Download Boundary Value Problems, Characteristic Functions

Chapter 4
Boundary Value Problems, Characteristic
Functions and Representations
1
Boundary Value Problems and Characteristic Functions
If a solution to an ordinary differential equation must satisfy certain specified conditions at
two or more boundaries then one has a boundary value problem (BVP). A BVP is homogeneous if the differential condition and the boundary conditions are all homogeneous. An
expression is homogeneoeus if when satisfied by a function F it is also satisfied by cF where
c is a constant. Recall that non-trivial solutions to a second order linear homogeneous ODE
can be expressed as a linear combinations of linearly independent functions with two constants. These must be determined from the boundary conditions resulting in an expression
in terms of only a single constant.
In many cases, the linearly independent solutions depend on a parameter λ which produces non-trivial solutions only for certain specific values λ1 , λ2 , .... Problems of this kind
are called characteristic-value or eigenvalue problems. The specific values of λ are called
characteristic values or eigenvalues and the solutions are called characteristic functions or
eigenfunctions.
2
The Rotating String
A taut string (length l, density ρ and tension T ) rotates about its length at angular velocity
ω. What is the shape y(x) of the rotating string? A force balance on an element of string
yields (assuming small deflections)
d2 y
+ λy = 0
dx2
where λ = ρω 2 /T . Since the ends are fixed the boundary conditions are y(0) = y(l) = 0.
The most general solution of the differential equation is
√
y(x) = C sin( λx) = Cφ(x)
1
√
where φ(x) = sin( λx).
√
In order to satisfy the condition y(l) = 0 it is necessary that λl = nπ where n = 1, 2, ...
producing the infinite set
λn =
n2 π 2
l2
n = 1, 2, ...
Therefore, there is an infinite set of φ(x)’s given by
q
φn (x) = sin( λn x) = sin(
nπ
x)
l
n = 1, 2, ...
are the characteristic functions of the problem and λn are the characteristic values corresponding to specific, physically meaningful critical values of the angular velocity
nπ
ωn =
l
s
T
ρ
n = 1, 2, ...
associated with characteristic deflection modes of the string.
A general expression capable of accounting for various types of boundary conditions in
the above problem is
T
Lyb0 = αLyb0 = −yb
kL
where yb is the displacement at the string boundary and k is the spring constant of an
elastic spring which may be attached to the the end of the string (i.e. a yielding support).
Specifically, if α = 0 (i.e. infinite restoring force) one gets yb = 0, the fixed end condition,
while if α = ∞ (i.e. zero restoring force) one obtains yb0 = 0, the free end condition.
3
The Rotating Shaft
Consider now a uniform shaft (length l, density ρ, Young’s modulus E and moment of inertia
I) rotating about its axis with angular velocity ω. From beam bending theory, the deflexion
of the shaft y(x) satisfies
d4 y
− λy = 0
dx4
where λ = ρω 2 /EI. This has the general solution
y(x) = c1 sinh(λ1/4 x) + c2 cosh(λ1/4 x) + c3 sin(λ1/4 x) + c4 cos(λ1/4 x)
If the shaft is regarded as hinged at both ends, the appropriate boundary conditions are
y(0) = y 00 (0) = 0
2
which gives c2 = c4 = 0, and
y(l) = y 00 (l) = 0
which gives the systems of equations
c1 sinh(µ) + c3 sin(µ) = 0
and
c1 sinh(µ) − c3 sin(µ) = 0
where µ = λ1/4 l.
From the basic theory of systems of linear algebraic equations, compatibility requires the
vanishing of the determinant of coefficients, i.e.
−2 sinh(µ) sin(µ) = 0
This can only happen if µ = 0 or if sin(µ) = 0. Nontrivial solutions result only in the second
case, that is if
µn = nπ
n = 1, 2, ...
Substituting into the first equation of the system above and recalling the properties of the
sinh function necessarily requires that c1 = 0 so that the eigenfunctions (deflection modes)
are
nπx
) n = 1, 2, ...
φn (x) = sin(
l
the eigenvalues are
λn =
n4 π 4
l4
n = 1, 2, ...
and the critical speeds are
n2 π 2
ωn = 2
l
4
s
EI
ρ
n = 1, 2, ...
