Green’s Function - Ordinary Differential Equations, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


21.6 Green Functions for First Order Equations
Consider the ?rst order inhomogeneous equation
L[y]= y
?
+p(x)y = f(x)? for x > a? (21.2)
subject to a homogeneous initial condition, B[y]= y(a) = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to G(a|?) = 0.
WecanrepresentthesolutiontotheinhomogeneousprobleminEquation21.2asanintegralinvolving
the Green function. To show that
y(x) =
?
8
a
G(x|?)f(?)d?
is the solution, we apply the linear operatorL to the integral. (Assume that the integral is uniformly
convergent.)
L
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
L[G(x|?)]f(?)d?
=
?
8
a
d(x-?)f(?)d?
= f(x)
The integral also satis?es the initial condition.
B
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
B[G(x|?)]f(?)d?
=
?
8
a
(0)f(?)d?
= 0
Now we consider the qualitiative behavior of the Green function. For x?= ?, the Green function
is simply a homogeneous solution of the di?erential equation, however at x = ? we expect some
singular behavior. G
?
(x|?) will have a Dirac delta function type singularity. This means that G(x|?)
will have a jump discontinuity at x = ?. We integrate the di?erential equation on the vanishing
interval (?
-
...?
+
) to determine this jump.
G
?
+p(x)G = d(x-?)
G(?
+
|?)-G(?
-
|?)+
?
?
+
?
?
p(x)G(x|?)dx = 1
G(?
+
|?)-G(?
-
|?) = 1 (21.3)
The homogeneous solution of the di?erential equation is
y
h
= e
-
?
p(x)dx
Since the Green function satis?es the homogeneous equation for x ?= ?, it will be a constant times
this homogeneous solution for x < ? and x > ?.
G(x|?) =
?
c
1
e
-
?
p(x)dx
a < x < ?
c
2
e
-
?
p(x)dx
? < x
Page 2


21.6 Green Functions for First Order Equations
Consider the ?rst order inhomogeneous equation
L[y]= y
?
+p(x)y = f(x)? for x > a? (21.2)
subject to a homogeneous initial condition, B[y]= y(a) = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to G(a|?) = 0.
WecanrepresentthesolutiontotheinhomogeneousprobleminEquation21.2asanintegralinvolving
the Green function. To show that
y(x) =
?
8
a
G(x|?)f(?)d?
is the solution, we apply the linear operatorL to the integral. (Assume that the integral is uniformly
convergent.)
L
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
L[G(x|?)]f(?)d?
=
?
8
a
d(x-?)f(?)d?
= f(x)
The integral also satis?es the initial condition.
B
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
B[G(x|?)]f(?)d?
=
?
8
a
(0)f(?)d?
= 0
Now we consider the qualitiative behavior of the Green function. For x?= ?, the Green function
is simply a homogeneous solution of the di?erential equation, however at x = ? we expect some
singular behavior. G
?
(x|?) will have a Dirac delta function type singularity. This means that G(x|?)
will have a jump discontinuity at x = ?. We integrate the di?erential equation on the vanishing
interval (?
-
...?
+
) to determine this jump.
G
?
+p(x)G = d(x-?)
G(?
+
|?)-G(?
-
|?)+
?
?
+
?
?
p(x)G(x|?)dx = 1
G(?
+
|?)-G(?
-
|?) = 1 (21.3)
The homogeneous solution of the di?erential equation is
y
h
= e
-
?
p(x)dx
Since the Green function satis?es the homogeneous equation for x ?= ?, it will be a constant times
this homogeneous solution for x < ? and x > ?.
G(x|?) =
?
c
1
e
-
?
p(x)dx
a < x < ?
c
2
e
-
?
p(x)dx
? < x
In order to satisfy the homogeneous initial condition G(a|?) = 0, the Green function must vanish on
the interval (a...?).
G(x|?) =
?
0 a < x < ?
ce
-
?
p(x)dx
? < x
The jump condition, (Equation 21.3), gives us the constraint G(?
+
|?) = 1. This determines the
constant in the homogeneous solution for x > ?.
G(x|?) =
?
0 a < x < ?
e
-
?
x
?
p(t)dt
? < x
We can use the Heaviside function to write the Green function without using a case statement.
