|z| eı arg(z)/2

=√

r eıθ/2.The Cauchy-Riemann equations for polar coordinates and the polar form f (z) = R(r, θ) eıΘ(r,θ) are

rRθ = 0, −RΘr = 0Since the Cauchy-Riemann equations are satisfied and the partial derivatives are continuous, f (z) is analytic in

|z| > 0, | arg(z)| < π The complex derivative in terms of polar coordinates is

We use this to differentiate f (z).

1 We consider the function

u = x Log r − y arctan(x, y) = r cos θ Log r − rθ sin θ

We compute the Laplacian

∆u = 1r

r2

∂2u

∂θ2

= 1r

vr = −1

ruθ, vθ = rur

vr = sin θ(1 + Log r) + θ cos θ, vθ = r (cos θ(1 + Log r) − θ sin θ)

We integrate the first equation with respect to r to determine v to within the constant of integration g(θ)

v = r(sin θ Log r + θ cos θ) + g(θ)

We differentiate this expression with respect to θ

vθ = r (cos θ(1 + Log r) − θ sin θ) + g0(θ)

We compare this to the second Cauchy-Riemann equation to see that g0(θ) = 0 Thus g(θ) = c We havedetermined the harmonic conjugate.

v = r(sin θ Log r + θ cos θ) + cThe corresponding analytic function is

f (z) = r cos θ Log r − rθ sin θ + ı(r sin θ Log r + rθ cos θ + c)

On the positive real axis, (θ = 0), the function has the value

r2

∂2u

∂θ2 = 0The function u is harmonic We find the harmonic conjugate v by solving the Cauchy-Riemann equations

We differentiate this expression with respect to θ.

vθ = g0(θ)

We compare this to the second Cauchy-Riemann equation to see that g0(θ) = 0 Thus g(θ) = c We havedetermined the harmonic conjugate

v = − Log r + cThe corresponding analytic function is

r2

∂2u

∂θ2

= 1r

The function u is harmonic We find the harmonic conjugate v by solving the Cauchy-Riemann equations.

We compute the Laplacian.

∆u = 1r

r2

∂2u

∂θ2

= 1r

∂

∂r



−sin θr

f (z) = sin θ

r + ı

cos θ

r + ıc

On the positive real axis, (θ = 0), the function has the value

2 We calculate the first partial derivatives of u and v.

ux= 2 ex2−y2(x cos(2xy) − y sin(2xy))

uy = −2 ex2−y2(y cos(2xy) + x sin(2xy))

vx= 2 ex2−y2(y cos(2xy) + x sin(2xy))

vy = 2 ex2−y2(x cos(2xy) − y sin(2xy))Since the Cauchy-Riemann equations, ux = vy and uy = −vx, are satisfied everywhere and the partial derivativesare continuous, f (z) is everywhere differentiable Since f (z) is differentiable in a neighborhood of every point, it

is analytic in the complex plane (f (z) is entire.)

Now to evaluate the derivative The complex derivative is the derivative in any direction We choose the xdirection

f0(z) = ux+ ıvx

f0(z) = 2 ex2−y2(x cos(2xy) − y sin(2xy)) + ı2 ex2−y2(y cos(2xy) + x sin(2xy))

f0(z) = 2 ex2−y2((x + ıy) cos(2xy) + (−y + ıx) sin(2xy))

Finding the derivative is easier if we first write f (z) in terms of the complex variable z and use complex tiation

differen-f (z) = ex2−y2(cos(2x, y) + ı sin(2xy))

f (z) = ex2−y2eı2xy

f (z) = e(x+ıy)2

f (z) = ez2

f0(z) = 2z ez2

Solution 8.19

1 Assume that the Cauchy-Riemann equations in Cartesian coordinates

ux = vy, uy = −vxare satisfied and these partial derivatives are continuous at a point z We write the derivatives in polar coordinates

in terms of derivatives in Cartesian coordinates to verify the Cauchy-Riemann equations in polar coordinates First

we calculate the derivatives

ur = cos θux+ sin θuy

= cos θvy − sin θvx

= 1

rvθ1

ruθ = − sin θux+ cos θuy

= − sin θvy− cos θvx

= −vrThis proves that the Cauchy-Riemann equations in Cartesian coordinates hold only if the Cauchy-Riemann equa-tions in polar coordinates hold (Given that the partial derivatives are continuous.) Next we prove the converse.Assume that the Cauchy-Riemann equations in polar coordinates

ur = 1

rvθ,

1

ruθ = −vr

are satisfied and these partial derivatives are continuous at a point z We write the derivatives in Cartesiancoordinates in terms of derivatives in polar coordinates to verify the Cauchy-Riemann equations in Cartesiancoordinates First we calculate the derivatives.