Buckling of Axially Loaded Columns
Consider now an uniform column, axially loaded at one end (length l, load P , density ρ,
Young’s modulus E and moment of inertia I). The deflection y(x) must satisfy the equation
d2 y
d4 y
+
λ
=0
dx4
dx2
3
where λ = P/EI. The general solution of this is of the form
y(x) = c1 sin(λ1/2 x) + c2 cos(λ1/2 x) + c3 x + c4
If both ends of the column are hinged the boundary conditions are
y(0) = y 00 (0) = 0
which gives c2 = c4 = 0 and
y(l) = y 00 (l) = 0
which gives non-trivial solutions only if
λn =
n2 π 2
l2
n = 1, 2, ...
with c3 = 0. The characteristic functions (deflection modes) in this case are then
φn (x) = sin(
nπx
)
l
n = 1, 2, ...
The critical buckling loads are
Pn = n2 π 2
EI
l2
n = 1, 2, ...
The unbent column is then stable only if
P < P1 = π 2
EI
l2
which is called the Euler load and the fundamental buckling mode y(x) = C sin(πx/l) occurs
when P = P1 .
5
Orthogonality of Characteristic Functions
Two arbitrary functions φm (x) and φn (x) for x ∈ [a, b] for which
Z
b
a
r(x)φm (x)φn (x)dx = 0
are called orthogonal with respect to the weight function r(x). If all pairs of functions in a
set are orthogonal, the set itself is called orthogonal.
The sets of characteristic functions corresponding to a wide variety of BVP encountered
in applications are orthogonal with respect to various specific weighing functions.
4
Consider the generic BVP for x ∈ [a, b] defined by the following ODE in standard form
dy
d
[p(x) ] + [q(x) + λr(x)]y = Ly + λr(x)y = 0
dx
dx
for which two eigenfunctions are φ1 (x) and φ2 (x). The characteristic functions must satisfy
dφ1
d
[p(x)
] + [q(x) + λ1 r(x)]φ1 = 0
dx
dx
and
dφ2
d
[p(x)
] + [q(x) + λ2 r(x)]φ2 = 0
dx
dx
Multiplying the first of these by φ2 , the second one by φ1 substracting and integrating results
in
(λ2 − λ1 )
Z
Z
b
rφ1 φ2 dx =
a
b
a
[φ2
d dφ1
d dφ2
(p
) − φ1 (p
)]dx
dx dx
dx dx
Integrating the RHS by parts gives
Z
(λ2 − λ1 )
b
a
rφ1 φ2 dx = [p(x){(φ2
dφ1
dφ2 b
) − (φ1
)}]
dx
dx a
Therefore, if the boundary conditions associated with the original problem are of the form
y = 0, or y 0 = 0, or y + γy 0 = 0, at each end point the RHS term equals zero. The term
can also be zero if p(x) = 0 when x = a or x = b; or if p(b) = p(a) and y(b) = y(a) and
y 0 (b) = y 0 (a). Finally, the RHS is also zero if the solutions are required to be periodic. In all
these cases one has then
Z
b
a
r(x)φ1 (x)φ2 (x)dx = 0
λ2 6= λ1
Boundary value problems of the above types are known as Sturm-Liouville problems.
Moreover, if p(x) > 0, r(x) > 0, q(x) ≤ 0 and if γ 6= 0 but is negative if the condition is
on the left boundary and positive otherwise, one has then a proper Sturm-Liouville problem.
Proper S-L problems are interesting because all their associated eigenvalues are real and ≥ 0
and their eigenfunctions are real too. Furthermore, the eigenvalues constitute an infinite
discrete set.
Finally, the integral
Z
Cn =
b
a
r(x)[φn (x)]2 dx
is called the norm of φn with respect to r and if the coefficient in φn is chosen so that Cn = 1,
φn is considered normalized with respect to r. A set of normalized orthogonal functions is
called orthonormal.