G(x|?) = e
-
?
x
?
p(t)dt
H(x-?)
Clearly the Green function is of little value in solving the inhomogeneous di?erential equation in
Equation 21.2, as we can solve that problem directly. However, we will encounter ?rst order Green
function problems in solving some partial di?erential equations.
Result 21.6.1 The ?rst order inhomogeneous di?erential equation with ho-
mogeneous initial condition
L[y]= y
?
+p(x)y = f(x)? for a < x? y(a) = 0?
has the solution
y =
?
8
a
G(x|?)f(?)d??
where G(x|?) satis?es the equation
L[G(x|?)] = d(x-?)? for a < x? G(a|?) = 0.
The Green function is
G(x|?) = e
-
?
x
?
p?t)dt
H(x-?)
21.7 Green Functions for Second Order Equations
Consider the second order inhomogeneous equation
L[y] = y
??
+p(x)y
?
+q(x)y = f(x)? for a < x < b? (21.4)
subject to the homogeneous boundary conditions
B
1
[y] = B
2
[y] = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to B
1
[G] = B
2
[G] = 0.
The Green function is useful because you can represent the solution to the inhomogeneous problem
in Equation 21.4 as an integral involving the Green function. To show that
y(x) =
?
b
a
G(x|?)f(?)d?
Page 3


21.6 Green Functions for First Order Equations
Consider the ?rst order inhomogeneous equation
L[y]= y
?
+p(x)y = f(x)? for x > a? (21.2)
subject to a homogeneous initial condition, B[y]= y(a) = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to G(a|?) = 0.
WecanrepresentthesolutiontotheinhomogeneousprobleminEquation21.2asanintegralinvolving
the Green function. To show that
y(x) =
?
8
a
G(x|?)f(?)d?
is the solution, we apply the linear operatorL to the integral. (Assume that the integral is uniformly
convergent.)
L
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
L[G(x|?)]f(?)d?
=
?
8
a
d(x-?)f(?)d?
= f(x)
The integral also satis?es the initial condition.
B
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
B[G(x|?)]f(?)d?
=
?
8
a
(0)f(?)d?
= 0
Now we consider the qualitiative behavior of the Green function. For x?= ?, the Green function
is simply a homogeneous solution of the di?erential equation, however at x = ? we expect some
singular behavior. G
?
(x|?) will have a Dirac delta function type singularity. This means that G(x|?)
will have a jump discontinuity at x = ?. We integrate the di?erential equation on the vanishing
interval (?
-
...?
+
) to determine this jump.
G
?
+p(x)G = d(x-?)
G(?
+
|?)-G(?
-
|?)+
?
?
+
?
?
p(x)G(x|?)dx = 1
G(?
+
|?)-G(?
-
|?) = 1 (21.3)
The homogeneous solution of the di?erential equation is
y
h
= e
-
?
p(x)dx
Since the Green function satis?es the homogeneous equation for x ?= ?, it will be a constant times
this homogeneous solution for x < ? and x > ?.
G(x|?) =
?
c
1
e
-
?
p(x)dx
a < x < ?
c
2
e
-
?
p(x)dx
? < x
In order to satisfy the homogeneous initial condition G(a|?) = 0, the Green function must vanish on
the interval (a...?).
G(x|?) =
?
0 a < x < ?
ce
-
?
p(x)dx
? < x
The jump condition, (Equation 21.3), gives us the constraint G(?
+
|?) = 1. This determines the
constant in the homogeneous solution for x > ?.
G(x|?) =
?
0 a < x < ?
e
-
?
x
?
p(t)dt
? < x
We can use the Heaviside function to write the Green function without using a case statement.
G(x|?) = e
-
?
x
?
p(t)dt
H(x-?)
Clearly the Green function is of little value in solving the inhomogeneous di?erential equation in
Equation 21.2, as we can solve that problem directly. However, we will encounter ?rst order Green
function problems in solving some partial di?erential equations.
Result 21.6.1 The ?rst order inhomogeneous di?erential equation with ho-
mogeneous initial condition
L[y]= y
?
