Free Style Problem Solver - octawian.ro
Transcript of Free Style Problem Solver - octawian.ro
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Octavian G. Mustafa
Free Style Problem Solver
Second order linear inhomogeneous ODEs
Publicatiile DAL
Craiova
Fisier prelucrat ın data de [December 15, 2017]
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To Lidia Astefanei †
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Disclaimer
This essay has not been submitted to a referee. Therefore its content must be taken
“as is.”
The author welcomes your comments to his e-mail address1 and thanks you in
advance for your effort.
Each project from Publicatiile DAL is to be considered “under construction” un-
less specified otherwise. Its version is given by the date stated on the title page.
Craiova, May 18, 2015 O.G.M.
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Preface
Many undergraduate problems in mathematics and mechanics reduce to finding
solutions to (in)homogeneous second order ordinary differential equations (aka
ODEs).
A set of computations, based on the simple Bellman’s estimate, is always useful
in this respect. In the author’s experience, it is a pretty good time saver.
Craiova, [December 15, 2017] O.G.M.
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Contents
1 Bellman’s estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 An integral inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Bellman’s estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Changing the variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.1 The case when a ∈C1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 The general case ⋆ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 The solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.1 The uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.2 The wronskian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.3 The general solution of the simplified equation . . . . . . . . . . . . . . . . . . 11
3.4 A particular solution of the inhomogeneous ordinary differential
equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.5 The solution of (3.4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4 Particular cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.1 Constant coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.2 Perturbations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
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Acronyms
ODE Ordinary differential equation
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Chapter 1
Bellman’s estimate
1.1 An integral inequality
Given the interval I ⊂ R, with [t0,T ]⊆ I, consider the next inequality
x(t)≤ x0 +∫ t
t0
a(s)x(s)ds, t ∈ [t0,T ], (1.1)
where x0 ≥ 0 is a constant and a,x are continuous, non–negative valued functions
defined in I.
Set ε > 0. The integral inequality still holds if we replace x0 with x0 + ε . The
difference is that now the right–hand member of the inequality is a positive quantity
and we can write
a(t)x(t)
x0 + ε +∫ t
t0a(s)x(s)ds
≤ a(t), t ∈ [t0,T ]. (1.2)
Since
d
dt
[
ln
(
x0 + ε +∫ t
t0
a(s)x(s)ds
)]
=a(t)x(t)
x0 + ε +∫ t
t0a(s)x(s)ds
,
an integration of (1.2) with respect to t over [t0,T ] leads to
ln
(
x0 + ε +∫ t
t0a(s)x(s)ds
x0 + ε
)
≤∫ t
t0
a(s)ds.
Now,
x0 + ε +∫ t
t0
a(s)x(s)ds ≤ (x0 + ε)e∫ tt0
a(s)ds, t ∈ [t0,T ].
Taking into account (1.1), we get
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2 1 Bellman’s estimate
x(t)≤ (x0 + ε)e∫ tt0
a(s)ds, t ∈ [t0,T ].
Finally, by making ε ց 0, we obtain
x(t)≤ x0e∫ tt0
a(s)ds, t ∈ [t0,T ]. (1.3)
1.2 Bellman’s estimate
Consider the second order linear homogeneous ODE below
y′′+a(t)y′+b(t)y = 0, t ∈ [t0,T ], (1.4)
where the functions a,b are defined in I and continuous. The solution y is always
assumed C2, that is twice continuously differentiable.
The equation (1.4) can be rewritten as a system of two first order ODEs by intro-
ducing the new variable v = y′. We have
{
y′ = v,
v′ =−b(t)y−a(t)v,t ∈ [t0,T ].
The initial data for the equation (1.4) read as
y(t0) = y0, y′(t0) = v(t0) = y1
for y0,y1 ∈ R.
An integration with respect to t over [t0,T ] leads to
y(t)− y0 =∫ t
t0
v(s)ds
and
v(t)− y1 =−∫ t
t0
[b(s)y(s)+a(s)v(s)]ds.
