MATH 215/255 Solutions to Additional Practice Problems April dy dt

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1 . For the nonlinear system MATH 5/55 Solutions to Additional Practice Problems April 08 dx dt = x( x y, dy dt = y(.5 y x, x 0, y 0, (a Show that if x(0 > 0 and y(0 = 0, then the solution (x(t, y(t of the nonlinear system with initial value (x(0, 0 at t = 0 satisfies y(t = 0 for all t. What differential equation does x(t satisfy? Sketch the phase portrait for this (one-dimensional differential equation in the x-axis. Give a similar analysis if x(0 = 0 and y(0 > 0. Solution: Substitute (x, y = (x(t, 0 into both differential equations in the system, and get dx(t dt = x(t( x(t, d dt 0 = 0, so the second equation is trivially satisfied, and the first equation is satisfied if x(t is a solution of x = x( x, a logistic equation of the type studied in Chapter. The phase portrait can be obtained by plotting F (x, 0 = x( x as a function of x, and interpreting the plot to obtain the phase portrait in the x-axis (x 0 only. Similarly, substituting (x, y = (0, y(t into the system, we get the first equation satified trivially, and the second equation satisfied if y(t is a solution of y = y(.5 y, with the phase portrait in the y-axis (y 0. For comparison to part (e later, it is helpful to plot the y-axis vertically. (b Find all equilibrium solutions (critical points (x 0, y 0 with x 0 0 and y 0 0. Solution: To find equilibrium solutions, set x = 0 and y = 0, so we must solve the algebraic system of equations 0 = x( x y, 0 = y(.5 y x. Thus (x = 0 or x y = 0 and (y = 0 or.5 y x = 0 and there could be as many as four possiblities: i. x = 0 and y = 0, ii. iii. iv. x y = 0 and y = 0, x = 0 and.5 x y = 0, x y = 0 and.5 y x = 0. In case iv however, there are no solutions (you can think of the algebraic system x + y =, x + y =.5 as finding the intersection points of two parallel lines in the xy-plane: there are none. The corresponding equilibrium solutions (x 0, y 0 are i. (0, 0, ii. (, 0, iii. (0,.5.

2 All three of these satisfy x 0 0 and y 0 0. (c For each equilibrium solution, examine the linearization (the corresponding linear system and when possible determine the type and stability of the equilibrium solution. Solution: We obtain the linearization at each equilibrium solution by evaluating the Jacobian matrix of the vector field at each point. The type and stability is determined using Theorem (you do not have to quote the theorem, just use it correctly. For any (x, y the vector field is ( F (x, y G(x, y ( x x xy = 3 y y xy and the Jacobian matrix is ( ( Fx (x, y F J(x, y = y (x, y x y x = G x (x, y G y (x, y 3 y y x which we need to evaluate at each of the equilibrium solutions to get the linearizations. i. At the equilibrium solution (0, 0, the linearization is diagonal: ( ( ( u 0 u v =, v and it has real unequal eigenvalues 0 3 r =, r = 3, with the same signs, both positive, so (by Theorem (0, 0 is a node and it is unstable. ii. At the equilibrium solution (, 0, the linearization is (upper-triangular: ( ( ( u u v =, v it has real unequal eigenvalues 0 r =, r =, with opposite signs, so (, 0 is a saddle point and it is unstable. iii. At the equilibrium solution ( 0, 3 the linearization is (lower-triangular: ( u and it has real unequal eigenvalues v = ( ( u v r =, r = 3, with the same signs, both negative, therefore ( 0, 3 is a node and it is asymptotically stable.,

