93 Analytical solution of differential equations

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1 1 93 Analytical solution of differential equations 1. Nonlinear differential equation The only kind of nonlinear differential equations that we solve analytically is the so-called separable differential equations including autonomous equations). See AMBS Ch Solution method: separate the variables and integrate Find analytical solution formulas for the following initial value problems. In each case sketch the graphs of the solutions and determine the half-life. See: P. Atkins and L. Jones, Chemical Principles. The Quest for Insight. Freeman, New York, second edition, 22, pp a) First order decay rate: b) Second order decay rate: c) Third order decay rate: u = ku, t >, k >, u > ) u = ku 2, t >, k >, u > ) u = ku 3, t >, k >, u > ) Find analytical solution formulas for the following initial value problems. Sketch the graphs. a) First order increase rate: b) Second order increase rate: c) Third order increase rate: d) The logistic equation: u = ku, t >, k >, u > ) u = ku 2, t >, k >, u > ) u = ku 3, t >, k >, u > ) u = ku1 u), t >, k >, u > ) e) u = tu 2, t >, u > )

2 2 2. Linear differential equation 2.1 Linear differential equation first order 93.1) u + at)u = ft). Here u = ut) is an unknown function of an independent variable t. The equation is called homogeneous if ft) and nonhomogeneous otherwise. The differential operator Lu = u +at)u has constant coefficient if at) = a is constant and it has variable coefficient otherwise. The equation is said to be a linear equation, because the operator L is a linear operator: Lαu + βv) = αlu + βlv, α, β R, u = ut), v = vt)) i.e., it preserves linear combinations of functions. Check this! Solution method: multiply by the integrating factor e At) with At) = as) ds, and integrate. See AMBS Ch constant coefficient, homogeneous) Solve the following. Sketch the graph of the solution. a) u + 2u =, t >, b) u 2u =, t >, constant coefficient, nonhomogeneous) Solve the following. a) u + 2u = ft), t >, b) u 2u = ft), t >, constant coefficient, nonhomogeneous) Solve the following. u + au = ft), t >, u) = u variable coefficient, nonhomogeneous) Solve the following. u + 2tu = ft), t >, 2.2 Linear differential equation second order constant coefficients 93.2) u + a 1 u + a u = ft). The equation is called homogeneous if ft) and nonhomogeneous otherwise. We assume that the differential operator Lu = u + a 1 u + a u has constant coefficients a 1 and a. Check that the operator L is linear! Variable coefficients: Linear differential equations of second order with variable coefficients u + a 1 t)u + a t)u = ft), cannot be solved analytically, except in some special cases. One such case can be found in AMBS Ch We do not discuss this here.

3 3 Homogeneous equation See AMBS Ch The homogeneous equation 93.2) may be written 93.3) D 2 u + a 1 Du + a u =, or P D)u =, where P r) = r 2 + a 1 r + a is the characteristic polynomial of the equation. The characteristic equation P r) = has two roots r 1 and r 2. Hence P r) = r r 1 )r r 2 ). All solutions of equation 93.2) are obtained as linear combinations 93.4) ut) = c 1 e r1t + c 2 e r2t, if r 1 r 2, ut) = c 1 + c 2 t)e r1t, if r 1 = r 2, where c 2, c 2 are arbitrary coefficients. The coefficients may be determined from an initial condition of the form u) = u, u ) = u 1. The formula 93.4) is called the general solution of homogeneous linear equation 93.3). Example We solve u + u 12u = ; u) = u, u ) = u 1. The equation is written D 2 D 12)u = and the characteristic equation is r 2 + r 12 = with roots r 1 = 3, r 2 = 4. The general solution is with the derivative The initial condition gives ut) = c 1 e 3t + c 2 e 4t u t) = 3c 1 e 3t 4c 2 e 4t. u = u) = c 1 + c 2 u 1 = u ) = 3c 1 4c 2 which implies c 1 = 4u + u 1 )/7, c 2 = 3u u 1 )/7. The solution is ut) = 4u + u 1 e 3t + 3u u 1 e 4t Prove the solution formula 93.4) by writing the equation as P D)u = D r 1 )D r 2 )u = and by solving two first order equations D r 1 )v = and D r 2 )u = v as in Problems 93.3 and Write the following equations as P D)u = and solve the initial value problem. Choose numerical values for the constants and sketch the graph of the solution. Solve the problem with your Matlab program my ode.m. a) u u 2u = ; u) = u, u ) = u 1. b) u k 2 u = ; u) = u, u ) = u 1. c) u + 4u + 4u = ; u) = u, u ) = u Solve the boundary value problem u x) k 2 ux) =, < x < L, u) =, ul) = u L.