in Cartesian coordinates hold We have demonstrated the equivalence of the two forms

2 We verify that log z is analytic for r > 0 and −π < θ < π using the polar form of the Cauchy-Riemann equations

r =

1

r1,1

r0 = −0

Since the Cauchy-Riemann equations are satisfied and the partial derivatives are continuous for r > 0, log z isanalytic there We calculate the value of the derivative using the polar differentiation formulas.

dz Log z =

−ız

3 Let {xi} denote rectangular coordinates in two dimensions and let {ξi} be an orthogonal coordinate system The distance metric coefficients hi are defined

Solution 8.20

1 We compute the Laplacian of u(x, y) = x3− y3

∇2u = 6x − 6ySince u is not harmonic, it is not the real part of on analytic function

2 We compute the Laplacian of u(x, y) = sinh x cos y + x

∇2u = sinh x cos y − sinh x cos y = 0Since u is harmonic, it is the real part of on analytic function We determine v by solving the Cauchy-Riemannequations

vx = −uy, vy = ux

vx = sinh x sin y, vy = cosh x cos y + 1

We integrate the first equation to determine v up to an arbitrary additive function of y

v = cosh x sin y + g(y)

We substitute this into the second Cauchy-Riemann equation This will determine v up to an additive constant

vy = cosh x cos y + 1cosh x cos y + g0(y) = cosh x cos y + 1

g0(y) = 1g(y) = y + a

v = cosh x sin y + y + a

f (z) = sinh x cos y + x + ı(cosh x sin y + y + a)Here a is a real constant We write the function in terms of z

f (z) = sinh z + z + ıa

3 We compute the Laplacian of u(r, θ) = rncos(nθ).

∇2u = n(n − 1)rn−2cos(nθ) + nrn−2cos(nθ) − n2rn−2cos(nθ) = 0Since u is harmonic, it is the real part of on analytic function We determine v by solving the Cauchy-Riemannequations

v = rnsin(nθ) + a

f (z) = rncos(nθ) + ı(rnsin(nθ) + a)Here a is a real constant We write the function in terms of z

f (z) = zn+ ıaSolution 8.21

1 We find the velocity potential φ and stream function ψ

Φ(z) = log z + ı log zΦ(z) = ln r + ıθ + ı(ln r + ıθ)

φ = ln r − θ, ψ = ln r + θ

Figure 8.7: The velocity potential φ and stream function ψ for Φ(z) = log z + ı log z.

A branch of these are plotted in Figure 8.7

Next we find the stream lines, ψ = c

ln r + θ = c

r = ec−θThese are spirals which go counter-clockwise as we follow them to the origin See Figure 8.8 Next we find thevelocity field

v = ∇φ

v = φrˆr + φθ

r

ˆθ

v = ˆr

r − θˆr

Figure 8.8: Streamlines for ψ = ln r + θ.

The velocity field is shown in the first plot of Figure 8.9 We see that the fluid flows out from the origin alongthe spiral paths of the streamlines The second plot shows the direction of the velocity field

2 We find the velocity potential φ and stream function ψ

Φ(z) = log(z − 1) + log(z + 1)Φ(z) = ln |z − 1| + ı arg(z − 1) + ln |z + 1| + ı arg(z + 1)

φ = ln |z2− 1|, ψ = arg(z − 1) + arg(z + 1)The velocity potential and a branch of the stream function are plotted in Figure8.10

The stream lines, arg(z − 1) + arg(z + 1) = c, are plotted in Figure8.11

Next we find the velocity field

Figure 8.9: Velocity field and velocity direction field for φ = ln r − θ.

The velocity field is shown in the first plot of Figure 8.12 The fluid is flowing out of sources at z = ±1 Thesecond plot shows the direction of the velocity field

Figure 8.10: The velocity potential φ and stream function ψ for Φ(z) = log(z − 1) + log(z + 1).

