13

B. Differential Equations A differential equation is an equation of the form 0]);(),.....(),(),(;[ )( =ʹ′ʹ′ʹ′ αtxtxtxtxtF n

where dtdxtx =ʹ′ )( ,

2

2)(dtxdtx =ʹ′ʹ′ , …….,

n

nn

dtxdtx =)()( .

A differential equation describes the behaviour of a state variable over continuous time which explains the use of derivatives (the time derivative measure the marginal changes at each instant of time). In general the above equation can be written explicitly in a more commonly used form ));(),.....,(),(),(;()( )1()( αtxtxtxtxtftx nn −ʹ′ʹ′ʹ′= The order of a differential equation is the order of the highest derivative that appears in the equation. A differential equation is said to be linear if f is linear function of the state variable and its derivatives. A differential equation is said to be autonomous if time does not enter directly in the function f as an argument. We shall be dealing only with differential equations of first order. It can be shown that any differential equation of higher order can be transformed into a system of first order differential equations by introducing new variables. Example: Consider the second-order linear differential equation )()()( tbxtxatx +ʹ′=ʹ′ʹ′ (10) Define a new variable )()( txty ʹ′= . Then )()( txty ʹ′ʹ′=ʹ′ . Putting these values in (10), we get the system of first order differential equations involving two variables x and y:

⎭⎬⎫

+=ʹ′

=ʹ′

)()()()()(

tbxtaytytytx

(10ʹ′)

The theory of differential equations is in many ways similar to theory of difference equations. Therefore in some cases we shall just state the results, leaving the derivations as exercises. B.1 Solving a Linear First Order Differential Equation (a) AUTONOMOUS EQUATIONS: Consider a linear autonomous equation of the following form:

btaxdttdx

+= )()( (11)

As in the case of difference equation, the generation solution to (11) is given by a sum of two terms cx and px , where cx solves the homogeneous part of (11) (complementary

function), and px is any particular solution to (11) (called a particular integral).First let us solve for the homogeneous component:

axdtdx

= (11ʹ′)

Equation (11ʹ′) can be written as

14

adtdxx

=1

Integrating both sides and simplifying, we can write the solution to equation (11ʹ′) as atctx exp)( =

where c is an arbitrary constant. Thus the complementary function is given by at

c cx exp= . For the particular integral we again proceed with a trial solution that ktx =)( (a

constant). Then 0=dtdx

. Putting these values in (11) we get

0=+ bak

abk −

=⇒ .

Hence the particular integral is given by abx p

−= .

Thus the general solution to the autonomous differential equation (11) is

abctx at )(exp)( −

+=

where c is an arbitrary constant whose value is to be determined from the given initial or any other boundary condition. (b) NON-AUTONOMOUS EQUATIONS: Let us consider a non-autonomous first order differential equation of the form:

)()()( tbtaxdttdx

+= (12)

The complementary function is once again given by atc ctx exp)( = . So we have to find

only a particular solution.

Suppose )(tb takes a particular functional form given by tbBtb exp )( = . Then the differential equation given in (12) takes the form:

tbBtaxdttdx exp )()(

+= (12ʹ′)

Let us now proceed with a trial solution, tbktx exp )( = . Then tbkbdtdx exp = .

Substituting these values to the above equation and simplifying, abBk−

= . Thus we get

a particular solution to (12ʹ′) as tbp ab

Btx exp )(−

= .

Hence the general solution to (12ʹ′) is

tbatabBctx expexp)(−

+= , ab ≠ ; c an arbitrary constant.

When ab = , the differential equation becomes

15

taBtxadttdx exp )( )(

+= (12ʺ″)

So the complementary function now is tac ctx exp)( = .

In order to find a particular solution, let us proceed with a trial solution takttx exp )( = .

Then tata ktakdtdx exp exp += . Putting these values back into the above equation and

simplifying, Bk = . Thus the particular integral is given by tap Bttx exp)( = .

