ODE2013_08_Operator_and_Laplace_Transform_Methods_v00-11-20
8.Operator and Laplace Transform Methods
There exist a panoply of operator methods that can be used to solve, among other things, differential equations. These form a whole branch of applied mathematics called operational calculus. Some of these operators are integral transforms, i.e., operators mapping functions to other functions via special integrals. This chapter will discuss the basics of the Laplace transform. It proves particularly useful when we are confronted with linear equations with a complicated inhomogeneous term. An example of a typical problem would be an electronic logic computer circuit driven by a clock signal which is a discontinuous function periodically jumping between two voltage values. Two other well known such integral transforms are the Fourier transform and the Mellin transform. Before dealing with Laplace transform we will take a look at polynomial differential operators (or vector fields). These are a subset of linear operators endowed with nice properties. We will see that there is a natiural way of defining an inverse for such an operator and that this thereby provide us with a means of computing particular solutions of equations.
8.1Differential Polynomial Operators
Roughly speaking, (functional) operators are transformations mapping functions into functions. Differentiation and integration are prime examples of operators. They map functions to other functions called their derivatives and integrals. Consider functions of a single function x. We denote the operator of differentiation with respect to x by D. In general, we define
D n y=d n y
d x n
.(1)
(D0y=y). We can think of D as an abbreviation for d/dx. We can construct polynomial differential operators of order n by taking linear combinations:
P(D)=a
0+a
1
D+a
2
D2+...+a
n
D n.(2)
This operator acts on a function f(x) in the expected manner:
P(D)f(x)=a
0f(x)+a
1
f'(x)+a
2
f''(x)+...+a
n
f(n)(x).(3)
Polynomial differential operators act linearly:
P(D)(af(x)+bg(x))=a P(D)f(x)+b P(D)g(x).(4) Example:
(D2?D+2)(e x+x3+cos(2x))
=D2(e x+x3+cos(2x))?D(e x+x3+cos(2x))+2(e x+x3+cos(2x))
=(e x+6x?4cos(2x))?(e x+3x2?2sin(2x))+(2e x+2x3+2cos(2x))
=2e x+6x?3x2+2x3?2cos(2x)+2sin(2x)
.(5) Principle of superposition. Consider a linear differential equation
P(D)y=R
1
(x)+R2(x)+...+R n(x).(6)
Let y
k
(x) be a solution of P(D)y=R k(x) for 1≤k≤n. Then
y(x)=y
1
(x)+y2(x)+...+y n(x) is a solution of (6).
P(D)y=R
1
(x)+R2(x)+...+R n(x).(7) Polynomial operators are linear:
(P1(D)+P2(D))y=P1(D)y+P2(D)y.(8) Example:
P
1
(D)=D3?2D2+3D?4
P
2
(D)=D2?4D+1
P 1(D)+P
2
(D)=D3?D2+2D?3
.(9)
One can also verify that polynomial operators are commutative and associative with respect to addition. Furthermore, there exists an identity operator: 0 defined by
0y(x)=0 and every polynomial operator P(D) has an inverse ?P(D). Thus the set of polynomial operators form s a commutative group under addition.
Given a function w(x) and a polynomial operator P(D) we define their product by
[w(x)P(D)]y=w(x)[P(D)y].(10) Example:
[x3(5D2?4)]e2x=x3[20e2x?4e2x]=16x3e2x.(11)
The product of two operators P 1(D ) and P 2(D ) is defined by
[P 1(D )P 2(D )]y =P 1(D )[P 2(D )y ].
(12)
One can verify that this product is commutative, associative and also distributive on addition:
P 1(D )P 2(D )=P 2(D )P 1(D )
[P 1(D )P 2(D )]P 3(D )=P 1(D )[P 2(D )P 3(D )]
P 1(D )[P 2(D )+P 3(D )]=P 1(D )P 2(D )+P 1(D )P 3(D )
.(13)
Exponential translation theorem. Consider a polynomial differential operator of order n:
P (D )=∑k =0n
a k D
k
(14)
With a n ≠0. Let u (x ) be a differentiable function up to order n, defined over some open interval. Then
P (D )(ue ax )=e ax P (D +a )u .
(15)
Corollary:
(D ?a )n (ue ax )=e ax D n u .
(16)Corollary: If u (x )=k , then
P (D )(ke ax )=ke ax P (a ).
(17)
Example:
?(x)=(2D2?3D?4)(e3x cos(2x)).(18)
Note that
P(D+3)=2(D+3)2?3(D+3)?4=2D2+12D+18?3D?9?4=2D2+9D+5.(19) Then
?(x)=e3x(2D2+9D+5)(cos(2x))
.(20) =e3x(?8cos(2x)?18sin(2x)+5cos(2x))
=?3e3x(cos(2x)+6sin(2x))
Example:
(D2?4)(e3x cos(2x))=e3x D2(cos(2x))=?4e3x cos(2x).(21)
Let's give an example to show how this can help us solve an ODE. The idea is to factorize the differential operator from which the equation arises as a product of order one operators and proceed to solve a sequence of first-order equations.