3 (c Plot the nullclines and direction field. Solution: The x = 0 nullcline consists of the y-axis x = 0 together with the line y = x +, and the y = 0 nullcline consists of the x-axis y = 0 together with the line y = x + 3. See Figure (only x 0 and y 0 is shown. Optional: If time permits, it can be useful to find stable and unstable eigenvector directions at any saddle points, and the weak and strong eigenvector directions at any nodes, to help plot the phase portrait more accurately. Otherwise, these directions can be guessed from the nullclines and direction field. Stable and unstable eigenvector directions at saddle points, or weak and strong eigenvector directions at nodes, must be consistent with the nullclines and direction field. Any inconsistencies would be due to mistakes in calculation, or in interpretation (e.g. arrows pointing in wrong directions, mistakes in plotting. At the unstable node (0, 0, for the weak stable eigenvalue (the eigenvalue with smaller absolute value r = the eigenvector direction is exactly horizontal, ( ξ ( = 0 and for the strong stable eigenvalue (the eigenvalue with larger absolute value r = 3 the eigenvector direction is exactly vertical, ( 0 ξ ( =. At the saddle point (, 0, for the negative eigenvalue r = the stable eigenvector direction is exactly horizontal, ( ξ ( = 0 and for the positive eigenvalue r = an unstable eigenvector is ( ξ ( = which has negative slope, steeper than 45 (the slope of the x = 0 nullcline at (, 0 is. At the asymptotically stable node ( 0, 3, for the weak unstable eigenvalue (the eigenvalue with smaller absolute value r = an eigenvector direction is ( ξ ( =, which has negative slope, steeper than the y = 0 nullcline at ( 0, 3, and for the strong unstable eigenvalue (the eigenvalue with larger absolute value r = 3 the eigenvector direction is exactly vertical, ( 0 ξ ( =

4 (e Sketch, by hand (without using any graphing device the phase portrait. (It must be consistent with the results of parts (a, (b, (c, (d. Solution: On the coordinate axes, the phase portraits should be as determined in part (a. The equilibrium solutions from part (b should be plotted in their correct positions. By Theorem 9.3.3, near a node or saddle point (or spiral point, but there are none in this problem x = x 0, y = y 0, the phase portrait for the nonlinear system should look like a small version of the phase portrait for the linearization from part (c, near the origin u = 0, v = 0, but here we only see the portions corresponding to x 0 and y 0. Trajectories should have slopes of their tangent lines consistent with the directions obtained in part (d, in particular they should intersect the x = 0 nullcline with vertical tangents in the correct direction (upwards or downwards, according to part (d, and they should intersect the y = 0 nullcline with horizontal tangents in the correct direction (leftwards or rightwards, according to part (d. Notice that on the coordinate axes, the nullclines and trajectories coincide, consistent with part (a. See Figure. [Optional: If you calculated eigenvectors, trajectories should leave saddle points in general tangent to the unstable eigenvector direction, and approach tangent to the stable eigenvector direction. Here trajectories approach exactly in the stable eigenvector direction because of part (a. At stable nodes in general, trajectories approach tangent to the weak eigenvector direction, except for two that approach tangent to the strong eigenvector direction. Again because of part (a, two trajectories approach exactly in the strong eigenvector direction. At unstable nodes in general, trajectories leave tangent to the weak eigenvector direction, except for two that leave tangent to the strong eigenvector direction. Again note part (a, and also note we only show x 0 and y 0.] Figure : Nullclines and direction field (left panel, and phase portrait (right panel for the coupled nonlinear system. (f What is the behaviour of solutions (x(t, y(t with x(0 0, y(0 0 as t (there 4

5 are several cases? Solution: [Note that you could answer some of part (f without having done part (e.] Case : If (x(0, y(0 is an equilibrium solution (x 0, y 0, then the solution of the initial value problem is constant for all t, it doesn t move, (x(t, y(t = (x 0, y 0 for all < t < therefore in each of the three subcases i., ii., iii., (in particular for the equilibrium solution at the origin we have lim (x(t, y(t = t (x0, y 0 (if x(0 = x 0, y(0 = y 0. Case : If y(0 = 0 then (see part (a y(t = 0 for all t, the solution (x(t, y(t = (x(t, 0 stays on the x-axis for all t, and the dynamics on the x-axis can is determined by x = x( x. The result should be consistent with the result of part (d. In this case (which overlaps with Case at the equilibrium solution (, 0 lim (x(t, y(t = (, 0 (if x(0 > 0, y(0 = 0. t Case 3 : Similarly, if x(0 = 0 then (see (a x(t = 0 for all t, the solution (x(t, y(t = (0, y(t stays on the y-axis for all t, and the dynamics on the y-axis are determined by y = y ( 3 y. In this case (which overlaps with Case at the equilibrium solution ( 0, 3 lim (x(t, y(t = ( 0, 3 (if x(0 = 0, y(0 > 0. t Case 4 : For almost all initial values (x(0, y(0, strictly inside the first quadrant, from part (e we see that all solutions will approach the asymptotically stable node, lim (x(t, y(t = ( 0, 3 (if x(0 > 0, y(0 > 0. t Instead, we could separate the cases differently so they do not overlap and write lim (x(t, y(t = (0, 0 (if x(0 = 0, y(0 = 0, t lim (x(t, y(t = (, 0 (if x(0 > 0, y(0 = 0, t lim (x(t, y(t = ( 0, 3 (if x(0 0, y(0 > 0. t [As a population model, the intepretation is that if any nonzero amount of the species represented by the y-variable is introduced at time t = 0, it will eventually completely take over, driving the other species (represented by x to extinction.]. For the initial value problem y = y t +, y(0 =. use the Euler method to estimate values of the solution at t = 0.. (Use a calculator and show 4 significant digits in your results. (a Use step size 0.. (b Use step size