4 4 Complex roots If the characteristic polynomial P r) has real coefficients, then its roots are either real numbers or a pair of conjugate complex numbers, see AMBS Ch In the latter case we have r 1 = α + iω and r 2 = α iω and the solution 93.4) becomes see AMBS Ch 33.2 for the definition of expz) with a complex variable z) ut) = c 1 e α+iω)t + c 2 e α iω)t with d 1 = c 1 + c 2, d 2 = ic 1 c 2 ). = e αt c 1 e iωt + c 2 e iωt) = e αt ) ) ) c 1 cosωt) + i sinωt) + c2 cosωt) i sinωt) = e αt ) d 1 cosωt) + d 2 sinωt), Write the equation as P D)u = and solve the initial value problem. Choose numerical values for the constants and sketch the graph of the solution. Solve the problem with your Matlab program my ode.m. a) u + 4u + 13u = ; u) = u, u ) = u 1. b) u + ω 2 u = ; u) = u, u ) = u 1. Nonhomogeneous equation See AMBS Ch The solution of the nonhomogeneous equation P D)u = ft) is given by 93.5) ut) = u h t) + u p t), where u h is the general solution 93.4) of the corresponding homogeneous equation, i.e., P D)u h =, and u p is a particular solution of the nonhomogeneous equation, i.e., P D)u p = ft). Proof: If u is given by 93.5), then Lu = Lu h + u p ) = Lu h + Lu p = + f = f, so that u solves the nonhomogeneous equation. On the other hand: if u p is a particular solution and u is any other solution of the nonhomogeneous equation, then Lu u p ) = Lu Lu p = f f =, i.e., u u p solves the homogeneous equation. Thus u u p = u h, which is 93.5). A particular solution can sometimes be found by guess-work: we make an Ansatz for u p of the same form as f. Example u u 2u = t. Here ft) = t is a polynomial of degree 1 and we make the Ansatz u p t) = At + B, i.e., a polynomial of degree 1. Substitution into the equation gives A 2At + B) = t. Identification of coefficients gives A = 1 2, B = 1 4, so that u pt) = t. The general solution of the homogeneous equation is u h t) = c 1 e t + c 2 e 2t, see Problem 93.8 a). Hence we get Solve the following. ut) = u h t) + u p t) = c 1 e t + c 2 e 2t t. a) u u 2u = e t Ansatz: u p t) = Ae t b) u u 2u = cost) Ansatz: u p t) = A cost) + B sint) c) u u 2u = t 3 d) u u 2u = e t Ansatz: u p t) = At 3 + Bt 2 + Ct + D Ansatz: u p t) = Ate t

5 5 Re-writing as a system of first order equations [ ] By setting w 1 = u, w 2 = u w1, w =, we can re-write 93.2) as a system of first order equations w 2 w t) = Awt) + F t); w) = w, where w = [ u u 1 ], A = [ ] [ ] 1, F t) =. a a 1 ft) To see this we compute w = [ ] [ ] u u u = a u a 1 u + ft) [ = w 2 a w 1 a 1 w 2 + ft) ] [ 1 = a ] [ ] w1 + a 1 w 2 [ ]. ft) It is necessary to do this re-writing before we can use our Matlab programs to solve 93.2) Write the equation in Problem 93.11a) as a system of first order. 2.3 System of linear differential equations of first order Constant coefficients homogeneous equations We finally mention 93.6) u + Au =, t >, u) = u, where ut), u R d, and A R d d is a constant matrix of coefficients. This kind of system will studied by means of eigenvalues and eigenvectors in the following course ALA-C. Answers and solutions Reaction of order 1 decay rate of order 1): { u = ku u) = u The half-life T 1/2 is given by ut) = u e kt ut 1/2 ) = u e kt 1/2 = 1 2 u, which leads to T 1/2 = log2). k