Example:
y''+y'?2y=xe2x.(22) We can rewrite this as
(D2+D?2)y=xe2x.(23) or
(D?1)(D+2)y=xe2x.(24)
Let
u=(D+2)y.(25) Then
(D?1)u=u'?u=xe2x.(26)
Using the method of undetermined coefficients, we find
u(x)=c
1
e x+(x?1)e2x.(27) But, from (25), this means
u(x)=c
1
e x+(x?1)e2x.(28) But, from (25), this means
y''+2y=c
1
e x+(x?1)e2x.(29) Again, using the method o
f undetermined coefficients, we obtain
y(x)=c
2e?2x+
c
1
2
e x+(
1
4
x?
5
16
)e2x.(30)
We define D?1 as the operator that gives the primitive of a function, i.e., that integrates a function and ignores the constant of integration. For example, D?1(cos(x))=sin(x). We can generalize this to the operator D?n=(D?1)n.
D?2(3x?4)=D?1∫(3x?4)dx=D?1(32x2?4x)=∫(32x2?4x)dx=12x3?2x2.(31) Let P(D) be a polynomial differential operator We define its inverse P?1(D) by
P(D)[P?1(D)y(x)]=y(x).(32) Let us see how we can use this to find particular solutions of ODEs. The method is rather simplistic: it consists in expressing the inverse of a first order differential operator as a series of powers of D:
(D?a)?1=?1
a
(1?
D
a
)
?1
=
?1
a[1+D a+(D a)2+(D a)3+...].(33)
Example:
(D?2)y(x)=x3.(34)
Then
y(x)=(D?2)?1x3.(35) From (33), we then have
y(x)=?1
2[1+D2+(D2)2+(D2)3+...]x3
=?1
2[x3+3x22+3x2+34
]=?18[4x3+6x2+6x+3].(36)
Example:
(D?a)y(x)=e kx.(37)
y(x)=?1
a[1+D a+(D a)2+(D a)3+...]e kx
=?1
a[1+k a+(k a)2+(k a)3+...]e kx=e kx k?a
.(38) Thus
(D?a)?1e kx=
e kx
k?a
.(39)
Since every polynomial differential operator can be factor as a product of first-order ones, then it directly follows that
P(D)?1e kx=
e kx
P(D)
.(40)
This can be used with equations whose inhomogeneous terms are trigonometric functions. Recall that e ix=cos(x)+isin(x). Then the equation P(D)y=A cos(bx) can be solved by first solving the equation P(D)y=A e ibx and then extracting its real part. Similarly,the equation P(D)y=A sin(bx) can be solved by first solving the equation P(D)y=A e ibx and then extracting its imaginary part.
The above result can be extended for product of functions by exponentials:
P(D)?1u(x)e kx=e kx1
P(D+k)
u(x).(41) Example:
y''?y'?2y=x3e2x.(42)
We can rewrite this as
(D2?D?2)y=x3e2x.(43) We have
y(x)=1
(D2?D?2)x3e2x=e2x1
P(D+2)
x3
=e2x
1
(D+2)2?(D+2)?2
x3=e2x
1
D2+3D
x3
=e2x
1
D(D+3)
x3=e2x
1
D(?13
)[1+D3+(D3)2+(D3)3+...]x3
=e2x 1
D(?13
)[x3+x23+2x3+29]=?13e2x[x44+x33+x23+2x9]
.(44)
Partial fraction expansions come in very handy for operator and Laplace transform methods. Such an expansion takes a rational expression whose denominator is polynomial and expresses it as a sum of rational fractions whose denominators are monomials.
Example:
F=3x+4
x2?1
.(45) The denominator is a difference of squares x2?1=(x+1)(x?1) and we want to write
F=A
x+1+
B
x?1.(46)
Bringing everything over a common denominator, we have
F=
A(x?1)
(x+1)(x?1)
+
B(x+1)
(x?1)(x+1)
=
(A+B)x+(B?A)
x2?1
.(47)
Comparing (45) with (47), we must have
A+B=3
B?A=4
.(48)
This implies A=?1/2 and B=7/2. Thus
F=7/2?1/2.(49)
The following example illustrates how to decompose a rational expression with a complex denominator. Let F=P(x)/Q(x) where
Q(x)=(x+α)(x+β)3(x2+γ)2(x4+δ).(50) Then
F=
A
x+α
+
B
1
x+β
+
B
2
(x+β)2
+
B
3
(x+β)3
+C
1
x+C
2
x2+γ
+
D
1
x+D
2
(x2+γ)2
+
E
1
x3+E
2
x2+E
3
x+E
4
x4+δ
.(51)
One particular case is easy to figure out. Let P(x)=1 and
Q(x)=(x?α
1
)(x?α2)...(x?αn).(52) Then
F =∑
k =1n
1
Q '(αk )(x ?αk
).
