Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 8 doc

We determine the coefficients from the inhomogeneous boundary conditions.. We determine the coefficients from the inhomogeneous boundary conditions... An alert student suggests that this

We have the implicit boundary conditions,

u(0, z) = u(2π, z), uθ(0, z) = uθ(0, z)and the boundedness condition,

f (θ) e−ınθ dθ ≡ fn

Thus the solution is

u(θ, z) → f0 = 1

2π

Z 2π 0

Figure 40.1: Decomposition of the problem

First we solve the problem for u

We have the eigenvalue problem,

X00 = −λ2X, X(0) = X(a) = 0,which has the solutions,

λn = nπ

a , Xn= sin

nπxa

, n ∈ N

The equation for Y (y) becomes,

nπya

, sinh

nπya

o

It will be convenient to choose solutions that satisfy the conditions, Y (b) = 0 and Y0(0) = 0, respectively

sinh nπ(b − y)

a

, coshnπy



αnsinh nπ(b − y)

a

+ βncoshnπy

a



We determine the coefficients from the inhomogeneous boundary conditions (Here we see how our choice of solutions

for Y (y) is convenient.)

cosh nπb

 2a

Z a 0

g1(x) sinnπx

a

dx

Z a 0

g2(x) sinnπx

a

dxNow we solve the problem for v

vxx+ vyy = 0, 0 < x < a, 0 < y < b,v(0, y) = f1(y), v(a, y) = f2(y),

λn= (2n − 1)π

2b , Yn= cos

 (2n − 1)πy2b

, n ∈ N.The equation for X(y) becomes,

Xn00 = (2n − 1)π

2b

2

Xn

We choose solutions that satisfy the conditions, X(a) = 0 and X(0) = 0, respectively.

sinh (2n − 1)π(a − x)

2b

, sinh (2n − 1)πx

2b



We determine the coefficients from the inhomogeneous boundary conditions

Z b 0

f1(y) cos (2n − 1)πy

2b

dyv(a, y) =

Z b 0

f2(y) cos (2n − 1)πy

2b

dyWith u and v determined, the solution of the original problem is w = u + v

Chapter 41

Waves

u(0, t) = u(4, t) = 02

Z x+ct x−ct

ut(τ, 0) dτ,for various values of t corresponding to the initial conditions:

1 u(x, 0) = 0, ut(x, 0) = sin ωx where ω is a constant,

−h(−x) for x < 0,and with initial velocity

ut(x, 0) = 0

Show that the solution of the wave equation satisfying these initial conditions also solves the following semi-infiniteproblem: Find u(x, t) satisfying the wave equation utt = c2uxx in 0 < x < ∞, t > 0, with initial conditionsu(x, 0) = h(x), ut(x, 0) = 0, and with the fixed end condition u(0, t) = 0 Here h(x) is any given function withh(0) = 0

2 Use a similar idea to explain how you could use the general solution of the wave equation to solve the finiteinterval problem (0 < x < l) in which u(0, t) = u(l, t) = 0 for all t, with u(x, 0) = h(x) and ut(x, 0) = 0 Takeh(0) = h(l) = 0

Exercise 41.4

The deflection u(x, T ) = φ(x) and velocity ut(x, T ) = ψ(x) for an infinite string (governed by utt = c2uxx) aremeasured at time T , and we are asked to determine what the initial displacement and velocity profiles u(x, 0) and

ut(x, 0) must have been An alert student suggests that this problem is equivalent to that of determining the solution

of the wave equation at time T when initial conditions u(x, 0) = φ(x), ut(x, 0) = −ψ(x) are prescribed Is she correct?

If not, can you rescue her idea?