The triangle inequality helps us estimate that
|y(t)| ≤ |y0|+∫ t
t0
|v(s)|ds (1.5)
and
|v(t)| ≤ |y1|+∫ t
t0
[|b(s)| · |y(s)|+ |a(s)| · |v(s)|]ds. (1.6)
The sum of (1.5) and (1.6) yields
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1.2 Bellman’s estimate 3
|y(t)|+ |v(t)| ≤ |y0|+ |y1|+∫ t
t0
(1+ |a(s)|+ |b(s)|) · (|y(s)|+ |v(s)|)ds
for every t ∈ [t0,T ].An application of formula (1.3), with
x(t) = |y(t)|+ |v(t)|, x0 = x(t0),
leads to the simple estimate
|y(t)|+ |y′(t)| ≤ (|y0|+ |y1|) · e∫ tt0(1+|a(s)|+|b(s)|)ds
, (1.7)
known as Bellman’s estimate. See also [1].
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Chapter 2
Changing the variables
2.1 The case when a ∈C1
Consider now the second order inhomogeneous ODE
y′′+a(t)y′+b(t)y = f (t), t ∈ [t0,T ], (2.1)
where the functions a,b, f : I → R are continuous and, as before, y ∈C2.
Our aim in this section is to simplify this equation by reducing it to the formula
d2z
ds2+ c(s)z = g(s), s ∈ [s0,S], (2.2)
where J ⊂ R is an interval, [s0,S]⊆ J and c,g are continuous in J.
To this end, let us assume first that a ∈C1. We look for a change of variables
y = v(t)z
that will make the term ”a(t)y′ ” from (2.1) vanish.
The formulas
y′ = v′z+ vz′, y′′ = v′′z+2v′z′+ vz′′,
once introduced into (2.1), yield
vz′′+[2v′+a(t)v]z′+[b(t)v+a(t)v′+ v′′]z = f (t), t ∈ [t0,T ].
We notice that, if 2v′+a(t)v = 0, we can get rid of z′ in the preceding equation.
In this way, by taking
v(t) = e−12
∫
a(t)dt = e− 1
2
∫ tt0
a(s)ds,
we reduce the equation (2.1) to the simpler form (2.2), that is
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6 2 Changing the variables
z′′+ c(t)z = g(t), t ∈ [t0,T ],
where
c(t) =b(t)v+a(t)v′+ v′′
v(t)= b(t)−
[a(t)]2
4−
a′(t)
2
and
g(t) =f (t)
v(t)= f (t) · e
12
∫ tt0
a(s)ds.
2.2 The general case ⋆
Let us notice that, by multiplying the equation (2.1) with
p(t) = e∫
a(t)dt = e∫ tt0
a(s)ds,
we can recast it as
[p(t)y′]′+ p(t)b(t)y = p(t) f (t), t ∈ [t0,T ].
We would like to have
p(t)y′ =dydt
p(t)
=dz
ds,
which means that we are interested in finding s = s(t) such that
z(s(t)) = y(t) anddz
ds(s(t)) = p(t)y′(t), t ∈ [t0,T ]. (2.3)
A differentiation of the first identity in (2.3) with respect to t yields
d
dt[z(s(t))] =
dz
ds(s(t)) · s′(t) = y′(t)
and, taking into account the second identity in (2.3), we obtain the initial value
problem
ds
dt=
1
p(t), s(t0) = s0. (2.4)
An integration with respect to t in [t0,T ] leads to
⋆ May be omitted at first reading.
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2.2 The general case ⋆ 7
s(t) = s0 +∫ t
t0
dτp(τ)
= s0 +∫ t
t0
e−∫ τt0
a(ξ )dξdτ .
Further,
[p(t)y′]′ =d
dt
[
dz
ds(s(t))
]
=d2z
ds2·
ds
dt=
1
p(t)·
d2z
ds2.
The equation (2.1) reads now as
d2z
ds2+[p(t)]2b(t)z = [p(t)]2 f (t), t ∈ [t0,T ]. (2.5)
The inverse of the function s = s(t), that is t = t(s), verifies the initial value
problem — recall (2.4) —
dt
ds= p(t), t(s0) = t0.
In conclusion, we get from (2.5) that
dz2
ds2+ c(s)z = g(s), s ∈ [s0,S],
where
c(s) = [p(t(s))]2b(t(s)), g(s) = [p(t(s))]2 f (t(s)).