Therefore the general solution to (12ʺ″) is

taBtctx exp)()( += , c an arbitrary constant. However it may not always be possible to find a suitable trial solution for a non-autonomous difference equation even when the functional form of )(tb is known. There is a more general method of solving any non-autonomous first order differential equation which requires knowledge of two concepts: (i) exact differential equation, (ii) integrating factor. Let us consider a first order non-autonomous differential equation of the form

0),(),( =+ dttxNdxtxM (13)

i.e., ),(),(txMtxN

dtdx

−=

This differential equation is said to be an exact differential equation if there exists a

function ),( txF such that ),( txMxF=

∂

∂ and ),( txN

tF=

∂

∂. The function ),( txF is

called the primitive function.

Remark. Since we know that for any function ),( txF , xtF

txF

∂∂

∂=

∂∂

∂ 22

, an easy criterion for

checking whether a given differential equation of the general form

0),(),( =+ dttxNdxtxM is exact or not is to verify if xN

tM

∂

∂=

∂

∂. If this holds then the

given differential equation is exact, otherwise not. It is easy to see that if a differential equation is exact and if we can identify the primitive function ),( txF , then such an exact differential equation can be readily solved in the following way:

0),(),( =+ dttxNdxtxM

0=∂

∂+

∂

∂⇒ dt

tFdx

xF

0),( =⇒ txdF Integrating, we get the general solution to this equation as ctxF =),( where c is an arbitrary constant (constant of integration).

16

Thus if we are given an exact difference equation to solve, all that we have to do is to find the corresponding primitive function ),( txF .

How to find the primitive function: we know that ),( txMxF=

∂

∂. Hence the function

),( txF must contain a partial integral of ),( txM . Let us start with a trial primitive function of the form ∫ +∂= )(),(),( txtxMtxF φ (14)

[Note that in (14), ),( txM is integrated partially with respect to x, i.e., in integrating the function we treat t as given.] If (14) is indeed the primitive function corresponding to (13), then its partial derivative with respect to t should be equal to ),( txN . Using this relation we can identify the )(tφ function and consequently identify the ),( txF function. Once ),( txF is found we can equate it with an arbitrary constant to get the general solution to (13). Example: Consider a differential equation of the form: 0)22()43( 22 =+++ dttxdxxtx (13ʹ′)

Here )43(),( 2 xtxtxM += and )22(),( 2 txtxN += . The first step is to check whether (13ʹ′) is an exact differential equation or not.

From the equation we find, xtM 4=∂

∂ and x

xN 4=∂

∂. Thus

xN

tM

∂

∂=

∂

∂ and the differential

equation is exact. So in order to solve (13ʹ′) we have to find the corresponding primitive function ),( txF . Let us proceed with the trial function: ∫ +∂= )(),(),( txtxMtxF φ

i.e., ∫ +∂+= )()43(),( 2 txxtxtxF φ

)()2(),( 23 ttxxtxF φ++=⇒

From above, dtdx

tF φ

+=∂

∂ 22 .

If the above trail function is indeed the primitive function corresponding to (13ʹ′), then

txdtdx 222 22 +=+φ

i.e., tdtd 2=φ

.

Integrating, 12)( ctt +=φ , where 1c is an arbitrary constant.

Thus we have derived the primitive function as 1

223 2),( cttxxtxF +++= . The general solution to (13ʹ′) is then given by, 2),( ctxF = , or

21223 2 ccttxx =+++ .

Combining the two arbitrary constants 1c and 2c we may write this solution as

cttxx =++ 223 2

17

Remark. In order to find the primitive function of the exact differential equation given in

(13), we can proceed in an alternative way. Since ),( txNtF=

∂

∂, the function ),( txF

must contain a partial integral of ),( txN as well. So we can write the trial solution as

∫ +∂= )(),(),( xttxNtxF ψ (14ʹ′)

Note once again that ),( txN is to integrated partially with respect to t, i.e., in integrating the function we have to treat x as given. If this is indeed the primitive function corresponding to (13), then its partial derivative with respect to x should be equal to ),( txM . Using this relation we can identify the

)(xψ function and consequently identify the ),( txF function. Both this procedures should give us the same primitive function. (Verify that for the above example, the alternative procedure described here will generate the same functional form of the primitive function.)

