Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 10 ppt
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The general solution of the differential equation is thus
y = c
1
cos x + c
2
sin x + x.
Applying the two initial conditions gives us the equations
c
1
= 1, c
2
+ 1 = 0.
The solution subject to the initial conditions is
y = cos x − sin x + x.
Solution 21.5
Solve
x
2
y
(x) − xy
(x) + y(x) = x.
The homogeneous equation is
x
2
y
(x) − xy
(x) + y(x) = 0.
Substituting y = x
λ
into the homogeneous differential equation yields
x
2
λ(λ − 1)x
λ−2
− xλx
λ
+ x
λ
= 0
λ
2
− 2λ + 1 = 0
(λ − 1)
2
= 0
λ = 1.
The homogeneous solutions are
y
1
= x, y
2
= x log x.
The Wronskian of the homogeneous solutions is
W [x, x log x] =
x x log x
1 1 + log x
= x + x log x −x log x
= x.
1134
Writing the inhomogeneous equation in the standard form:
y
(x) −
1
x
y
(x) +
1
x
2
y(x) =
1
x
.
Using variation of parameters to find the particular solution,
y
p
= −x
log x
x
dx + x log x
1
x
dx
= −x
1
2
log
2
x + x log x log x
=
1
2
x log
2
x.
Thus the general solution of the inhomogeneous differential equation is
y = c
1
x + c
2
x log x +
1
2
x log
2
x.
Solution 21.6
1. First we find the homogeneous solutions. We substitute y =
e
λx
into the homogeneous differential equation.
y
+ y = 0
λ
2
+ 1 = 0
λ = ±ı
y =
e
ıx
,
e
−ıx
We can also write the solutions in terms of real-valued functions.
y = {cos x, sin x}
1135
The Wronskian of the homogeneous solutions is
W [cos x, sin x] =
cos x sin x
−sin x cos x
= cos
2
x + sin
2
x = 1.
We obtain a particular solution with the variation of parameters formula.
y
p
= −cos x
e
x
sin x dx + sin x
e
x
cos x dx
y
p
= −cos x
1
2
e
x
(sin x − cos x) + sin x
1
2
e
x
(sin x + cos x)
y
p
=
1
2
e
x
The general solution is the particular solution plus a linear combination of the homogeneous solutions.
y =
1
2
e
x
+ cos x + sin x
2.
y
+ λ
2
y = sin x, y(0) = y
(0) = 0
Assume that λ is positive. First we find the homogeneous solutions by substituting y =
e
αx
into the homogeneous
differential equation.
y
+ λ
2
y = 0
α
2
+ λ
2
= 0
α = ±ıλ
y =
e
ıλx
,
e
−ıλx
y = {cos(λx), sin(λx)}
1136
The Wronskian of these homogeneous solution is
W [cos(λx), sin(λx)] =
cos(λx) sin(λx)
−λ sin(λx) λ cos(λx)
= λ cos
2
(λx) + λ sin
2
(λx) = λ.
We obtain a particular solution with the variation of parameters formula.
y
p
= −cos(λx)
sin(λx) sin x
λ
dx + sin(λx)
cos(λx) sin x
λ
dx
We evaluate the integrals for λ = 1.
y
p
= −cos(λx)
cos(x) sin(λx) − λ sin x cos(λx)
λ(λ
2
− 1)
+ sin(λx)
cos(x) cos(λx) + λ sin x sin(λx)
λ(λ
2
− 1)
y
p
=
sin x
λ
2
− 1
The general solution for λ = 1 is
y =
sin x
λ
2
− 1
+ c
1
cos(λx) + c
2
sin(λx).
The initial conditions give us the constraints:
c
1
= 0,
1
λ
2
− 1
+ λc
2
= 0,
For λ = 1, (non-resonant forcing), the solution subject to the initial conditions is
y =
λ sin(x) − sin(λx)
λ(λ
2
− 1)
.
