Module 13 - Differential Equations 2

FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 13

DIFFERENTIAL EQUATIONS III

1. Differential operators 2. General solution of linear inhomogeneous ODEs 3. Free and forced oscillations

1. Differential operators First let us introduce some general notation which is useful in stating later results. Operators are functions that transform a function to a different function. e.g. φ[f (t)] = (f (t))2 so φ[t + 1] = (t + 1)2 = t2 + 2t + 1 . d denotes a differential operator. dt d2 f d2 f + f = t can be expressed L[f (t)] = t . If L[f (t)] = 2 + f , then the differential equation (DE) dt dt2 A DE L[f (t)] = g(t) is linear if the operator satisfies The symbol

L[a f1 + b f2 ] = a L[f1 ] + b L[f2 ] , for all constants a and b . Linearity principle: if x1 and x2 are both solutions of the homogeneous linear DE L[x] = 0 then so is the quantity a x1 + b x2 , where a and b are both arbitrary constants. (This important result was used in module 6, DEs I). 2. General solution of linear inhomogeneous ODEs Linearly independent functions are all distinct from each other (e.g. 1 , sin t , et ). Linearly dependent functions are not all distinct - i.e. at least one function is a linear combination of some, or all, of the others. For example, the functions in the set 1, t , 2 + 3t are dependent, since the final member can be written in terms of the other two: 2 + 3t = 2(1) + 3(t) .) With the above notation some general results are stated below: Result 1. If x1 , x2 ...xp are linearly independent solutions of L[x] = 0 , then the general solution is

(pth order, homogeneous, linear ODE)

x = A1 x1 + A2 x2 + ... + Ap xp .

∗

Result 2. If x is any solution of L[x] = f (t), (pth order, inhomogeneous, linear ODE) and xc is the general solution of L[x] = 0 , then x∗ + xc is the general solution of L[x] = f (t) . [Proof: Result 2 is easily proved since L[x∗ + xc ] = L[x∗ ] + L[xc ] = f (t) + 0 = f (t) .] 1

In the above x∗ is called a particular integral, xc is called the complementary function. In this module we investigate the particular case of second order ODEs with constant coefficients, i.e. the equation dx d2 x + c x = f (t) . a 2 +b dt dt How do we find x∗ ? The procedure is usually called the method of undetermined coefficients. It involves using a trial function, which includes some unknown constants, for the particular integral and then determining these constants after substituting into the DE. The three standard types are listed below: trial function

f (t) polynomial of degree p

most general polynomial function of degree p

αt

Ceαt C sin(αt) + D cos(αt)

e sin(αt)

Let us illustrate the method of solution with three examples. d2 x dx + 2x = t . −3 2 dt dt First find the complementary function (CF), i.e. the solution of

Ex 1.

Find the general solution of

auxiliary equation

dx d2 x + 2x = 0 , −3 dt2 dt m2 − 3m + 2 = 0 ,

(recall DEs I, module 6)

(m − 2)(m − 1) = 0 , → m = 1, 2 t

(roots real and distinct) 2t

→ xc = A e + B e ,

(from DEs I)

For the particular integral (PI) x∗ try x = Ct + D , since the RHS of the given ODE is a polynomial of this order. d2 x dx =C, Differentiating gives = 0 , and substituting into the original equation leads to dt dt2 0 − 3C + 2(Ct + D) = t , 2Ct + (−3C + 2D) = t . Equating the coefficients of the corresponding terms in the polynomials on both sides of the above equation (i.e. the coefficients of t and constant terms in turn) leads to 2C = 1 , −3C + 2D = 0 , Hence x∗ =

→ C = 1/2 → 2D = 3C = 3/2 ,

3 t + and the general solution to the given ODE is 2 4 x = A et + B e2t +

2

3 t + . 2 4

D = 3/4 .

Ex 2.

d2 x dx + 4x = e3t . +4 dt2 dt dx d2 x + 4x = 0 , +4 2 dt dt m2 + 4m + 4 = 0 ,

Find the general solution of CF: aux. eqn.

(m + 2)2 = 0 , → m = −2 (twice) , xc = (At + B) e PI: try leads to

x = Ce3t ,

dx = 3C e3t , dt

→

−2t

(roots real and equal) ,

(from DEs I)

d2 x = 9C e3t , and substituting into the original equation dt2

9Ce3t + 4(3Ce3t ) + 4(Ce3t ) = e3t , (9C + 12C + 4C) e3t = e3t .