If the given differential equation is not exact, then it is possible to transform it into an exact differential equation by multiplying both sides with a common factor. This common factor that transforms a non-exact differential equation to an exact one is called an integrating factor. We shall only discuss the procedure for solving a linear non-autonomous first order differential equation which is not exact. Consider a first order linear differential equation of the form

)(tbaxdtdx

+= (15)

Writing this in the 0),(),( =+ dttxNdxtxM format, we get 0)}({ =−−+ dttbaxdx (15ʹ′) Thus 1),( =txM and )}({),( tbaxtxN +−= . Obviously, this differential equation is not exact as long as 0≠a . Suppose I is the integrating factor (which is yet unknown). Then multiplying both sides of (15ʹ′) by I should convert it to an exact differential equation. Hence the following is an exact differential equation:

0)}({ =−−+ dttbaxIdxI (15ʺ″) The new M and N function corresponding to (15ʺ″) are given by ItxM =),( and

)}({),( tbaxItxN +−= . Since (15ʺ″) is an exact differential equation (by construction),

xN

tM

∂

∂=

∂

∂. That is, aI

dtdI

−= . Integrating, we can identify the integrating factor as,

) ( log nintegratio ofconstant thebeingKKatI +−=

or, )exp( exp Kat AAI == − . Any function of the above form will act as the integrating factor for (15ʹ′), whatever be the value of A. Hence for simplicity we set 1=A . Thus using the integrating factor

atI −= exp , we can write (15) as an exact differential equation in the following way:

0)}({expexp =+− −− dttbaxdx atat (15ʹ′ʺ″) To find the primitive function corresponding to (15ʹ′ʺ″), we proceed with the trial function

18

)(exp)(exp)(),( txtxtxMtxF atat φφφ +=+∂=+∂= −−∫∫

Hence dtdax

tF at φ

+−=∂

∂ −exp . Again from (15ʹ′ʺ″) )}({exp),( tbaxtxN at +−= − . Equating

the two,

)(exp tbdtd at−−=φ

dttbat∫ −−=⇒ )(expφ .

Thus the primitive function is given by dttbxtxF atat ∫ −− −= )(expexp),( .

Hence the general solution to a linear non-autonomous difference equation of the form given in (15) is cdttbx atat =− ∫ −− )(expexp

Rearranging, we can write the general solution as ( )∫ −+= dttbctx atat )(expexp)( (16)

where c is an arbitrary constant whose value is to be determined from the given boundary condition. B.2 Autonomous First Order Differential Equation: Steady States and Stability Consider an autonomous first order differential equation of the form ));(()( αtxftx =ʹ′ (17) The steady state and stability of this autonomous differential equation is defines as follows. Steady state: A point Xx ∈ is a steady state of the dynamic system represented by the differential equation (17) if it satisfies the following condition: 0);( =αxf . In other words, the steady state can be found by setting 0)( =ʹ′ tx . Stability: A dynamical system with a steady state Xx ∈ is said to be asymptotically stable if xtx

t=

∞→)(lim .

Example: Let us consider a linear difference equation of the form

btaxdttdx

+= )()(

We have seen that this function has a general solution abctx at )(exp)( −

+= .

Setting 0)( =ʹ′ tx , i.e., 0=dtdx

, we can derive the steady state as abx −= , which is

nothing but the particular integral as derived earlier. Hence the general solution to the above equation can be written as

xctx at+= exp)( . Thus no matter what the value of c is (i.e., no matter what the initial condition is), the system will approach its steady state as t tends to infinity if and only if 0<a . Thus the stability condition here is given by 0<a .