1137
Now consider the case λ = 1. We obtain a particular solution with the variation of parameters formula.
y
p
= −cos(x)
sin
2
(x) dx + sin(x)
cos(x) sin x dx
y
p
= −cos(x)
1
2
(x − cos(x) sin(x)) + sin(x)
−
1
2
cos
2
(x)
y
p
= −
1
2
x cos(x)
The general solution for λ = 1 is
y = −
1
2
x cos(x) + c
1
cos(x) + c
2
sin(x).
The initial conditions give us the constraints:
c
1
= 0
−
1
2
+ c
2
= 0
For λ = 1, (resonant forcing), the solution subject to the initial conditions is
y =
1
2
(sin(x) − x cos x).
Solution 21.7
1. A set of linearly independent, homogeneous solutions is {cos t, sin t}. The Wronskian of these solutions is
W (t) =
cos t sin t
−sin t cos t
= cos
2
t + sin
2
t = 1.
We use variation of parameters to find a particular solution.
y
p
= −cos t
g(t) sin t dt + sin t
g(t) cos t dt
1138
The general solution can be written in the form,
y(t) =
c
1
−
t
a
g(τ) sin τ dτ
cos t +
c
2
+
t
b
g(τ) cos τ dτ
sin t.
2. Since the initial conditions are given at t = 0 we choose the lower bounds of integration in the general solution
to be that point.
y =
c
1
−
t
0
g(τ) sin τ dτ
cos t +
c
2
+
t
0
g(τ) cos τ dτ
sin t
The initial condition y(0) = 0 gives the constraint, c
1
= 0. The derivative of y(t) is then,
y
(t) = −g(t) sin t cos t +
t
0
g(τ) sin τ dτ sin t + g(t) cos t sin t +
c
2
+
t
0
g(τ) cos τ dτ
cos t,
y
(t) =
t
0
g(τ) sin τ dτ sin t +
c
2
+
t
0
g(τ) cos τ dτ
cos t.
The initial condition y
(0) = 0 gives the constraint c
2
= 0. The solution subject to the initial conditions is
y =
t
0
g(τ)(sin t cos τ − cos t sin τ) dτ
y =
t
0
g(τ) sin(t − τ) dτ
3. The solution of the initial value problem
y
+ y = sin(λt), y(0) = 0, y
(0) = 0,
is
y =
t
0
sin(λτ) sin(t − τ) dτ.
1139
Figure 21.5: Non-resonant Forcing
For λ = 1, this is
y =
1
2
t
0
cos(t − τ − λτ) − cos(t − τ + λτ )
dτ
=
1
2
−
sin(t − τ − λτ)
1 + λ
+
sin(t − τ + λτ)
1 − λ
t
0
=
1
2
sin(t) − sin(−λt)
1 + λ
+
−sin(t) + sin(λt)
1 − λ
y = −
λ sin t
1 − λ
2
+
sin(λt)
1 − λ
2
. (21.6)
The solution is the sum of two periodic functions of period 2π and 2π/λ. This solution is plotted in Figure 21.5
on the interval t ∈ [0, 16π] for the values λ = 1/4, 7/8, 5/2.
1140
Figure 21.6: Resonant Forcing
For λ = 1, we have
y =
1
2
t
0
cos(t − 2τ) − cos(tau)
dτ
=
1
2
−
1
2
sin(t − 2τ) − τ cos t
t
0
y =
1
2
(sin t − t cos t) . (21.7)
The solution has both a periodic and a transient term. This solution is plotted in Figure 21.5 on the interval
t ∈ [0, 16π].
Note that we can derive (21.7) from (21.6) by taking the limit as λ → 0.
lim
λ→1
sin(λt) − λ sin t
1 − λ
2
= lim
λ→1
t cos(λt) − sin t
−2λ
=
1
2
(sin t − t cos t)
1141
Solution 21.8
Let y
1
, y
2
and y
3
be linearly independ ent homogeneous solutions to the differential equation
L[y] = y
+ p
2
y
+ p
1
y
+ p
0
y = f(x).
We will look for a particular solution of the form
y
p
= u
1
y
1
+ u
2
y
2
+ u
3
y
3
.