+p(x)y = f(x)? for a < x? y(a) = 0?
has the solution
y =
?
8
a
G(x|?)f(?)d??
where G(x|?) satis?es the equation
L[G(x|?)] = d(x-?)? for a < x? G(a|?) = 0.
The Green function is
G(x|?) = e
-
?
x
?
p?t)dt
H(x-?)
21.7 Green Functions for Second Order Equations
Consider the second order inhomogeneous equation
L[y] = y
??
+p(x)y
?
+q(x)y = f(x)? for a < x < b? (21.4)
subject to the homogeneous boundary conditions
B
1
[y] = B
2
[y] = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to B
1
[G] = B
2
[G] = 0.
The Green function is useful because you can represent the solution to the inhomogeneous problem
in Equation 21.4 as an integral involving the Green function. To show that
y(x) =
?
b
a
G(x|?)f(?)d?
is the solution, we apply the linear operatorL to the integral. (Assume that the integral is uniformly
convergent.)
L
?
?
b
a
G(x|?)f(?)d?
?
=
?
b
a
L[G(x|?)]f(?)d?
=
?
b
a
d(x-?)f(?)d?
= f(x)
The integral also satis?es the boundary conditions.
B
i
?
?
b
a
G(x|?)f(?)d?
?
=
?
b
a
B
i
[G(x|?)]f(?)d?
=
?
b
a
[0]f(?)d?
= 0
One of the advantages of using Green functions is that once you ?nd the Green function for a
linear operator and certain homogeneous boundary conditions,
L[G] = d(x-?)? B
1
[G] = B
2
[G] = 0?
you can write the solution for any inhomogeneity, f(x).
L[f] = f(x)? B
1
[y] = B
2
[y] = 0
You do not need to do any extra work to obtain the solution for a di?erent inhomogeneous term.
Qualitatively, what kind of behavior will the Green function for a second order di?erential equa-
tion have? Will it have a delta function singularity; will it be continuous? To answer these questions
we will ?rst look at the behavior of integrals and derivatives of d(x).
The integral of d(x) is the Heaviside function, H(x).
H(x) =
?
x
-8
d(t)dt =
?
0 for x < 0
1 for x > 0
The integral of the Heaviside function is the ramp function, r(x).
r(x) =
?
x
-8
H(t)dt =
?
0 for x < 0
x for x > 0
The derivative of the delta function is zero for x?= 0. At x = 0 it goes from 0 up to +8, down to
-8 and then back up to 0.
In Figure 21.2 we see conceptually the behavior of the ramp function, the Heaviside function,
the delta function, and the derivative of the delta function.
We write the di?erential equation for the Green function.
G
??
(x|?)+p(x)G
?
(x|?)+q(x)G(x|?) = d(x-?)
weseethatonlytheG
??
(x|?)termcanhaveadeltafunctiontypesingularity. Ifoneoftheotherterms
had a delta function type singularity then G
??
(x|?) would be more singular than a delta function
and there would be nothing in the right hand side of the equation to match this kind of singularity.
Analogous to the progression from a delta function to a Heaviside function to a ramp function, we
see that G
?
(x|?) will have a jump discontinuity and G(x|?) will be continuous.
Page 4


21.6 Green Functions for First Order Equations
Consider the ?rst order inhomogeneous equation
L[y]= y
?
+p(x)y = f(x)? for x > a? (21.2)
subject to a homogeneous initial condition, B[y]= y(a) = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to G(a|?) = 0.
WecanrepresentthesolutiontotheinhomogeneousprobleminEquation21.2asanintegralinvolving
the Green function. To show that
y(x) =
?
8
a
G(x|?)f(?)d?
is the solution, we apply the linear operatorL to the integral. (Assume that the integral is uniformly
convergent.)
L
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
L[G(x|?)]f(?)d?
=
?
8
a
d(x-?)f(?)d?
= f(x)
The integral also satis?es the initial condition.
B
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
B[G(x|?)]f(?)d?
=
?
8
a
(0)f(?)d?
= 0
Now we consider the qualitiative behavior of the Green function. For x?= ?, the Green function
is simply a homogeneous solution of the di?erential equation, however at x = ? we expect some
singular behavior. G
?