5
Although the above statements referred to the given ODE in standard form they are
widely applicable since that the generic ODE
a0 (x)
d2 y
dy
+ a1 (x) + [a2 (x) + λa3 (x)]y = 0
2
dx
dx
can always be transformed to the standard form by setting
Z
a1
dx)
a0
p = exp(
q=
a2
p
a0
r=
a3
p
a0
and
Exercise. Investigate the eigenvalues and eigenfunctions associated with the BVP in
x ∈ [0, l],
y 00 + λy = 0
subject to
y(0) = y(l) = 0
6
Characteristic Function Representations
Linear combinations of eigenfunctions from a complete orthogonal set constitute useful formal representations of arbitrary functions. This is reminiscent of the representation of arbitrary vectors in Euclidean space by means of linear combination of the complete orthogonal
set of unit vectors.
Assume that an arbitrary function f (x) can be represented as a linear combination of
characteristic functions forming an orthogonal set with respect to a certain weighting function
r(x), {φn (x)}∞
0
f (x) = A0 φ0 (x) + A1 φ1 (x) + A2 φ2 (x) + ... =
∞
X
An φn (x)
n=0
Multiplication of the above by r(x)φk (x), followed by integration from x = a to x = b and
taking into account the orthogonality of the eigenfunctions leads to
Rb
r(x)f (x)φn (x)dx
An = aR b
2
a r(x)[φn (x)] dx
6
which determines the formal series representation of f (x).
If the associated Sturm-Liouville problem is proper and p(x), q(x) and r(x) are regular
in (a, b), the formal representation given above of any piecewise differentiable function f (x)
converges to f (x) inside (a, b) wherever f (x) is continuous and converges to the mean value
wherever finite jumps occur. Furthermore, if f (x) is continuous and its derivative is piecewise
differentiable, the term by term derivative of the series converges to f 0 (x) wherever the
derivative is continuous.
Note the great generality of series representation in terms of eigenfunctions in comparison
with power series expansions. While Taylor series representations require the existence of
derivatives of all orders and even in such case the representation may not exist, this is not
the case at all with Fourier series.
The characteristic function representation can be operated upon by the operator L =
dy
d
(p(x)
) + q to give
dx
dx
Lf (x) ≈
∞
X
An Lφn (x) = −r(x)
n=0
∞
X
λn An φn (x)
n=0
and if Lf (x)/r(x) is piecewise differentiable
∞
Lf (x) X
=
Bn φn (x)
r(x)
n=0
with Bn = −λn An . Therefore, in these conditions the operator can by applied term by term
to the series.
7
Boundary Value Problems involving Nonhomogeneous
Equations
Consider the nonhomogeneous differential equation
Ly + Λry = h(x)
and its associated homogeneous problem when h(x) = 0 with corresponding eigenfunctions
P
φn (x) and eigenvalues λn with the series representation of its solution y = an φn Since
P
Lφn = −λn rφn , substitution into the original ODE gives r (Λ − λn )an φn = h, i.e.
X
X
h(x)
= f (x) = (Λ − λn )an φn (x) =
An φ(x)
r(x)
where h/r is assumed to be piecewise differentiable. The solution of the nonhomogeneous
problem is then
y(x) =
X
An
φn (x)
Λ − λn
7
8
Fourier Series
The characteristic functions φn (x) = sin(nπx/l) are eigenfunctions of the problem
d2 y
+ λy = 0,
dx2
y(0) = 0,
y(l) = 0
with the characteristic numbers λn = n2 π 2 /l2 .
If a function f (x) can be represented as
f (x) =
∞
X
An φn (x) =
n=1
∞
X
An sin(
n=1
nπx
)
l
in x ∈ [0, l] then, because of orthogonality,
Rl
An =
f (x) sin( nπx
)dx
2Z l
nπx
l
)dx
=
f (x) sin(
Rl
nπx 2
l 0
l
0 [sin( l )] dx
0
The series is the Fourier sine series representation of f (x) in terms of individual harmonics and An are called the Fourier coefficients and are equal to twice the average value of
f sin(nπx/l) in [0, l]. Since the sign of each harmonic is reversed when x is replaced by
−x, the Fourier sine series is a representation of an odd function f (x) (i.e. one for which
f (−x) = −f (x)) in the interval [−l, l]. If f (x) is odd and periodic, the series represents f (x)
everywhere.