Since the u
j
’s are undetermined functions, we are free to impose two constraints. We choose the constraints to simplify
the algebra.
u
1
y
1
+ u
2
y
2
+ u
3
y
3
= 0
u
1
y
1
+ u
2
y
2
+ u
3
y
3
= 0
Differentiating the expression for y
p
,
y
p
= u
1
y
1
+ u
1
y
1
+ u
2
y
2
+ u
2
y
2
+ u
3
y
3
+ u
3
y
3
= u
1
y
1
+ u
2
y
2
+ u
3
y
3
y
p
= u
1
y
1
+ u
1
y
1
+ u
2
y
2
+ u
2
y
2
+ u
3
y
3
+ u
3
y
3
= u
1
y
1
+ u
2
y
2
+ u
3
y
3
y
p
= u
1
y
1
+ u
1
y
1
+ u
2
y
2
+ u
2
y
2
+ u
3
y
3
+ u
3
y
3
Substituting the expressions for y
p
and its derivatives into the differential equation,
u
1
y
1
+ u
1
y
1
+ u
2
y
2
+ u
2
y
2
+ u
3
y
3
+ u
3
y
3
+ p
2
(u
1
y
1
+ u
2
y
2
+ u
3
y
3
) + p
1
(u
1
y
1
+ u
2
y
2
+ u
3
y
3
)
+ p
0
(u
1
y
1
+ u
2
y
2
+ u
3
y
3
) = f(x)
u
1
y
1
+ u
2
y
2
+ u
3
y
3
+ u
1
L[y
1
] + u
2
L[y
2
] + u
3
L[y
3
] = f(x)
u
1
y
1
+ u
2
y
2
+ u
3
y
3
= f(x).
1142
With the two constraints, we have the system of equations,
u
1
y
1
+ u
2
y
2
+ u
3
y
3
= 0
u
1
y
1
+ u
2
y
2
+ u
3
y
3
= 0
u
1
y
1
+ u
2
y
2
+ u
3
y
3
= f(x)
We solve for the u
j
using Kramer’s rule.
u
1
=
(y
2
y
3
− y
2
y
3
)f(x)
W (x)
, u
2
= −
(y
1
y
3
− y
1
y
3
)f(x)
W (x)
, u
3
=
(y
1
y
2
− y
1
y
2
)f(x)
W (x)
Here W (x) is the Wronskian of {y
1
, y
2
, y
3
}. Integrating the expressions for u
j
, the particular solution is
y
p
= y
1
(y
2
y
3
− y
2
y
3
)f(x)
W (x)
dx + y
2
(y
3
y
1
− y
3
y
1
)f(x)
W (x)
dx + y
3
(y
1
y
2
− y
1
y
2
)f(x)
W (x)
dx.
Green Functions
Solution 21.9
We consider the Green function problem
G
= f(x), G(−∞|ξ) = G
(−∞|ξ) = 0.
The homogeneous solution is y = c
1
+ c
2
x. The homogeneous solution that satisfies the boundary conditions is y = 0.
Thus the Green function has the form
G(x|ξ) =
0 x < ξ,
c
1
+ c
2
x x > ξ.
The continuity and jump conditions are then
G(ξ
+
|ξ) = 0, G
(ξ
+
|ξ) = 1.
1143
Thus the Green function is
G(x|ξ) =
0 x < ξ,
x − ξ x > ξ
= (x − ξ)H(x − ξ).
The solution of the problem
y
= f(x), y(−∞) = y
(−∞) = 0.
is
y =
∞
−∞
f(ξ)G(x|ξ) dξ
y =
∞
−∞
f(ξ)(x − ξ)H(x −ξ) dξ
y =
x
−∞
f(ξ)(x − ξ) dξ
We differentiate this solution to verify that it satisfies the differential equation.
y
= [f(ξ)(x − ξ)]
ξ=x
+
x
−∞
∂
∂x
(f(ξ)(x − ξ)) dξ =
x
−∞
f(ξ) dξ
y
= [f(ξ)]
ξ=x
= f(x)
Solution 21.10
Since we are dealing with an Euler equation, we substitute y = x
λ
to find the homogeneous solutions.
