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0: Ordinary Differential Equations 6 Some definitions: Consider an ODE which has solution φ. I Order of ODE: Highest derivative involved in equation. I Linear: Equation does not involved powers of solution φ or its derivatives I Homogenous: Every term involves φ or its derivatives. (Similarly for PDEs.) 7 First order, linear ODEs df dx = −αf f is a function of one variable, x. α: constant Physics: Radioactive decay Maths: rate of change proportional to function itself. 1 - Trial solution. Guess f (x) = Aeλx . 2 - Separate variables and integrate. λ = −α 8 Second-order, linear ODEs a d 2f df + cf +b 2 dx dx = 0 (0.1) Again f is only dependent on one variable, x, so f (x) Constant coefficients (a, b, c) Homogeneous (RHS=0) - no driving force Try exponential solution f = Aeλx 9 Second derivative - need to integrate twice - two constants of integration f df dx d 2f dx 2 Substitute in to eqn. 0.1 Can only be true for all x if Auxiliary eqn., quadratic, 2 roots. So 2 solutions f1 (x) = Aeλx = λAeλx = λ2 Aeλx aλ2 + bλ + c Aeλx = 0 aλ2 + bλ + c = 0 √ −b ± b2 − 4ac λ= 2a = A1 eλ1 x and f2 (x) = A2 eλ2 x with two constants to be determined from the boundary conditions. 10 Principle of Superposition: A linear superposition (sum) of solutions to a linear equation is also a solution to the equation. Prove by differentiation and subsitution of f (x) = f1 (x) + f2 (x) = A1 eλ1 x + A2 eλ2 x in to eqn. 0.1. aλ21 + bλ1 + c A1 eλ1 x + aλ22 + bλ2 + c A2 eλ2 x = 0 Terms in parenthesis are zero. So the general solution is f (x) = A1 eλ1 x + A2 eλ2 x - 2 constants to be determined from boundary conditions. 11 A Simple (But Important) Second Order ODE d 2f = α2 f dx 2 Auxillary equation General solution (0.2) λ2 = α2 → λ = ±α f (x) = Ceαx + De−αx Growing or decaying not oscillatory or f (x) = C 0 cosh αx + D 0 sinh αx Hyperbolic functions 12 Simple Harmonic Motion (SHM) d 2f dx 2 = −k 2 f (0.3) sin and cos solutions. The general solution is a linear superposition: f (x) = A cos(kx) + B sin(kx) (0.4) 13 Try an exponential solution: f (x) = eλx Exponentials are easier to differentiate ! Try f = eλx df = λeλx dx d 2f = λ2 eλx dx 2 Compare with initial ODE and they can be identical if the auxillary equation is λ2 = −k 2 !!! So λ is complex! λ = ±ik where i 2 = −1 14 Complex exponentials Terms like eikx Easy to differentiate. Essential for quantum mechanics. d ikx e dx d 2 ikx e dx 2 = ikeikx = (ik)2 eikx = −k 2 eikx Equation of SHM with f (x) = eikx as a solution. 15 Complex Solutions SHM Showed that f = Ceλx where λ = ±ik d 2f = −k 2 f is a solution of dx 2 So general solution is f (x) = Ceikx + De−ikx (0.5) These are wave-like/oscillatory solutions. 16 Just showed the general solution to the equation of SHM is f (x) = A cos(kx) + B sin(kx) But complex exponentials are also a solution. How can these be equivalent? Look at the f (x) = eikx term at x = 0 and equate to the solution eqn. 0.4 at x = 0. eikx = A cos(kx) + B sin(kx) 0 At x = 0, e = 1 = A cos 0 + B sin 0 = A Differentiate f 0 (0) = ike0 = ik → ik = −Ak sin 0 + Bk cos 0 = kB So A = 1 and B = i and eikx = cos kx + i sin kx -Euler’s formula. Similarly for the e−ikx term or just subsitute x = −x in to Euler’s formula, noting that cos(−x) = cos(x) and sin(−x) = − sin(x) Linear combinations of complex exp. equivalent to linear combinations of sines and cosines. 17