Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 9 pot

Chapter 43Method of Characteristics Consider the following first order wave equation.. By comparing Equations 43.3 and 43.4 we obtain ordinary differential equations for the characterist

2√ν

+ c2

We write this solution in terms of x and t.

u(x, t) = c1erf

x

2√νt

+ c2

This is not the general solution of the heat equation There are many other solutions Note that since x and t do notexplicitly appear in the heat equation,

u(x, t) = c1erf x − x0

2pν(t − t0)

!+ c2

differential equation for f (ξ).

f0 = 4ξ2f00+ 6ξf0

f00

f0 = 14ξ2 − 32ξlog(f0) = − 1

2√ξ

+ c2

Chapter 43

Method of Characteristics

Consider the following first order wave equation

Let x(t) be some path in the phase plane Perhaps x(t) describes the position of an observer who is noting the value

of the solution u(x(t), t) at their current location We differentiate with respect to t to see how the solution varies forthe observer

Let the observer start at the position x0 Then we have an initial value problem for x(t).

dx

dt = c, x(0) = x0x(t) = x0 + ct

These lines x(t) are called characteristics of Equation43.1

Let the initial condition be u(x, 0) = f (x) We have an initial value problem for u(x(t), t)

du

dt = 0, u(0) = f (x0)u(x(t), t) = f (x0)

Again we see that the solution is constant along the characteristics We substitute the equation for the characteristicsinto this expression

u(x0+ ct, t) = f (x0)u(x, t) = f (x − ct)Now we see that the solution of Equation 43.1 is a wave moving with velocity c The solution at time t is the initialcondition translated a distance of ct

Consider the following quasi-linear equation

By comparing Equations 43.3 and 43.4 we obtain ordinary differential equations for the characteristics x(t) and thesolution along the characteristics u(x(t), t).

dx

dt = a(x, t, u),

du

dt = 0Suppose an initial condition is specified, u(x, 0) = f (x) Then we have ordinary differential equation, initial valueproblems

dx

dt = a(x, t, u), x(0) = x0du

dt = 0, u(0) = f (x0)

First we solve the equation for u u = f (x0) Then we solve for x x = x0+ f (x0)t This gives us an implicit solution

of the Burger equation

u(x0+ f (x0)t, t) = f (x0)

43.3 The Method of Characteristics and the Wave Equation

Consider the one dimensional wave equation,



x

= 0The eigenvalues and eigenvectors of the matrix are

λ1 = −c, λ2 = c, ξ1 =1

c

, ξ2 = 1

−c

.The matrix is diagonalized by a similarity transformation



c −c

 αβ



c −c

 αβ



x

= 0

Now we left multiply by the inverse of the matrix of eigenvectors to obtain an uncoupled system that we can solvedirectly.

αβ



x

= 0

α(x, t) = p(x + ct), β(x, t) = q(x − ct),Here p, q ∈ C2 are arbitrary functions We change variables back to a and b

Consider the Cauchy problem for the wave equation on −∞ < x < ∞

utt = c2uxx, −∞ < x < ∞, t > 0u(x, 0) = f (x), ut(x, 0) = g(x)

We know that the solution is the sum of right-moving and left-moving waves

The initial conditions give us two constraints on F and G.

F (x) = 1

2f (x) −

12c

Zg(x) dx − k, G(x) = 1

2f (x) +

12c

Zg(x) dx + k

Note that the value of the constant k does not affect the solution, u(x, t) For simplicity we take k = 0 We substitute

F and G into Equation 43.5 to determine the solution

u(x, t) = 1

2(f (x − ct) + f (x + ct)) +

12c

Z x+ct x−ct

g(ξ) dξ

u(x, t) = 1

2(u(x − ct, 0) + u(x + ct, 0)) +

12c

Z x+ct x−ct

ut(ξ, 0) dξ

Consider the wave equation for a semi-infinite domain

utt = c2uxx, 0 < x < ∞, t > 0u(x, 0) = f (x), ut(x, 0) = g(x), u(0, t) = h(t)

Again the solution is the sum of a right-moving and a left-moving wave.

u(x, t) = F (x − ct) + G(x + ct)For x > ct, the boundary condition at x = 0 does not affect the solution Thus we know the solution in this domainfrom our work on the wave equation in the infinite domain

u(x, t) = 1

2(f (x − ct) + f (x + ct)) +

12c

Z x+ct x−ct

g(ξ) dξ, x > ctFrom this, F (ξ) and G(ξ) are determined for ξ > 0

F (ξ) = 1

2f (ξ) −

12c

Zg(ξ) dξ, ξ > 0

G(ξ) = 1

2f (ξ) +

12c

Zg(ξ) dξ, ξ > 0

In order to determine the solution u(x, t) for x, t > 0 we also need to determine F (ξ) for ξ < 0 To do this, wesubstitute the form of the solution into the boundary condition at x = 0

Z −ξ

g(ψ) dψ + h(−ξ/c), ξ < 0

We determine the solution of the wave equation for x < ct

u(x, t) = F (x − ct) + G(x + ct)u(x, t) = −1

2f (−x + ct) −

12c

Z −x+ct

g(ξ) dξ + h(t − x/c) + 1

2f (x + ct) +

12c

Z x+ct

−x+ct

g(ξ) dξ + h(t − x/c), x < ct

Finally, we collect the solutions in the two domains.