(x|?) will have a Dirac delta function type singularity. This means that G(x|?)
will have a jump discontinuity at x = ?. We integrate the di?erential equation on the vanishing
interval (?
-
...?
+
) to determine this jump.
G
?
+p(x)G = d(x-?)
G(?
+
|?)-G(?
-
|?)+
?
?
+
?
?
p(x)G(x|?)dx = 1
G(?
+
|?)-G(?
-
|?) = 1 (21.3)
The homogeneous solution of the di?erential equation is
y
h
= e
-
?
p(x)dx
Since the Green function satis?es the homogeneous equation for x ?= ?, it will be a constant times
this homogeneous solution for x < ? and x > ?.
G(x|?) =
?
c
1
e
-
?
p(x)dx
a < x < ?
c
2
e
-
?
p(x)dx
? < x
In order to satisfy the homogeneous initial condition G(a|?) = 0, the Green function must vanish on
the interval (a...?).
G(x|?) =
?
0 a < x < ?
ce
-
?
p(x)dx
? < x
The jump condition, (Equation 21.3), gives us the constraint G(?
+
|?) = 1. This determines the
constant in the homogeneous solution for x > ?.
G(x|?) =
?
0 a < x < ?
e
-
?
x
?
p(t)dt
? < x
We can use the Heaviside function to write the Green function without using a case statement.
G(x|?) = e
-
?
x
?
p(t)dt
H(x-?)
Clearly the Green function is of little value in solving the inhomogeneous di?erential equation in
Equation 21.2, as we can solve that problem directly. However, we will encounter ?rst order Green
function problems in solving some partial di?erential equations.
Result 21.6.1 The ?rst order inhomogeneous di?erential equation with ho-
mogeneous initial condition
L[y]= y
?
+p(x)y = f(x)? for a < x? y(a) = 0?
has the solution
y =
?
8
a
G(x|?)f(?)d??
where G(x|?) satis?es the equation
L[G(x|?)] = d(x-?)? for a < x? G(a|?) = 0.
The Green function is
G(x|?) = e
-
?
x
?
p?t)dt
H(x-?)
21.7 Green Functions for Second Order Equations
Consider the second order inhomogeneous equation
L[y] = y
??
+p(x)y
?
+q(x)y = f(x)? for a < x < b? (21.4)
subject to the homogeneous boundary conditions
B
1
[y] = B
2
[y] = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to B
1
[G] = B
2
[G] = 0.
The Green function is useful because you can represent the solution to the inhomogeneous problem
in Equation 21.4 as an integral involving the Green function. To show that
y(x) =
?
b
a
G(x|?)f(?)d?
is the solution, we apply the linear operatorL to the integral. (Assume that the integral is uniformly
convergent.)
L
?
?
b
a
G(x|?)f(?)d?
?
=
?
b
a
L[G(x|?)]f(?)d?
=
?
b
a
d(x-?)f(?)d?
= f(x)
The integral also satis?es the boundary conditions.
B
i
?
?
b
a
G(x|?)f(?)d?
?
=
?
b
a
B
i
[G(x|?)]f(?)d?
=
?
b
a
[0]f(?)d?
= 0
One of the advantages of using Green functions is that once you ?nd the Green function for a
linear operator and certain homogeneous boundary conditions,
L[G] = d(x-?)? B
1
[G] = B
2
[G] = 0?
you can write the solution for any inhomogeneity, f(x).
L[f] = f(x)? B
1
[y] = B
2
[y] = 0
You do not need to do any extra work to obtain the solution for a di?erent inhomogeneous term.
Qualitatively, what kind of behavior will the Green function for a second order di?erential equa-
tion have? Will it have a delta function singularity; will it be continuous? To answer these questions
we will ?rst look at the behavior of integrals and derivatives of d(x).
The integral of d(x) is the Heaviside function, H(x).
H(x) =
?
x
-8
d(t)dt =
?
0 for x < 0
1 for x > 0
The integral of the Heaviside function is the ramp function, r(x).
r(x) =
?
x
-8
H(t)dt =
?
0 for x < 0
x for x > 0
The derivative of the delta function is zero for x?= 0. At x = 0 it goes from 0 up to +8, down to
-8 and then back up to 0.
