Chapter 2
Retrospection of Di↵erential
Equations
2.1
Introduction
In this chapter, we will revisited the basic concepts and techniques of solving ordinary di↵erential equations that are often encountered in structural dynamics. Therefore, only few essential categories are
only considered in this chapter to maintain its brevity. For complete knowledge reader may refer to any
standard or dedicate di↵erential equation book. This chapter contains the essential concepts which will
be used in the later chapters.
2.2
First order di↵erential equations
The most general forms of a first order ordinary di↵erential equation (ODE) can be expressed as:
F
✓
dy
, y, x
dx
◆
dy
= F1 (y, x)
dx
= 0,
(2.1)
However, in all cases, conversion between the equations are not feasible.
2.2.1
Separable ODE
While the function of Equation 3.1 is separable in the product of functions of only x and y, then the
di↵erential equation is separable. A separabele di↵erential equation can be written as:
dy
= f (x)g(y)
dx
!
dy
= f (x)dx
g(y)
(2.2)
Integrating the left and right hand side of both the equation we can get:
Z
dy
=
g(y)
Z
15
f (x)dx + c
(2.3)
Example
Q. Find the solution of the following 1st order ODE:
dy
x2
=
dx
1 y2
(2.4)
A. First we can rearrange and then integrate in both the side as:
✓
◆
x2
y3
2
2
(1 y )dy = x dx
!
+ y
=C
3
3
(2.5)
The value of C can be calculated from the boundary condition.
2.2.2
Standard Linearized form
The most general form of a first order ODE can be expressed as:
dy
= p(x)y + Q(x)
dx
After setting Q(x) = 0 in 2.6 it becomes:
Z
Z
dy
dy
= p(x)y !
= p(x)dx
dx
y
! ln y =
Z
(2.6)
p(x)dx + c
R
y = c(x)e p(x)dx
(2.7)
further it can be di↵erentiated as:
R
dy
dc(x) R p(x)dx
dc(x) R p(x)dx
=
e
+ c(x)p(x)e p(x)dx =
e
+ p(x)y
dx
dx
dx
(2.8)
Substituting Eq.2.8 into Eq.2.6, we can obtain:
R
dc(x)
= Q(x)e p(x)dx
dx
(2.9)
The solution of c(x) and subsequently y(x) can be obtained by integrating Eq.2.9 as:
✓Z
◆
Z
R
R
R
c(x) = Q(x)e p(x)dx dx + C
! y = e p(x)dx
Q(x)e p(x)dx dx + C
(2.10)
Many ODEs can be expressed in terms of standard linear forms.
Example
2.2.3
Solution by Taylor series expansion
If the boundary of initial conditions are given then the solution can be obtained by Taylor series expansion. Lets solve Eq.3.1 using Taylor series expansion. In that case the higher order derivatives with
respect to x become:
dy
= F1 (x, y)
dx
d2 y
@F1 (x, y) @F1 (x, y) dy
@F1 (x, y) @F1 (x, y)
=
+
=
+
F1 = F2 (x, y)
2
dx
@x
@y
dx
@x
@y
dn y
@Fn 1 (x, y) @Fn 1 (x, y) dy
@Fn 1 (x, y) @Fn 1 (x, y)
=
+
=
+
Fn 1 = Fn (x, y)
dxn
@x
@y
dx
@x
@y
16
(2.11)
Using Taylor series expansion about a point x0 , we can obtain:
y = y0 + F1 (x0 , y0 )(x
x0 ) + F2 (x0 , y0 )
(x
x0 ) 2
+ ...
2!
(2.12)
Now, the initial condition is at x = 0, y = c. Then the approximate solution is:
y(x) = c + F1 (0, c)x + F2 (0, c)
x2
+ ....
2
(2.13)
For infinite number of terms, the solution approaches to the exact solution. This method can give fast
first approximation of the result.
Several other forms of first order ODEs can be found in the book entitled ”Theory of Di↵erential Equations in Engineering and Mechanics” by K.T. Chau.
