Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 5 pdf

Thus the solution of the integral equation is... Since L1e−t 2 /4 = 0 the result of this problem does not produce any negative eigenvalues.. Since this function is square integrable it i

Thus the solution of the integral equation is



2 For x < 1 the integral equation reduces to

φ(x) = x

For x ≥ 1 the integral equation becomes,

φ(x) = x + λ

Z 1 0

coefficients of the eigenfunctions.

φ(x) = x + λ

Z 1 0

L1L2− L2L1 = d

dt +

t2

 

−d

dt +

t2

  d

dt +

t2



= −d

dt +

t2

d

dt − 1

2I +

t2

We note that e−t2/4 is an eigenfunction corresponding to the eigenvalue λ = 1/2 Since L1e−t 2 /4 = 0 the result

of this problem does not produce any negative eigenvalues However, Ln2 e−t 2 /4 is the product of e−t2/4 and apolynomial of degree n in t Since this function is square integrable it is and eigenfunction Thus we have theeigenvalues and eigenfunctions,

λn= n − 1

2, φn=

 t

2− ddt

for all x Now we apply the definition of the adjoint.

hx, (T − λ1I)∗yi = 0, ∀x

hx, (T∗− λ1I)yi = 0, ∀x(T∗− λ1I)y = 0

y is an eigenfunction of T∗ corresponding to the eigenvalue λ1

Solution 48.26

1

u00(t) +

Z 1 0

sin(k(s − t))u(s) ds = f (t), u(0) = u0(0) = 0

u00(t) + cos(kt)

Z 1 0

sin(ks)u(s) ds − sin(kt)

Z 1 0

cos(ks)u(s) ds = f (t)

u00(t) + c1cos(kt) − c2sin(kt) = f (t)

u00(t) = f (t) − c1cos(kt) + c2sin(kt)The solution of

u00(t) = g(t), u(0) = u0(0) = 0using Green functions is

u(t) =

Z t 0

(t − τ )g(τ ) dτ

Thus the solution of our problem has the form,

u(t) =

Z t 0

(t − τ )f (τ ) dτ − c1

Z t 0

(t − τ ) cos(kτ ) dτ + c2

Z t 0

(t − τ ) sin(kτ ) dτ

u(t) =

Z t 0

(t − τ )f (τ ) dτ − c11 − cos(kt)

k2 + c2kt − sin(kt)

k2

We could determine the constants by multiplying in turn by cos(kt) and sin(kt) and integrating from 0 to 1 Thiswould yields a set of two linear equations for c1 and c2.

2

u(x) = λ

Z π 0

∞

X

n=1

sin nx sin nsn

∞

X

m=1

Z π 0

12π

1 − r2

1 − 2r cos(θ − t) + r2φ(t) dt, |r| < 1

We use Poisson’s formula

φ(θ) = λu(r, θ),

where u(r, θ) is harmonic in the unit disk and satisfies, u(1, θ) = φ(θ) For a solution we need λ = 1 and thatu(r, θ) is independent of r In this case u(θ) satisfies

(eınx+ e−ınx) +n

1

(eı(n−2)x+ e−ı(n−2)x) + · · ·

+

n(n − 1)/2

(eıx+ e−ıx)

#

= 1

2n

n0

cos((n − 2m)x)

n(n − k)/2

cos(kx)

(eınx+ e−ınx) +n

1

(eı(n−2)x+ e−ı(n−2)x) + · · ·

+

nn/2 − 1

(ei2x+ e−i2x) +

nn/2

#

= 1

2n

n0



2 cos(2x) +

nn/2



= 1

2n

nn/2

cos((n − 2m)x)

= 1

2n

nn/2

n(n − k)/2

cos(kx)



We substitute this into the integral equation

φ(x) = λ

Z π

−π

a02

For even n, substituting φ(x) = 1 yields λ = 1

πa 0 For n and m both even or odd, substituting φ(x) = cos(mx)

or φ(x) = sin(mx) yields λ = πa1

m For even n we have the eigenvalues and eigenvectors,

2π2− 6π|x − s| + 3(x − s)2
φ(s) dsφ(x) = λ

From this we obtain the eigenvalues and eigenfunctions,

K(x, s)φ(s) ds = 0

Z 2π 0

This has the solutions φ = const The set of eigenfunctions

3 We consider the problem

u − λT u = f

For λ 6= λ, (λ not an eigenvalue), we can obtain a unique solution for u

u(x) = f (x) +

Z 2π 0

Γ(x, s, λ)f (s) dsSince K(x, s) is self-adjoint and L2(0, 2π), we have

p(λ) is an eigenvalue of p(T ).