λ(λ − 1) + λ − 1 = 0
(λ − 1)(λ + 1) = 0
y
1
= x, y
2
=
1
x
1144
Variation of Parameters. The Wronskian of the homogeneous solutions is
W (x) =
x 1/x
1 −1/x
2
= −
1
x
−
1
x
= −
2
x
.
A particular solution is
y
p
= −x
x
2
(1/x)
−2/x
dx +
1
x
x
2
x
−2/x
dx
= −x
−
x
2
2
dx +
1
x
−
x
4
2
dx
=
x
4
6
−
x
4
10
=
x
4
15
.
The general solution is
y =
x
4
15
+ c
1
x + c
2
1
x
.
Applying the initial conditions,
y(0) = 0 → c
2
= 0
y
(0) = 0 → c
1
= 1.
Thus we have the solution
y =
x
4
15
+ x.
1145
Green Function. Since this problem has both an inhomogeneous term in the differential equation and inhomoge-
neous boundary conditions, we separate it into the two problems
u
+
1
x
u
−
1
x
2
u = x
2
, u(0) = u
(0) = 0,
v
+
1
x
v
−
1
x
2
v = 0, v(0) = 0, v
(0) = 1.
First we solve the inhomogeneous differential equation with the homogeneous boundary conditions. The Green
function for this problem satisfies
L[G(x|ξ)] = δ(x − ξ), G(0|ξ) = G
(0|ξ) = 0.
Since the Green function must satisfy the homogeneous boundary conditions, it has the form
G(x|ξ) =
0 for x < ξ
cx + d/x for x > ξ.
From the continuity condition,
0 = cξ + d/ξ.
The jump condition yields
c − d/ξ
2
= 1.
Solving these two equations, we obtain
G(x|ξ) =
0 for x < ξ
1
2
x −
ξ
2
2x
for x > ξ
1146
Thus the solution is
u(x) =
∞
0
G(x|ξ)ξ
2
dξ
=
x
0
1
2
x −
ξ
2
2x
ξ
2
dξ
=
1
6
x
4
−
1
10
x
4
=
x
4
15
.
Now to solve the homogeneous differential equation with inhomogeneous boundary conditions. The general solution
for v is
v = cx + d/x.
Applying the two boundary conditions gives
v = x.
Thus the solution for y is
y = x +
x
4
15
.
Solution 21.11
The Green function satisfies
G
(x|ξ) + p
2
(x)G
(x|ξ) + p
1
(x)G
(x|ξ) + p
0
(x)G(x|ξ) = δ(x − ξ).
First note that only the G
(x|ξ) term can have a delta function singularity. If a lower derivative had a delta function
type singularity, then G
(x|ξ) would be more singular than a delta function and there would be no other term in the
equation to balance that behavior. Thus we see that G
(x|ξ) will have a delta function singularity; G
(x|ξ) will have
a jump discontinuity; G
(x|ξ) will be continuous at x = ξ. Integrating the differential equation from ξ
−
to ξ
+
yields
ξ
+
ξ
−
G
(x|ξ) dx =
ξ
+
ξ
−
δ(x − ξ) dx
1147
G
(ξ
+
|ξ) − G
(ξ
−
|ξ) = 1.
Thus we have the three continuity conditions:
G
(ξ
+
|ξ) = G
(ξ
−
|ξ) + 1
G
(ξ
+
|ξ) = G
(ξ
−
|ξ)
G(ξ
+
|ξ) = G(ξ
−
|ξ)
Solution 21.12
Variation of Parameters. Consider the problem
x
2
y
− 2xy
+ 2y =
e
−x
, y(1) = 0, y
(1) = 1.
Previously we showed that two homogeneous solutions are
y
1
= x, y
2
= x
2
.
The Wronskian of these solutions is
W (x) =
x x
2
1 2x
= 2x
2
− x
2
= x
2
.