The general case is, obviously, more difficult than the first case discussed here.
Fortunately, in most problems we encounter a continuously differentiable coefficient
a. See also [3].
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Chapter 3
The solution
3.1 The uniqueness
We are interested in this essay in solving the initial value problem
y′′+a(t)y′+b(t)y = f (t), t ∈ [t0,T ],y(t0) = y0,
y′(t0) = y1.
(3.1)
According to the previous section, we first simplify the formula of the equation.
Since in most undergraduate applications the continuous coefficients a,b are con-
stant — meaning that a is always C1 —, we shall keep the notation t for the argument
of the unknown function z.
Before writing down the initial value problem for the simplified equation, let us
notice that, given the change of variables
y(t) = e− 1
2
∫ tt0
a(s)dsz(t), t ∈ [t0,T ], (3.2)
we have
y′(t) =−1
2a(t) · y(t)+ e
− 12
∫ tt0
a(s)ds· z′(t). (3.3)
The new initial value problem reads as
z′′+ c(t)z = g(t), t ∈ [t0,T ],z(t0) = z0,
z′(t0) = z1,
(3.4)
where, taking into account (3.2), (3.3) for t = t0, we have
{
z0 = y0,
z1 =a(t0)
2· y0 + y1.
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10 3 The solution
Let us establish now that the initial value problem (3.4) has a unique solution.
To this end, assume that z1,z2 verify both the problem (3.4).
Their difference, namely Y = z1 − z2, verifies the next initial value problem
Y ′′+ c(t)Y = 0, t ∈ [t0,T ],Y (t0) = 0,
Y ′(t0) = 0.
Now, according to Bellman’s estimate (1.7), we deduce that
|Y (t)|+ |Y ′(t)| ≤ (|Y (t0)|+ |Y ′(t0)|) · e∫ tt0(1+|c(s)|)ds
= 0, t ∈ [t0,T ].
In conclusion, z1 = z2 throughout [t0,T ].
3.2 The wronskian
Given the linear homogeneous ODE
y′′+a(t)y′+b(t)y = 0, t ∈ [t0,T ], (3.5)
assume that we know a particular solution y1 which is positive everywhere in [t0,T ].The wronskian of the pair (y1,y) of solutions — here, y stands for the general
solution of (3.5) — is defined by the formula
W [y1,y](t) =
∣
∣
∣
∣
y1(t) y(t)y′1(t) y′(t)
∣
∣
∣
∣
, t ∈ [t0,T ].
We make the following computations
dW [y1,y]
dt=
∣
∣
∣
∣
y′1(t) y′(t)y′1(t) y′(t)
∣
∣
∣
∣
+
∣
∣
∣
∣
y1(t) y(t)y′′1(t) y′′(t)
∣
∣
∣
∣
=
∣
∣
∣
∣
y1(t) y(t)y′′1(t) y′′(t)
∣
∣
∣
∣
=
∣
∣
∣
∣
y1(t) y(t)−a(t)y′1(t)−b(t)y1(t) −a(t)y′(t)−b(t)y(t)
∣
∣
∣
∣
.
Recalling that, if we add the first row multiplied by ”b(t)” to the second row, the
determinant does not change, we get
dW [y1,y]
dt=
∣
∣
∣
∣
y1(t) y(t)−a(t)y′1(t) −a(t)y(t)
∣
∣
∣
∣
=−a(t) ·W [y1,y], t ∈ [t0,T ].
In conclusion, the wronskian verifies the formula
W [y1,y](t) =C · e−∫
a(t)dt =W [y1,y](t0) · e−∫ tt0
a(s)ds(3.6)
throughout [t0,T ].
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3.4 A particular solution of the inhomogeneous ordinary differential equation 11
3.3 The general solution of the simplified equation
Given the simplified ODE below
z′′+ c(t)z = 0, t ∈ [t0,T ],
assume that it has a positive valued solution z1.
The wronskian reads in this case as
W [z1,z](t) =
∣
∣
∣
∣
z1(t) z(t)z′1(t) z′(t)
∣
∣
∣
∣
=C, C ∈ R.