Now assume that µ is an eigenvalues of p(T ), p(T )φ = µφ We assume that T has a complete, orthonormal set ofeigenfunctions, {φn} corresponding to the set of eigenvalues {λn} We expand φ in these eigenfunctions

p(T )φ = µφp(T )Xcnφn= µXcnφnX

cnp(λn)φn=Xcnµφnp(λn) = µ, ∀n such that cn 6= 0

Thus all eigenvalues of p(T ) are of the form p(λ) with λ an eigenvalue of T

f (x) cos(ωx) dx,

f (x) = 2

Z ∞ 0

cos(2xs)φ(s) ds

We multiply the integral equation by 1

π cos(2xs) and integrate

1π

Z ∞ 0

cos(2xs)φ(x) dx = 1

π

Z ∞ 0

cos(2xs)f (x) dx + λ

Z ∞ 0

cos(2xs) ˆφ(2x) dxˆ

φ(2s) = ˆf (2s) + λ

2

Z ∞ 0

cos(xs) ˆφ(x) dxˆ

φ(x) = f (x) + λ

R∞

0 f (s) cos(2xs) ds

1 − πλ2/4Solution 48.30

D

(v∇2u + avux+ bvuy + cuv) dx dy

=Z

Thus we see that

Let G be the harmonic Green function, which satisfies,

We expand the operators to obtain the first form.

u +Z

D

((c − ax− by)G − aGx− bGy)u dx dy = U

Here U is the harmonic function that satisfies U = f on C

We use integration by parts to obtain the second form

u +Z

D

(cuG − axuG − byuG + axuG + auxG + byuG + buyG) dx dy = U

u +Z

min(x, s)φ(s) ds = λ

Z x 0

esφ(s) ds +

Z 1 x

exφ(s) ds



φ0(x) = λ

xφ(x) +

Z 1 x

φ(s) ds − xφ(x)



= λ

Z 1 x

φ(s) ds

φ00(x) = −λφ(x)

We note that that φ(x) satisfies the constraints,

φ(0) = λ

Z 1 0

0 · φ(s) ds = 0,

φ0(1) = λ

Z 1 1

2 First we differentiate the integral equation.

φ(x) = λ

Z x 0

esφ(s) ds +

Z 1 x

φ(s) ds − exφ(x)



= λ ex

Z 1 x

Thus we see that there are only positive eigenvalues The differential equation has the general solution

φn(x) = v0(1; λn)u(x; λn) − u0(1; λn)v(x; λn)

Solution 48.32

1 First note that

sin(kx) sin(lx) = sign(kl) sin(ax) sin(bx)where

G−(t) = ı−√1 − t2+ t(log(−t − 1) − ıπ) ,

G+(t) − G−(t) = −2π√

t2− 1 + t.For x ∈ (−1, 1) we have

C

f (t) dt

t − zFrom the Plemelj formula we have,

F+(t0) + F−(t0) = 1

ıπ −Z

C

f (t)

t − t0 dt,

f (t0) = F+(t0) − F−(t0)

With W (z) defined as above, we have

W+(t0) + W−(t0) = F+(t0) − F−(t0) = f (t0),

and also

W+(t0) + W−(t0) = 1

ıπ −Z

C

g(t)

t − t0

dt

(1 − e−ı2πγ) dτ(τ − β)1−γ(τ − α)γ(τ − ζ)1

ıπ −Z

C

dτ(τ − β)1−γ(τ − α)γ(τ − ζ) = −ı cot(πγ)

that tends to unity as z → ∞ We find a series expansion of this function about infinity.

γ



1 −αz

j − k

−γk

j − k

−γk

iπ −Z

iπ −Z

iπ −Z

Z 1

−1

φ(y) + φ(−y)

y − x dy

12x −

Z 1

−1

φ(y) + φ(−y)

y − x dy = f (x)1

we seek to determine F to within a multiplicative constant.

log F+(x) − log F−(x) = log λ + 1

We have left off the additive term of ı2πn in the above equation, which will introduce factors of zk and (z − 1)m in