In the variation of parameters formula, we will choose 1 as the lower bound of integration. (This will simplify the
algebra in applying the initial conditions.)
y
p
= −x
x
1
e
−ξ
ξ
2
ξ
4
dξ + x
2
x
1
e
−ξ
ξ
ξ
4
dξ
= −x
x
1
e
−ξ
ξ
2
dξ + x
2
x
1
e
−ξ
ξ
3
dξ
= −x
e
−1
−
e
−x
x
−
x
1
e
−ξ
ξ
dξ
+ x
2
e
−x
2x
−
e
−x
2x
2
+
1
2
x
1
e
−ξ
ξ
dξ
= −x
e
−1
+
1
2
(1 + x)
e
−x
+
x + x
2
2
x
1
e
−ξ
ξ
dξ
1148
If you wanted to, you could write the last integral in terms of exponential integral functions.
The general solution is
y = c
1
x + c
2
x
2
− x
e
−1
+
1
2
(1 + x)
e
−x
+
x +
x
2
2
x
1
e
−ξ
ξ
dξ
Applying the boundary conditions,
y(1) = 0 → c
1
+ c
2
= 0
y
(1) = 1 → c
1
+ 2c
2
= 1,
we find that c
1
= −1, c
2
= 1.
Thus the solution subject to the initial conditions is
y = −(1 +
e
−1
)x + x
2
+
1
2
(1 + x)
e
−x
+
x +
x
2
2
x
1
e
−ξ
ξ
dξ
Green Functions. The solution to the problem is y = u + v where
u
−
2
x
u
+
2
x
2
u =
e
−x
x
2
, u(1) = 0, u
(1) = 0,
and
v
−
2
x
v
+
2
x
2
v = 0, v(1) = 0, v
(1) = 1.
The problem for v has the solution
v = −x + x
2
.
The Green function for u is
G(x|ξ) = H(x −ξ)u
ξ
(x)
where
u
ξ
(ξ) = 0, and u
ξ
(ξ) = 1.
1149
Thus the Green function is
G(x|ξ) = H(x −ξ)
−x +
x
2
ξ
.
The solution for u is then
u =
∞
1
G(x|ξ)
e
−ξ
ξ
2
dξ
=
x
1
−x +
x
2
ξ
e
−ξ
ξ
2
dξ
= −x
e
−1
+
1
2
(1 + x)
e
−x
+
x +
x
2
2
x
1
e
−ξ
ξ
dξ.
Thus we find the solution for y is
y = −(1 +
e
−1
)x + x
2
+
1
2
(1 + x)
e
−x
+
x +
x
2
2
x
1
e
−ξ
ξ
dξ
Solution 21.13
The differential equation for the Green function is
G
− G = δ(x − ξ), G
x
(0|ξ) = G(1|ξ) = 0.
Note that cosh(x) and sinh(x − 1) are homogeneous solutions that satisfy the left and right boundary conditions,
respectively. The Wronskian of these two solutions is
W (x) =
cosh(x) sinh(x − 1)
sinh(x) cosh(x − 1)
= cosh(x) cosh(x − 1) − sinh(x) sinh(x − 1)
=
1
4
e
x
+
e
−x
e
x−1
+
e
−x+1
−
e
x
−
e
−x
e
x−1
−
e
−x+1
=
1
2
e
1
+
e
−1
= cosh(1).
1150
The Green function for the problem is then
G(x|ξ) =
cosh(x
<
) sinh(x
>
− 1)
cosh(1)
,
G(x|ξ) =
cosh(x) sinh(ξ−1)
cosh(1)
for 0 ≤ x ≤ ξ,
cosh(ξ) sinh(x−1)
cosh(1)
for ξ ≤ x ≤ 1.
Solution 21.14
The differential equation for the Green function is
G
− G = δ(x − ξ), G(0|ξ) = G(∞|ξ) = 0.
Note that sinh(x) and
e
−x
are homogeneous solutions that satisfy the left and right boundary conditions, respectively.