In Figure 21.2 we see conceptually the behavior of the ramp function, the Heaviside function,
the delta function, and the derivative of the delta function.
We write the di?erential equation for the Green function.
G
??
(x|?)+p(x)G
?
(x|?)+q(x)G(x|?) = d(x-?)
weseethatonlytheG
??
(x|?)termcanhaveadeltafunctiontypesingularity. Ifoneoftheotherterms
had a delta function type singularity then G
??
(x|?) would be more singular than a delta function
and there would be nothing in the right hand side of the equation to match this kind of singularity.
Analogous to the progression from a delta function to a Heaviside function to a ramp function, we
see that G
?
(x|?) will have a jump discontinuity and G(x|?) will be continuous.
Figure 21.2: r(x), H(x), d(x) and
d
dx
d(x)
Lety
1
andy
2
betwolinearlyindependentsolutionstothehomogeneousequation,L[y] = 0. Since
the Green function satis?es the homogeneous equation for x?= ?, it will be a linear combination of
the homogeneous solutions.
G(x|?) =
?
c
1
y
1
+c
2
y
2
for x < ?
d
1
y
1
+d
2
y
2
for x > ?
We require that G(x|?) be continuous.
G(x|?)
?
?
x??
?
= G(x|?)
?
?
x??
+
We can write this in terms of the homogeneous solutions.
c
1
y
1
(?)+c
2
y
2
(?) = d
1
y
1
(?)+d
2
y
2
(?)
We integrate L[G(x|?)] = d(x-?) from ?
-
to ?+.
?
?
+
?
?
[G
??
(x|?)+p(x)G
?
(x|?)+q(x)G(x|?)] dx =
?
?
+
?
?
d(x-?)dx.
Since G(x|?) is continuous and G
?
(x|?) has only a jump discontinuity two of the terms vanish.
?
?
+
?
?
p(x)G
?
(x|?)dx = 0 and
?
?
+
?
?
q(x)G(x|?)dx = 0
?
?
+
?
?
G
??
(x|?)dx =
?
?
+
?
?
d(x-?)dx
?
G
?
(x|?)
?
?
+
?
?
=
?
H(x-?)
?
?
+
?
?
G
?
(?
+
|?)-G
?
(?
-
|?) = 1
We write this jump condition in terms of the homogeneous solutions.
d
1
y
?
1
(?)+d
2
y
?
2
(?)-c
1
y
?
1
(?)-c
2
y
?
2
(?) = 1
Combined with the two boundary conditions, this gives us a total of four equations to determine
our four constants, c
1
, c
2
, d
1
, and d
2
.
Page 5


21.6 Green Functions for First Order Equations
Consider the ?rst order inhomogeneous equation
L[y]= y
?
+p(x)y = f(x)? for x > a? (21.2)
subject to a homogeneous initial condition, B[y]= y(a) = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to G(a|?) = 0.
WecanrepresentthesolutiontotheinhomogeneousprobleminEquation21.2asanintegralinvolving
the Green function. To show that
y(x) =
?
8
a
G(x|?)f(?)d?
is the solution, we apply the linear operatorL to the integral. (Assume that the integral is uniformly
convergent.)
L
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
L[G(x|?)]f(?)d?
=
?
8
a
d(x-?)f(?)d?
= f(x)
The integral also satis?es the initial condition.
B
? ?
8
a
G(x|?)f(?)d?
?
=
?
8
a
B[G(x|?)]f(?)d?
=
?
8
a
(0)f(?)d?
= 0
Now we consider the qualitiative behavior of the Green function. For x?= ?, the Green function
is simply a homogeneous solution of the di?erential equation, however at x = ? we expect some
singular behavior. G
?
(x|?) will have a Dirac delta function type singularity. This means that G(x|?)
will have a jump discontinuity at x = ?. We integrate the di?erential equation on the vanishing
interval (?
-
...?
+
) to determine this jump.
G
?
+p(x)G = d(x-?)
G(?
+
|?)-G(?
-
|?)+
?
?
+
?
?
p(x)G(x|?)dx = 1
G(?
+
|?)-G(?
