Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps
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What would happen if we continued this method? Since y = x +
1
c−x
is a solution of the Ricatti equation we can
make the substitution,
y = x +
1
c − x
+
1
u(x)
, (18.7)
which will lead to a solution for y which has two constants of integration. Then we could repeat the process,
substituting the sum of that solution and 1/u(x) into the Ricatti equation to find a solution with three constants
of integration. We know that the general solution of a first order, ordinary differential equation has only one
constant of integration. Does this method for Ricatti equations violate this theorem? There’s only one way to
find out. We substitute Equation 18.7 into the Ricatti equation.
u
= −
−2x + 2
x +
1
c − x
u − 1
u
= −
2
c − x
u − 1
u
+
2
c − x
u = −1
The integrating factor is
I(x) =
e
2/(c−x)
=
e
−2 log(c−x)
=
1
(c − x)
2
.
Upon multiplying by the integrating factor, the equation b ecome s exact.
d
dx
1
(c − x)
2
u
= −
1
(c − x)
2
u = (c − x)
2
−1
c − x
+ b(c − x)
2
u = x − c + b(c − x)
2
Thus the Ricatti equation has the solution,
y = x +
1
c − x
+
1
x − c + b(c − x)
2
.
1014
It appears that we we have found a solution that has two constants of integration, but appearances can be
deceptive. We do a little algebraic simplification of the solution.
y = x +
1
c − x
+
1
(b(c − x) −1)(c − x)
y = x +
(b(c − x) −1) + 1
(b(c − x) −1)(c − x)
y = x +
b
b(c − x) −1
y = x +
1
(c − 1/b) −x
This is actually a solution, (namely the solution we had before), with one constant of integration, (namely c−1/b).
Thus we see that repeated applications of the procedure will not produce more general solutions.
3. The substitution
y = −
u
au
gives us the second order, linear, homogeneous differential equation,
u
−
a
a
+ b
u
+ acu = 0.
The solution to this linear equation is a linear combination of two homogeneous s olutions, u
1
and u
2
.
u = c
1
u
1
(x) + c
2
u
2
(x)
The solution of the Ricatti equation is then
y = −
c
1
u
1
(x) + c
2
u
2
(x)
a(x)(c
1
u
1
(x) + c
2
u
2
(x))
.
1015
Since we can divide the numerator and denominator by either c
1
or c
2
, this answer has only one constant of
integration, (namely c
1
/c
2
or c
2
/c
1
).
Exchanging the Dependent and Independent Variables
Solution 18.6
Exchanging the dependent and independent variables in the differential equation,
y
=
√
y
xy + y
,
yields
x
(y) = y
1/2
x + y
1/2
.
1016
This is a first order differential equation for x(y).
x
− y
1/2
x = y
1/2
d
dy
x exp
−
2y
3/2
3
= y
1/2
exp
−
2y
3/2
3
x exp
−
2y
3/2
3
= −exp
−
2y
3/2
3
+ c
1
x = −1 + c
1
exp
2y
3/2
3
x + 1
c
1
= exp
2y
3/2
3
log
x + 1
c
1
=
2
3
y
3/2
y =
3
2
log
x + 1
c
1
2/3
y =
c +
3
2
log(x + 1)
2/3
Autonomous Equations
*Equidimensional-in-x Equations
*Equidimensional-in-y Equations
*Scale-Invariant Equations
1017
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 the
misfortune of universal geniuses that in attempting to be at home everywhere are so nowhere. Intensity gives eminence
and rises to the heroic in matters sublime.
-Balthasar Gracian
19.1 The Constant Coefficient Equation
The solution of any second order linear homogeneous diffe rential equation can be written in terms of the solutions to
either
y
= 0, or y
− y = 0
Consider the general equation
y
+ ay
+ by = 0.
1018
We can solve this differential equation by making the substitution y =
e
λx
. This yields the algebraic equation
λ
2
+ aλ + b = 0.