6 6 Reaction of order n > 1 decay rate of order n > 1): { u = ku n u) = u du = k dt un ut ) u u n du = T k dt [ u n+1 ] ut ) = kt n + 1 u ut ) n+1 u n+1 = n 1)kT 1 ut ) n 1 = 1 u n 1 + n 1)kT = ut ) = u 1 + n 1)u n 1 kt ) 1/n 1) 1 + n 1)un 1 kt u n 1 The half-life T 1/2 is given by which implies a) ut) = u e kt b c) order n > 1: ut) = d) ut) = e) ut) = ut 1/2 ) = u u + 1 u )e kt u 1 + t 2 u /2 a) ut) = e 2t u b) ut) = e 2t u a) ut) = e 2t u + e 2t s) fs) ds b) ut) = e 2t u + e2t s) fs) ds u 1 + n 1)u n 1 kt 1/2 ) 1/n 1) = 1 2 u T 1/2 = 2n 1 1 n 1)u n 1 k u 1 n 1)u n 1 kt ), t < 1 1/n 1) n 1)u n 1 k ut) = e at u + e at s) fs) ds Integrating factor: e t2. Solution ut) = e t2 u + e t2 s 2) fs) ds.

7 The equation for v is D r 1 )v =, or v r 1 v =, v) = v, with unique solution vt) = v e r1t. The equation for u is D r 2 )u = v, or u r 2 u = v e r1t, u) = u, with unique solution, see Problem 93.5, If r 1 = r 2, then we have instead ut) = u e r2t + e r2t e r2s v e r1s ds = u e r2t + v e r2t e r1 r2)s ds = u e r2t + v e r2t[ e r1 r2)s ] t r 1 r 2 s= = u e r2t + v e r 1t e r2t) r 1 r 2 = v e r1t + u v ) r 1 r 2 r 1 r 2 = c 1 e r1t + c 2 e r2t, if r 1 r 2. ut) = u e r2t + e r2t e r2s v e r1s ds = u e r1t + v e r1t ds = u e r1t + v te r1t = c 1 + c 2 t)e r1t a) ut) = 1 3 2u u 1 )e t u + u 1 )e 2t. b) ut) = c 1 e kt + c 2 e kt = d 1 coshkt) + d 2 sinhkt), d 1 = c 1 + c 2, d 2 = c 1 c 2. The initial condition gives ut) = 1 2 u +u 1 /k)e kt u u 1 /k)e kt or alternatively ut) = u coshkt)+ u 1 /k) sinhkt). c) ut) = u + 2u + u 1 )t)e 2t ux) = u L sinhkx)/ sinhkl) ) a) ut) = e u 2t cos3t) u + u 1 ) sin3t). e r2t b) ut) = u cosωt) + u 1 /ω) sinωt). Compare to Problem 93.8 b) a) ut) = c 1 e t + c 2 e 2t 1 2 et. b) ut) = c 1 e t + c 2 e 2t 3 1 cost) 1 1 sint). c) ut) = c 1 e t + c 2 e 2t 1 2 t t2 9 4 t d) ut) = c 1 e t + c 2 e 2t 1 3 te t. Note: the Ansatz u p t) = Ae t does not work, because e t is a solution of the homogeneous equation, P D)e t =, i.e., e t is contained in u h. [ ] w = w /stig [ ] [ ] [ ] 1 w w 2 e t.

Example. Mathematics 255: Lecture 17. Example. Example (cont d) Consider the equation. d 2 y dt 2 + dy

Example. Mathematics 255: Lecture 17. Example. Example (cont d) Consider the equation. d 2 y dt 2 + dy Mathematics 255: Lecture 17 Undetermined Coefficients Dan Sloughter Furman University October 10, 2008 6y = 5e 4t. so the general solution of 0 = r 2 + r 6 = (r + 3)(r 2), 6y = 0 y(t) = c 1 e 3t + c 2