-
|?) = 1 (21.3)
The homogeneous solution of the di?erential equation is
y
h
= e
-
?
p(x)dx
Since the Green function satis?es the homogeneous equation for x ?= ?, it will be a constant times
this homogeneous solution for x < ? and x > ?.
G(x|?) =
?
c
1
e
-
?
p(x)dx
a < x < ?
c
2
e
-
?
p(x)dx
? < x
In order to satisfy the homogeneous initial condition G(a|?) = 0, the Green function must vanish on
the interval (a...?).
G(x|?) =
?
0 a < x < ?
ce
-
?
p(x)dx
? < x
The jump condition, (Equation 21.3), gives us the constraint G(?
+
|?) = 1. This determines the
constant in the homogeneous solution for x > ?.
G(x|?) =
?
0 a < x < ?
e
-
?
x
?
p(t)dt
? < x
We can use the Heaviside function to write the Green function without using a case statement.
G(x|?) = e
-
?
x
?
p(t)dt
H(x-?)
Clearly the Green function is of little value in solving the inhomogeneous di?erential equation in
Equation 21.2, as we can solve that problem directly. However, we will encounter ?rst order Green
function problems in solving some partial di?erential equations.
Result 21.6.1 The ?rst order inhomogeneous di?erential equation with ho-
mogeneous initial condition
L[y]= y
?
+p(x)y = f(x)? for a < x? y(a) = 0?
has the solution
y =
?
8
a
G(x|?)f(?)d??
where G(x|?) satis?es the equation
L[G(x|?)] = d(x-?)? for a < x? G(a|?) = 0.
The Green function is
G(x|?) = e
-
?
x
?
p?t)dt
H(x-?)
21.7 Green Functions for Second Order Equations
Consider the second order inhomogeneous equation
L[y] = y
??
+p(x)y
?
+q(x)y = f(x)? for a < x < b? (21.4)
subject to the homogeneous boundary conditions
B
1
[y] = B
2
[y] = 0.
The Green function G(x|?) is de?ned as the solution to
L[G(x|?)] = d(x-?) subject to B
1
[G] = B
2
[G] = 0.
The Green function is useful because you can represent the solution to the inhomogeneous problem
in Equation 21.4 as an integral involving the Green function. To show that
y(x) =
?
b
a
G(x|?)f(?)d?
is the solution, we apply the linear operatorL to the integral. (Assume that the integral is uniformly
convergent.)
L
?
?
b
a
G(x|?)f(?)d?
?
=
?
b
a
L[G(x|?)]f(?)d?
=
?
b
a
d(x-?)f(?)d?
= f(x)
The integral also satis?es the boundary conditions.
B
i
?
?
b
a
G(x|?)f(?)d?
?
=
?
b
a
B
i
[G(x|?)]f(?)d?
=
?
b
a
[0]f(?)d?
= 0
One of the advantages of using Green functions is that once you ?nd the Green function for a
linear operator and certain homogeneous boundary conditions,
L[G] = d(x-?)? B
1
[G] = B
2
[G] = 0?
you can write the solution for any inhomogeneity, f(x).
L[f] = f(x)? B
1
[y] = B
2
[y] = 0
You do not need to do any extra work to obtain the solution for a di?erent inhomogeneous term.
Qualitatively, what kind of behavior will the Green function for a second order di?erential equa-
tion have? Will it have a delta function singularity; will it be continuous? To answer these questions
we will ?rst look at the behavior of integrals and derivatives of d(x).
The integral of d(x) is the Heaviside function, H(x).
H(x) =
?
x
-8
d(t)dt =
?
0 for x < 0
1 for x > 0
The integral of the Heaviside function is the ramp function, r(x).
r(x) =
?
x
-8
H(t)dt =
?
0 for x < 0
x for x > 0
The derivative of the delta function is zero for x?= 0. At x = 0 it goes from 0 up to +8, down to
-8 and then back up to 0.
In Figure 21.2 we see conceptually the behavior of the ramp function, the Heaviside function,
the delta function, and the derivative of the delta function.
We write the di?erential equation for the Green function.
G
??
(x|?)+p(x)G
?
(x|?)+q(x)G(x|?) = d(x-?)
weseethatonlytheG
??