λ =
1
2
−a ±
√
a
2
− 4b
There are two cases to consider. If a
2
= 4b then the solutions are
y
1
=
e
(−a+
√
a
2
−4b)x/2
, y
2
=
e
(−a−
√
a
2
−4b)x/2
If a
2
= 4b then we have
y
1
=
e
−ax/2
, y
2
= x
e
−ax/2
Note that regardless of the values of a and b the solutions are of the form
y =
e
−ax/2
u(x)
We would like to write the solutions to the general differential equation in terms of the solutions to simpler differential
equations. We make the substitution
y =
e
λx
u
The derivatives of y are
y
=
e
λx
(u
+ λu)
y
=
e
λx
(u
+ 2λu
+ λ
2
u)
Substituting these into the differential equation yields
u
+ (2λ + a)u
+ (λ
2
+ aλ + b)u = 0
In order to get rid of the u
term we choose
λ = −
a
2
.
The equation is then
u
+
b −
a
2
4
u = 0.
There are now two cases to consider.
1019
Case 1. If b = a
2
/4 then the differential equation is
u
= 0
which has solutions 1 and x. The general solution for y is then
y =
e
−ax/2
(c
1
+ c
2
x).
Case 2. If b = a
2
/4 then the differential equation is
u
−
a
2
4
− b
u = 0.
We make the change variables
u(x) = v(ξ), x = µξ.
The derivatives in terms of ξ are
d
dx
=
dξ
dx
d
dξ
=
1
µ
d
dξ
d
2
dx
2
=
1
µ
d
dξ
1
µ
d
dξ
=
1
µ
2
d
2
dξ
2
.
The differential equation for v is
1
µ
2
v
−
a
2
4
− b
v = 0
v
− µ
2
a
2
4
− b
v = 0
We choose
µ =
a
2
4
− b
−1/2
1020
to obtain
v
− v = 0
which has solutions
e
±ξ
. The solution for y is
y =
e
λx
c
1
e
x/µ
+c
2
e
−x/µ
y =
e
−ax/2
c
1
e
√
a
2
/4−b x
+c
2
e
−
√
a
2
/4−b x
19.2 Normal Form
19.2.1 Second Orde r Equations
Consider the second order equation
y
+ p(x)y
+ q(x)y = 0. (19.1)
Through a change of dependent variable, this equation can be transformed to
u
+ 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).
au
+ 2a
u
+ a
u + p(au
+ a
u) + qau = 0
u
+
2
a
a
+ p
u
+
a
a
+
pa
a
+ q
u = 0
To eliminate the u
term, a(x) must satisfy
2
a
a
+ p = 0
a
+
1
2
pa = 0
1021
a = c exp
−
1
2
p(x) dx
.
For this choice of a, our differential equation for u becomes
u
+
q −
p
2
4
−
p
2
u = 0.
Two differential equations having the same normal form are called equivalent.
Result 19.2.1 The change of variables
y(x) = exp
−
1
2
p(x) dx
u(x)
transforms the differential equation
y
+ p(x)y
+ q(x)y = 0
into its normal form
u
+ I(x)u = 0
where the invariant of the equation, I(x), is
I(x) = q −
p
2
4
−
p
2
.
19.2.2 Higher Order Differential Equations
Consider the third order differential equation
y
+ p(x)y
+ q(x)y
+ r(x)y = 0.
1022
We can eliminate the y
term. Making the change of dependent variable
y = u exp
−
1
3
p(x) dx
y
=
u
−
1
3
pu
exp
−
1
3
p(x) dx
y
=
u
−
2
3
pu
+
1
9
(p
2
− 3p
)u
exp
−
1
3
p(x) dx
y
=
u
− pu
+
1
3
(p
2
− 3p
)u
+
1
27
(9p
− 9p
− p
3
)u
exp
−
1
3
p(x) dx
yields the differential equation
u
+
1
3
(3q − 3p
− p
2
)u
+
1
27
(27r −9pq − 9p
+ 2p
3
)u = 0.
Result 19.2.2 The change of variables
y(x) = exp
−
1
n
p
n−1
(x) dx
u(x)
transforms the differential equation
y
(n)
+ p
n−1
(x)y
(n−1)
+ p
n−2
(x)y
(n−2)
+ ···+ p
0
(x)y = 0
into the form
u
(n)
+ a
n−2
(x)u
(n−2)
+ a
n−3
(x)u
(n−3)
+ ···+ a
0
(x)u = 0.
1023
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
y
+ p(x)y
+ q(x)y = 0.
We make the change of independent variable
ξ = f (x), u(ξ) = y(x).
