Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 6 ppsx

49.3 Solutions Solution 49.1 1. When β = γ = 0 the equations are dx dt = rxy, dy dt = −rxy. Adding these two equations we see that dx dt = − dy dt . Integrating and applying the initial conditions x(0) = x 0 and y(0) = y 0 we obtain x = x 0 + y 0 − y Substituting this into the differential equation for y, dy dt = −r(x 0 + y 0 − y)y dy dt = −r(x 0 + y 0 )y + ry 2 . 2174 We recognize this as a Bernoulli equation and make the substitution u = y −1 . −y −2 dy dt = r(x 0 + y 0 )y −1 − r du dt = r(x 0 + y 0 )u − r d dt  e −r(x 0 +y 0 )t u  = −re −r(x 0 +y 0 )t u = e r(x 0 +y 0 )t  t −re −r(x 0 +y 0 )t dt + ce r(x 0 +y 0 )t u = 1 x 0 + y 0 + ce r(x 0 +y 0 )t y =  1 x 0 + y 0 + ce r(x 0 +y 0 )t  −1 Applying the initial condition for y,  1 x 0 + y 0 + c  −1 = y 0 c = 1 y 0 − 1 x 0 + y 0 . The solution for y is then y =  1 x 0 + y 0 +  1 y 0 − 1 x 0 + y 0  e r(x 0 +y 0 )t  −1 Since x = x 0 + y 0 − y, the solution to the system of differential equations is x = x 0 + y 0 −  1 y 0 + 1 x 0 + y 0  1 − e r(x 0 +y 0 )t   −1 , y =  1 y 0 + 1 x 0 + y 0  1 − e r(x 0 +y 0 )t   −1 . 2175 2. For β = 0, γ = 0, the equation for x is ˙x = rxy −γx. At t = 0, ˙x(0) = x 0 (ry 0 − γ). Thus we see that if ry 0 < γ, x is initially decreasing. If ry 0 > γ, x is initially increasing. Now to show that x(t) → 0 as t → ∞. First note that if the initial conditions satisfy x 0 , y 0 > 0 then x(t), y(t) > 0 for all t ≥ 0 because the axes are a seqaratrix. y(t) is is a strictly dec reasing function of time. Thus we see that at some time the quantity x(ry − γ) will become negative. Since y is decreasing, this quantity will remain negative. Thus after some time, x will become a strictly decreasing quantity. Finally we see that regardless of the initial conditions, (as long as they are positive), x(t) → 0 as t → ∞. Taking the ratio of the two differential equations, dx dy = −1 + γ ry . x = −y + γ r ln y + c Applying the intial condition, x 0 = −y 0 + γ r ln y 0 + c c = x 0 + y 0 − γ r ln y 0 . Thus the solution for x is x = x 0 + (y 0 − y) + γ r ln  y y 0  . 3. When β > 0 and γ > 0 the system of equations is ˙x = rxy − γx ˙y = −rxy + β. 2176 The equilibrium solutions occur when x(ry − γ) = 0 β −rxy = 0. Thus the singular point is x = β γ , y = γ r . Now to classify the point. We make the substitution u = (x − β γ ), v = (y − γ r ). ˙u = r  u + β γ   v + γ r  − γ  u + β γ  ˙v = −r  u + β γ   v + γ r  + β ˙u = rβ γ v + ruv ˙v = −γu − rβ γ v − ruv The linearized system is ˙u = rβ γ v ˙v = −γu − rβ γ v Finding the eigenvalues of the linearized system,      λ − rβ γ γ λ + rβ γ      = λ 2 + rβ γ λ + rβ = 0 2177 λ = − rβ γ ±  ( rβ γ ) 2 − 4rβ 2 Since both eigenvalues have negative real part, we see that the singular point is asymptotically stable. A plot of the vector field for r = γ = β = 1 is attached. We note that there appears to be a stable singular point at x = y = 1 which corroborates the previous results. Solution 49.2 The singular points are u = 0, v = 0, u = 0, v = 1, u = 0, v = p. The point u = 0, v = 0. The linearized system about u = 0, v = 0 is du dx = ru dv dx = u. The eigenvalues are     λ − r 0 −1 λ     = λ 2 − rλ = 0. λ = 0, r. Since there are positive eigenvalues, this point is a source. The critical point is unstable. The point u = 0, v = 1. Linearizing the system about u = 0, v = 1, we make the substitution w = v − 1. du dx = ru + (w + 1)(−w)(p −1 −w) dw dx = u du dx = ru + (1 − p)w