(x|?)termcanhaveadeltafunctiontypesingularity. Ifoneoftheotherterms
had a delta function type singularity then G
??
(x|?) would be more singular than a delta function
and there would be nothing in the right hand side of the equation to match this kind of singularity.
Analogous to the progression from a delta function to a Heaviside function to a ramp function, we
see that G
?
(x|?) will have a jump discontinuity and G(x|?) will be continuous.
Figure 21.2: r(x), H(x), d(x) and
d
dx
d(x)
Lety
1
andy
2
betwolinearlyindependentsolutionstothehomogeneousequation,L[y] = 0. Since
the Green function satis?es the homogeneous equation for x?= ?, it will be a linear combination of
the homogeneous solutions.
G(x|?) =
?
c
1
y
1
+c
2
y
2
for x < ?
d
1
y
1
+d
2
y
2
for x > ?
We require that G(x|?) be continuous.
G(x|?)
?
?
x??
?
= G(x|?)
?
?
x??
+
We can write this in terms of the homogeneous solutions.
c
1
y
1
(?)+c
2
y
2
(?) = d
1
y
1
(?)+d
2
y
2
(?)
We integrate L[G(x|?)] = d(x-?) from ?
-
to ?+.
?
?
+
?
?
[G
??
(x|?)+p(x)G
?
(x|?)+q(x)G(x|?)] dx =
?
?
+
?
?
d(x-?)dx.
Since G(x|?) is continuous and G
?
(x|?) has only a jump discontinuity two of the terms vanish.
?
?
+
?
?
p(x)G
?
(x|?)dx = 0 and
?
?
+
?
?
q(x)G(x|?)dx = 0
?
?
+
?
?
G
??
(x|?)dx =
?
?
+
?
?
d(x-?)dx
?
G
?
(x|?)
?
?
+
?
?
=
?
H(x-?)
?
?
+
?
?
G
?
(?
+
|?)-G
?
(?
-
|?) = 1
We write this jump condition in terms of the homogeneous solutions.
d
1
y
?
1
(?)+d
2
y
?
2
(?)-c
1
y
?
1
(?)-c
2
y
?
2
(?) = 1
Combined with the two boundary conditions, this gives us a total of four equations to determine
our four constants, c
1
, c
2
, d
1
, and d
2
.
Result 21.7.1 The second order inhomogeneous di?erential equation with
homogeneous boundary conditions
L[y] = y
??
+p(x)y
?
+q(x)y = f(x)? for a < x < b? B
1
[y] = B
2
[y] = 0?
has the solution
y =
?
b
a
G(x|?)f(?)d??
where G(x|?) satis?es the equation
L[G(x|?)] = d(x-?)? for a < x < b? B
1
[G(x|?)] = B
2
[G(x|?)] = 0.
G(x|?) is continuous andG
?
(x|?) has a jump discontinuity of height 1 atx = ?.
Example 21.7.1 Solve the boundary value problem
y
??
= f(x)? y(0) = y(1) = 0?
using a Green function.
A pair of solutions to the homogeneous equation are y
1
= 1 and y
2
= x. First note that only the
trivial solution to the homogeneous equation satis?es the homogeneous boundary conditions. Thus
there is a unique solution to this problem.
The Green function satis?es
G
??
(x|?) = d(x-?)? G(0|?) = G(1|?) = 0.
The Green function has the form
G(x|?) =
?
c
1
+c
2
x for x < ?
d
1
+d
2
x for x > ?.
Applying the two boundary conditions, we see that c
1
= 0 and d
1
=-d
2
. The Green function now
has the form
G(x|?) =
?
cx for x < ?
d(x-1) for x > ?.
Since the Green function must be continuous,
c? = d(?-1) ? d = c
?
?-1
.
From the jump condition,
d
dx
c
?
?-1
(x-1)
?
?
?
x=?
-
d
dx
cx
?
?
?
x=?
= 1
c
?
?-1
-c = 1
c = ?-1.
Thus the Green function is
G(x|?) =
?
(?-1)x for x < ?
?(x-1) for x > ?.
The Green function is plotted in Figure 21.3 for various values of ?. The solution to y
??
= f(x) is