2.3
Second order ODE
The most general case of 2nd order ODE can be considered as:
d2 y
=f
dx2
✓
y, x,
dy
dx
◆
! y 00 + p(x)y 0 + q(x)y = g(x)
(2.14)
Due to the presence of the non-homogeneous part g(x) in Eq.2.14, the solution need to be written as the
summation of homogeneous solution and particular solution as:
y(x) = yh (x) + yp (x)
(2.15)
yh is the solution of the homogeneous part of Eq.2.14, i.e., while g(x) = 0 and yp satisfies Eq.2.14 as it
is. The proof of the separate solution can be given as follows:
y 00 + p(x)y 0 + q(x)y = (yh00 + yp00 ) + p(x)(yh0 + yp0 ) + q(x)(yh + yp ) =
yp00 + p(x)yp0 + q(x)yp + yh00 + p(x)yh0 + q(x)yh = g(x) + 0 = g(x)
(2.16)
For a homogeneous di↵erential equation, we can write the solution as:
y = C1 y1 (x) + C2 y2 (x)
(2.17)
The boundary or initial conditions are as follows:
y(x0 ) = y0 , y 0 (x0 ) = y⇤
(2.18)
Substituting, Eq.2.18 into Eq.2.17 yields:
9
C y (x ) + C y (x ) = y =
1 1
0
2 2
0
0
C1 y10 (x0 ) + C2 y20 (x0 ) = y⇤;
! C1 =
y0 y20 (x0 ) y ⇤ y2 (x0 )
y ⇤ y1 (x0 ) y0 y10 (x0 )
, C2 =
0
0
y1 (x0 )y2 (x0 ) y1 (x0 )y2 (x0 )
y1 (x0 )y20 (x0 ) y10 (x0 )y2 (x0 )
(2.19)
The denominator is called Wronskian. For a solvable system, Wornskian must be non-zero.
17
2.3.1
Homogeneous ODE with constant Coefficients
The simplest case of second order ODE is a ODE with constant coefficient and non-homogeneous term
0 as shown below.
y 00 + py 0 + qy = 0
(2.20)
An exponential function are assumed as solution:
y = erx
(2.21)
substituting Eq.2.21 into Eq.2.20 we can obtain:
2
! r1,2 =
r + pr + q = 0
where,
= p2
Case-1
>0
4q. Based on the value of
p±
2
p
(2.22)
three di↵erent case can be possible.
. In this case, r1 and r2 are di↵erent value,
r1 =
p+
2
p
,
r2 =
p
p
2
(2.23)
and the general solution is:
y = C 1 e r1 x + C 2 e r 2 x
(2.24)
These C1 and C2 can be computed form the boundary condition.
Case-2
=0
. In this case, r1 = r2 =
p
2 and the general solution is:
y = (C1 + C2 x)e
p
2x
(2.25)
These C1 and C2 can be computed form the boundary condition.
Case-3
<0
. In this case, r1 and r2 are becomes complex conjugate. Thus, these can be written as:
p
p
p+
p
r1 =
= ↵ + i , r2 =
=↵ i
2
2
(2.26)
and the general solution is:
y = D1 e(↵+i )x + D2 e(↵ i )x = e↵x (D1 ei x + D2 e i x ) = e↵x (C1 cos x + C2 sin x)
These C1 and C2 can be computed form the boundary condition.
18
(2.27)
2.3.2
Non-homogeneous ODE with constant coefficient
The non-homogeneous part of Eq.2.14 can be solved adopting the following process. Lets assume g(x)
can be expressed as a product of a polynomial and exponential function as:
g(x) = e x Pm (x),
Pm (x) = a0 xm + a1 xm 1 + a2 xm 2 + a3 xm 3 .... + am
(2.28)
As the exponential terms cannot be vanished by di↵erentiation; thus, we should assume the solution
proportional with the same exponential function. The assumed particular solution is:
y = Q(x)e x
! y 0 = Q0 e x + Qe x
! y 00 = Q00 e x + 2 Q0 e x +
2
Qe x
(2.29)
Substituting, Eq.2.29 into Eq.2.28, we can obtain:
Q00 + 2 Q0 +
2
Q + p (Q0 + Q) + qQ = Pm
! Q00 + (2 + p)Q0 + ( 2 + p + q)Q = Pm
(2.30)
We can see, the coefficient of Q and Q0 in Eq.2.30 is the characteristics equation of homogeneous case.
