Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 9 doc

In this case the homogeneous equation subject to homogeneousboundary conditions has only the trivial solution.. Since the completely homogeneous problem has no solutions, we know that B1

We solve this system with Kramer’s rule.

c1(ξ) = − y2(ξ)

p(ξ)(−W (ξ)), c2(ξ) = −

y1(ξ)p(ξ)(−W (ξ))Here W (x) is the Wronskian of y1(x) and y2(x) The Green function is

G(x|ξ) =

(y 1 (x)y 2 (ξ) p(ξ)W (ξ) for a ≤ x ≤ ξ,

y 2 (x)y 1 (ξ) p(ξ)W (ξ) for ξ ≤ x ≤ b

The solution of the Sturm-Liouville problem is

y =

Z b a

G(x|ξ)f (ξ) dξ

Result 21.7.2 The problem

L[y] = (p(x)y0)0+ q(x)y = f (x), subject to

B1[y] = α1y(a) + α2y0(a) = 0, B2[y] = β1y(b) + β2y0(b) = 0.

has the Green function

G(x|ξ) =

(y 1 (x)y 2 (ξ) p(ξ)W (ξ) for a ≤ x ≤ ξ,

y2(x)y1(ξ) p(ξ)W (ξ) for ξ ≤ x ≤ b, where y1 and y2 are non-trivial homogeneous solutions that satisfy B1[y1] = B2[y2] = 0, and

W (x) is the Wronskian of y1 and y2.

Example 21.7.5 Consider the equation

sinh x sinh(x − 1)cosh x cosh(x − 1)

= sinh x cosh(x − 1) − cosh x sinh(x − 1)

sinh(x−1) sinh ξ sinh(1) for ξ ≤ x ≤ 1

The solution to the problem is

y = sinh(x − 1)

sinh(1)

Z x 0

sinh(ξ)f (ξ) dξ + sinh(x)

sinh(1)

Z 1 x

subject the the initial conditions

y(a) = γ1, y0(a) = γ2.The solution is y = u + v where

u00+ p(x)u0+ q(x)u = f (x), u(a) = 0, u0(a) = 0,and

v00+ p(x)v0+ q(x)v = 0, v(a) = γ1, v0(a) = γ2.Since the Wronskian

W (x) = c exp



−

Zp(x) dx

Let u1and u2be two linearly independent solutions of the differential equation For x < ξ, G(x|ξ) is a linear combination

of these solutions Since the Wronskian is non-vanishing, only the trivial solution satisfies the homogeneous initialconditions The Green function must be

G(x|ξ) =

(

0 for x < ξ

uξ(x) for x > ξ,where uξ(x) is the linear combination of u1 and u2 that satisfies

uξ(ξ) = 0, u0ξ(ξ) = 1

Note that the non-vanishing Wronskian ensures a unique solution for uξ We can write the Green function in the form

G(x|ξ) = H(x − ξ)uξ(x)

This is known as the causal solution The solution for u is

u =

Z b a

G(x|ξ)f (ξ) dξ

=

Z b a

H(x − ξ)uξ(x)f (ξ) dξ

=

Z x a

uξ(x)f (ξ) dξ

Now we have the solution for y,

y = v +

Z x a

uξ(x)f (ξ) dξ

Result 21.7.3 The solution of the problem

y00 + p(x)y0+ q(x)y = f (x), y(a) = γ1, y0(a) = γ2, is

y = yh+

Z x

a

yξ(x)f (ξ) dξ where yh is the combination of the homogeneous solutions of the equation that satisfy the initial conditions and yξ(x) is the linear combination of homogeneous solutions that satisfy

yξ(ξ) = 0, yξ0(ξ) = 1.

21.7.3 Problems with Unmixed Boundary Conditions

The Green function for u satisfies

of u1 and u2 Since the homogeneous equation with homogeneous boundary conditions has only the trivial solution,

W (x) is nonzero on [a, b] The Green function has the form

G(x|ξ)f (ξ) dξ

Thus if there is a unique solution for v, the solution for y is

y = v +

Z b a

G(x|ξ)f (ξ) dξ

Result 21.7.4 Consider the problem

y00 + p(x)y0 + q(x)y = f (x),

α1y(a) + α2y0(a) = γ1, β1y(b) + β2y0(b) = γ2.

