Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps

Since we can divide the numerator and denominator by either c1 or c2, this answer has only one constant ofintegration, namely c1/c2 or c2/c1.Exchanging the Dependent and Independent Vari

What would happen if we continued this method? Since y = x + 1

c−x is a solution of the Ricatti equation we canmake the substitution,

I(x) = e2/(c−x) = e−2 log(c−x)= 1

(c − x)2.Upon multiplying by the integrating factor, the equation becomes exact

ddx

1(c − x)2u

y = x + 1

c − x +

1

x − c + b(c − x)2

It appears that we we have found a solution that has two constants of integration, but appearances can bedeceptive We do a little algebraic simplification of the solution.

y = x + 1

c − x +

1(b(c − x) − 1)(c − x)

y = − c1u

0

1(x) + c2u02(x)a(x)(c1u1(x) + c2u2(x)).

Since we can divide the numerator and denominator by either c1 or c2, this answer has only one constant ofintegration, (namely c1/c2 or c2/c1).

Exchanging the Dependent and Independent Variables

Solution 18.6

Exchanging the dependent and independent variables in the differential equation,

y0 =

√y

xy + y,

yields

x0(y) = y1/2x + y1/2

This is a first order differential equation for x(y).

x0− y1/2x = y1/2d

Chapter 19

Transformations and Canonical Forms

Prize intensity more than extent Excellence resides in quality not in quantity The best is always few and rare abundance lowers value Even among men, the giants are usually really dwarfs Some reckon books by the thickness,

-as if they were written to exercise the brawn more than the brain Extent alone never rises above mediocrity; it is themisfortune of universal geniuses that in attempting to be at home everywhere are so nowhere Intensity gives eminenceand rises to the heroic in matters sublime

-Balthasar Gracian

The solution of any second order linear homogeneous differential equation can be written in terms of the solutions toeither

y00= 0, or y00− y = 0Consider the general equation

y00+ ay0+ by = 0

We can solve this differential equation by making the substitution y = eλx This yields the algebraic equation

λ2+ aλ + b = 0

λ = 12



−a ±√a2− 4bThere are two cases to consider If a2 6= 4b then the solutions are

y = e−ax/2u(x)

We would like to write the solutions to the general differential equation in terms of the solutions to simpler differentialequations We make the substitution

y = eλxuThe derivatives of y are

y0 = eλx(u0+ λu)

y00= eλx(u00+ 2λu0 + λ2u)Substituting these into the differential equation yields

u00+ (2λ + a)u0+ (λ2+ aλ + b)u = 0

In order to get rid of the u0 term we choose

λ = −a

2.The equation is then

Case 1 If b = a2/4 then the differential equation is

u00 = 0which has solutions 1 and x The general solution for y is then

y = e−ax/2(c1+ c2x).Case 2 If b 6= a2/4 then the differential equation is

d

dx =

dξdx

d

dξ =

1µ

ddξ

d2

dx2 = 1

µ

ddξ

1µ

to obtain

v00− v = 0which has solutions e±ξ The solution for y is

19.2.1 Second Order Equations

Consider the second order equation

Through a change of dependent variable, this equation can be transformed to

u00+ I(x)y = 0

This is known as the normal form of (19.1) The function I(x) is known as the invariant of the equation

Now to find the change of variables that will accomplish this transformation We make the substitution y(x) =a(x)u(x) in (19.1)

a = c exp −1

2 p(x) dx .For this choice of a, our differential equation for u becomes

Two differential equations having the same normal form are called equivalent

Result 19.2.1 The change of variables

y(x) = exp



− 1 2

Z p(x) dx

 u(x) transforms the differential equation

y00 + p(x)y0 + q(x)y = 0 into its normal form

u00 + I(x)u = 0 where the invariant of the equation, I(x), is

19.2.2 Higher Order Differential Equations

Consider the third order differential equation

y000+ p(x)y00+ q(x)y0+ r(x)y = 0

We can eliminate the y00 term Making the change of dependent variable

y = u exp



−13

Zp(x) dx



−13

Zp(x) dx



−13

Zp(x) dx



−13

Zp(x) dx

Z

pn−1(x) dx

 u(x) transforms the differential equation

y(n)+ pn−1(x)y(n−1)+ pn−2(x)y(n−2)+ · · · + p0(x)y = 0 into the form

u(n)+ an−2(x)u(n−2) + an−3(x)u(n−3)+ · · · + a0(x)u = 0.

19.3 Transformations of the Independent Variable

19.3.1 Transformation to the form u” + a(x) u = 0

Consider the second order linear differential equation

d

dξ = f

0 ddξ



f =

Zexp



−

Zp(x) dx

dx

The differential equation for u is then

u00+ q(f0)2u = 0

u00(ξ) + q(x) exp 2 p(x) dx u(ξ) = 0.

Result 19.3.1 The change of variables

ξ =

Z exp



−

Z p(x) dx



dx, u(ξ) = y(x) transforms the differential equation

y00 + p(x)y0 + q(x)y = 0 into

u00(ξ) + q(x) exp

 2

Z p(x) dx

 u(ξ) = 0.