This arises three cases based on the value of .
Case-1:
is not a characteristic root
That condition implies neither of the coefficient of Q and Q0 in Eq.2.28 will be cancelled out. Thus,
highest power of Q in right hand side must be same as polynomial Pm . Therefore, we can consider:
Q(x) = Qm (x) = b0 xm + b1 xm 1 + ... + bm ;
yp = Qm (x)e x
(2.31)
Substituting Eq.2.31 into Eq.2.14 we can obtain:
[m(m
1)b0 xm 2 + ... + 2bm 2 ] + (2 + p)[mb0 xm 1 + ... + bm 1 ]+
( 2 + p + q)[b0 xm + b1 xm 1 + ... + bm ] = a0 xm + a1 xm 1 + ... + am
(2.32)
Matching the coefficients for di↵erent powers of x, we can found m equations in ascending order as:
( 2 + p + q)b0 = a0
(2 + p)mb0 + ( 2 + p + q)b1 = a1
(2.33)
:::
2bm 2 + (2 + p)bm 1 + ( 2 + p + q)bm = am
Case-2:
If
is a simple characteristic root
matches with characteristics root of the homogeneous di↵erential equation, we have:
2
2 + p 6= 0
(2.34)
Q00 (x) + (2 + p)Q0 (x) = Pm (x)
(2.35)
+ p + q = 0,
Thus, Eq.2.30 can be reduced to:
As the highest order term of the power series expansion will come from Q0 (x) which is xm 1 ; thus, to
perform the matching similar to the case-1, we need to assume:
Q(x) = xQm (x)
!
19
yp = xQm (x)e x
(2.36)
Case-3:
If
is a simple characteristic root
matches with the double or repetitive characteristics roots of the homogeneous equation then we
have:
2
+ p + q = 0,
2 +p=0
(2.37)
Thus, Eq.2.30 can be reduced to:
Q00 (x) = Pm (x)
(2.38)
As the highest order term of the power series expansion will come from Q00 (x) which is xm 2 ; thus, to
perform the matching similar to the case-1, we need to assume:
Q(x) = x2 Qm (x)
2.4
yp = x2 Qm (x)e x
!
(2.39)
Higher order ODE
The general form of linear ODE of nth order can be written as:
L[y] =
dn y
dn 1 y
dy
+
P
(x)
+ ... + Pn 1 (x)
+ Pn (x)y = g(x)
1
n
n
1
dx
dx
dx
(2.40)
and corresponding boundary conditions are: y(x0 ) = y0 , y 0 (x0 ) = y00 , y 00 (x0 ) = y000 , ..., and y (n 1) (x0 ) =
(n 1)
y0
.
The solution of y can be assumed as:
y(x) = c1 y1 (x) + c2 y2 (x) + ... + cn yn (x)
(2.41)
Substituting Eq.2.41 into Eq.2.40 we can obtain:
⇥
⇤
c1 y1n + p1 (x)y1n 1 + ... + Pn 1 (x)y1 + pn (x)y1
⇥
⇤
+c2 y2n + p1 (x)y2n 1 + ... + Pn 1 (x)y2 + pn (x)y2
(2.42)
...
⇥
⇤
+cn ynn + p1 (x)ynn 1 + ... + Pn 1 (x)yn + pn (x)yn = 0
Substituting the assumed solutions in boundary conditions we can get:
c1 y1 (x0 )+...+Cn yn (x0 ) = y0 ,
c1 y10 (x0 )+...+Cn yn0 (x0 ) = y00 ,
...
c1 y1n 1 (x0 )+...+Cn ynn 1 (x0 ) = y0n 1
(2.43)
Clearly, for the system to have unique solution, we must have the Wornskian to be non-zero, which
implies:
W (y1 , y2 , ..., yn )(x0 ) =
y1 (x0 )
y2 (x0 )
...
yn (x0 )
y10 (x0 )
y20 (x0 )
...
..
.
..
.
...
yn0 (x0 )
..
.
(n 1)
y1
(x0 )
(n 1)
y2
(x0 )
...