If the homogeneous differential equation subject to the inhomogeneous boundary conditions has the unique solution yh, then the problem has the unique solution

y = yh+

Z b

a

G(x|ξ)f (ξ) dξ where

B1[y] = α11y(a) + α12y0(a) + β11y(b) + β12y0(b) = γ1,

B2[y] = α21y(a) + α22y0(a) + β21y(b) + β22y0(b) = γ2

The solution is y = u + v where

u00+ p(x)u0+ q(x)u = f (x), B1[u] = 0, B2[u] = 0,and

v00+ p(x)v0 + q(x)v = 0, B1[v] = γ1, B2[v] = γ2

The problem for v may have no solution, a unique solution or an infinite number of solutions Again we consideronly the case that there is a unique solution for v In this case the homogeneous equation subject to homogeneousboundary conditions has only the trivial solution

Let y1 and y2 be two solutions of the homogeneous equation that satisfy the boundary conditions B1[y1] = 0 and

B2[y2] = 0 Since the completely homogeneous problem has no solutions, we know that B1[y2] and B2[y1] are nonzero.The solution for v has the form

v = c1y1+ c2y2.Applying the two boundary conditions yields

v = γ2

B2[y1]y1+

γ1

B1[y2]y2.The Green function for u satisfies

With this form, the continuity and jump conditions are automatically satisfied Applying the boundary conditions yields

G(x|ξ)f (ξ) dξ

Thus if there is a unique solution for v, the solution for y is

y =

Z b a

G(x|ξ)f (ξ) dξ + γ2

B2[y1]y1+

γ1

B1[y2]y2.

Result 21.7.5 Consider the problem

y00 + p(x)y0 + q(x)y = f (x),

B1[y] = α11y(a) + α12y0(a) + β11y(b) + β12y0(b) = γ1,

B2[y] = α21y(a) + α22y0(a) + β21y(b) + β22y0(b) = γ2.

If the homogeneous differential equation subject to the homogeneous boundary conditions has no solution, then the problem has the unique solution

sec-21.8 Green Functions for Higher Order Problems

Consider the nth order differential equation

L[y] = y(n)+ pn−1(x)y(n−1)+ · · · + p1(x)y0 + p0y = f (x) on a < x < b,subject to the n independent boundary conditions

Bj[y] = γjwhere the boundary conditions are of the form

L[w] = 0, with Bj[w] = 0,has only the trivial solution, then the solution for y exists and is unique We will construct this solution using Greenfunctions

First we consider the problem for v Let {y1, , yn} be a set of linearly independent solutions The solution for vhas the form

To solve the problem for u we consider the Green function satisfying

L[G(x|ξ)] = δ(x − ξ), with Bj[G] = 0

Let yξ(x) be the linear combination of the homogeneous solutions that satisfy the conditions

yξ(ξ) = 0

yξ0(ξ) = 0

The constants are determined by the matrix equation

The solution for u then is

u =

Z b a

G(x|ξ)f (ξ) dξ

Result 21.8.1 Consider the nth order differential equation

L[y] = y(n)+ pn−1(x)y(n−1) + · · · + p1(x)y0 + p0y = f (x) on a < x < b,

subject to the n independent boundary conditions

cj and dj can be determined by solving sets of linear equations.

Example 21.8.1 Consider the problem

y000− y00+ y0− y = f (x),y(0) = 1, y0(0) = 2, y(1) = 3

The completely homogeneous associated problem is

w000− w00+ w0− w = 0, w(0) = w0(0) = w(1) = 0

The solution of the differential equation is

w = c1cos x + c2sin x + c2ex.The boundary conditions give us the equation

v = c1cos x + c2sin x + c2ex.The boundary conditions yields the equation





The solution for v is

e − cos 1 − sin 1(e + sin 1 − 3) cos x + (2e − cos 1 − 3) sin x + (3 − cos 1 − 2 sin 1) ex

Now we find the Green function for the problem in u The causal solution is

H(x − ξ)uξ(x) = H(x − ξ)1

2(sin ξ − cos ξ) cos x − (sin ξ + cos ξ) sin ξ + e−ξex,

H(x − ξ)uξ(x) = 1

2H(x − ξ) ex−ξ− cos(x − ξ) − sin(x − ξ)

The Green function has the form

G(x|ξ) = H(x − ξ)uξ(x) + c1cos x + c2sin x + c3ex.The constants are determined by the three conditions

G(x|ξ)f (ξ) dξ

Thus the solution for y is

y =

Z 1 0

G(x|ξ)f (ξ) dξ + 1

e − cos 1 − sin 1(e + sin 1 − 3) cos x+ (2e − cos 1 − 3) sin x + (3 − cos 1 − 2 sin 1) ex

Orthogonality Two real vectors, u and v are orthogonal if u · v = 0 Consider two functions, u(x) and v(x),defined in [a, b] The dot product in vector space is analogous to the integral

Z b a

u(x)v(x) dx

in function space Thus two real functions are orthogonal if

Z b a

u(x)v(x) dx = 0

Consider the nth order linear inhomogeneous differential equation

L[y] = f (x) on [a, b],subject to the linear inhomogeneous boundary conditions

Bj[y] = 0, for j = 1, 2, n

The Fredholm alternative theorem tells us if the problem has a unique solution, an infinite number of solutions, or

no solution Before presenting the theorem, we will consider a few motivating examples

No Nontrivial Homogeneous Solutions In the section on Green functions we showed that if the completelyhomogeneous problem has only the trivial solution then the inhomogeneous problem has a unique solution.