The Wronskian of these two solutions is
W (x) =
sinh(x)
e
−x
cosh(x) −
e
−x
= −sinh(x)
e
−x
−cosh(x)
e
−x
= −
1
2
e
x
−
e
−x
e
−x
−
1
2
e
x
+
e
−x
e
−x
= −1
The Green function for the problem is then
G(x|ξ) = −sinh(x
<
)
e
−x
>
G(x|ξ) =
−sinh(x)
e
−ξ
for 0 ≤ x ≤ ξ,
−sinh(ξ)
e
−x
for ξ ≤ x ≤ ∞.
1151
Solution 21.15
a) The Green function problem is
xG
(x|ξ) + G
(x|ξ) = δ(x − ξ), G(0|ξ) bounded, G(1|ξ) = 0.
First we find the homogeneous solutions of the differential equation.
xy
+ y
= 0
This is an exact equation.
d
dx
[xy
] = 0
y
=
c
1
x
y = c
1
log x + c
2
The homogeneous solutions y
1
= 1 and y
2
= log x satisfy the left and right boundary conditions, respectively.
The Wronskian of these solutions is
W (x) =
1 log x
0 1/x
=
1
x
.
The Green function is
G(x|ξ) =
1 · log x
>
ξ(1/ξ)
,
G(x|ξ) = log x
>
.
b) The Green function problem is
G
(x|ξ) − G(x|ξ) = δ(x − ξ), G(−a|ξ) = G(a|ξ) = 0.
1152
{
e
x
,
e
−x
} and {cosh x, sinh x} are both linearly independent sets of homogeneous solutions. sinh(x + a) and
sinh(x − a) are homogeneous solutions that satisfy the left and right boundary conditions, respectively. The
Wronskian of these two solutions is,
W (x) =
sinh(x + a) sinh(x −a)
cosh(x + a) cosh(x − a)
= sinh(x + a) cosh(x − a) − sinh(x − a) cosh(x + a)
= sinh(2a)
The Green function is
G(x|ξ) =
sinh(x
<
+ a) sinh(x
>
− a)
sinh(2a)
.
c) The Green function problem is
G
(x|ξ) − G(x|ξ) = δ(x − ξ), G(x|ξ) bounded as |x| → ∞.
e
x
and
e
−x
are homogeneous solutions that satisfy the left and right boundary conditions, respectively. The
Wronskian of these solutions is
W (x) =
e
x
e
−x
e
x
−
e
−x
= −2.
The Green function is
G(x|ξ) =
e
x
<
e
−x
>
−2
,
G(x|ξ) = −
1
2
e
x
<
−x
>
.
d) The Green function from part (b) is,
G(x|ξ) =
sinh(x
<
+ a) sinh(x
>
− a)
sinh(2a)
.
1153
We take the limit as a → ∞.
lim
a→∞
sinh(x
<
+ a) sinh(x
>
− a)
sinh(2a)
= lim
a→∞
(
e
x
<
+a
−
e
−x
<
−a
) (
e
x
>
−a
−
e
−x
>
+a
)
2 (
e
2a
−
e
−2a
)
= lim
a→∞
−
e
x
<
−x
>
+
e
x
<
+x
>
−2a
+
e
−x
<
−x
>
−2a
−
e
−x
<
+x
>
−4a
2 − 2
e
−4a
= −
e
x
<
−x
>
2
Thus we see that the solution from part (b) approaches the solution from part (c) as a → ∞.
Solution 21.16
1. The problem,
y
+ λy = f(x), y(0) = y(π) = 0,
has a Green function if and only if it has a unique solution. This inhomogeneous problem has a unique solution
if and only if the homogeneous problem has only the trivial solution.
First consider the case λ = 0. We find the general solution of the homogeneous differential equation.
y = c
1
+ c
2
x
Only the trivial solution satisfies the boundary conditions. The problem has a unique solution for λ = 0.
Now consider non-zero λ. We find the general solution of the homogeneous differential equation.
y = c
1
cos
√
λx
+ c
2
sin
√
λx
.
The solution that satisfies the left boundary condition is
y = c sin
√
λx
.