Recasting this identity as a first order inhomogeneous ODE with respect to z, we
obtain
z′ =z′1(t)
z1(t)· z+
C
z1(t). (3.7)
The general solution of the equation (3.7) reads as
z(t) = C(t) · e∫ z′
1(t)
z1(t)dt=C(t) · e
∫ tt0
z′1(τ)
z1(τ)dτ
= C(t)z1(t), t ∈ [t0,T ],
where C is an yet unknown smooth function.
To find C, we introduce the preceding formula into the equation and get
C′(t) =C
[z1(t)]2
everywhere in [t0,T ].By an integration with respect to t, we conclude that the general solution of the
simplified ODE reads as
z(t) = C(t)z1(t) =∫
C
[z1(t)]2dt · z1(t)
= C1 · z1(t)+C2 · z1(t)
∫ t
t0
dτ[z1(τ)]2
,
where the numbers C1,C2 ∈ R are the integration constants.
3.4 A particular solution of the inhomogeneous ordinary
differential equation
Assuming that the homogeneous ODE
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12 3 The solution
z′′+ c(t)z = 0, t ∈ [t0,T ], (3.8)
has a positive valued particular solution z1, we would like to find a particular solu-
tion zp of the inhomogeneous ODE
z′′+ c(t)z = g(t), t ∈ [t0,T ]. (3.9)
Let us try the following formula
zp = D(t)z1, (3.10)
where D is an yet unknown smooth function with real values.
Inserting (3.10) into the equation (3.9), we get
z1(t)D′′+2z′1(t)D
′+[z′′1(t)+ c(t)z1(t)]D = g(t), t ∈ [t0,T ].
Notice that the sum between the brackets is zero, z1 being a solution of the equation
(3.8).
Introduce the function E = E(t) with the formula E = D′. We can now recast the
latter formula as an inhomogeneous first order ODE, that is
E ′ =−2z′1(t)
z1(t)·E +
g(t)
z1(t), t ∈ [t0,T ]. (3.11)
As before, the general solution of this equation reads as
E(t) =C(t) · e−2∫ z′
1(t)
z1(t)dt=
C(t)
[z1(t)]2(3.12)
for every t ∈ [t0,T ], where C is an yet unknown smooth function.
The formula (3.12), once inserted into (3.11), yields
C′(t) = z1(t)g(t).
Since we are looking for a particular solution of (3.11), we take
C(t) =∫ t
t0
z1(s)g(s)ds.
Now, because of
D′(t) =C(t)
[z1(t)]2=
1
[z1(t)]2
∫ t
t0
z1(s)g(s)ds, t ∈ [t0,T ], (3.13)
a variant of D is given by
D(t) =
∫ t
t0
1
[z1(s)]2
∫ s
t0
z1(τ)g(τ)dτds. (3.14)
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3.5 The solution of (3.4) 13
Via an integration by parts, we have that
D(t)
=∫ t
t0
d
ds
{
∫ s
t0
dτ[z1(τ)]2
}
·
[
∫ s
t0
z1(τ)g(τ)dτ]
ds
=∫ t
t0
dτ[z1(τ)]2
·∫ t
t0
z1(τ)g(τ)dτ −∫ t
t0
{
z1(s)∫ s
t0
dτ[z1(τ)]2
}
g(s)ds.
Notice that, for our choice of D, — recall (3.13), (3.14) for t = t0 —
D(t0) = D′(t0) = 0. (3.15)
Finally, collecting all the details into (3.10), we obtain
zp(t) = z1(t)
∫ t
t0
dτ[z1(τ)]2
·
∫ t
t0
z1(s)g(s)dτ
−z1(t)∫ t
t0
z1(s)∫ s
t0
dτ[z1(τ)]2
·g(s)ds
= z1(t)∫ t
t0
z1(s)
{
∫ t
t0
dτ[z1(τ)]2
−∫ s
t0
dτ[z1(τ)]2
}
g(s)ds
= z1(t)∫ t
t0
{
z1(s)∫ t
s
dτ[z1(τ)]2
}
g(s)ds, t ∈ [t0,T ]. (3.16)
Taking into account (3.15), we have also that
zp(t0) = D(t0)z1(t0) = 0 (3.17)
and
z′p(t0) = D′(t0)z1(t0)+D(t0)z′1(t0) = 0. (3.18)
3.5 The solution of (3.4)
Since the problem (3.4) has a unique solution, we look for constants C1,C2 ∈ R
so that the solution can be written as
z(t) = a particular solution of the equation (3.8) + zp(t)
= C1z1(t)+C2z1(t)∫ t
t0
ds
[z1(s)]2+ zp(t), t ∈ [t0,T ].