Since the function behaves like 1/z at infinity, it is analytic there

(b) The denominator of 1/ sin z has first order zeros at z = nπ, n ∈ Z Thus the function has first order poles

at these locations Now we examine the point at infinity with the change of variables z = 1/ζ

1sin z =

1sin(1/ζ) =

ı2

eı/ζ− e−ı/ζ

We see that the point at infinity is a singularity of the function Since the denominator grows exponentially,there is no multiplicative factor of ζn that will make the function analytic at ζ = 0 We conclude that thepoint at infinity is an essential singularity Since there is no deleted neighborhood of the point at infinitythat does contain first order poles at the locations z = nπ, the point at infinity is a non-isolated singularity.(c)

log 1 + z2
= log(z + ı) + log(z − ı)There are branch points at z = ±ı Since the argument of the logarithm is unbounded as z → ∞ there is

a branch point at infinity as well Branch points are non-isolated singularities

Figure 8.11: Streamlines for ψ = arg(z − 1) + arg(z + 1).

(d)

z sin(1/z) = 1

2z e

ı/z+ eı/z
The point z = 0 is a singularity Since the function grows exponentially at z = 0 There is no multiplicativefactor of zn that will make the function analytic Thus z = 0 is an essential singularity

There are no other singularities in the finite complex plane We examine the point at infinity

Figure 8.12: Velocity field and velocity direction field for φ = ln |z2− 1|.

tan−1(z)

z sinh2(πz) =

ı log ı+zı−z
2z sinh2(πz)

There are branch points at z = ±ı due to the logarithm These are non-isolated singularities Note thatsinh(z) has first order zeros at z = ınπ, n ∈ Z The arctangent has a first order zero at z = 0 Thus there

is a second order pole at z = 0 There are second order poles at z = ın, n ∈ Z \ {0} due to the hyperbolicsine Since the hyperbolic sine has an essential singularity at infinity, the function has an essential singularity

at infinity as well The point at infinity is a non-isolated singularity because there is no neighborhood ofinfinity that does not contain second order poles

2 (a) (z − ı) e1/(z−1) has a simple zero at z = ı and an isolated essential singularity at z = 1

(b)

sin(z − 3)(z − 3)(z + ı)6

has a removable singularity at z = 3, a pole of order 6 at z = −ı and an essential singularity at z∞

With more overlapping domains D3, D4, we could perhaps extend f1(z) to more of the complex plane Sometimes

it is impossible to extend a function beyond the boundary of a domain This is known as a natural boundary If a

Figure 9.1: Overlapping Domains

function f1(z) is analytically continued to a domain Dn along two different paths, (See Figure 9.2.), then the twoanalytic continuations are identical as long as the paths do not enclose a branch point of the function This is theuniqueness theorem of analytic continuation

D1

Dn

Figure 9.2: Two Paths of Analytic Continuation

Consider an analytic function f (z) defined in the domain D Suppose that f (z) = 0 on the arc AB, (see Figure9.3.)Then f (z) = 0 in all of D

Consider a point ζ on AB The Taylor series expansion of f (z) about the point z = ζ converges in a circle C at

B

ζ

CA

Figure 9.3: Domain Containing Arc Along Which f (z) Vanishes

least up to the boundary of D The derivative of f (z) at the point z = ζ is

To prove Result 9.1.1, we define the analytic function g(z) = f1(z) − f2(z) Since g(z) vanishes in the region or

on the arc, then g(z) = 0 and hence f1(z) = f2(z) for all points in D

Result 9.1.2 Consider analytic functions f1(z) and f2(z) defined on the domains D1 and

D2, respectively Suppose that D1 ∩ D2 is a region or an arc and that f1(z) = f2(z) for all

z ∈ D1 ∩ D2 (See Figure 9.4 ) Then the function

Example 9.2.1 Consider the function

9.3 Analytic Functions Defined in Terms of Real Variables

Result 9.3.1 An analytic function, u(x, y) + ıv(x, y) can be written in terms of a function

of a complex variable, f (z) = u(x, y) + ıv(x, y).

Result 9.3.1is proved in Exercise 9.1

Example 9.3.1

f (z) = cosh y sin x (x excos y − y exsin y) − cos x sinh y (y excos y + x exsin y)

+ ı cosh y sin x (y excos y + x exsin y) + cos x sinh y (x excos y − y exsin y)

is an analytic function Express f (z) in terms of z

On the real line, y = 0, f (z) is

f (z = x) = x exsin x(Recall that cos(0) = cosh(0) = 1 and sin(0) = sinh(0) = 0.)