19

Remark. By comparing this example with corresponding difference equation, one can readily see the similarity between the two cases. However, one should note that in the case of difference equation stability depends on whether the parameter a is greater than unity or less than unity. On the other hand, in the case of differential equation stability depends on whether the parameter a is positive or negative. B.3 Solving a System of First Order Differential Equations (Linear and Autonomous) (a) n dimensional system: Consider a system of linear and autonomous differential equations of the form bxx +=ʹ′ )()( tAt (18) where A is an nn× matrix of constant coefficients. As in the case of system of difference equations, we first identify the particular solution

(integral) as xbx =−= − )( 1Atp (the steady state vector). Next, to find the complementary function, let us consider the homogenous system )()( tAt xx =ʹ′ (18ʹ′) If the n dimensional square matrix A has n distinct roots then we can use the corresponding matrix of the eigenvectors M to diagonalize A and the transform the homogeneous system given by (18ʹ′) to a simpler system of the form: )()( tt yy Λ=ʹ′ (18ʺ″) where Λ is a diagonal matrix with all the distinct eigenvalues of A as its diagonal elements. Equation (18ʺ″) represents a system of n independent linear and autonomous differential equations of the following kind:

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

=

=

=

nnn ydtdy

ydtdy

ydtdy

λ

λ

λ

.............

222

111

The solution to this system is given by tcty 1exp)( 11λ= , tcty 2exp)( 22

λ= ,..., tn

nn cty λexp)( = , where 1c , 2c , ..., nc are all arbitrary constants. We can then find the general solution to the original system given by (18) as

byx 1)()( −−= AtMt . The system will be asymptotically stable if all the eigenvalues are negative; will be unstable if all the eigenvalues are positive; and will be saddle point stable if some of the eigenvalues are positive and others are negative. If A matrix is not digonizable, then we can transform it to its Jordan canonical form by using a similarity transformation and solve the resulting simplified system to arrive at the general solution to (18). We shall however discuss this procedure only for the 2 dimensional case.

20

(b) 2 dimensional system: Consider a two dimensional system of the form

⎪⎪⎭

⎪⎪⎬

⎫

++=

++=

22221212

12121111

bxaxadtdx

bxaxadtdx

(19)

The co-efficient matrix is given by

⎥⎦

⎤⎢⎣

⎡=

2221

1211

aaaa

A .

The particular integral can be found as 1

2( )p xt

x

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

x , where 1

2

x

x

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

is obtained by solving

the equations: 11 1 12 2 1

21 1 22 2 2

00

a x a x ba x a x b

+ + = ⎫⎬

+ + = ⎭. This particular solution is also the steady state

solution of the dynamic system. In order to find the complementary function, we derive the characteristic roots and the corresponding characteristic vectors of A. Case (i): the roots are real and distinct

Let the two roots be 1λ and 2λ with the associated characteristic vectors ⎟⎟⎠

⎞⎜⎜⎝

⎛=

21

111 e

ee

and ⎟⎟⎠

⎞⎜⎜⎝

⎛=

22

122 e

ee respectively. Then the transformation matrix that diagonalizes A is given

by ),( 21 ee=M , which has the two eigenvectors 1e and 2e as its columns. As in the difference equation case, the general solution to the system is then given by:

⎪⎭

⎪⎬

⎫

++=

++=

22

2221

1212

12

2121

111

expexp)(

expexp)(

xcecetx

xcecetxtt

tti

λλ

λλ

(20)

where 21 and xx are the two steady state values. The arbitrary constraints 1c and

2c will be determined by the given boundary conditions The system will be stable if both 1λ and 2λ are negative; will be unstable if 1λ and 2λ both positive; and will be saddle point stable if one of the eigenvalues is negative and the other is not. As in the case of difference equation, we can find the equation of the saddle path in the following way. Let 01 >λ and 02 <λ . Choosing the initial values such that 1c (the

21

coefficient associated with the unstable root 1λ ) vanishes, we can derive corresponding stable time paths of 1x and 2x as:

12

2121 exp)( xcetxt+=

λ

22

2222 exp)( xcetxt+=

λ

Eliminating t

c 22 exp

λ, we can write the equation of the saddle path as:

2112

222 ))(()( xxtx

eetx i +−=

Case (ii): the roots are real and repeated Let the repeated root be denoted by λ. In this case the matrix A is not diagonaizable. However, as in the case of difference equations, by a similarity transformation, we can

still convert A to its canonical form ⎥⎦

⎤⎢⎣

⎡=

λ

λ

01

C

If the eigenvector associated with λ be ⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

ee

e , then the invertible transformation

matrix that converts A to C is given by ⎥⎦

⎤⎢⎣

⎡=

22

11veve

M , where ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡−

2

1

2

1)(ee

vv

IA λ .