(n 1)
yn
=0
(2.44)
(x0 )
For constant coefficients and homogeneous ODE, Eq.2.40 can be reduced and its solution can be written
in exponential function as:
y (n) + P1 y (n 1) + ... + Pn 1 y 0 + Pn y = 0
20
! y = erx
(2.45)
where, r can be computed from the characteristic equation as:
Z(r) = rn + P1 rn 1 + ... + Pn 1 r + Pn = 0
(2.46)
If all the roots are di↵erent then the solution can be written as:
Z(r) = (r
r1 )(r
r2 )(r
r3 )...(r
rn ) = 0
! y = C1 er1 x + C2 er2 x + ... + Cn ern x
(2.47)
If first k number of roots are equal then:
Z(r) = (r r1 )k (r rk+1 )(r rk+2 )...(r rn ) = 0
! y = C0 + C1 x + ... + CK 1 xk 1 er1 x +Ck+1 erk+1 x +...+Cn ern x
(2.48)
2.4.1
D-operator method
D-operator method is one of the most popular method to solve Non-homogeneous ODE. In this method,
di↵erentiation (D) is defined as:
D=
d
d2
dn
, D2 = 2 , ...Dn = n
dx
dx
dx
(2.49)
The equation of 2.40 can be expressed in the form of D-operator as:
L(D)y(x) = g(x) ! y(x) =
g(x)
,
L(D)
L(D) = Dn + P1 (x)Dn 1 + P2 (x)Dn 2 + ... + Pn 1 (x)D + Pn
(2.50)
We can notice that equation 2.50 is an algebraic polynomial of variable D.
Properties of L(D) operator
1. Operator L(D) is Linear, i.e. L(D)[C1 y1 (x) + C2 y2 (x)] = C1 L(D)y1 (x) + C2 L(D)y2 (x) .
2. Dm Dn = Dm+n .
3. Commutative law of addition: L1 (D) + L2 (D) = L2 (D) + L1 (D)
4. Associate law of addition: [L1 (D) + L2 (D)] + L3 (D) = L1 (D) + [L2 (D) + L3 (D)].
5. Commutative law of multiplication: Lc1 (D).Lc2 (D) = Lc2 (D).Lc1 (D).
6. Associative law of multiplication: Lc1 (D)[Lc2 (D)Lc3 (D)] = [Lc1 (D)Lc2 (D)]Lc3 (D)
7. Distributive law of multiplication over addition: Lc1 (D)[Lc2 (D) + Lc3 (D)] = Lc1 (D)Lc2 (D) +
Lc1 (D)Lc3 (D).
where, L1 (D), L2 (D), ... represents the di↵erent polynomial of D. Lc (D) denotes the polynomial
having constant coefficients.
21
Example
with Exponential functions: We can substitute D = a, where a is the coefficient of the x in the power
of exponential (eax ).
Q1. Solve a) (D + 3)2 y = 50e2x and b) (D
2)2 y = 50e2x .
2x
50e
2x
A1. Solution of part a) is very straight forward: yp = (D+3)
, after substitution of D = 2;
2 = 2e
and the complete solution is y = (A + Bx)e 3x + 2e2x .
Whereas, solution for part b) is bit complicated as the root of the L(D) is same with the power of
the exponential series, thus substitution of D = 2 leads to infinity. The particular integral (yp ), could
2
x
2x
be computed as: yp = D
= 25x2 e2x and complete solution is y = (A + Bx + 25x2 )e2x .
2 50e
with sine and cosine function: We can substitute, D2 =
a2 , where a is the coefficient of the
x as cos(ax) or sin(ax).
Q1. Solve (D2 + 3D + 2)y = cos 2x.
2x
cos 2x
A1. Particular solution can be written as: yp = (D2cos
+3D+2) = (3D 2) , as we can only substitute for
D2 not for D.
3D+2
3D+2
1
After rewriting the solution, we can get yp = 9D
2 4 cos 2x =
40 cos 2x = 20 (3 sin 2x
cos 2x).
with polynomial of x Use binomial theorem to simplify.
Q1. Solve (D2
4D + 3)y = x3
A1. The particular solution can be written as:
✓
◆
x2
1
1
1
=
x3 =
(D2 4D + 3)
2 1 D 3 D
◆
1
D D2
x3
4
26
80
(1 +
+
+) x3 =
+ x2 + x +
3
3
9
3
3
9
27
yp =
1
2
✓
2
3
(1 + D + D + D + ...)