Nontrivial Homogeneous Solutions Exist If there are nonzero solutions to the homogeneous problem L[y] = 0that satisfy the homogeneous boundary conditions Bj[y] = 0 then the inhomogeneous problem L[y] = f (x) subject tothe same boundary conditions either has no solution or an infinite number of solutions

Suppose there is a particular solution yp that satisfies the boundary conditions If there is a solution yh to thehomogeneous equation that satisfies the boundary conditions then there will be an infinite number of solutions since

yp+ cyh is also a particular solution

The question now remains: Given that there are homogeneous solutions that satisfy the boundary conditions, how

do we know if a particular solution that satisfies the boundary conditions exists? Before we address this question wewill consider a few examples

Example 21.9.1 Consider the problem

y00+ y = cos x, y(0) = y(π) = 0

The two homogeneous solutions of the differential equation are

y1 = cos x, and y2 = sin x

y2 = sin x satisfies the boundary conditions Thus we know that there are either no solutions or an infinite number of

solutions A particular solution is

y = 1

2x sin x + c1cos x + c2sin x.

Applying the two boundary conditions yields

y = 1

2x sin x + c sin x.

Thus there are an infinite number of solutions

Example 21.9.2 Consider the differential equation

y00+ y = sin x, y(0) = y(π) = 0

The general solution is

y = −1

2x cos x + c1cos x + c2sin x.

Applying the boundary conditions,

y(0) = 0 → c1 = 0y(π) = 0 → − 1

2π cos(π) + c2sin(π) = 0

2 = 0.

Since this equation has no solution, there are no solutions to the inhomogeneous problem

In both of the above examples there is a homogeneous solution y = sin x that satisfies the boundary conditions

In Example 21.9.1, the inhomogeneous term is cos x and there are an infinite number of solutions In Example 21.9.2,the inhomogeneity is sin x and there are no solutions In general, if the inhomogeneous term is orthogonal to all thehomogeneous solutions that satisfy the boundary conditions then there are an infinite number of solutions If not, thereare no inhomogeneous solutions

Result 21.9.1 Fredholm Alternative Theorem Consider the nth order inhomogeneous problem

L[y] = f (x) on [a, b] subject to Bj[y] = 0 for j = 1, 2, , n, and the associated homogeneous problem

L[y] = 0 on [a, b] subject to Bj[y] = 0 for j = 1, 2, , n.

If the homogeneous problem has only the trivial solution then the inhomogeneous problem has

a unique solution If the homogeneous problem has m independent solutions, {y1, y2, , ym}, then there are two possibilities:

• If f (x) is orthogonal to each of the homogeneous solutions then there are an infinite number of solutions of the form

inho-Example 21.9.3 Consider the problem

y00+ y = cos 2x, y(0) = 1, y(π) = 2

cos x and sin x are two linearly independent solutions to the homogeneous equation sin x satisfies the homogeneousboundary conditions Thus there are either an infinite number of solutions, or no solution

To transform this problem to one with homogeneous boundary conditions, we note that g(x) = x

π + 1 and makethe change of variables y = u + g to obtain

u00+ u = cos 2x − x

π − 1, y(0) = 0, y(π) = 0

Since cos 2x −πx − 1 is not orthogonal to sin x, there is no solution to the inhomogeneous problem

To check this, the general solution is

y = −1

3cos 2x + c1cos x + c2sin x.

Applying the boundary conditions,

From the Fredholm Alternative Theorem we see that the inhomogeneous problem has a unique solution

To find the solution, start with

y = −1

3cos 2x + c1cos x + c2sin x.

y0(0) = 1 → c2 = 1

3− c1 = 1Thus the solution is

u00+ u = cos 2x + 1

3, y(0) = 0, y(π) = 0.

Now we check if sin x is orthogonal to cos 2x + 13

Z π 0

sin x

cos 2x + 1

3



dx =

Z π 0

As a check, then general solution for y is

y = −1

3cos 2x + c1cos x + c2sin x.

Applying the boundary conditions,

y = −1

3cos 2x + cos x + c sin x.