Equating the coefficients implies to the given ODE is

25C = 1 ,

→C=

1 3t 1 . Hence x∗ = e and the general solution 25 25

x = (At + B) e−2t +

d2 x dx + 2x = sin t . +2 2 dt dt dx d2 x + 2x = 0 , +2 CF: dt2 dt aux. eqn. m2 + 2m + 2 = 0 , p p −2 ± (−4) −2 ± (22 − 4(2)(1)) −2 ± 2j = = = −1 ± j , (roots complex) m= 2 2 2 → xc = e−t (A cos t+B sin t) , (again using module 6)

Ex 3.

PI:

1 3t e . 25

Find the general solution of

try

x = C cos t + D sin t , →

dx = −C sin t + D cos t , dt

d2 x = −C cos t − D sin t . dt2

Substituting into the original equation leads to (−C cos t − D sin t) + 2(−C sin t + D cos t) + 2(C cos t + D sin t) = sin t , (−C + 2D + 2C) cos t + (−D − 2C + 2D) sin t = sin t , (C + 2D) cos t + (−2C + D) sin t = sin t , Equating the coefficients gives

C + 2D = 0 ,

(1)

−2C + D = 1 .

(2)

eqn. (1) − 2 × eqn.(2) then eqn.(1) Hence x∗ = −

→

→ 5C = −2 , C = −2/5   1 2 1 C − = D=− =− 2 2 5 5

1 2 cos t + sin t , and the general solution to the ODE is 5 5 x = e−t (A cos t + B sin t) − 3

1 2 cos t + sin t . 5 5

Examples 1, 2 and 3 above show the three different types of complementary function that were considered in more detail in module 6, and the three different types of f (t) on the RHS of the ODE. Clearly one can “mix and match” the different cases. Extra conditions must be prescribed to obtain the constants A and B. Note that these conditions must be applied to the general solution, not to the complementary function on its own. Two additional complications in the above method can arise and we shall look at these in turn. First, suppose f (t) is a sum of functions of the types considered above. Then we can use the result that PI for full equation = sum of PIs for each separate function (or trial function for full equation = sum of trial functions for each separate function) d2 x dx + 4x = t + e3t . +4 dt2 dt This is the same as Ex 2, except for the extra term on the RHS. The above general result implies Find the general solution of

Ex 4.

general solution = xc + (x∗ for t) + (x∗ for e3t ) . 1 3t e From the earlier question we know the first and third terms on the RHS are (At + B) e−2t and 25 respectively. Hence to complete the general solution it is necessary to find the PI corresponding to t , i.e. find the PI of the ODE dx d2 x + 4x = t . +4 2 dt dt d2 x dx =C, = 0. dt dt2 Substituting into the original ODE gives 0 + 4C + 4(Ct + D) = t , and equating coefficients leads to PI:

try

→

x = Ct + D ,

4C = 1 , 4C + 4D = 0 , x∗ =

Hence

→ C = 1/4 → D = −C = −1/4 .

1 t − and the general solution is 4 4 x = (At + B)e−2t +

1 1 t − + e3t . 4 4 25

The second difficulty is illustrated by the following example. Ex 5.

Find the general solution of

CF:

same as Ex 1

d2 x dx + 2x = et . −3 dt2 dt

xc = Aet + Be2t .

dx = Cet , dt Substituting into the original ODE gives PI:

try

x = Cet ,

→

d2 x = Cet . dt2

Cet − 3(Cet ) + 2(Cet ) = et , 0Cet = et , 0 = et , The standard approach therefore fails:

WHY? 4

NOT possible

We tried to find a PI of the same form as a term which arises in the CF. This trial function will never work (since it appears in the CF and therefore satisfies the ODE with RHS zero). When this situation arises, then use a new trial function which is the original guess multiplied by t . This new trial function will work, provided it does not appear explicitly in the CF. If it does appear in the latter then you multiply it by another t and keep going until you obtain the first expression which does not appear in the CF. In the current example, therefore, we would normally argue as follows. Looking at the RHS the normal choice for the PI is x = Cet . This function appears in the CF, however, so we should try x = t(Cet ) = Ctet . This term does not appear in the CF so the new trial function should work. Differentiating the new trial function, which is a product of functions of t , dx = C((1)et + tet ) = C(1 + t)et , dt d2 x = C((1)et + (1 + t)et ) = C(2 + t)et . dt2 Hence substituting into the original ODE C(2 + t)et − 3C(1 + t)et + 2Ctet = et , tet (C − 3C + 2C) + et (2C − 3C) = et , −Cet = et , C = −1 . Hence the PI is

x∗ = −tet and the general solution is x = A et + B e2t − t et .