1154
We apply the right boundary condition and find nontrivial solutions.
sin
√
λπ
= 0
λ = n
2
, n ∈ Z
+
Thus the problem has a unique solution for all complex λ except λ = n
2
, n ∈ Z
+
.
Consider the case λ = 0. We find solutions of the homogeneous equation that satisf y the lef t and right boundary
conditions, respectively.
y
1
= x, y
2
= x − π.
We compute the Wronskian of these functions.
W (x) =
x x −π
1 1
= π.
The Green function for this case is
G(x|ξ) =
x
<
(x
>
− π)
π
.
We consider the case λ = n
2
, λ = 0. We find the solutions of the homogeneous equation that satisfy the left
and right boundary conditions, respectively.
y
1
= sin
√
λx
, y
2
= sin
√
λ(x − π)
.
We compute the Wronskian of these functions.
W (x) =
sin
√
λx
sin
√
λ(x − π)
√
λ cos
√
λx
√
λ cos
√
λ(x − π)
=
√
λ sin
√
λπ
The Green function for this case is
G(x|ξ) =
sin
√
λx
<
sin
√
λ(x
>
− π)
√
λ sin
√
λπ
.
1155
2. Now we consider the problem
y
+ 9y = 1 + αx, y(0) = y(π) = 0.
The homogeneous solutions of the problem are constant multiples of sin(3x). Thus for each value of α, the
problem either has no solution or an infinite number of solutions. There will be an infinite number of solutions if
the inhomogeneity 1 + αx is orthogonal to the homogeneous solution sin(3x) and no solution otherwise.
π
0
(1 + αx) sin(3x) dx =
πα + 2
3
The problem has a solution only for α = −2/π. For this case the general solution of the inhomogeneous differential
equation is
y =
1
9
1 −
2x
π
+ c
1
cos(3x) + c
2
sin(3x).
The one-parameter family of solutions that satisfies the boundary conditions is
y =
1
9
1 −
2x
π
− cos(3x)
+ c sin(3x).
3. For λ = n
2
, n ∈ Z
+
, y = sin(nx) is a solution of the homogeneous equation that satisfies the boundary
conditions. Equation 21.5 has a (non-unique) solution only if f is orthogonal to sin(nx).
π
0
f(x) sin(nx) dx = 0
The modified Green function satisfies
G
+ n
2
G = δ(x − ξ) −
sin(nx) sin(nξ)
π/2
.
We expand G in a series of the eigenfunctions.
G(x|ξ) =
∞
k=1
g
k
sin(kx)
1156
We substitute the expansion into the differential equation to determine the coefficients. This will not determine
g
n
. We choose g
n
= 0, which is one of the choices that will make the modified Green function symmetric in x
and ξ.
∞
k=1
g
k
n
2
− k
2
sin(kx) =
2
π
∞
k=1
k=n
sin(kx) sin(kξ)
G(x|ξ) =
2
π
∞
k=1
k=n
sin(kx) sin(kξ)
n
2
− k
2
The solution of the inhomogeneous problem is
y(x) =
π
0
f(ξ)G(x|ξ) dξ.
Solution 21.17
We separate the problem for u into the two problems:
Lv ≡ (pv
)
+ qv = f(x), a < x < b, v(a) = 0, v(b) = 0
Lw ≡ (pw
)
+ qw = 0, a < x < b, w(a) = α, w(b) = β
and note that the solution for u is u = v + w.
The problem for v has the solution,
v =
b
a
g(x; ξ)f(ξ) dξ,
with the Green function,
g(x; ξ) =
v
1
(x
<
)v
2
(x
>
)
p(ξ)W (ξ)
≡
v
1
(x)v
2
(ξ)
p(ξ)W (ξ)
for a ≤ x ≤ ξ,
v
1
(ξ)v
2
(x)
p(ξ)W (ξ)
for ξ ≤ x ≤ b.
Here v
1
and v
2
are homogeneous solutions that respectively satisfy the left and right homogeneous boundary conditions.
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