The derivatives in terms of ξ are
d
dx
=
dξ
dx
d
dξ
= f
d
dξ
d
2
dx
2
= f
d
dξ
f
d
dξ
= (f
)
2
d
2
dξ
2
+ f
d
dξ
The differential equation becomes
(f
)
2
u
+ f
u
+ pf
u
+ qu = 0.
In order to eliminate the u
term, f must satisfy
f
+ pf
= 0
f
= exp
−
p(x) dx
f =
exp
−
p(x) dx
dx.
The differential equation for u is then
u
+
q
(f
)
2
u = 0
1024
u
(ξ) + q(x) exp
2
p(x) dx
u(ξ) = 0.
Result 19.3.1 The change of variables
ξ =
exp
−
p(x) dx
dx, u(ξ) = y(x)
transforms the differential equation
y
+ p(x)y
+ q(x)y = 0
into
u
(ξ) + q(x) exp
2
p(x) dx
u(ξ) = 0.
19.3.2 Transformation to a Constant Coefficient Equation
Consider the second order linear differential equation
y
+ p(x)y
+ q(x)y = 0.
With the change of independent variable
ξ = f (x), u(ξ) = y(x),
the differential equation becomes
(f
)
2
u
+ (f
+ pf
)u
+ qu = 0.
For this to be a constant coefficient equation we must have
(f
)
2
= c
1
q, and f
+ pf
= c
2
q,
1025
for some constants c
1
and c
2
. Solving the first condition,
f
= c
√
q,
f = c
q(x) dx.
The second constraint becomes
f
+ pf
q
= const
1
2
cq
−1/2
q
+ pcq
1/2
q
= const
q
+ 2pq
q
3/2
= const.
Result 19.3.2 Consider the differential equation
y
+ p(x)y
+ q(x)y = 0.
If the expression
q
+ 2pq
q
3/2
is a constant then the change of variables
ξ = c
q(x) dx, u(ξ) = y(x),
will yield a constant coefficient differential equation. (Here c is an arbitrary constant.)
1026
19.4 Integral Equations
Volterra’s Equations. Volterra’s integral equation of the first kind has the form
x
a
N(x, ξ)f(ξ) dξ = f(x).
The Volterra equation of the second kind is
y(x) = f(x) + λ
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
b
a
N(x, ξ)f(ξ) dξ = f(x),
y(x) = f(x) + λ
b
a
N(x, ξ)y(ξ) dξ.
19.4.1 Initial Value Problems
Consider the initial value problem
y
+ p(x)y
+ q(x)y = f(x), y(a) = α, y
(a) = β.
Integrating this equation twice yields
x
a
η
a
y
(ξ) + p(ξ)y
(ξ) + q(ξ)y(ξ) d ξ dη =
x
a
η
a
f(ξ) dξ dη
1027
x
a
(x − ξ)[y
(ξ) + p(ξ)y
(ξ) + q(ξ)y(ξ)] d ξ =
x
a
(x − ξ)f(ξ) dξ.
Now we use integration by parts.
(x − ξ)y
(ξ)
x
a
−
x
a
−y
(ξ) dξ +
(x − ξ)p(ξ)y(ξ)
x
a
−
x
a
[(x − ξ)p
(ξ) −p(ξ)]y(ξ) dξ
+
x
a
(x − ξ)q(ξ)y(ξ) dξ =
x
a
(x − ξ)f(ξ) dξ.
− (x − a)y
(a) + y(x) − y(a) − (x −a)p(a)y(a) −
x
a
[(x − ξ)p
(ξ) −p(ξ)]y(ξ) dξ
+
x
a
(x − ξ)q(ξ)y(ξ) dξ =
x
a
(x − ξ)f(ξ) dξ.
We obtain a Volterra integral equation of the second kind for y(x).
y(x) =
x
a
(x − ξ)f(ξ) dξ + (x − a)(αp(a) + β) + α +
x
a
(x − ξ)[p
(ξ) −q(ξ)] −p(ξ)
y(ξ) dξ.
Note that the initial conditions for the differential equation are “built into” the Volterra equation. Setting x = a in
the Volterra equation yields y(a) = α. Differentiating the Volterra equation,
y
(x) =
x
a
f(ξ) dξ + (αp(a) + β) −p(x)y(x) +
x
a
[p
(ξ) −q(ξ)] −p(ξ)y(ξ) dξ
and setting x = a yields
y
(a) = αp(a) + β − p(a)α = β.
