If the characteristic equation 12.4.1.2 has complex conjugate roots, then in the above solution, one should extract the real part on the basis of the relation expα iβ = e α cos β i sin β
Trang 1Example 1 The eigenvalue problem for the equation
y xx + λ(1+ x2)–2y= 0
with the boundary conditions (12.3.7.8) admits an exact analytic solution and has eigenvalues λ1= 15, λ2= 63 ,
, λ n= 16n 2 – 1
According to the Rayleigh–Ritz principle, formula (12.2.5.6) for z = sin(πx) yields the approximate value
λ0= 15 33728 The solution of the Cauchy problem (12.3.7.9)–(12.3.7.10) with f (x) =1, g(x) = λ(1+ x2) – 2 ,
h (x) =0yields x0 = 0 983848 , 1– x0 = 0 016152 , y2 = 0 024585, y x (x0 ) = – 0 70622822
The first iteration for the first eigenvalue is determined by (12.3.7.11) and results in the value λ1= 14 99245
with the relative error Δλ/λ1 = 5 × 10 – 4
The second iteration results in λ2= 14 999986 with the relative error Δλ/λ2 < 10 – 6
Example 2 Consider the eigenvalue problem for the equation
(√
1+ x y x) x + λy =0
with the boundary conditions (12.3.7.8).
The Rayleigh–Ritz principle yields λ0 = 11 995576 The next two iterations result in the values λ1 =
11 898578and λ2 = 11 898458 For the relative error we haveΔλ/λ2 < 10 – 5
For more details about finite-difference methods and other numerical methods, see, for instance, the books by Lambert (1973), Keller (1976), Schiesser (1993), and Zwillinger (1997)
12.4 Linear Equations of Arbitrary Order
12.4.1 Linear Equations with Constant Coefficients
12.4.1-1 Homogeneous linear equations
An nth-order homogeneous linear equation with constant coefficients has the general form
y(n)
x + a n–1y(n–1 )
x +· · · + a1y
The general solution of this equation is determined by the roots of the characteristic equation
P (λ) =0, where P (λ) = λ n + a n–1λ n–1+· · · + a1λ + a0 (12.4.1.2) The following cases are possible:
1◦ All roots λ1, λ2, , λ nof the characteristic equation (12.4.1.2) are real and distinct.
Then the general solution of the homogeneous linear differential equation (12.4.1.1) has the form
y = C1exp(λ1x ) + C2exp(λ2x) +· · · + C n exp(λ n x)
2◦ There are m equal real roots λ1= λ2=· · · = λ m (m≤n), and the other roots are real and distinct In this case, the general solution is given by
y = exp(λ1x )(C1+ C2x+· · · + C m x m–1)
+ C m+1exp(λ m+1x ) + C m+2exp(λ m+2x) +· · · + C n exp(λ n x)
3◦ There are m equal complex conjugate roots λ = α iβ (2m≤n), and the other roots are real and distinct In this case, the general solution is
y = exp(αx) cos(βx)(A1+ A2x+· · · + A m x m–1)
+ exp(αx) sin(βx)(B1+ B2x+· · · + B m x m–1)
+ C2m+1exp(λ2m+1x ) + C2m+2exp(λ2m+2x) +· · · + C n exp(λ n x),
where A1, , A m , B1, , B m , C2m+1, , C n are arbitrary constants
Trang 24◦ In the general case, where there are r different roots λ1, λ2, , λ r of multiplicities
m1, m2, , m r, respectively, the left-hand side of the characteristic equation (12.4.1.2)
can be represented as the product
P (λ) = (λ – λ1)m1(λ – λ2)m2 (λ – λ r)m r,
where m1+ m2+· · · + m r = n The general solution of the original equation is given by
the formula
r
k=1
exp(λ k x )(C k,0+ C k,1x+· · · + C k,m k 1x m k 1),
where C k,lare arbitrary constants
If the characteristic equation (12.4.1.2) has complex conjugate roots, then in the above
solution, one should extract the real part on the basis of the relation exp(α iβ) = e α (cos β
i sin β).
Example 1 Find the general solution of the linear third-order equation
y + ay – y – ay =0
Its characteristic equation is λ3+ aλ2– λ – a =0 , or, in factorized form,
(λ + a)(λ –1)(λ +1 ) = 0
Depending on the value of the parameter a, three cases are possible.
1 Case a≠ 1 There are three different roots, λ1= –a, λ2 = – 1, and λ3 = 1 The general solution of the
differential equation is expressed as y = C1e–ax + C2e–x + C3e .