dw dx = u 2178     λ − r (p − 1) −1 λ     = λ 2 − rλ + p −1 = 0 λ = r ±  r 2 − 4(p − 1) 2 Thus we see that this point is a saddle point. The critical point is unstable. The point u = 0, v = p. Linearizing the system about u = 0, v = p, we make the substitution w = v −p. du dx = ru + (w + p)(1 − p − w)(−w) dw dx = u du dx = ru + p(p − 1)w dw dx = u     λ − r p(1 − p) −1 λ     = λ 2 − rλ + p(1 −p) = 0 λ = r ±  r 2 − 4p(1 − p) 2 Thus we see that this point is a source. The critical point is unstable. The solution of for special values of α and r. Differentiating u = αv(1 − v), du dv = α −2αv. 2179 Taking the ratio of the two differential equations, du dv = r + v(1 −v)(p − v) u = r + v(1 −v)(p − v) αv(1 − v) = r + (p − v) α Equating these two expressions, α − 2αv = r + p α − v α . Equating coefficients of v, we see that α = 1 √ 2 . 1 √ 2 = r + √ 2p Thus we have the solution u = 1 √ 2 v(1 −v) when r = 1 √ 2 − √ 2p. In this case, the differential equation for v is dv dx = 1 √ 2 v(1 −v) −v −2 dv dx = − 1 √ 2 v −1 + 1 √ 2 2180 We make the change of variablles y = v −1 . dy dx = − 1 √ 2 y + 1 √ 2 d dx  e x/ √ 2 y  = e x/ √ 2 √ 2 y = e −x/ √ 2  x e x/ √ 2 √ 2 dx + ce −x/ √ 2 y = 1 + ce −x/ √ 2 The solution for v is v(x) = 1 1 + ce −x/ √ 2 . Solution 49.3 We make the change of variables x = r cos θ y = r sin θ. Differentiating these expressions with respect to time, ˙x = ˙r cos θ − r ˙ θ sin θ ˙y = ˙r sin θ + r ˙ θ cos θ. Substituting the new variables into the pair of differential equations, ˙r cos θ − r ˙ θ sin θ = −r sin θ + r cos θ(1 − r 2 ) ˙r sin θ + r ˙ θ cos θ = r cos θ + r sin θ(1 − r 2 ). 2181 Multiplying the equations by cos θ and sin θ and taking their sum and difference yields ˙r = r(1 − r 2 ) r ˙ θ = r. We can integrate the second equation. ˙r = r(1 − r 2 ) θ = t + θ 0 At this point we could note that ˙r > 0 in (0, 1) and ˙r < 0 in (1, ∞). Thus if r is not initially zero, then the solution tends to r = 1. Alternatively, we can solve the equation for r exactly. ˙r = r − r 3 ˙r r 3 = 1 r 2 − 1 We make the change of variables u = 1/r 2 . − 1 2 ˙u = u − 1 ˙u + 2u = 2 u = e −2t  t 2e 2t dt + ce −2t u = 1 + ce −2t r = 1 √ 1 + ce −2t Thus we see that if r is in itiall nonzero, the solution tends to 1 as t → ∞. 2182 [...]... = (1 + A ) + 2 (1 + ω1 )A cos θ + A2 cos 2θ 2 2 To avoid secular terms, we must have ω1 = −1 A particular solution for u is 1 1 u = 1 + A2 − A2 cos 2θ 2 6 The the solution for v is v(φ) = 1 + A cos((1 − )φ) + 1 1 1 + A2 − A2 cos(2(1 − )φ) + O( 2 ) 2 6 2188 Solution 49.7 Substituting the expressions for x and ω into the differential equations yields 2 a2 ω0 d2 x2 + x2 dθ2 2 + α cos2 θ + a3 ω0 d2 x3 +... cos 2θ) 2 6 0 Equating the coefficent of a3 gives us the differential equation 2 ω0 d2 x3 + x3 dθ2 α2 5α2 + 2 − 2ω0 ω2 + 2 3ω0 6 0 α2 α2 cos θ + 2 cos 2θ + 2 cos 3θ = 0 3ω0 6 0 To avoid secular terms we must have 5α2 12ω0 Solving the differential equation for x3 subject to the intial conditions x3 (0) = x3 (0) = 0, ω2 = − x3 = α2 (−48 + 29 cos θ + 16 cos 2θ + 3 cos 3θ) 4 144ω0 Thus our solution for x(t)... Assuming intersections do occur, what is the time of the first intersection? You may assume that f is everywhere continuous and differentiable 2 3 Apply this to the case where f (x) = 1 − e−x to indicate where and when a shock will form and sketch (roughly) the solution both before and after this time Exercise 50.2 Solve the equation φt + (1 + x)φx + φ = 0 in − ∞ < x < ∞, t > 0, with initial