(53)
Example:
F =
3x +4
(x ?1)(x +2)
.(54)
We have
F =A x ?1+B x +2+C
(x +2)2=A (x +2)2+B (x ?1)(x +2)+C (x ?1)(x ?1)(x +2)
2
.(55)
Expanding and comparing with (54), we get
(A +B )x 2+(4A +B +C )x +(4?2B ?C )=3x +4.
(56)The solution is A =3/5, B =?3/5 and C =6/5. Thus
F =
3/5x ?1?3/5x +2+6/5
(x +2)
.(57)The above result can be used to find particular solutions of ODEs with constant coefficients.Example:
y (4)+4y '''?3y ''?4y ?4=x 2e 3x .
(58)
We can rewrite this as
(D 4+4D 3?3D 2?4D ?4)y =x 2e 3x .
(59)
or
(D?1)(D+2)2(D+1)y=x2e3x.(60) Thus
y=1
(D?1)(D+2)2(D+1)
x2e3x.(61) Let
F=1
(D?1)(D+2)(D+1)=
a
D?1
+
b
D+1
+
c
D+2
+
d
(D+2)
=a(D+2)2(D+1)
D?1
+
b(D+2)2(D?1)
D+1
+
c(D?1)(D+1)(D+2)
D+2
+
d(D?1)(D+1)
(D+2)2
.(62)
Expanding and comparing with (60), we have
a+b+c=0
5a+3b+2c+d=0
8a?c=0
4a?4b?2c?d=1
.(63) The solution is a=1/18, b=?1/2. c=4/9 and d=1/3. Thus (61) becomes
y=(1/18D?1?1/2D+1+4/9D+2+1/3(D+2)2)x2e3x.(64) Using (41), we get
y=e3x(1/18D+2?1/2D+4+4/9D+5+1/3(D+5)2)x2.(65) From (33) we have
(D?a)?1x2=?1
a[1+D a+(D a)2+...]x2=?1a [x2+2x a+2a]
(D?a)?2x2=?1
a[1+D a+(D a)2+...]
2
x2=
1
a2[x2+4x a+6a2
].(66)
Using this, (65) yields
y=e3x[136(x2?x+12)?18(x2?12x+18)+445(x2?25x+225)+175(x2?45x+625)] =e3x(1200x2?232000x+34340000)
.(67)
8.2The Laplace Transform and its Properties
We consider the space Γ of piecewise continuous functions. We define a mapping ￡:Γ→Γ:f →￡[f ]￡ by
￡[f ](s )=∫0∞
e ?sx
f (x )dx .
(1)
￡[f ] Is called the Laplace transform of f . We will adopt the convention to denote an ordinary function by a lower-case letter, e.g., h (x ) and its transform by an upper-case letter; thus ￡[h ](s )=H (s ).
From the definition, it is clear that the Laplace transform is linear:
￡[af (x )+bg (x )](s )=a ￡[f (x )]+b ￡[g (x )].
(1)
Let us calculate some examples.
￡[k ](s )=∫0
∞e
?sx
k dx =ke ?sx ?s |0∞=k s
.
(1)
￡[e ax
](s )=∫0
∞
e
?sx
e ax
dx =∫0
∞
e
?(s ?a )x
dx =e ?(s ?a )x ?(s ?a )|0∞=1s ?a
.
(2)
￡[e iax
](s )=∫0
∞
e
?sx
e iax
dx =∫0
∞
e (ia ?s )x dx
=e (ia ?s )x ia ?s |0∞
==(C ax +iS ax )e ?sx ia ?s =e (ia ?s )x ia ?s
|0
∞=(C ax +iS ax )(?ia ?s )e ?sx
(ia ?s )(?ia ?s )|0∞=?iaC ax +aS ax ?sC ax +isS ax s 2+a 2
e ?sx |0
∞=(aS ax ?sC ax )+i (sS ax ?aC ax )s 2+a
2e ?sx |0
∞.(3)
where C ax =cos (ax ) and S ax =sin (ax ). Since e iax =C ax +iS ax , it then follows that
￡[cos (ax )](s )=
(aS ax ?sC ax )
s 2+a 2e ?sx |0∞=
s
s 2+a 2￡[sin (ax )](s )=(sS ax ?aC ax )s 2+a 2e ?sx |0
∞=a
s 2+a 2
.(4)
The Laplace transform of the derivative of a function can easily be obtained, using
integration by parts:
￡[f '(x )](s )=∫0∞
e ?sx
f '(x )dx =e
?sx
f (x )|0
∞?∫0
∞
f (x )(?s )e ?sx dx
=e
?sx
f (x )|0
∞+s ∫0
∞f (x )e
?sx
dx =?f (0)+s ￡[f (x )](s )=s ￡[f (x )](s )?f (0)
.