2 Case a =1 There is a double root, λ1= λ2 = – 1, and a simple root, λ3 = 1 The general solution of the
differential equation has the form y = (C1+ C2x )e–x + C3e .
3 Case a = –1 There is a double root, λ1= λ2= 1, and a simple root, λ3 = – 1 The general solution of
the differential equation is expressed as y = (C1+ C2x )e x + C3e–x.
Example 2 Consider the linear fourth-order equation
y xxxx – y =0
Its characteristic equation, λ4– 1 = 0 , has four distinct roots, two real and two pure imaginary,
λ1= 1 , λ2= – 1 , λ3= i, λ4= –i.
Therefore the general solution of the equation in question has the form (see Item 3◦)
y = C1e + C2e–x + C3sin x + C4cos x.
12.4.1-2 Nonhomogeneous linear equations Forms of particular solutions
1◦ An nth-order nonhomogeneous linear equation with constant coefficients has the
gen-eral form
y(n)
x + a n–1y(n–1 )
x +· · · + a1y
x + a0y = f (x). (12.4.1.3) The general solution of this equation is the sum of the general solution of the
corre-sponding homogeneous equation with f (x)≡ 0(see Paragraph 12.4.1-1) and any particular solution of the nonhomogeneous equation (12.4.1.3)
If all the roots λ1, λ2, , λ n of the characteristic equation (12.4.1.2) are different, equation (12.4.1.3) has the general solution:
n
ν=1
C ν e λ ν x+n
ν=1
e λ ν x
P
λ (λ ν)
(for complex roots, the real part should be taken)
In the general case, if the characteristic equation (12.4.1.2) has multiple roots, the solution to equation (12.4.1.3) can be constructed using formula (12.4.2.5)
Trang 3TABLE 12.4 Forms of particular solutions of the constant coefficient nonhomogeneous linear equation
y(x n) + a n–1 y x(n–1)+· · · + a1y x + a0y = f (x) that correspond to some special forms of the function f (x)
Form of the
function f (x)
Roots of the characteristic equation
λ n + a n–1 λ n–1+· · · + a1λ + a0= 0 Form of a particularsolution y = 2y(x)
Zero is not a root of the
characteristic equation (i.e., a0≠ 0 ) P2m (x)
P m (x)
Zero is a root of the
characteristic equation (multiplicity r) x r P2m (x)
αis not a root of the characteristic equation P2m (x)e αx
P m (x)e αx
(α is a real constant) αis a root of the
characteristic equation (multiplicity r) x r P2m (x)e αx
iβis not a root of the characteristic equation
2
P ν (x) cos βx
+ 2Q ν (x) sin βx
P m (x) cos βx
+ Q n (x) sin βx
iβis a root of the
characteristic equation (multiplicity r)
x r[ 2P ν (x) cos βx
+ 2Q ν (x) sin βx]
α + iβ is not a root of the
characteristic equation
[ 2P ν (x) cos βx
+ 2Q ν (x) sin βx]e αx [P m (x) cos βx
+ Q n (x) sin βx]e αx
α + iβ is a root of the characteristic equation (multiplicity r)
x r[ 2P ν (x) cos βx
+ 2Q ν (x) sin βx]e αx
Notation: P m and Q n are polynomials of degrees m and n with given coefficients; 2 P m, 2P ν, and 2Q νare
polynomials of degrees m and ν whose coefficients are determined by substituting the particular solution into the basic equation; ν = max(m, n); and α and β are real numbers, i2= – 1
2◦ Table 12.4 lists the forms of particular solutions corresponding to some special forms
of functions on the right-hand side of the linear nonhomogeneous equation
3◦ Consider the Cauchy problem for equation (12.4.1.3) subject to the homogeneous initial
conditions
y(0) = y x (0) = = y(x n–1)(0) =0 (12.4.1.5)
Let y(x) be the solution of problem (12.4.1.3), (12.4.1.5) for arbitrary f (x) and let u(x) be the solution of the auxiliary, simpler problem (12.4.1.3), (12.4.1.5) with f (x)≡ 1, so that
u (x) = y(x)|f(x)≡1 Then the formula
y (x) =
x
0 f (t)u
x (x – t) dt holds It is called the Duhamel integral.
12.4.1-3 Solution of the Cauchy problem using the Laplace transform
Consider the Cauchy problem for equation (12.4.1.3) with arbitrary initial conditions
y(0) = y0, y
x(0) = y1, ., y(n–1 )
x (0) = y n–1, (12.4.1.6)
where y0, y1, , y n–1are given constants
Trang 4Problem (12.4.1.3), (12.4.1.6) can be solved using the Laplace transform based on the formulas (for details, see Section 11.2)
2y(p) = L5y (x)6
, f2(p) =L5f (x)6, where L5f (x)6≡ ∞
0 e
–px f (x) dx.