If instead the following BVP is considered
d2 y
+ λy = 0,
dx2
y 0 (0) = 0,
y 0 (l) = 0
with characteristic functions
φn (x) = cos(
nπx
),
l
(n = 0, 1, 2, ...)
and eigenvalues λn = n2 π 2 /l2 and this is used to produce a (Fourier cosine series) representation of the function f (x)
f (x) = A0 +
∞
X
An cos(
n=1
nπx
)
l
the corresponding Fourier coefficients are
A0 =
1
l
Z
0
l
f (x)dx
and
2Z l
nπx
)dx,
f (x) cos(
An =
l 0
l
8
(n = 1, 2, 3, ...)
The Fourier cosine series is a representation of an even function f (x) (i.e. one for which
f (−x) = f (x)) in the interval [−l, l]. If f (x) is even and periodic, the series represents
f (x) everywhere. Since all harmonics are even and periodic, they constitute convergent
representations to piecewise differentiable even functions f (x) everywhere.
Since any function f (x) can be expressed as the sum of an even fe (x) and an odd function
fo (x) with
fe (x) = A0 +
∞
X
An cos(
n=1
fo (x) =
∞
X
Bn sin(
n=1
nπx
)
l
nπx
)
l
such that , over the interval [−l, l],
f (x) = A0 +
∞
X
(An cos(
n=1
nπx
nπx
) + Bn sin(
))
l
l
where
1 Zl
f (x)dx
2l −l
A0 =
1
An =
l
Z
l
−l
f (x) cos(
nπx
)dx
l
f (x) sin(
nπx
)dx
l
and
1
Bn =
l
Z
l
−l
The series representation of f (x) is called the complete Fourier sine-cosine series representation. This can be expressed in more compact form by using Euler’s formula:
f (x) =
∞
X
Ck eikx
k=−∞
where
Ck =
1 Zl
1
i
f (x)e−ikx dx = Ak − Bk
2π −l
2
2
The complete Fourier series representation of f (x) over any interval [a, a + P ] is easily
obtained.
9
Fourier series representations of f (x) can be differentiated term by term and the result
becomes a convergent representation of f 0 (x) as long as f 0 is piecewise differentiable.
Exercise. Obtain Fourier sine series and cosine series representations of f (x) = ex in
x ∈ [0, π].
For the sine series
An =
2
π
Z
0
π
ex sin(nx)dx =
2 n
(1 − eπ cos(nπ))
π n2 + 1
And for the cosine series
1
A0 =
π
Z
0
π
ex dx =
eπ − 1
π
2Zπ x
2 1
(eπ cos(nπ) − 1)
An =
e cos(nx)dx =
2
π 0
πn +1
9
Fourier-Bessel Series
The BVP
p2
d dy
(x ) + (− + µ2 x)y = 0
dx dx
x
is of Sturm-Liouville form with p(x) = x, q(x) = −p2 /x, r(x) = x and λ = µ2 . Its general
solution is of the form y(x) = Zp (µx) where
Zp (µx) = c1 Jp (µx) + c2 J−p (µx)
when p is not integer and
Zp (µx) = c1 Jp (µx) + c2 Yp (µx)
when p is integer. The functions Jp and Yp constitute and orthogonal set of characteristic
functions.
If at x = 0 y is finite and xy 0 = 0 and at x = l, y(l) = 0, the eigenvalues µn s are the
roots of Jp (µn l) = 0.
If instead at x = l, y 0 (l) = 0, the µn s are the roots of Jp0 (µn l) = 0.
Finally if at x = l, ky(l) + ly 0 (l) = 0 (with k ≥ 0), the µs are the roots of kJp (µn l) +
µn lJp0 (µn l) = 0.
In the first case the characteristic functions are
φn (x) = Jp (µn x)
which are orthogonal on [0, l] with respect to r(x) = x.
10
So, the representation
f (x) =
∞
X
An Jp (µn x)
n=1
with
An =
and
Z
Cn =
0
l
1
Cn
Z
l
0
xf (x)Jp (µn x)dx
1
dJp (µn x)
|x=l )2 ]
[(µ2n l2 − p2 )Jp (µn l)2 + l2 (
2
2µn
dx
x[Jp (µn x)]2 dx =
is called the Fourier-Bessel series representation of f (x).