The constants C1,C2 are found by means of the data from problem (3.4). To this
end, we have — recall (3.17), (3.18) —
z(t0) =C1z1(t0)+ zp(t0) = z0
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14 3 The solution
and
z′1(t0) =C1z′1(t0)+C2 ·1
z1(t0)+ z′p(t0) = z1.
We have obtained
C1 =z0
z1(t0), C2 = z1 · z1(t0)− z0 · z
′1(t0).
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Chapter 4
Particular cases
4.1 Constant coefficients
If the coefficients a,b of the general linear inhomogeneous ODE (1.4) are con-
stant then the coefficient c of the simplified ODE (3.8) is also a constant,
c(t) = b−a2
4= c.
We recall the fundamental formula (3.16),
zp(t) =∫ t
t0
G(t,s)g(s)ds, G(t,s) = z1(t)z1(s)∫ t
s
dτ[z1(τ)]2
,
where t0 ≤ s ≤ t ≤ T .
We have three cases. In the first one, c =−ω2 < 0 for some constant ω > 0. We
take z1(t) = eωt . Here, the quantity G(t,s) reads as
G(t,s) = eω(t+s)∫ t
s
dτe2ωτ =
1
ω· sinhω(t − s).
The solution of (3.4) is given by
z(t) = z0 coshω(t − t0)+z1
ωsinhω(t − t0)+
1
ω
∫ t
t0
sinhω(t − s) ·g(s)ds.
In the second case, c = 0. We take z1(t) = 1. Here, the quantity G(t,s) reads as
G(t,s) = t − s.
The solution of (3.4) is given by
z(t) = z0 + z1(t − t0)+∫ t
t0
(t − s)g(s)ds.
15
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16 4 Particular cases
In the third case, c = ω2 for some constant ω > 0. We take z1(t) = cosωt — the
interval [t0,T ] must be chosen appropriately! —. Here, the quantity G(t,s) reads as
G(t,s) = sinωt sinωs
∫ t
s
dτcos2 ωτ
=1
ω· sinω(t − s).
The solution of (3.4) is given by
z(t) = z0 cosω(t − t0)+z1
ωsinω(t − t0)+
1
ω
∫ t
t0
sinω(t − s) ·g(s)ds.
4.2 Perturbations
In many undergraduate textbooks, e.g. [2, pages 168, 176], one can find presentations of the
so-called method of undetermined coefficients as some sort of independent enterprise and not as
of a less-obvious application of the fundamental variation of parameters procedure. The computa-
tions in this section show that the method of undetermined coefficients is nothing but a disguised
particular case of the method of variation of parameters.
Given the simplified inhomogeneous equation (3.9), assume that
c(t) =±ω2, g(t) = tneαt cosβ t, (4.1)
where ω > 0, n ≥ 1 is an integer and α,β ∈ R.
The particular solution zp can be computed in this case by taking into account
formula (3.14), that is
zp(t) = z1(t)∫ t
t0
1
[z1(s)]2
∫ s
t0
z1(τ)g(τ)dτds, t ∈ [t0,T ].
In fact, we shall work loosely and compute the solution without caring about the
integration constants, namely
zp(t) = z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt,
since the integration constants will join the homogeneous part of the general solution
(the expense will be on the initial data).
Now, given γ ∈ C−{0} and m ≥ 0 an integer, the following formula
∫
tmeγtdt = eγt
[
1
γ· tm
+ (sign m)m
∑k=1
(−1)k m(m−1) · · ·(m− k+1)
γk+1· tm−k
]
(4.2)
can be established by mathematical induction.
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4.2 Perturbations 17
Notice that the perturbation from (4.1) is the real part ”Re ” of the quantity
g(t) = tneηt, η = α + iβ .