F (z) We will choose these factors so that F (z) has integrable algebraic singularites and vanishes at infinity Note that

we have defined γ to be the real parameter,

γ = log 1 + λ

1 − λ



By the discontinuity theorem,

log F (z) = 1

ı2π

Z 1 0

γ + ıπ

t − z dz

= 1

2 − ı γ2π

log 1 − z

eγ/2− e−γ/2
f (t)

t − xdt = e

γ/2+ e−γ/2
f (x)1

ıπ −

Z 1 0

f (t)

t − xdt = tanh

γ2



f (x)Thus we see that the eigenfunctions are

px(1 − x)

 1 − xx

−ı tanh−1(λ)/π

for −1 < λ < 1

The method used in this problem cannot be used to construct eigenfunctions for λ > 1 For this case we cannotfind an F (z) that has integrable algebraic singularities and vanishes at infinity

Solution 48.38

1

ıπ −

Z 1 0

f (t)

t − xdt = −

ıtan(x)f (x)

We define the function,

F (z) = 1

ı2π −

Z 1 0

The ı2πn term will give us the factors (z − 1)k and zm in the solution for F (z) We will choose the integers k and m

so that F (z) has only algebraic singularities and vanishes at infinity We drop the ı2πn term for now

log F (z) = 1

ı2π

Z 1 0

We replace e1/π by a multiplicative constant and multiply by (z − 1)1 to give F (z) the desired properties.

Since the integral and g(z) are analytic functions inside C we know that f (z) is analytic inside C We use Cauchy’stheorem to evaluate the integral and obtain a differential equation for f (x).

C

1

t − x + P (t − x)



f (t) dt = g(x)1

ıπ −Z

C

f (τ )

τ − ζ dτ = g(ζ)then

f (ζ) = 1

ıπ −Z

C

g(τ )

τ − ζ dτ.

We apply this theorem to the integral equation.

f (x) = − 1

π2 −Z

t − xdt

= − 1

π2 −Z

C

g(t)

t − xdt −

1ıπZ

C

g(t)

t − xdt −

1ıπZ

C

g(τ )

τ − tdτ

dt

= − 1

π2 −Z

C

P (t − x)

τ − t dt

g(τ ) dτ

= − 1

π2 −Z

Part VII Nonlinear Differential Equations

Chapter 49

Nonlinear Ordinary Differential Equations

1 Find the solution, x(t), y(t), in the case β = γ = 0.

2 Show for the case β = 0, γ 6= 0 that x(t) first decreases or increases according as ry0 < γ or ry0 > γ Showthat x(t) → 0 as t → ∞ in both cases Find x as a function of y

3 In the phase plane: Find the position of the singular point and its type when β > 0, γ > 0

Exercise 49.2

Find the singular points and their types for the system

du

dx = ru + v(1 − v)(p − v), r > 0, 0 < p < 1,dv

dx = u,which comes from one of our nonlinear diffusion problems Note that there is a solution with

u = α(1 − v)for special values of α and r Find v(x) for this special case

dt = x + y(1 − r

2)(r = x2+ y2), and that all solution curves spiral into it

Exercise 49.4

Consider

 ˙y = f (y) − x

˙x = yIntroduce new coordinates, R, θ given by

x = R cos θ

y = √1

R sin θand obtain the exact differential equations for R(t), θ(t) Show that R(t) continually increases with t when R 6= 0.Show that θ(t) continually decreases when R > 1

1 Invistigate the nature of the sigular point at (0, 0, 0) by finding the eigenvalues and their behavior for all 0 < R <

∞

2 Find the other singular points when R > 1

3 Show that the appropriate eigenvalues for these other singular points satisfy the cubic

x = a cos θ + a2x2(θ) + a3x3(θ) + · · · , θ = ωt

ω = ω0+ a2ω2+ · · · ,

to find a periodic solution and its natural frequency ω.

Note that, with the expansions given, there are no “secular term” troubles in the determination of x2(θ), but x2(θ)

is needed in the subsequent determination of x3(θ) and ω

Show that a term aω1 in the expansion for ω would have caused trouble, so ω1 would have to be taken equal tozero

(We take the imaginary part of pn(t) in the final answers.)

1 Find p1(t) directly from the equation for n = 1 and note the behavior as t → ∞

2 Find the generating function

Exercise 49.9

1 For the equation modified with a reaction time, namely

d

dtpn(t + τ ) = α[pn−1(t) − pn(t)] n ≥ 1,find a solution of the form in 1(c) by direct substitution in the equation Again take its imaginary part

2 Find a condition that the disturbance is stable, i.e pn(t) remains bounded as n → ∞.

3 In the stable case show that the disturbance is wave-like and find the wave velocity