Therefore, if y(l) = 0, Jp (µn l) = 0 and
Z
Cn =
0
If y (l) = 0,
Jp0 (µn l)
l
0
x[Jp (µn x)]2 dx =
l2
[Jp+1 (µn l)]2
2
= 0 and
Cn =
µ2n l2 − p2
[Jp (µn l)]2
2
2µn
And if ky(l) + ly 0 (l) = 0, kJp (µn l) + µn lJp0 (µn l) = 0
Cn =
µ2n l2 − p2 + k 2
[Jp (µn l)]2
2µ2n l2
Note that in all the above cases the characteristic functions are of the form
φn (x) = Jp (µn x)
The integral involved in the calculation of An can be readily obtained by numerical
methods.
Exercise. Determine the Fourier-Bessel series expansion of the function f (x) = 1 for
x ∈ [0, l] in terms of J0 .
Let
1=
∞
X
An J0 (µn x)
n=1
where J0 (µn l) = 0. Thus
An =
with
Z
Cn =
0
l
1 Zl
xJ0 (µn x)dx
Cn 0
x[J0 (µn x)]2 dx =
11
l2
[J1 (µn l)]2
2
10
Legendre Series
Legendre’s differential equation
d
dy
[(1 − x2 ) ] + p(p + 1)y = 0
dx
dx
has the form of the differential equation of a Sturm-Liouville problem with p(x) = 1 − x2 ,
q(x) = 0, r(x) = 1 and λ = p(p + 1). Finite solutions at x = ±1 require p = 0, 1, 2, ... and
are
φn (x) = Pn (x)
where Pn (x) are the Legendre polynomials. Since these are orthogonal, any piecewise differentiable function in the interval (−1, 1) can be represented as
∞
X
f (x) =
An Pn (x)
n=0
where
R1
f (x)Pn (x)dx
=
An = −1
R1
2
−1 [Pn (x)] dx
(
2n+1 R 1
−1 f (x)Pn (x)dx
2
n
2n+1 R 1
(1 − x2 )n d dxf (x)
n dx
2n+1 n! −1
Because of the properties of Pn , if f (x) is even, An = 0 if n is odd and viceversa.
Any polynomial of degree k can be expressed as a linear combination of the first k + 1
Legendre polynomials.
11
Fourier Integral
R
Consider the expression 0∞ A(u) sin(ux)du where u is any positive number. Could this be
a representation of a well behaved function f (x) for 0 < x < ∞? Assume that is the case,
therefore,
Z
f (x) =
∞
0
A(u) sin(ux)du
Multiplying both sides by sin(u0 x), integrating from 0 to l, taking the limit as l → ∞ and
rearranging gives
2
A(u) =
π
Z
0
∞
f (x) sin(ux)dx
Switching the dummy integration variable from x to t and substituting in the original representation
Z ∞
2Z∞
f (x) =
sin(ux)
f (t) sin(ut)dtdu
π 0
0
12
This is the FourierR sine integral representation of f (x) and is valid in ∀x if f is odd, piecewise
differentiable and 0∞ |f (x)|dx exists. Similarly, the Fourier cosine integral representation can
be defined as
Z
Z ∞
2 ∞
f (x) =
cos(ux)
f (t) cos(ut)dtdu
π 0
0
valid ∀x if f is even. For a function f (x) = fe (x) + fo (x), the complete Fourier integral
representation in −∞ < x < ∞
Z
f (x) =
∞
0
[A(u) cos(ux) + B(u) sin(ux)]du
with
Z
∞
A(u) =
1
π
B(u) =
1Z∞
f (t) sin(ut)dt
π 0
0
f (t) cos(ut)dt
and
Substituting and using complex notation
f (x) =
The integral
transform is
R∞
−∞
1
2π
Z
Z
∞
−∞
eiux
∞
−∞
e−iut f (t)dtdu
¯ of f . The inverse
e−ixt f (x)dx is called the Fourier transform f (u)
1 Z ∞ iux ¯
e f (u)du
f (x) =
2π −∞
Fourier sine and cosine integral transforms can also be defined as
Z
fS (x) =
∞
0
f (x) sin(ux)dx
with
2
f (x) =
π
and
Z
0
Z
fC (x) =
∞
0
∞
fS (u) sin(ux)du
f (x) cos(ux)dx
with
2Z∞
f (x) =
fC (u) cos(ux)du
π 0
13