As in the difference equation case, we can then define a new set of variables

)()( 1 tMt xy −= and transform the homogeneous system )()( tAt xx =ʹ′ to a simplified system given by: (t)Ct yy =ʹ′ )(

i.e., ⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡ʹ′

ʹ′

2

1

2

1 0

1yy

yy

λ

λ

The solution for the second equation is given by tcty λexp)( 22 = . Putting this solution back in the first equation, we get a non-autonomous differential equation of the form

tcyy λλ exp211 +=ʹ′ , which has a solution ttccty 211 exp)()( λ+= .

Given the solutions to y, once again the general solutions for 1tx and 2

tx can be derived as xyx += tt M . Stability of the system depends on the value of λ. The system is stable if 0<λ and is unstable if 0>λ . Case (iii): the roots are complex conjugate When the eigenvalues of A are complex, (20) would still be a solution to the system. However, as in the difference equation case, we can simplify the system to get a solution in terms of real numbers so that the stability property of the system becomes visible.

22

If the Matrix A has a specific form such that aaa == 1122 (say) and baa =−= 1221 (say)

so that A looks like this: ⎟⎟⎠

⎞⎜⎜⎝

⎛ −

abba

, then as in the difference equation case, we can

derive the eigen values and eigen vectors of A as iba ± and ⎟⎟⎠

⎞⎜⎜⎝

⎛

− i1

, ⎟⎟⎠

⎞⎜⎜⎝

⎛

i1

respectively.

Then directly using (20), we can obtain the general solution for this special case as: 1)(

2)(

11 expexp xccx tibatibat ++= −+

2)(2

)(1

2 expexp xicicx tibatibat ++−= −+ (20ʹ′)

Again applying the second part of De Moivre’s Theorem, the solution can be simplified to

1 13 4

2 23 4

exp ( cos sin )

exp ( sin cos )

att

att

x c t c t x

x c t c t x

θ θ

θ θ

= + +

= − +

where )( 213 ccc += and iccc )( 214 +−= are two arbitrary constants (not necessarily real). To determine the stability property, observe that as in the difference equation case, the two terms cos tθ and sin tθ move in a periodic manner, taking values between –1 and

+1, and returning to the same value after every θπ2

period. Hence 1x and 2x will also

move in cyclically fashion. However, whether they approach the steady state over time or not depends crucially on the term expat . If 0a < , the system will approach the steady state over time, albeit cyclically. If, 0a > the system will move away from the steady state. If 0a = , the system will exhibit limit cycles, moving in the same orbit period after period, neither approaching the steady state, nor moving away from it. Remark. Note that though the methods of solving difference and differential equations are very similar, and often the stability properties are also somewhat similar, in the complex root case the stability condition for a system of differential equations is substantially different from that of difference equation. Recall that in a 22× system of

difference equations with complex roots a ib± , stability requires that 2 2 1r a b= + < . But in a 22× system of differential equations with complex roots a ib± , stability requires that 0a < . Thus in the latter case, only the real part of the complex root is important for stability, the imaginary part does not play any role.

If the matrix A does not follow the specific form given by ⎟⎟⎠

⎞⎜⎜⎝

⎛ −

abba

, we can construct a

matrix M and transform the x-system into a system of new variables y such that the new co-efficient matrix of the y-system has the specific form. Once we solve the y-system, we can derive the general solution for the x-system as: t tM= +x y x . The stability property of the x-system will be the same as the y-system.