1 Find the general solution of y00+ y = ex.

2 Solve y00+ λ2y = sin x, y(0) = y0(0) = 0 λ is an arbitrary real constant Is there anything special about λ = 1?

g(τ ) sin τ dτ

cos t +



c2+

Z t b

g(τ ) cos τ dτ

sin t,where c1 and c2 are arbitrary constants and a and b are any conveniently chosen points

2 Using the result of part (a) show that the solution satisfying the initial conditions y(0) = 0 and y0(0) = 0 is givenby

y(t) =

Z t 0

g(τ ) sin(t − τ ) dτ

Notice that this equation gives a formula for computing the solution of the original initial value problem for anygiven inhomogeneous term g(t) The integral is referred to as the convolution of g(t) with sin t.

3 Use the result of part (b) to solve the initial value problem,

y00+ y = sin(λt), y(0) = 0, y0(0) = 0,where λ is a real constant How does the solution for λ = 1 differ from that for λ 6= 1? The λ = 1 case provides

an example of resonant forcing Plot the solution for resonant and non-resonant forcing

Hint, Solution

Exercise 21.8

Find the variation of parameters solution for the third order differential equation

y000+ p2(x)y00+ p1(x)y0+ p0(x)y = f (x)

Exercise 21.11

What are the continuity conditions at x = ξ for the Green function for the problem

y000+ p2(x)y00+ p1(x)y0+ p0(x)y = f (x)

Find the Green function for each of the following:

a) xu00+ u0 = f (x), u(0+) bounded, u(1) = 0

b) u00− u = f (x), u(−a) = u(a) = 0

c) u00− u = f (x), u(x) bounded as |x| → ∞

d) Show that the Green function for (b) approaches that for (c) as a → ∞.

Hint, Solution

Exercise 21.16

1 For what values of λ does the problem

have a unique solution? Find the Green functions for these cases

2 For what values of α does the problem

y00+ 9y = 1 + αx, y(0) = y(π) = 0,have a solution? Find the solution

3 For λ = n2, n ∈ Z+ state in general the conditions on f in Equation 21.5 so that a solution will exist What isthe appropriate modified Green function (in terms of eigenfunctions)?

Hint, Solution

Exercise 21.17

Show that the inhomogeneous boundary value problem:

Lu ≡ (pu0)0 + qu = f (x), a < x < b, u(a) = α, u(b) = βhas the solution:

u(x) =

Z b a

i) u(0) = 0 |u(∞)| < ∞,ii) u0(0) = 0 |u(∞)| < ∞.

Express these results in simplified forms without absolute values

Hint, Solution

Exercise 21.19

1 Determine the Green function for solving:

y00− a2y = f (x), y(0) = y0(L) = 0

2 Take the limit as L → ∞ to find the Green function on (0, ∞) for the boundary conditions: y(0) = 0, y0(∞) = 0

We assume here that a > 0 Use the limiting Green function to solve:

y00− a2y = e−x, y(0) = 0, y0(∞) = 0

Check that your solution satisfies all the conditions of the problem

Hint, Solution

where the yj’s are homogeneous solutions Impose the constraints

subject to unmixed, homogeneous boundary conditions is

G(x|ξ) = y1(x<)y2(x>)

p(ξ)W (ξ) ,

G(x|ξ) =

(y 1 (x)y 2 (ξ) p(ξ)W (ξ) for a ≤ x ≤ ξ,

y 1 (ξ)y 2 (x) p(ξ)W (ξ) for ξ ≤ x ≤ b,where y1 and y2 are homogeneous solutions that satisfy the left and right boundary conditions, respectively

Recall that if y(x) is a solution of a homogeneous, constant coefficient differential equation then y(x + c) is also asolution

A particular solution is

yp = 3

17(sin(2t) − 4 cos(2t)).

The general solution of the differential equation is

2(2a − d cos(t) − e sin(t)) + 3(2at + b − d sin(t) + e cos(t))

+ at2+ bt + c + d cos(t) + e sin(t) = t2+ 3 sin(t)

(a − 1)t2+ (6a + b)t + (4a + 3b + c) + (−d + 3e) cos(t) − (3 + 3d + e) sin(t) = 0

a − 1 = 0, 6a + b = 0, 4a + 3b + c = 0, −d + 3e = 0, 3 + 3d + e = 0

a = 1, b = −6, c = 14, d = − 9

10, e = −

310

λ = 1The homogeneous solution is

(6a − 1)t + 2b = 06a − 1 = 0, 2b = 0

a = 1

6, b = 0

λ2+ 2λ + 5 = 0

λ = −1 ±√

1 − 5

λ = −1 ± ı2The homogeneous solution is

yh = c1e−tcos(2t) + c2e−tsin(2t)

B2[y]... α21y(a) + α22y0(a) + β21y(b) + β22y0(b) = γ2.

If... p(x)y0+ q(x)y = f (x), y(a) = γ1, y0(a) = γ2, is

y = yh+

Z x