19.3.2 Transformation to a Constant Coefficient Equation

Consider the second order linear differential equation

y00+ p(x)y0+ q(x)y = 0

With the change of independent variable

ξ = f (x), u(ξ) = y(x),the differential equation becomes

(f0)2u00+ (f00+ pf0)u0+ qu = 0

For this to be a constant coefficient equation we must have

(f0)2 = c1q, and f00+ pf0 = c2q,

for some constants c1 and c2 Solving the first condition,

f0 = c√

q,

f = c

Zpq(x) dx

The second constraint becomes

19.4 Integral Equations

Volterra’s Equations Volterra’s integral equation of the first kind has the form

Z x a

N (x, ξ)f (ξ) dξ = f (x)

The Volterra equation of the second kind is

y(x) = f (x) + λ

Z x a

N (x, ξ)y(ξ) dξ

N (x, ξ) is known as the kernel of the equation

Fredholm’s Equations Fredholm’s integral equations of the first and second kinds are

Z b a

N (x, ξ)f (ξ) dξ = f (x),

y(x) = f (x) + λ

Z b a

N (x, ξ)y(ξ) dξ

19.4.1 Initial Value Problems

Consider the initial value problem

y00+ p(x)y0+ q(x)y = f (x), y(a) = α, y0(a) = β.Integrating this equation twice yields

Z x a

Z η a

y00(ξ) + p(ξ)y0(ξ) + q(ξ)y(ξ) dξ dη =

Z x a

Z η a

f (ξ) dξ dη

Z x a

(x − ξ)[y00(ξ) + p(ξ)y0(ξ) + q(ξ)y(ξ)] dξ =

Z x a

−y0(ξ) dξ +(x − ξ)p(ξ)y(ξ)xa−

Z x a

[(x − ξ)p0(ξ) − p(ξ)]y(ξ) dξ+

Z x a

(x − ξ)q(ξ)y(ξ) dξ =

Z x a

(x − ξ)f (ξ) dξ

− (x − a)y0(a) + y(x) − y(a) − (x − a)p(a)y(a) −

Z x a

[(x − ξ)p0(ξ) − p(ξ)]y(ξ) dξ+

Z x a

(x − ξ)q(ξ)y(ξ) dξ =

Z x a

(x − ξ)f (ξ) dξ

We obtain a Volterra integral equation of the second kind for y(x)

y(x) =

Z x a

(x − ξ)f (ξ) dξ + (x − a)(αp(a) + β) + α +

Z x a

f (ξ) dξ + (αp(a) + β) − p(x)y(x) +

Z x a

[p0(ξ) − q(ξ)] − p(ξ)y(ξ) dξ

and setting x = a yields

y0(a) = αp(a) + β − p(a)α = β

(Recall from calculus that

ddx

Z x

g(x, ξ) dξ = g(x, x) +

Z x ∂

∂x[g(x, ξ)] dξ.)

Result 19.4.1 The initial value problem

y00+ p(x)y0 + q(x)y = f (x), y(a) = α, y0(a) = β.

is equivalent to the Volterra equation of the second kind

y(x) = F (x) +

Z x

a

N (x, ξ)y(ξ) dξ where

19.4.2 Boundary Value Problems

Consider the boundary value problem

The homogeneous solutions of the differential equation that satisfy the left and right boundary conditions are

G(x|ξ) =

((x−a)(ξ−b) b−a , for x ≤ ξ

(x−b)(ξ−a) b−a , for x ≥ ξThus the solution of the (19.2) is

y(x) = α + β − α

b − a(x − a) +

Z b a

G(x|ξ)f (ξ) dξ

Now consider the boundary value problem

y00+ p(x)y0+ q(x)y = 0, y(a) = α, y(b) = β

From the above result we can see that the solution satisfies

y(x) = α + β − α

b − a(x − a) +

Z b a

G(x|ξ)[f (ξ) − p(ξ)y0(ξ) − q(ξ)y(ξ)] dξ

Using integration by parts, we can write

−

Z b a

G(x|ξ)p(ξ)y0(ξ) dξ = −G(x|ξ)p(ξ)y(ξ)ba+

Z b a

=

Z b a

Substituting this into our expression for y(x),

y(x) = α + β − α

b − a(x − a) +

Z b a

G(x|ξ)f (ξ) dξ +

Z b a

we obtain a Fredholm integral equation of the second kind

Result 19.4.2 The boundary value problem

y00 + p(x)y0 + q(x)y = f (x), y(a) = α, y(b) = β.

is equivalent to the Fredholm equation of the second kind

y(x) = F (x) +

Z b

a

N (x, ξ)y(ξ) dξ where

(x−b)(ξ−a) b−a , for x ≥ ξ, H(x|ξ) =

((x−a) b−a p(ξ) + (x−a)(ξ−b)b−a [p0(ξ) − q(ξ)] for x ≤ ξ

(x−b) b−a p(ξ) + (x−b)(ξ−a)b−a [p0(ξ) − q(ξ)] for x ≥ ξ.