(Recall from calculus that
d
dx
x
g(x, ξ) dξ = g(x, x) +
x
∂
∂x
[g(x, ξ)] dξ.)
1028
Result 19.4.1 The initial value problem
y
+ p(x)y
+ q(x)y = f(x), y(a) = α, y
(a) = β.
is equivalent to the Volterra equation of the second kind
y(x) = F (x) +
x
a
N(x, ξ)y(ξ) dξ
where
F (x) =
x
a
(x − ξ)f(ξ) dξ + (x −a)(αp(a) + β) + α
N(x, ξ) = (x −ξ)[p
(ξ) −q(ξ)] − p(ξ).
19.4.2 Boundary Value Problems
Consider the boundary value problem
y
= f(x), y(a) = α, y(b) = β. (19.2)
To obtain a problem with homogeneous boundary conditions, we make the change of variable
y(x) = u(x) + α +
β −α
b − a
(x − a)
to obtain the problem
u
= f(x), u(a) = u(b) = 0.
Now we will use Green’s functions to write the solution as an integral. First we solve the problem
G
= δ(x −ξ), G(a|ξ) = G(b|ξ) = 0.
1029
The homogeneous solutions of the differential equation that satisfy the left and right boundary conditions are
c
1
(x − a) and c
2
(x − b).
Thus the Green’s function has the form
G(x|ξ) =
c
1
(x − a), for x ≤ ξ
c
2
(x − b), for x ≥ ξ
Imposing continuity of G(x|ξ) at x = ξ and a unit jump of G(x|ξ) at x = ξ, we obtain
G(x|ξ) =
(x−a)(ξ−b)
b−a
, for x ≤ ξ
(x−b)(ξ−a)
b−a
, for x ≥ ξ
Thus the solution of the (19.2) is
y(x) = α +
β −α
b − a
(x − a) +
b
a
G(x|ξ)f (ξ) dξ.
Now consider the boundary value problem
y
+ p(x)y
+ 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) +
b
a
G(x|ξ)[f (ξ) −p(ξ)y
(ξ) −q(ξ)y(ξ)] d ξ.
Using integration by parts, we can write
−
b
a
G(x|ξ)p(ξ)y
(ξ) dξ = −
G(x|ξ)p(ξ)y(ξ)
b
a
+
b
a
∂G(x|ξ)
∂ξ
p(ξ) + G(x|ξ)p
(ξ)
y(ξ) dξ
=
b
a
∂G(x|ξ)
∂ξ
p(ξ) + G(x|ξ)p
(ξ)
y(ξ) dξ.
1030
Substituting this into our expression for y(x),
y(x) = α +
β −α
b − a
(x − a) +
b
a
G(x|ξ)f (ξ) dξ +
b
a
∂G(x|ξ)
∂ξ
p(ξ) + G(x|ξ)[p
(ξ) −q(ξ)]
y(ξ) dξ,
we obtain a Fredholm integral equation of the second kind.
Result 19.4.2 The boundary value problem
y
+ p(x)y
+ 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) +
b
a
N(x, ξ)y(ξ) dξ
where
F (x) = α +
β − α
b − a
(x − a) +
b
a
G(x|ξ)f(ξ) dξ,
N(x, ξ) =
b
a
H(x|ξ)y(ξ) dξ,
G(x|ξ) =
(x−a)(ξ−b)
b−a
, for x ≤ ξ
(x−b)(ξ−a)
b−a
, for x ≥ ξ,
H(x|ξ) =
(x−a)
b−a
p(ξ) +
(x−a)(ξ−b)
b−a
[p
(ξ) −q(ξ)] for x ≤ ξ
(x−b)
b−a
p(ξ) +
(x−b)(ξ−a)
b−a
[p
(ξ) −q(ξ)] for x ≥ ξ.
1031
19.5 Exercises
The Constant Coefficient Equation
Normal Form
Exercise 19.1
Solve the differential equation
y
+
2 +
4
3
x
y
+
1
9
24 + 12x + 4x
2
y = 0.