The analytic continuation of f (z) into the complex plane is

f (z) = z ezsin z

Alternatively, for x = 0 we have

f (z = ıy) = y sinh y(cos y − ı sin y)

The analytic continuation from the imaginary axis to the complex plane is

f (z) = −ız sinh(−ız)(cos(−ız) − ı sin(−ız))

= ız sinh(ız)(cos(ız) + ı sin(ız))

= z sin z ez

Example 9.3.2 Consider u = e−x(x sin y − y cos y) Find v such that f (z) = u + ıv is analytic.

From the Cauchy-Riemann equations,

F (x) is an arbitrary function of x Substitute this expression for v into the equation for ∂v/∂x

−y e−xsin y − x e−xcos y + e−xcos y + F0(x) = −y e−xsin y − x e−xcos y + e−xcos yThus F0(x) = 0 and F (x) = c

v = e−x(y sin y + x cos y) + cExample 9.3.3 Find f (z) in the previous example (Up to the additive constant.)

2



= ı(x + ıy) e−(x+ıy)

= ız e−z

Method 2 f (z) = f (x + ıy) = u(x, y) + ıv(x, y) is an analytic function.

On the real axis, y = 0, f (z) is

f (z = x) = u(x, 0) + ıv(x, 0)

= e−x(x sin 0 − 0 cos 0) + ı e−x(0 sin 0 + x cos 0)

= ıx e−xSuppose there is an analytic continuation of f (z) into the complex plane If such a continuation, f (z), exists, then itmust be equal to f (z = x) on the real axis An obvious choice for the analytic continuation is

f (z) = u(z, 0) + ıv(z, 0)since this is clearly equal to u(x, 0) + ıv(x, 0) when z is real Thus we obtain

f (z) = ız e−zExample 9.3.4 Consider f (z) = u(x, y) + ıv(x, y) Show that f0(z) = ux(z, 0) − ıuy(z, 0)

f0(z) = ux+ ıvx

= ux− ıuy

f0(z) is an analytic function On the real axis, z = x, f0(z) is

f0(z = x) = ux(x, 0) − ıuy(x, 0)Now f0(z = x) is defined on the real line An analytic continuation of f0(z = x) into the complex plane is

f0(z) = ux(z, 0) − ıuy(z, 0)

= 0 − ı z e−z− e−z

= ı −z e−z+ e−z
Integration yields the result

f (z) = ız e−z+cExample 9.3.6 Find f (z) given that

u(x, y) = cos x cosh2y sin x + cos x sin x sinh2yv(x, y) = cos2x cosh y sinh y − cosh y sin2x sinh y

f (z) = u(x, y) + ıv(x, y) is an analytic function On the real line, f (z) is

f (z = x) = u(x, 0) + ıv(x, 0)

= cos x cosh20 sin x + cos x sin x sinh20 + ı cos2x cosh 0 sinh 0 − cosh 0 sin2x sinh 0

= cos x sin xNow we know the definition of f (z) on the real line We would like to find an analytic continuation of f (z) into thecomplex plane An obvious choice for f (z) is

f (z) = cos z sin zUsing trig identities we can write this as

f (z) = sin(2z)

2 .Example 9.3.7 Find f (z) given only that

u(x, y) = cos x cosh2y sin x + cos x sin x sinh2y

Recall that

f0(z) = ux+ ıvx

= ux− ıuyDifferentiating u(x, y),

ux = cos2x cosh2y − cosh2y sin2x + cos2x sinh2y − sin2x sinh2y

uy = 4 cos x cosh y sin x sinh y

f0(z) is an analytic function On the real axis, f0(z) is

f0(z = x) = cos2x − sin2xUsing trig identities we can write this as

f0(z = x) = cos(2x)Now we find an analytic continuation of f0(z = x) into the complex plane

f0(z) = cos(2z)Integration yields the result

f0(z) = ex2< /sup>−y2< /sup>((x... sin(2xy))

uy = ? ?2 ex2< /sup>−y2< /sup>(y cos(2xy) + x sin(2xy))

vx= ex2< /sup>−y2< /sup>(y cos(2xy)... 12< /span>

2 We calculate the first partial derivatives of u and v.

ux= ex2< /sup>−y2< /sup>(x cos(2xy)