Equations involving an unknown function and its derivatives. Physical laws encoded in differential equations INTRODUCTION TO DIFFERENTIAL EQUATIONS

Transcription

1 INTRODUCTION TO DIFFERENTIAL EQUATIONS Equations involving an unknown function and its derivatives e. : +f = e solution for f specified by equation + initial data [e.g., value of f at a point] Physical laws encoded in differential equations

2 In this course we will talk of ordinary differential equations: involving derivatives with respect to one variable to be contrasted with partial differential equations: involving derivatives with respect to more than one variable Partial DE will occur in Hilary Term course Normal modes, wave motion and the wave equation

3 B. Ordinary differential equations (ODE) I. Generalities Differential operators. Initial Conditions. Linearity. General solutions and particular integrals. II. First order ODEs The linear case. Non-linear equations. III. Second order linear ODEs General structure of solutions. Equations with constant coefficients. Application: the mathematics of oscillations. IV. Systems of first-order differential equations Solution methods for linear systems. Application: electrical circuits.

4 Differential Equations e.g. f sin + = + Initial conditions Differential operators. Functions : map numbers numbers Operators : map functions functions Differential operators : e f f; f 1/ f; f f + f f f f f ; f ; f + f ; ) d d d d d d d d Convenient to name operator eg.. L( f) : f

5 Order of a differential operator L1( f) + 3f isfirstorder, d f L( f) + 3f is second order, d f L3( f) + 4 is second order., real or comple numbers Linear operator If L( f + g) = L( f) + L( g), then L is a linear operator e.g. f and f f are linear 1 f and f f f + are not linear

6 The principle of superposition Suppose f and g are solutions to L( y ) = 0 for different initial conditions ie.. L( f) = 0, L( g) = 0 Consider the Linear Combination : f + g If L linear then L( f + g) = L( f) + L( g) = 0 i.e. a linear combination of solutions of a linear operator is also a solution principle of superposition

7 Inhomogeneous terms e.g. f sin + = L( f) = h( ) Sometimes called driving term Lf = 0 homogeneous differential equation e. : +f = 0 Lf = h() 0 e. : inhomogeneous differential equation +f = sin

8 Solution to differential equations L( f) = h( ) The number of independent complementary functions is the number of integration constants equal to the order of the differential equation Complementary function 1) Construct f the general solution to the homogeneous equation Lf = ) Find a solution, f, to the inhomogeneous equation Lf 1 1 = h Particular integral General solution : f0 + f1 For a nth order differential equation need n independent solutions to Lf=0 to specify the complementary function

9 First order linear equations General form : q( ) f h( ) + =. Easy to solve Integrating factor Look for a function I() such that I() +I()q()f dif =I()h() Solution: CF PI dif =0 is If =const I() f()= I(')h(')' Solution : f()= 1 I() 0 I(')h(')'

10 First order linear equations General form : q( ) f h( ) + =. Easy to solve Integrating factor dif Look for a function I() such that I() + I( q ) ( ) f = I( h ) ( ) Solution : 1 f( ) = I( ') h( ') ' I( ) 0 We have I()q()= di First order : 1 integration constant (CF) ln(i())= q(')' I()=e q(')' Integrating factor

11 E1 Solve d f =. Writing in standard form : f = 1 so q 1 1 ln 1 = and I = e =. Pluggingthisintothe form of the solutionwehave : f = d = ( ) 0 3/ 1/ 0

12 Eample : Solve f +f = with f(3) = 0. Integrating factor I() = ep( ) = ep() f() = ep( )[ ep()+const.] = ep( )[ep()( +1/)/+const.] Initial condition f(3) = 0 ep( 6)[ep(6)13/4+const.] = 0 const. = ep(6)13/4 Thus f() = ( +1/)/ (13/4)ep(6 )

13 Eample 3: Solve f +sin f = with f(0) =. Integrating factor I() = ep( sin ) = ep( cos) f() = ep(cos)[ ep( cos)+const.] Initial condition f(0) = f() = ep(cos)[/e+ 0 dt tep( cost)]

14 The discussion of 1st-order linear ODE shows that the general solution depends on one arbitrary constant, and one initial condition is needed to specify the solution uniquely. What is the number of initial conditions necessary to specify solution uniquely for a linear ODE of order n Consider general form (f (k) = d k f/ k ) Lf = a n f (n) +a n 1 f (n 1) +...+a 0 f = 0 Rearrange this as f (n) = a n 1 a n f (n 1)... a 0 a n f Now differentiate both sides wrt and re-epress f (n) in rhs using above equation: f (n+1) = in terms of f and its derivatives up to f (n 1)

15 Differentiate once more, and by the same reasoning f (n+) = in terms of f and its derivatives up to f (n 1) So any derivative f (k) can be epressed in terms of f and derivatives up to f (n 1). Now use Taylor series epansion f() = f( 0 )+( 0 )f (1) ( 0 )+ 1 ( 0) f () ( 0 )+... Any of the derivatives at 0 is epressed in terms of f( 0 ),...,f (n 1) ( 0 ). Therefore f at is determined by original n-th order DE + n conditions giving the n constants f (r) ( 0 ), r = 0,...,n 1 (or equivalently an alternative set of n constants).

16 Summary Superposition principle for linear homogeneous ODEs. If f 1 and f are solution of Lf = 0, then any linear combination αf 1 +βf is also solution. General solution of linear inhomogeneous ODE Lf = h is sum of the particular integral (PI) and complementary function (CF). First-order linear ODE solvable by general method: f +qf = h f = e q [ e q h+c]. n initial conditions needed to specify solution of linear ODE of order n