Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Derivations of the Lorentz transformations wikipedia , lookup
Dynamical system wikipedia , lookup
Wave packet wikipedia , lookup
Classical central-force problem wikipedia , lookup
Numerical continuation wikipedia , lookup
Differential (mechanical device) wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Spinodal decomposition wikipedia , lookup
Routhian mechanics wikipedia , lookup
We have considered so far first order differential equations both linear and non-linear. Tonight we begin considering differential equations of higher order. dv m f( t ) which describes the velocity. dt As an example consider the example of motion: Once we solve this DE for v(t) we use the fact that dx/dt = v(t) to solve for position. But dv d 2 x( t ) so it is natural to ask can we solve for the position directly without having to first solve dt d t 2 for the velocity via the second order linear differential equation : 2 d m x( t ) f( t ) ? d t2 The answer is yes. Tonight we are going to concentrate on the general second order homogeneous linear differential equation with constant coefficients : a 2 d y b dy cy 0 . dx 2 dx We will then consider in great detail the application of this DE to the motion of a mass on a spring. However, before we begin we need to discuss some basic notions concerning nth order linear differential equations in general. The general nth order homogeneous linear differential equation is an equation of the form: a n( x) n d y an n dx n d y d n 1 n dx n an ( x) 1 y a 1( x) ................... 1 d n 1 y dy dx n ................... 1 a 1( x) a 0( x) y 0 dy a 0( x) y f( x) then it is not homogeneous dx dx dx (because of f(x) on the right hand side). If any of the derivatives are raised to a power or if there are any terms which are the products of the If a n( x) 1 ( x) various derivatives and or y it is not linear. For example : y 2 d y dy 2 dx dx 2 0 is not linear for two reasons, what are they? For the IVP : a n( x) n d y n dx an 1 ( x) d n 1 n dx y 1 ................... a 1( x) dy dx a 0( x) y 0 with initial conditions for y(0),y ' (0)................yn-1(0) has a unique solution of the form : y( x) A 1 y 1( x) A 2 y 2( x) ......................................... A n y n( x) where the y i (x) am linearly independent solutions to the DE and the Ai are constants determined from the initial conditions. The question is what do we mean by linearly independent functions ? Definition Two functions are linearly dependent if there exist non-zero constants c1 and c2 such that c1f1(x) +c2f2(x) = 0 i.e. f1(x) is a scalar multiple of f2(x). Functions which are not linearly dependent are called linearly independent. Eg: sin(x) and cos(x) are linearly independent. Suppose not then Then c2 sin ( x) c1 c1sin(x) +c2cos(x) = 0 for some c1 and c2. cos ( x) but if x = /2 we obtain 1 = 0. therefore sin(x) and cos(x) must be linearly independent. Eg : ex and ex are linearly independent unless = . Suppose c1ex + c2ex = 0 for some c1 and c2. Then ex = c ex where c = c2 . c1 taking the natural log of both sides x = ln(c) + x x(- ) = ln(c) if x = 1 then (- ) = ln(c) if x = -1 then (- ) = - ln(c) adding the two equations we have - = 0 or = . A little later we'll develop a much more direct way of establishing the linear independence of any number of functions (the Wronskian) but as regards second order linear differential equations with constant coefficients we have everything we need as regards linear independence of functions. Let's return to the general second order homogeneous linear differential equation with constant coefficients : a (I) 2 d y 2 dx b dy cy 0 . dx We start by looking for solutions of the type y (x) = e x . Then dy dx e x 2 and d y 2 x e . 2 dx Putting these into our DE we obtain : 2 x x x a e b e c e 0 Dividing through by e x we obtain : a 2 b c 0. This is called the characteristic polynomial or the characteristic equation of the differential equation. The roots of this equation lead us to The general solution to (I) above. This is of course just a quadratic whose solutions are given by the quadratic formula : 1 b b 2 4 ac b and 2 b 2 4 ac 2a 2a There are 3 possibilities depending on the number and type of roots. Case 1 Real Distinct Roots b Then 1 y( x) A e b 1 x b 2 2a B e 4 ac 2 4 ac 0 b and 2 b 4 ac and the general solution to (I) is : 2a 2 x 2 As an example consider the IVP : 2 d y 5 dy 6 y 0 with y(0) = 1 and y ' (0) = 0 dx 2 dx Here note that a = 1, b = - 5, and c = 6 so the characteristic equation is : 2 - 5 + 6 = 0 . - 3) ( -2) = 0 = 3 and = 2 Therefore the general solution is : y(x) = A e 2x + B e 3x. Applying the initial condition y(0) = 1 we obtain A+B=1 Applying the initial condition y'(0) = 0 we obtain 2A + 3 B = 0. Solving this system we obtain : A = 3 and B = - 2. 