can be written in terms of one of the following canonical forms:



y = 0

Hint, Solution

19.6 Hints

The Constant Coefficient Equation

Normal Form

Hint 19.1

Transform the equation to normal form

Transformations of the Independent Variable

dx = e

−t ddt

x d

dx =

ddtHint 19.5

Transform the equation to normal form

Z 

2 + 4

3x

dx

u

= e−x−x2/3uThe invariant of the equation is

I(x) = 1

9 24 + 12x + 4x

2
− 14

ddx

u = c1cos x + c2sin xThus the equation for y has the general solution

y = c1e−x−x2/3cos x + c2e−x−x2/3sin x

Transformations of the Independent Variable

Z2(a + bx) dx

u

β 6= 0 We have the equation

v00 = ξv

γ 6= 0 The scale transformation x = λξ + µ yields

v00+ λ2[α + β(λξ + µ) + γ(λξ + µ)2]v = 0

v00+ λ2[α + βµ + γµ2+ λ(β + 2γµ)ξ + λ2γξ2]v = 0.Choosing

λ = γ−1/4, µ = − β

2γyields the differential equation

v00+ (ξ2+ A)v = 0where

A = γ−1/2− 1

4βγ

−3/2



a + bx

dx

u

2



a + bx

2

−12

ddx

2



a + bx

The scale transformation u(x) = v(ξ), x = λξ yields

dx = e

−t ddt

x d

dx =

ddtNow we express x2 d2

Thus under the change of variables, x = et, y(x) = u(t), the Euler equation becomes

Z1

xdx



= x−1/2uwill put the equation in normal form The invariant is

I(x) =



1 − 14x2



− 14

 1

x2



−12

−1

x2 = 1.Thus we have the differential equation

u00+ u = 0,with the solution

u = c1cos x + c2sin x

The solution of Bessel’s equation of order 1/2 is

y = c1x−1/2cos x + c2x−1/2sin x

Chapter 20

The Dirac Delta Function

I do not know what I appear to the world; but to myself I seem to have been only like a boy playing on a seashore,and diverting myself now and then by finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean

of truth lay all undiscovered before me

- Sir Issac Newton

The Heaviside function H(x) is defined

of the word One can derive the properties of the delta function rigorously, but the treatment in this text will be almostentirely heuristic

The Dirac delta function is defined by the properties

Figure 20.1: The Dirac Delta Function

Let f (x) be a continuous function that vanishes at infinity Consider the integral

Z ∞

−∞

f (x)δ(x) dx

We use integration by parts to evaluate the integral.

at infinity should not affect the value of the integral Thus it is reasonable that f (0) =R∞

−∞f (x)δ(x) dx holds for allcontinuous functions By changing variables and noting that δ(x) is symmetric we can derive a more general formula

Consider a function b(x, ) defined by

The graph of b(x, 1/10) is shown in Figure 20.2.

5 10

Let f (x) be a continuous function and let F0(x) = f (x) We compute the integral of f (x)δ(x).

20.4 Non-Rectangular Coordinate Systems

We can derive Dirac delta functions in non-rectangular coordinate systems by making a change of variables in therelation,

Z

Rn

δn(x) dx = 1Where the transformation is non-singular, one merely divides the Dirac delta function by the Jacobian of the transfor-mation to the coordinate system

Example 20.4.1 Consider the Dirac delta function in cylindrical coordinates, (r, θ, z) The Jacobian is J = r

Z ∞

−∞

Z 2π 0

Z ∞ 0

δ3(x − x0) r dr dθ dz = 1For r0 6= 0, the Dirac Delta function is

δ3(x − x0) = 1

rδ (r − r0) δ (θ − θ0) δ (z − z0)since it satisfies the two defining properties

δ (r − r0) dr

Z 2π 0

δ (θ − θ0) dθ

Z ∞

−∞

δ (z − z0) dz = 1For r0 = 0, we have

δ3(x − x0) = 1

2πrδ (r) δ (z − z0)

since this again satisfies the two defining properties.

12πrδ (r) δ (z − z0) = 0 for (r, z) 6= (0, z0)

Z ∞

−∞

Z 2π 0

Z ∞ 0

12πrδ (r) δ (z − z0) r dr dθ dz =

12π

Z ∞ 0

δ (r) dr

Z 2π 0

dθ

Z ∞

−∞

δ (z − z0) dz = 1

Determine the Dirac delta function in spherical coordinates, (r, θ, φ).

x = r cos θ sin φ, y = r sin θ sin φ, z = r cos φ

Hint, Solution

δ(y(x)) dx =

(Rβ α

δ(y)

|y 0 (x n )|dy if y(xn) = 0 for a < xn< b

for α = min(y(a), y(b)) and β = max(y(a), y(b)) Now consider the integral on the interval (−∞ ∞) as the sum

of integrals on the intervals {(ξm ξm+1)}

Hint 20.6

The Dirac delta function is defined by the following two properties

δ(x − a) = 0 for x 6= aZ

Rn

δ(x − a) dx = 1Verify that δ(ξ − α)/|J | satisfies these properties in the ξ coordinate system.Hint 20.7

Consider the special cases φ0 = 0, π and r0 = 0

= f (0

−) + f (0+)2