(5)
This can be generalized for the derivative of any order:
￡[f
(n )
(x )](s )=s n
￡[f (x )]?∑k =0
n ?1
s k
f
(n ?k ?1)
(0).
(6)
￡[xf (x )](s )=∫0∞?d ds (e ?sx )f (x )dx =?d ds ∫0∞
e ?sx
f (x )dx =?d
ds ￡[f (x )](s ).
(7)
This can be generalized to
￡[x n
f (x )](s )=(?1)n
d n
ds
n ￡[f (x )](s ).
(8)
￡[e ax
f (x )](s )=∫0
∞
e
?sx
e ax
f (x )dx =∫0
∞
e ?(s ?a )x
f (x )dx =￡[f (x )](s ?a ).
(9)
￡[x n
e ax
](s )=(?1)n
d n ds n ￡[
e ax ](s )=?(?1)n d n
ds n 1s ?a =n !(s ?a )
n +1
.(10)
￡[x ?1f (x )](s )=∫0
∞e ?sx x ?1f (x )dx =?∫0
∞
dx ∫0
s
ds f (x )e ?sx
=?∫0
s
ds ∫0
∞
dx f (x )e
?sx
=?∫0
s
￡[f ](s )ds
.
(11)
￡[∫
x
f (t )dt ](s )=1
s
￡[f ](s ).
(12)
The following table shows some properties of Laplace transforms
f (x )
F (s )
af (x )+bg (x )aF (s )+bG (s )af (ax )F (s /a )e ax f (x )
F (s ?a )
The following table lists some special Laplace transforms
If the function F is the Laplace transform of the function f , i.e., if ￡[f ], then f is the inverse Laplace transform of F , which we denote by f =￡?1[F ].The inverse Laplace transform is a linear operator:
￡?1[a 1F 1+a 2F 2]=a 1￡?1[F 1]+a 2￡?1[F 2].
(13)
There is no canonical recipe to compute inverse Laplace transforms. Typically, Laplace transforms have the form of quotients of functions. In many applications, their
denominators are algebraic. Thus one would try to bring a given function F into that form. Generically we have to deal with ratios of the form p (s )/q (s ). where both functions are polynomials and the degree of the numerator is less than that of the
denominator (if not, then one reexpresses the numerator as p (s )=n q (s )+r (s ) and then consider the ratio r (s )/q (s )).
Partial fraction decomposition is a helpful. The denominator will be a product of linear and quadratic terms (in case of complex conjugate roots). Each term in p (s ) of the form (s ?a )k yields a sum
∑j =1
k
A j
(s ?a )
j
=A 1(s ?a )+A 2(s ?a )2+...+A k
(s ?a )
k
.(14)
Similarly, each term in p (s ) of the form (s 2+bs +c )k yields a sum
∑j =1
k
B j s +
C j (s 2+bs +c )j =B 1s +C 1(s 2+bs +c )+B 2s +C 2(s 2+bs +c )2+...+B k s +C k
(s 2+bs +c )k .
(15)
Example:
￡
?1
[
s +1s 2?16]=￡?1[s s 2?16]+￡
?1[1s 2?16]
=cosh (4x )+￡?1[144s 2?42
]
=cosh (4x )+1
4
sinh (4x ).
(16)
Example:
￡?1
[
s s ?4s +13]=￡?1[s (s ?2)+9]=￡?1[(s ?2)+2(s ?2)+9]
=￡?1[
s ?2(s ?2)2+9]+￡?1
[2(s ?2)2
+9
]=e 2x cos (3x )+￡?1[233(s ?2)2
+9]
=e 2x
cos (3x )+23
e 2x sin (3x ).(17)
Example:
￡
?1
[
s +3
(s ?2)(s +1)
]
.
(18)
Using a partial fraction expansion, this can be rewritten as
53￡?1[1s ?2]
?23￡?1[1s +1]
=53e 2x ?23
e ?x .(19)
Example:
￡
?1
[
1
(s 2+1)(s 2
+4s +8)
]
.(20)
Using a partial fraction expansion, this can be rewritten as
765￡?1[1s 2+1]?465￡?1[s s 2+1
]
+￡?1
[
465s +7
65
s 2
+4s +8
]
.(21)
Note that s 2+4s +8=(s +2)2+4. Reorganizing the numerator in the last, we get