Take z1(t) = e−λ t , where λ ∈ {ω, iω}. Then, taking into account (4.2) for γ =η −λ 6= 0, we obtain
1
[z1(t)]2
∫
z1(t)g(t)dt
= e(2λ+γ)t
[
1
γ· tn +
n
∑k=1
(−1)k n(n−1) · · ·(n− k+1)
γk+1· tn−k
]
.
Further, by assuming that 2λ + γ = η +λ 6= 0, we obtain
z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt
= e−λ t
[
∫
1
γ· tne(2λ+γ)tdt
+n
∑k=1
(−1)k n(n−1) · · ·(n− k+1)
γk+1
∫
tn−ke(2λ+γ)tdt
]
= e−λ t
{
e(2λ+γ)t
γ
[
tn
2λ + γ+
n
∑p=1
(−1)p n(n−1) · · ·(n− p+1)tn−p
(2λ + γ)p+1
]
+n−1
∑k=1
(−1)k n(n−1) · · ·(n− k+1)
γk+1
×e(2λ+γ)t
[
tn−k
2λ + γ+
n−k
∑p=1
(−1)p (n− k) · · ·(n− k− p+1)tn−k−p
(2λ + γ)p+1
]
+(−1)n n!
γn+1·
e(2λ+γ)t
2λ + γ
}
= e(λ+γ)t{
1
γ(2λ + γ)· tn
+n
∑p=1
(−1)pn · · ·(n− p+1)
[
1
γ(2λ + γ)p+1+
1
γ p+1(2λ + γ)
]
· tn−p
+n−1
∑k=1
n−k
∑p=1
(−1)k+p n · · ·(n− k− p+1)
γk+1(2λ + γ)p+1· tn−k−p
}
= eηt · (n–th order polynomial in t).
If γ = η −λ 6= 0 and η +λ = 0 then
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18 4 Particular cases
z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt
= e−λ t ·
12γ · t
2 − 1γ2 · t, when n = 1,
1(n+1)γ · t
n+1 +n
∑k=1
(−1)k n···(n−k+2)
γk+1 · tn−k+1, when n ≥ 2,
= e−λ tt ·
12γ · t −
1γ2 , when n = 1,
1(n+1)γ · t
n +n
∑k=1
(−1)k n···(n−k+2)
γk+1 · tn−k, when n ≥ 2,
= teηt · (n–th order polynomial in t).
If γ = η −λ = 0 then
z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt
= eλ t ·
14λ · t2 − 1
4λ 2 · t +1
8λ 3 , when n = 1,
1(n+1)(2λ ) · t
n+1 +n+1
∑k=1
(−1)k n···(n−k+2)
(2λ )k+1 · tn−k+1, when n ≥ 2,
= eλ t · [(n+1)–th order polynomial in t]
= teηt · (n–th order polynomial in t)+ constant · eλ t.
Since the latter term of this sum is a solution of the homogeneous (part of the)
equation, we can neglect it.
In all of these three situations, the particular solution reads as
zp(t) = Re
{
z1(t)∫
1
[z1(t)]2
[
∫
z1(t)g(t)dt
]
dt
}
, t ∈ [t0,T ].
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References
1. Bellman, R.: Stability theory of differential equations. McGraw-Hill, London (1953)
2. Boyce, W.E., DiPrima, R.C.: Elementary differential equations and boundary value problems,
Sixth edition. J. Wiley & Sons, New York (1997)
3. Hartman, P., Wintner, A.: On the assignment of asymptotic values for the solutions of linear
differential equations of second order. Amer. J. Math. 77, 475–483 (1955)
19
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Index
Astefanei, Lidia, v
Bellman’s estimate, ix, 3, 10
Bellman, Richard, ix
continuously differentiable, 7
determinant, 10
homogeneous, ix, 2, 11, 16, 18
inhomogeneous, ix, 11, 15, 16
initial data, 16
initial value problem, 6, 9
integral inequality, 1
integration by parts, 13
integration constant, 11, 16
inverse of a function, 7
mathematical induction, 16
method of undetermined coefficients, 16
ODE, ix, 2, 5, 10–12, 15
polynomial, 17
real part, 17
smooth, 11, 12
triangle inequality, 2
variation of parameters, 16
wronskian, 10, 11
21