2.5
System of ODEs
2.5.1
Reduction of Higher order ODE to a system of ODEs
(2.51)
A higher order ODE can be reduced to a system of ODEs in the following manner. Lets assume the
higher order given in Eq.2.40 need to be expressed as a system of 1st order ODEs. In that case, we can
add, n
1 number of auxiliary equations as:
y (n) =
P1 (x)y (n 1)
...
y (n 1) = y (n 1)
Pn 1 (x)y 0
Pn (x)y + g(x)
(2.52)
(2.53)
..
.
(2.54)
y 00 = y 00
(2.55)
y0 = y0
(2.56)
22
The set of above equations can be expressed in matrix form as:
8
9 2
9 8
9
38
>
>
P1 (x) ...
Pn 2 (x)
Pn 1 (x)
Pn (x) >
g(x)>
> y (n) >
>
>y (n 1) >
>
>
>
>
>
>
>
>
>
>
> 6
7>
> >
>
>
.. >
> (n 1) >
> 6
7>
> .. >
> >
>
>
>
>
>
>
>
>
y
1
...
0
0
0
7
>
>
6
>
>
>
>
.
.
>
>
>
>
<
= 6
7>
<
= >
<
=
..
..
..
..
6 ..
7
00
=
+
6
7
...
.
.
.
> . >
> 6 .
7>
> y >
> >
> 0 >
>
>
>
>
6
7>
>
>
>
>
>
>
>
00
0
> y >
> 6 0
7>
> y >
> >
> 0 >
>
...
1
0
0
>
>
>
>
>
>
>
>
>
5
4
>
>
>
> >
>
>
>
>
>
>
: y0 >
;
:
;
:
0
...
0
1
0
y
0 ;
| {z } |
{z
} | {z } | {z }
X0
A
2.5.2
X
! X 0 = AX + b
b
(2.57)
Constant coefficient equations
The solution of Eq.2.57 can be written as:
XI = AX
After substituting the assumed solution of X =
e t=A e t
which implies that
and
e t into Equation 2.58
!
A )=0
1 and the eigenvectors are . Thus the solution is:
8
9
2
< 0.4472=
0.4472
1
3t
t
= A1
e + A2
e =4
:x ;
:0.8944;
: 0.8944 ;
0.8944
2
(2.59)
2
The eigenvalue solution of the matrix is 3 and
2.6
(
are the eigenvectors and eigenfrequencies of matrix A. For an example
8 9 2
38 9
<ẋ =
1 1 < x1 =
1
5
=4
(2.60)
:ẋ ;
4 1 :x ;
2
8 9
<x =
(2.58)
8
9
<0.4472=
2
63t
6
4
3
0.4472
5e 0
0.8944
3
0 72
3
7
5
t 4 A1 5
A2
(2.61)
Fourier analysis
Fourier analysis on a single degree of freedom system can be performed in the following manner. Lets
assume the equation of the system is:
M ẍ + C ẋ + Kx = f (t)
(2.62)
The solution of above equation can be expressed as:
N
x(t) =
a0 X
+
(ai sin(i!t) + bi cos(i!t))
2
i
(2.63)
Further the time derivaties can be written as:
ẋ = !
N
X
( iai cos(i!t) + ibi sin(i!t))
i
ẍ =
!2
N
X
(2.64)
i (ai sin(i!t) + bi cos(i!t))
i
23
Substituting Eq2.63 into 2.62 we can get:
M !2
N
X
i (ai sin(i!t) + bi cos(i!t)) + C!
i
N
X
( iai cos(i!t) + ibi sin(i!t))
i
N
N
X
a0
f0 X
+K
+K
(ai sin(i!t) + bi cos(i!t)) =
+
(fsi sin(i!t) + fci cos(i!t))
2
2
i
i
2.7
PDE
2.8
weighted residual
2.8.1
Galerkine
2.9
Di↵erential equation with Non-constant coefficients
2.9.1
Series expansion
2.10
Conclusion
24
(2.65)