condition... in the form ∂ρ ∂σ + =0 ∂t ∂x ∂ρ = ασ − βρ − γρσ; ∂t α, β, γ = positive constants (50.3) (50.4) 1 Investigate wave solutions in which ρ = ρ(X), σ = σ(X), X = x − U t, U = constant, and show that ρ(X) must satisfy an ordinary differential equation of the form dρ = quadratic in ρ dX 2 Discuss ths “smooth shock” solution as we did for a different example in class In particular find the expression for U in... φ(x(t), t) = ce−t , φ(ξ, 0) = f (ξ), ξ = (x + 1)e−t − 1 φ(x, t) = f ((x + 1)e−t − 1)e−t 2202 1 0.8 0 .6 0.4 0.2 -3 -2 -1 1 0.8 0 .6 0.4 0.2 -3 -2 -1 1 2 3 1 0.8 0 .6 0.4 0.2 -3 -2 -1 1 2 3 1 2 3 1 0.8 0 .6 0.4 0.2 -3 -2 -1 1 2 3 Figure 50.1: The solution at t = 0, 1/2, 1, 1. 165 82 Thus the solution to the partial differential equation is φ(x, t) = f ((x + 1)e−t − 1)e−t Solution 50.3 αφ dφ = φt + x (t)φx... characteristic polynomial has two pure imaginary roots ±ıµ and one real root, then it has the form (λ − r)(λ2 + µ2 ) = λ3 − rλ2 + µ2 λ − rµ2 Equating the λ2 and the λ term with the characteristic polynomial yields r=− 41 , 3 µ= 8 (10 + R) 3 Equating the constant term gives us the equation 160 41 8 (10 + Rc ) = (Rc − 1) 3 3 3 which has the solution 470 19 For this critical value of R the characteristic polynomial... values of ρ as X → ±∞, and find the sign of dρ/dX Check that U= σ2 − σ1 ρ2 − ρ1 in agreement with the “discontinuous theory.” Exercise 50.8 Find solitary wave solutions for the following equations: 1 ηt + ηx + 6 ηx − ηxxt = 0 (Regularized long wave or B.B.M equation) 2 utt − uxx − 3 2 u xx 2 − uxxxx = 0 (“Boussinesq”) 3 φtt − φxx + 2φx φxt + φxx φt − φxxxx = 0 (The solitary wave form is for u = φx ) 4 ut... between the h equation obtained from (1) and the h equation obtained from (2) There will be two possible choices for V (h) depending on a choice of sign Consider each case separately In each case fix the arbitrary constant that arises in V (h) by stipulating that before the waves arrive, h is equal to the undisturbed depth h0 and V (h0 ) = 0 Find the h equation and the wave speed c(h) in each case 2198... 8(10 + R)λ + 160 (R − 1) = 0 Linearizing about the point − 8 (R − 1), − 3 8 (R − 1), R − 1 3 yields    ˙ X  ˙ Y  =   ˙ Z − −10 10 1 −1 8 (R 3 − 1) − 8 (R 3 0 − 1)    X 8 (R − 1)  Y  3  8 Z − 3 The characteristic polynomial of the matrix is λ3 + 160 41 2 8(10 + R) λ + λ+ (R − 1) 3 3 3 21 86  Thus the eigenvalues of the matrix satisfy the polynomial, 3λ3 + 41λ2 + 8(10 + R)λ + 160 (R − 1) =... method of characteristics to solve the problem with c = F (x) at t = 0 (µ is a positive constant.) 2 Find equations for the envelope of characteristics in the case F (x) < 0 3 Deduce an inequality relating max |F (x)| and µ which decides whether breaking does or does not occur Exercise 50 .6 For water waves in a channel the so-called shallow water equations are ht + (hv)x = 0 1 (hv)t + hv 2 + gh2 = 0, g . terms, we must have ω 1 = −1. A particular solution for u is u = 1 + 1 2 A 2 − 1 6 A 2 cos 2θ. The the solution for v is v(φ) = 1 + A cos((1 − )φ) +   1 + 1 2 A 2 − 1 6 A 2 cos(2(1 − )φ)  + O( 2 ). 2188 Solution. θ  . We make the assumptions that 0 <  < 1 and that f(y) is an odd function that is nonnegative for positive y and satisfies |f(y)| ≤ 1 for all y. Since sin θ is odd, sin θf  1 √  R sin. 0, x 3 = α 2 144ω 4 0 (−48 + 29 cos θ + 16 cos 2θ + 3 cos 3θ). Thus our solution for x(t) is x(t) = a cos θ + a 2  α 6 2 0 (−3 + 2 cos θ + cos 2θ)  + a 3  α 2 144ω 4 0 (−48 + 29 cos θ + 16 cos 2θ + 3 cos 3θ)  +