“whip-cracking” problem,

utt = c2uxx,(with c a constant) in the domain x > 0, t > 0 with initial conditions

u(x, 0) = ut(x, 0) = 0 x > 0,and boundary conditions

u(0, t) = γ(t)prescribed for all t > 0 Here γ(0) = 0 Find α and β so as to determine u for x > 0, t > 0

Hint: (From physical considerations conclude that you can take α(ξ) = 0 Your solution will corroborate this.)Use the initial conditions to determine α(ξ) and β(ξ) for ξ > 0 Then use the initial condition to determine β(ξ) for

ξ < 0

Exercise 41.6

Let u(x, t) satisfy the equation

utt = c2uxx;(with c a constant) in some region of the (x, t) plane

1 Show that the quantity (ut− cux) is constant along each straight line defined by x − ct = constant, and that(ut+ cux) is constant along each straight line of the form x + ct = constant These straight lines are calledcharacteristics; we will refer to typical members of the two families as C+ and C− characteristics, respectively.Thus the line x − ct = constant is a C+ characteristic

2 Let u(x, 0) and ut(x, 0) be prescribed for all values of x in −∞ < x < ∞, and let (x0, t0) be some point in the(x, t) plane, with t0 > 0 Draw the C+ and C− characteristics through (x0, t0) and let them intersect the x-axis

at the points A,B Use the properties of these curves derived in part (a) to determine ut(x0, t0) in terms of initialdata at points A and B Using a similar technique to obtain ut(x0, τ ) with 0 < τ < t, determine u(x0, t0) byintegration with respect to τ , and compare this with the solution derived in class:

u(x, t) = 1

2(u(x + ct, 0) + u(x − ct, 0)) +

12c

Z x+ct x−ct

u(0, t) = T cos(ωt)

Seek a solution of the form

u(x, t) = < A eıωt−αx
a) Find u(x, t) satisfying u → 0 as x → +∞, (i.e deep into the Earth)

b) Find the temperature variation at a fixed depth, h, below the surface

c) Find the phase lag δ(x) such that when the maximum temperature occurs at t0 on the surface, the maximum atdepth x occurs at t0+ δ(x)

d) Show that the seasonal, (i.e yearly), temperature changes and daily temperature changes penetrate to depths inthe ratio:

xyear

xday =

√365,where xyear and xday are the depths of same temperature variation caused by the different periods of the source

u(r, θ, t) = v(r, θ) eıωt.Exercise 41.9

Plane waves are incident on a “soft” cylinder of radius a whose axis is parallel to the plane of the waves Find thefield scattered by the cylinder In particular, examine the leading term of the solution when a is much smaller than thewavelength of the incident waves If v(x, y, t) is the scattered field it must satisfy:

Wave Equation: vtt = c2∆v, x2 + y2 > a2;Soft Cylinder: v(x, y, t) = − eı(ka cos θ−ωt), on r = a, 0 ≤ θ < 2π;

where C is the capacitance, G is the conductance, L is the inductance and R is the resistance.

a) Show that both I and V satisfy a damped wave equation

b) Find the relationship between the physical constants, C, G, L and R such that there exist damped travelingwave solutions of the form:

V (x, t) = e−γt(f (x − at) + g(x + at))

What is the wave speed?

Solve the Helmholtz equation for v with a Fourier series expansion,

x − 32



We extend the domain of the problem to (−∞ ∞) and add image sources in the initial condition so thatu(x, 0) is odd about x = 0 and x = 4 This enforces the boundary conditions at these two points

utt− uxx = 0, x ∈ (−∞ ∞), t ∈ (0 ∞)u(x, 0) =

x − 3

2 − 8n



− H 1

2−



−H 1

2 −

x −13

2 − 8n − t



− H 1

2 −

...

We extend the domain of the problem to (−∞ ∞) and add image sources in the initial condition so thatu(x, 0) is odd about x = and x = This enforces the boundary conditions at these two points... 3

2 − 8n



− H 1

2−

x − 13

2 − 8n

x − 3

2 − 8n − t

+...

Figure 41.1: The solution at t = 1/2, 1, 3/2, for the boundary conditions u(0, t) = u(4, t) =

u(x, 0) is even about x = and x = This enforces the boundary conditions at these two points