Consider the wave equation for the infinite domain

utt = c2uxx, −∞ < x < ∞, t > 0u(x, 0) = f (x), ut(x, 0) = g(x)

If f (x) and g(x) are odd about x = 0, (f (x) = −f (−x), g(x) = −g(−x)), then u(x, t) is also odd about x = 0 Wecan demonstrate this with D’Alembert’s solution

u(x, t) = 1

2(f (x − ct) + f (x + ct)) +

12c

Z x+ct x−ct

g(ξ) dξ

−u(−x, t) = −1

2(f (−x − ct) + f (−x + ct)) −

12c

Z x−ct x+ct

g(−ξ) (−dξ)

= 1

2(f (x − ct) + f (x + ct)) +

12c

Z x+ct x−ct

g(ξ) dξ

= u(x, t)Thus if the initial conditions f (x) and g(x) are odd about a point then the solution of the wave equation u(x, t) isalso odd about that point The analogous result holds if the initial conditions are even about a point These resultsare useful in solving the wave equation on a finite domain

Consider a string of length L with fixed ends.

utt = c2uxx, 0 < x < L, t > 0u(x, 0) = f (x), ut(x, 0) = g(x), u(0, t) = u(L, t) = 0

We extend the domain of the problem to x ∈ (−∞ ∞) We form the odd periodic extensions ˜f and ˜g which areodd about the points x = 0, L

If a function h(x) is defined for positive x, then sign(x)h(|x|) is the odd extension of the function If h(x) is definedfor x ∈ (−L L) then its periodic extension is

x − 2L x + L

2L





˜g(x) = sign

x − 2L x + L

2L



Z x+ct x−ct

˜g(ξ) dξ

Consider the tangent lines to the parabola y = x2 The slope of the tangent at the point (x, x2) is 2x The set oftangents form a one parameter family of lines,

f (x, t) = t2+ (x − t)2t = 2tx − t2

-1 1

-1

Figure 43.1: A parabola and its tangents

The parabola and some of its tangents are plotted in Figure 43.1

The parabola is the envelope of the family of tangent lines Each point on the parabola is tangent to one of thelines Given a curve, we can generate a family of lines that envelope the curve We can also do the opposite, given

a family of lines, we can determine the curve that they envelope More generally, given a family of curves, we candetermine the curve that they envelope Let the one parameter family of curves be given by the equation F (x, y, t) = 0.For the example of the tangents to the parabola this equation would be y − 2tx + t2 = 0

Let y(x) be the envelope of F (x, y, t) = 0 Then the points on y(x) must lie on the family of curves Thus y(x)must satisfy the equation F (x, y, t) = 0 The points that lie on the envelope have the property,

∂

∂tF (x, y, t) = 0.

We can solve this equation for t in terms of x and y, t = t(x, y) The equation for the envelope is then

F (x, y, t(x, y)) = 0

Consider the example of the tangents to the parabola The equation of the one-parameter family of curves is

These are circles of unit radius and center (t, t) To determine the envelope of the family, we first use the constraint

Ft(x, y, t) to solve for t(x, y)

Ft(x, y, t) = −2(x − t) − 2(y − t) = 0

t(x, y) = x + y

2Now we substitute this into the equation F (x, y, t) = 0 to determine the envelope

2

+



y − x + y2

2

− 1 = 0

 x − y2

2

+ y − x2

2

− 1 = 0(x − y)2 = 2

y = x ±√

2The one parameter family of curves and its envelope is shown in Figure 43.2

-3 -2 -1 1 2 3

-3 -2 -1

1 2

Figure 43.2: The envelope of (x − t)2+ (y − t)2− 1 = 0

Exercise 43.1

Consider the small transverse vibrations of a composite string of infinite extent, made up of two homogeneous strings

of different densities joined at x = 0 In each region 1) x < 0, 2) x > 0 we have

utt− c2

juxx = 0 j = 1, 2 c1 6= c2,and we require continuity of u and ux at x = 0 Suppose for t < 0 a wave approaches the junction x = 0 from theleft, i.e as t approaches 0 from negative values:

1 Formulate the appropriate initial values for u at t = 0.

2 Solve the initial-value problem for −∞ < x < ∞ , t > 0

3 Identify the incident, reflected and transmitted waves in your solution and determine the reflection and transmissioncoefficients for the junction in terms of c1 and c2 Comment also on their values in the limit c1 → c2

Exercise 43.2

Consider a semi-infinite string, x > 0 For all time the end of the string is displaced according to u(0, t) = f (t) Findthe motion of the string, u(x, t) with the method of characteristics and then with a Fourier transform in time Thewave speed is c

Solve using characteristics:

(y + u)ux+ yuy = x − y, u ... satisfy y(s) ≥ for all real s Figure43.4 shows some characteristics in the (x, y)plane with starting points from (? ?5, 1) to (5, 1) and a plot of the solution

0. 250 .50 . 751 1. 251 .51 . 75

-1012x... class="page_container" data-page="20">

We solve for f for x < c2t and for g for x > −c1t.

The reflection and transmission coefficients are, respectively,

For x < 0, the right-moving wave is F (x − c1t) and the left-moving wave is zero for x < −c1t For x >