3. Free and forced oscillations (a) Free oscillation - no damping Consider the simplest situation where a mass m is attached to one end of a light spring of stiffness k . The other end of the spring is attached to a fixed point on the table. The mass and spring lie on a smooth horizontal table.

R T

m

mg x When the spring is extended by a distance x and T is the tension in the spring, the equation of motion, using Newton’s second law, is d2 x −T = m 2 . dt The tension and extension are related by Hooke’s law T = kx and hence elimination of T gives −kx = m or

d2 x + ω2 x = 0 dt2

d2 x , dt2 where

ω2 =

k . m

As you probably know the above equation governs simple harmonic motion (SHM). This homogeneous equation is of the type discussed in module 6, and the associated auxiliary equation is with solution

m2 + ω 2 = 0 , → m = ±jω , x = A cos(ωt)+B sin(ωt) , where ω is the natural frequency. 5

(b) Free oscillation - with damping In real situations damping is often proportional to speed, and in these situations the above equation of motion for the mass on the table becomes −T − λ

d2 x dx =m 2 , dt dt

where λ is a positive constant. Using Hooke’s law, dividing by m and rearranging leads to λ d2 x + dt2 m d2 x + 2ζω dt2

or

dx + ω2x = 0 , dt dx + ω2x = 0 . dt

The equation is rewritten in the above form to simplify the following algebra. The quantity ζ is positive and is called the damping parameter. To determine the solution note that the auxiliary equation is m2 + 2ζωm + ω 2 = 0 with solution m=−

2ζω ±

p

p (4ζ 2 ω 2 − 4ω 2 ) = −ζω ± ω (ζ 2 − 1) . 2

In the usual way there are three different types of solution depending on the sign of the discriminant (ζ 2 −1) : ζ > 1 , the system is overdamped (figure 1); ζ = 1 , the system is critically damped (figure 1); ζ < 1 , the system is underdamped (figure 2)

Figure 1

Figure 2

The precise form of the solution depends on the initial conditions but for all overdamped and critically damped cases there is at most one turning point. (c) Forced oscillation (This is important in applications, but is non-examinable). When a forcing term F cos(Ωt) is applied to the system in (b) then the equation becomes d2 x dx + ω 2 x = F cos(Ωt) . + 2ζω 2 dt dt The general form of the solution of this equation is x = xc + x∗ , where xc , the complementary function, is called the transient solution which is shown in section (b) to tend to zero as t → ∞ and x∗ is called the steady state solution, since this is the only part of the solution which remains at large time. The method for solving the differential equation has been covered in the earlier sections of this module. The details are algebraically complicated, however, and therefore only the results are stated here. 6

Using the trial function x = C cos(Ωt) + D sin(Ωt) for the particular integral it can be shown, after differentiation and substitution into the original differential equation, that x∗ =

(ω 2 − Ω2 )F cos(Ωt) + 2ζωΩ F sin(Ωt) . (ω 2 − Ω2 )2 + 4ζ 2 ω 2 Ω2

Using trigonometric identities this can be rewritten x∗ =

F cos(Ωt − δ) , [(ω 2 − Ω2 )2 + 4ζ 2 ω 2 Ω2 ]1/2

which is the steady state solution. Comparing this solution with the forcing term F cos(Ωt) it is clear that: (1) (2)

the frequencies of the forcing term and solution are the same; the forcing term and solution are out of phase (since δ is usually non-zero);

(3) the amplitudes are different and the maximum amplitude of the solution occurs when the denominator ( (ω 2 − Ω2 )2 + 4ζ 2 ω 2 Ω2 ) is a minimum. The phenomenon of a large response from a small forcing term is called resonance, and gives rise to many practical occurrences (e.g. tuning receivers of radio signals, collapse of bridges by marching soldiers, breaking of wine glasses by singers).

rec/01lode3

7