Hint, Solution
Transformations of the Independent Variable
Integral Equations
Exercise 19.2
Show that the solution of the differential equation
y
+ 2(a + bx)y
+ (c + dx + ex
2
)y = 0
can be written in terms of one of the foll owing canonical forms:
v
+ (ξ
2
+ A)v = 0
v
= ξv
v
+ v = 0
v
= 0.
Hint, Solution
Exercise 19.3
Show that the solution of the differential equation
y
+ 2
a +
b
x
y
+
c +
d
x
+
e
x
2
y = 0
1032
can be written in terms of one of the foll owing canonical forms:
v
+
1 +
A
ξ
+
B
ξ
2
v = 0
v
+
1
ξ
+
A
ξ
2
v = 0
v
+
A
ξ
2
v = 0
Hint, Solution
Exercise 19.4
Show that the second order Euler equation
x
2
d
2
y
d
2
x
+ a
1
x
dy
dx
+ a
0
y = 0
can be transformed to a constant co effici ent equation.
Hint, Solution
Exercise 19.5
Solve Bessel’s equation of order 1/2,
y
+
1
x
y
+
1 −
1
4x
2
y = 0.
Hint, Solution
1033
19.6 Hints
The Constant Coefficient Equation
Normal Form
Hint 19.1
Transform the equation to normal form.
Transformations of the Independent Variable
Integral Equations
Hint 19.2
Transform the equation to normal form an d then apply the scale transformation x = λξ + µ.
Hint 19.3
Transform the equation to normal form an d then apply the scale transformation x = λξ.
Hint 19.4
Make the change of variables x =
e
t
, y(x) = u(t). Write the derivatives with respect to x in terms of t.
x =
e
t
dx =
e
t
dt
d
dx
=
e
−t
d
dt
x
d
dx
=
d
dt
Hint 19.5
Transform the equation to normal form.
1034
19.7 Solutions
The Constant Coefficient Equation
Normal Form
Solution 19.1
y
+
2 +
4
3
x
y
+
1
9
24 + 12x + 4x
2
y = 0
To transform the equation to normal form we make the substitution
y = exp
−
1
2
2 +
4
3
x
dx
u
=
e
−x−x
2
/3
u
The invariant of the equation is
I(x) =
1
9
24 + 12x + 4x
2
−
1
4
2 +
4
3
x
2
−
1
2
d
dx
2 +
4
3
x
= 1.
The normal form of the differential equation is then
u
+ u = 0
which has the general solution
u = c
1
cos x + c
2
sin x
Thus the equation for y has the general solution
y = c
1
e
−x−x
2
/3
cos x + c
2
e
−x−x
2
/3
sin x.
1035
Transformations of the Independent Variable
Integral Equations
Solution 19.2
The substitution that will transform the equation to normal form i s
y = exp
−
1
2
2(a + bx) dx
u
=
e
−ax−bx
2
/2
u.
The invariant of the equation is
I(x) = c + dx + ex
2
−
1
4
(2(a + bx))
2
−
1
2
d
dx
(2(a + bx))
= c −b − a
2
+ (d − 2ab)x + (e − b
2
)x
2
≡ α + βx + γx
2
The normal form of the differential equation is
u
+ (α + βx + γx
2
)u = 0
We consider the following cases:
γ = 0.
β = 0.
α = 0. We immediately have the equation
u
= 0.
α = 0. With the change of variables
v(ξ) = u(x), x = α
−1/2
ξ,
we obtain
v
+ v = 0.
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β = 0. We have the equation
y
+ (α + βx)y = 0.
The scale transformation x = λξ + µ yields
v
+ λ
2
(α + β(λξ + µ))y = 0
v
= [βλ
3
ξ + λ
2
(βµ + α)]v.
Choosing
λ = (−β)
−1/3
, µ = −
α
β
yields the differential equation
v
= ξv.
γ = 0. The scale transformation x = λξ + µ yields
v
+ λ
2
[α + β(λξ + µ) + γ(λξ + µ)
2
]v = 0
v
+ λ
2
[α + βµ + γµ
2
+ λ(β + 2γµ)ξ + λ
2
γξ
2
]v = 0.
Choosing
λ = γ
−1/4
, µ = −
β
2γ
yields the differential equation
v
+ (ξ
2
+ A)v = 0
where
A = γ
−1/2
−
1
4
βγ
−3/2
.
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