To this end, let us multiply equation (12.4.1.3) by e–px and then integrate with respect to x
from zero to infinity Taking into account the differentiation rule
L5y(n)
= p n 2y(p) –n
k=1
p n–k y(k–1 )
and the initial conditions (12.4.1.6), we arrive at a linear algebraic equation for the trans-form2y(p):
P (p) 2y(p) – Q(p) = 2 f (p), (12.4.1.7) where
P (p) = p n + a n–1p n–1+· · · + a1p + a0, Q (p) = b n–1p n–1+· · · + b1p + b0,
b k = y n–k–1+ a n–1y n–k–2+· · · + a k+2y1+ a k+1y0, k=0, 1, , n –1
The polynomial P (p) coincides with the characteristic polynomial (12.4.1.2) at λ = p.
The solution of equation (12.4.1.7) is given by the formula
2y(p) = f2(p) + Q(p)2
On applying the Laplace inversion formula (see in Section 11.2) to (12.4.1.8), we obtain a solution to problem (12.4.1.3), (12.4.1.6) in the form
y (x) = 1
2πi
c+i∞
c–i∞
2
f (p) + Q(p)
2
Since the transform 2y(p) (12.4.1.8) is a rational function, the inverse Laplace transform
(12.4.1.9) can be obtained using the formulas from Paragraph 11.2.2-2 or the tables of Section T3.2
Remark In practice, the solution method for the Cauchy problem based on the Laplace transform leads
to the solution faster than the direct application of general formulas like (12.4.1.4), where one has to determine
the coefficients C1, , C n.
Example 3 Consider the following Cauchy problem for a homogeneous fourth-order equation:
y xxxx + a4y= 0 ; y( 0) = y x ( 0) = y xxx( 0 ) = 0 , y xx( 0) = b.
Using the Laplace transform reduces this problem to a linear algebraic equation for2y(p): (p4+ a4)2y(p) –
bp= 0 It follows that
2y(p) = bp
p4+ a4.
In order to invert this expression, let us use the table of inverse Laplace transforms T3.2.2 (see row 52) and take into account that a constant multiplier can be taken outside the transform operator to obtain the solution to the original Cauchy problem in the form
y (x) = b
a2 sin
ax
√
2
sinh
ax
√
2
.
Trang 512.4.2 Linear Equations with Variable Coefficients
12.4.2-1 Homogeneous linear equations Structure of the general solution
The general solution of the nth-order homogeneous linear differential equation
f n (x)y(x n) + f n–1(x)y(x n–1)+· · · + f1(x)y x + f0(x)y =0 (12.4.2.1) has the form
y = C1y1(x) + C2y2(x) + · · · + C n y n (x). (12.4.2.2)
Here, y1(x), y2(x), , y n (x) is a fundamental system of solutions (the y k are linearly
independent particular solutions, y k0); C1, C2, , C nare arbitrary constants
12.4.2-2 Utilization of particular solutions for reducing the order of the equation
1◦ Let y1= y1(x) be a nontrivial particular solution of equation (12.4.2.1) The substitution
y = y1(x)
z (x) dx results in a linear equation of order n –1for the function z(x).
2◦ Let y1 = y1(x) and y2 = y2(x) be two nontrivial linearly independent solutions of
equation (12.4.2.1) The substitution
y = y1
y2w dx – y2
y1w dx
results in a linear equation of order n –2for w(x).
3◦ Suppose that m linearly independent solutions y1(x), y2(x), , y
m (x) of equation (12.4.2.1) are known Then one can reduce the order of the equation to n – m by successive application of the following procedure The substitution y = y m (x)
z (x) dx leads to an equation of order n –1for the function z(x) with known linearly independent solutions:
z1=
y1
y m
y2
y m
x, ., z m–1 =
y
m–1
y m
x. The substitution z = z m–1(x)
w (x) dx yields an equation of order n –2 Repeating this
procedure m times, we arrive at a homogeneous linear equation of order n – m.
12.4.2-3 Wronskian determinant and Liouville formula
The Wronskian determinant (or simply, Wronskian) is the function defined as
W (x) =
y
n (x)
y(n–1 )
where y1(x), , y n (x) is a fundamental system of solutions of the homogeneous equa-tion (12.4.2.1); y(k m) (x) = d m y k
dx m , m =1, , n –1; k =1, , n.