2 Therefore the solution to the IVP : d y 5 dy 6 y 0 with y(0) = 1 and y ' (0) = 0 is y(x) = 3 e 2x - 2 dx 2 dx e 3x. Case 2 Complex Roots b 2 4 ac 0 Here we'll use Euler's Identity : e i = cos() + i sin() where i = In this case both possibilities 1 b i b 2 2a result as you should verify so let's work with 1. 4 ac and 2 1 b i b 2 2a 4 ac yield the same For ease of calculation let's write 1 = + i where Then the solution to a 2 d y b dy b b and 2a 2 4 ac . 2a is y(x) = e ( + i) = e ei = e ( cos() + isin() ) . cy 0 dx 2 dx The only way this can solve the DE is if the real part of the solution, e cos(), and the imaginary part of the solution, e sin(), independently are solutions. Therefore the general solution is : y(x) = e ( Acos() +Bsin() ) . 2 Example Solve d y dy 2 dx dx 2 y 0 with y(0) = 1 and y ' (0) = -1 . Here the characteristic equation is : 2 + + 2 = 0 . The solutions are 1 i 2 7 and 1 2 7 i 2 2 x The general solution to the differential equation is : y( x) e A cos 2 7 x 2 B sin 7 x 2 Now we'll apply the initial conditions: y( 0) 1 A 1 1 1 1 y ' (0) = -1 = A 7B = 7B 2 2 2 2 1 Therefore A =1 B = 7 7 2 The solution to the IVP d y dy 2 dx dx 2 y 0 with y(0) = 1 and y ' (0) = -1 is : x y( x) e 2 cos 7 x 2 1 7 7 sin 7 x 2 Case 3 Repeated Real Roots b b In this case 1 = 2 = . 2a 2 4 ac 0 b 2 Therefore one solution to the DE a d y 2 dx b dy dx cy 0 is y = e 2 a . However, we know we need two independent solutions. In this case it turns out the general solution is y e (Convince yourself that e x and x e x x (A B x) where the second solution is y e are indeed linearly independent). x B x. The reason for this deals with obtaining the general solution to a second order DE from a known solution by a method called reduction of order i.e. : If y1(x) is a solution to a second order DE then we can obtain the general solution by setting y(x) = y1(x)v(x) and solving for v(x). Let's illustrate this for the case where we have a repeated root in the characteristic equation. 2 d y Eg Solve 2 dy 1 0 . dx 2 dx 2 The characteristic equation is 2 -2 + 1 = ( 1) = 0 . Here we have the double root = 1. Therefore y = ex is a solution. We claim the general solution is y(x) = ex (A +Bx) is the general solution. To obtain the general solution let y = ex v(x). Then dy x e v dx 2 d y x dv e dx x e v 2 dx (Verify this ) x dv 2e dx 2 x d v e 2 dx 2 d y Putting these results in 2 dy dx x e v x dv 2 e dx 2 x d v e 2 dx 1 0 we obtain : dx 2 x 2 e v x dv e dx x e v( x) 0 2 x d v 0 Which simplifies to : e 2 dx 2 Which simplifies further to d v 0 2 dx Therefore dv B and v Bx A dx x x Therefore y( x) e v e ( A Bx) as advertised. The point is we don't have to go through this process each time : if there is a single root, to the characteristic equation then the general solution to a 2 d y 2 dx Differential Operator Notation b dy dx cy 0 is y(x) = ex ( A+Bx ) . For the differential equation a 2 d y b dy cy in particular and for higher order differential equations dx 2 dx in general obtaining the characteristic equation is made easier if we use what is called differential operator notation. Define D = d dy then dx Similarly D2 = d 2 2 then 2 And in general Dn = 2 2 dx = D2 y dx d n d y n n dx d y d y 2 dx Therefore a can be written as simply Dy. D is called a differential operator. dx b dy n = Dn y. dx cy = 0 becomes (aD2 + bD +c)y = 0 . Compare this with the characteristic dx equation a2 + b + c = 0 The point being that if an nth order differential equation with constant coefficients is written using differential operators then the characteristic equation is obtained simply by replacing D with . As an example consider the 3d order eqn : 7 3 d y 3 15 2 d y 2 dx dx 3 2 (7 D - 15D +4) y = 0 therefore the characteristic eqn is : 73 -152 + 4 = 0. 4 y 0 . Written using operator notation: