ENGI 5432 Advanced Calculus

Faculty of Engineering and Applied Science
2009 Winter

Problem Set 8 - Solutions

[Sections 1.7, 2.6, 3]

A point charge Q at the origin gives rise to the electrostatic potential field

Find
1. the potential V at the point (x, y, z ) = (3, -4, 12).
  
  (x, y, z) = (3, -4, 12) Þ r² = 9 + 16 + 144 = 169
  Þ r = 13 . Therefore
2. the electric field vector E at the point (x, y, z ) = (3, -4, 12).
3. the volume charge density r_V at the point (x, y, z ) = (3, -4, 12).
  [Hint: use Poisson’s equation div D = r_V .]
  
  Therefore
4. the magnitude Q of the point charge at the origin.
  
  $Q = Integral {D dot dS} = e Integral {E dot dS} = 13 e Integral {(1/r^2) dS}$
  where S is any sphere, centre O, radius r.
  $Q = 13e Int_0^2pi {Int_0^pi { r^2 sin theta / r^2 dtheta} dphi} (r not= 0)$
  
  Therefore
  
  or

In spherical polar coordinates, the scalar field V is defined by
V(x, y, z) = r cos²q .
1. The vector field F is defined by F = ÑV . Find F.
2. Find the divergence of F.
  
  $div F = (1/r^2) d/dr(r^2 (cos^2 theta)) + + (1/(r sin theta)) {d/dtheta(sin theta (-2 sin theta cos theta)) + d/dphi (0)}$
  $= (2/r) cos^2 theta - (2/(r sin theta)) {d/dtheta(sin^2 theta cos theta)) + 0}$
  
  Therefore
  
  Note that the divergence of F is also the Laplacian of V :
3. Find the curl of F.
  
  [Long method:]
  
  Therefore (except possibly on the z axis)
  
  [It can be shown that the curl remains the zero vector as one approaches any point on the z axis in the limit from any direction.]
  
  [Short method:]
  
  F is the gradient of a scalar function V.
  A quotable identity is curl grad V º 0 for all twice-differentiable scalar fields V.
  Therefore curl F º 0 .

For the scalar field $V(x,y,z) = z * sqrt{x^2 + y^2}$
1. Find ÑV in the Cartesian coordinate system.
  
  Note that
  $d/dx (x^2 + y^2)^(1/2) = x / sqrt{x^2 + y^2}$
  Using symmetry, it then follows that $d/dy (x^2 + y^2)^(1/2) = y / sqrt{x^2 + y^2}$
  
  $grad V = < xz/sqrt{x^2 + y^2}, yz/sqrt{x^2 + y^2}, sqrt{x^2 + y^2} >$
2. Find the Laplacian Ñ²V in the Cartesian coordinate system in simplified form.
  
  Using the quotient rule:
  $= (zr - x^2 z/r)/r^2 + (zr - y^2 z/r)/r^2 + 0 where r = sqrt{x^2 + y^2}$
  $= (zr^2 - x^2 z)/r^3 + (zr^2 - y^2 z)/r^3 where r = sqrt{x^2 + y^2}$
  Therefore
  
  $del^2 V = z/sqrt{x^2 + y^2}$
  
  [Also note that (Laplacian of V) = div grad V .]
3. Convert V(x, y, z) to spherical polar coordinates.
  
  x = r sin q cos f
  y = r sin q sin f
  z = r cos q
  Þ x² + y² = r² sin²q (cos²f + sin²f) = (r sinq)²
  $==> z sqrt{x^2 + y^2} = r cos theta r sin theta$
  Therefore
4. Find ÑV in the spherical polar coordinate system in simplified form.
  
  Therefore
  
  This is equivalent to
5. Show that the Laplacian Ñ²V in the spherical polar coordinate system is cot q.
  
  Either use (Laplacian of V) = div grad V, or quote the spherical polar form for the Laplacian:
  $del^2 V = (1/r^2) d/dr(r^2 dV/dr)) + (1/(r^2 sin theta)) {d/dtheta(sin theta dV/dtheta) + (1/sin theta) d^2 V/dphi^2}$
  $= (1/r^2) d/dr(r^3 sin 2theta)) + (1/(r^2 sin theta)) {d/dtheta(r^2 sin theta cos 2theta) + 0}$
  $= (3 r^2/r^2) sin 2theta + r^2 /(r^2 sin theta)) {cos theta cos 2theta - 2 sin theta sin 2theta)}$
  
  Therefore
  
  Ñ²V = cot q
6. Hence show that the value of the Laplacian of V is the same in both coordinate systems (except possibly on the z-axis).
  
  From part (c), z = r cos q and x² + y² = (r sinq)²
  From part (b), $del^2 V = z/sqrt{x^2 + y^2}$
  Therefore
  
  which is the spherical polar form found in part (e).

A set of functions {f_n(x) } is said to be orthonormal on the interval (a, b) if

$Integral{phi_m(x)*phi_n(x) dx = 0 (for m not= n) ; = 1 (for m=n)$ where m and n are positive integers.
Show that { sin npx }, { cos npx } are both orthonormal on the interval (-1, 1).

[Note: It then follows that if $f(x) = Sum_k=1 ^oo {a_k phi_k(x)}$ then $a_n = Integral {f(x) phi_n(x)} dx / Integral {(phi_n(x))^2} dx$ which leads to methods of series solution of differential equations, including Fourier series.]

The following trigonometric identities are quotable:

sin A sin B º ( cos(A-B) - cos(A+B) ) / 2
and
cos A cos B º ( cos(A+B) + cos(A-B) ) / 2
Therefore

If m ¹ n then

[Note that m + n cannot be zero because both m and n are positive.]

If m = n then

Therefore { sin np x } is an orthonormal set of functions on (-1, 1).

If m ¹ n then

If m = n then

Therefore { cos np x } is an orthonormal set of functions on (-1, 1).

[One can also show that every member of the set { cos mp x } is orthogonal to every member of the set { sin np x } on (-1, 1).]

Find the Fourier series for the function f (x) defined on the interval [–1, 1] by

L = 1

The Fourier series of f (x) on [–L, +L] in general is

Therefore

$f(x) = 1/3 + Sum { 2(-1)^(n+1) cos(n pi x)/(n pi)^2 + (2((-1)^n - 1)/(n pi)^3 - 1/(n pi)) sin(n pi x) }$

The first few terms of this series are

The convergence is quite slow, due to the discontinuity at x = 0 and the abrupt change in slope at the ends of the interval, as this graph of the fourth partial sum S₄ illustrates:

y = f (x) is in red
y = S₄ is in cyan.

Find the Fourier series for the function f (x) defined on the interval [–1, 1] by

L = 1
f (x) is even Þ b_n = 0 for all n.

Therefore the Fourier series is

$f(x) = 1/2 + (4/pi^2) Sum { cos((2k-1) pi x)/((2k-1) pi)^2 }$

The first few terms of this series are

The partial sum of just the first three non-zero terms yields an impressively good approximation everywhere except near the points where f (x) is not differentiable:

y = f (x) is in red
y = S₃ is in blue.

Find the Fourier cosine series for the function f (x) defined on the interval [0, 2] by

L = 2
An even extension of f (x) is required
b_n = 0 for all n.

Therefore the Fourier series is

$f(x) = 8/3 + (16/pi^2) Sum { (-1)^(n+1) cos(n pi x/2)/(n pi)^2 }$

The first few terms of this series are

The partial sum of the first four terms yields a fair approximation everywhere except near the endpoints, where f (x) is not differentiable:

y = f (x) is in red
y = S₃ is in blue.

Find the Fourier sine series for the function f (x) defined on the interval [0, 2] by

Also comment on why the rate of convergence of this Fourier series everywhere in the interval [0, 2] is much more rapid than in question 7.

[graph of y = f(x)]
L = 2
An odd extension of f (x) is required.
a_n = 0 for all n.

Note how the periodic extension of f (x) is not only continuous everywhere but differentiable everywhere, even at the endpoints. We therefore anticipate a more rapid convergence than in question 7.

Integral for b_n
b_n = 32 / (n pi)^3 for odd n only [tabular integration by parts]

Therefore the Fourier series is

$f(x) = (32/pi^3) Sum { sin((2k-1) pi x/2)/(2k-1)^3 }$

The first few terms of this series are

The partial sum of just the first two non-zero terms yields an excellent approximation everywhere:

[graph of y = f(x)]
y = f (x) is in red
y = S₃ is in blue.

[A more challenging question:]
An incompressible viscous fluid is in steady-state flow around the z-axis, trapped between a pair of coaxial rotating cylinders, aligned along the z-axis. The inner cylinder has radius a and rotates at angular velocity w k. The outer cylinder has radius b (> a) and rotates at angular velocity u k.
In response, the velocity vector at any point in the fluid, in cylindrical polar coordinates (r, f, z), is and v(r) obeys the ordinary differential equation

Where it is in contact with a cylinder, the fluid must move with the same velocity as that cylinder.
Solve the ODE to find the velocity at all points in the fluid.

The Laplacian of any scalar function V in cylindrical polar coordinates is, in general,

However, the velocity vector has neither radial nor vertical components.
The f component of the velocity vector is a function of r only, so that the ODE becomes

so that the ODE becomes

This is a linear first order ODE and can be solved by that method,
or one may use a similar shortcut to an exact form again:

The two arbitrary constants are determined by the two boundary conditions.
The linear speed of each cylinder is just the product of the radius of that cylinder with its angular speed:

The complete solution for the velocity vector at any point (r, f, z) in the fluid is therefore

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Created 2008 01 29 and most recently modified 2008 12 27 by Dr. G.H. George