The following Liouville formula holds:
W (x) = W (x0) exp
– x
x0
f n–1(t)
f n (t) dt
Trang 6
12.4.2-4 Nonhomogeneous linear equations Construction of the general solution.
1◦ The general nonhomogeneous nth-order linear differential equation has the form
f n (x)y x(n) + f n–1(x)y(x n–1)+· · · + f1(x)y x + f0(x)y = g(x). (12.4.2.4) The general solution of the nonhomogeneous equation (12.4.2.4) can be represented as the sum of its particular solution and the general solution of the corresponding homogeneous equation (12.4.2.1)
2◦ Let y1(x), , y
n (x) be a fundamental system of solutions of the homogeneous equa-tion (12.4.2.1), and let W (x) be the Wronskian determinant (12.4.2.3) Then the general
solution of the nonhomogeneous linear equation (12.4.2.4) can be represented as
n
ν=1
C ν y ν (x) +
n
ν=1
y ν (x)
ν (x) dx
f n (x)W (x), (12.4.2.5)
where W ν (x) is the determinant obtained by replacing the νth column of the matrix (12.4.2.3) by the column vector with the elements 0, 0, , 0, g.
3◦ Let y(x, σ) be the solution to the Cauchy problem for the homogeneous equation (12.4.2.1) with nonhomogeneous initial conditions at x = σ:
y (σ) = y x (σ) = · · · = y(n–2 )
x (σ) =0, y(n–1 )
x (σ) =1,
where σ is an arbitrary parameter Then a particular solution of the nonhomogeneous linear
equation (12.4.2.4) with homogeneous boundary conditions
y (x0) = y x (x0) =· · · = y(n–1 )
is given by the Cauchy formula
¯y(x) =
x
x0
y (x, σ) g (σ)
f n (σ) dσ.
4◦ Superposition principle The solution of a nonhomogeneous linear equation
L[y] =
m
k=1
g k (x), L[y]≡f n (x)y x(n) + f n–1(x)y x(n–1)+· · · + f1(x)y x + f0(x)y
is determined by adding together the solutions,
m
k=1
y k,
of m (simpler) equations,
L[y k ] = g k (x), k=1, 2, , m,
corresponding to respective nonhomogeneous terms in the original equation
Trang 712.4.2-5 Euler equation.
1◦ The nonhomogeneous Euler equation has the form
x n y(n)
x + a n–1x n–1y(n–1 )
x +· · · + a1xy
x + a0y = f (x).
The substitution x = be t (b ≠ 0) leads to a constant coefficient linear equation of the form (12.4.1.3)
2◦ Particular solutions of the homogeneous Euler equation [with f (x)≡ 0] are sought in
the form y = x k If all k are real and distinct, its general solution is expressed as
y (x) = C1|x|k1 + C2|x|k2 +· · · + C n|x|k n
Remark. To a pair of complex conjugate values k = α iβ there corresponds a pair of particular solutions:
y= |x|α sin(β|x|) and y =|x|α cos(β|x| ).
12.4.2-6 Solution of equations using the Laplace transform Laplace equation
1◦ Some classes of equations (12.4.2.1) or (12.4.2.4) with polynomial coefficients
f k (x) =
s k
m=0
a km x m
may be solved using the Laplace transform (see Paragraph 12.4.1-3 and Section 11.2) To this end, one uses the following formula for the Laplace transform of the product of a power function and a derivative of the unknown function:
L5x m y(n)
6
= (–1)m d m
p n 2y(p) –
n
k=1
p n–k y(k–1 )
(12.4.2.6)
The right-hand side contains initial data y(m)
x (+0), m =0, 1, , n –1(specified in the Cauchy problem) As a result, one arrives at a linear ordinary differential equation, with
respect to p, for the transform 2y(p); the order of this equation is equal to max
1≤k n{s k}, the highest degree of the polynomials that determine the equation coefficients In some cases, the equation for2y(p) turns out to be simpler than the initial equation for y(x) and can be solved in closed form The desired function y(x) is found by inverting the transform 2y(p)
using the formulas from Paragraph 11.2.2-2 of the tables from Section T3.2
2◦ Consider the Laplace equation
(a n + b n x )y(x n) + (a n–1+ b n–1x )y(x n–1)+· · · + (a1+ b1x )y x + (a0+ b0x )y =0, (12.4.2.7)
whose coefficients are linear functions of the independent variable x The application of the
Laplace transform, in view of formulas (12.4.2.6), brings it to a linear first-order ordinary differential equation for the transform2y(p).