6 Solution: The Euler method formula y n+ = y n + h f(t n, y n applied to this problem is y n+ = y n + h [ y n t n + ]. With step size h = 0. we need only one step to reach t = 0.: y 0 = t 0 = 0 y = + (0.[ ( ] =.50 t = = 0. so we have the estimate φ(0..50 using the Euler method with one step of size 0.. (b With step size h = 0.05 we need two steps to reach t = 0.: y 0 = t 0 = 0 y = + (0.05[ ( ] =.5 t = = 0.05 y =.5 + (0.05[ ( ] =.60 t = = 0. so we have the estimate φ(0..60 using the Euler method with two steps of size Repeat parts (a, (b for y = 5t 3 y, y(0 =. Solution: (a The Euler method formula y n+ = y n + h f(t n, y n applied to this problem is y n+ = y n + h ( 5t n 3 y n. With step size h = 0. we need one step to reach t = 0.: y 0 = t 0 = 0 y = + (0.[ 5(0 3 ] =.576 t = = 0. so we have the estimate φ( using the Euler method with one step of size 0.. (b With step size h = 0.05 we need two steps to reach t = 0.: y 0 = t 0 = 0 y = + (0.05[ 5(0 3 ] =.788 t = = 0.05 y = (0.05[ 5( ] =.600 t = = 0. 6

7 so we have the estimate φ( using the Euler method with two steps of size For the initial value problem y = t + y, y(0 = obtain a formula for the local truncation error for the Euler method in terms of t and the solution φ(t. Solution: The formula for the local truncation error for the Euler method with step size h is e n+ = φ ( t n h, where φ(t is the solution of the initial value problem and t n is some value between t n and t n+ = t n + h. Since φ(t is the solution of the differential equation, it must satisfy φ (t = t + [φ(t], and we have φ (t = d dt φ (t = d dt {t + [φ(t] } = t + φ(tφ (t and we express e n+ in terms of t and φ(t as = t + φ(t{t + [φ(t] } = [t + t φ(t + [φ(t] 3 } e n+ = { t n + t nφ( t n + [φ( t n ] 3 }h, where t n is some value between t n and t n+ = t n + h. 5. For the initial value problem y = y t +, y(0 = (a Obtain a formula for the local truncation error for the Euler method in terms of t and the solution φ(t. Solution: The general formula for the local truncation error for the Euler method with step size h is e n+ = φ ( t n h, where φ(t is the solution of the initial value problem, and t n is some number, generally unknown, between t n and t n+ = t n + h. Since φ(t is a solution of the differential equation, it satisfies φ (t = φ(t t, and therefore φ (t = d dt φ (t = d dt [ φ(t t + ] = φ (t = [ φ(t t + = 4φ(t t ] 7

8 and we can express e n+ in terms of t and φ(t as where t n is some value between t n and t n+. e n+ = [ φ( t n t n ]h, (b Using the result of part (a, obtain an upper bound for the absolute value of the local truncation error that is valid for 0 t 0.. Solution: Using the triangle inequality, we get an upper bound valid for 0 t 0., e n+ = φ( t n t n h φ( t n + t n h ( max φ(t + 0. h. 0 t 0. (c Solve the initial value problem exactly to find a formula for φ(t. Solution: To solve the initial value problem, observe that the differential equation is linear, so it can be solved with an integrating factor. Write the differential equation in standard form as y y = t, multiply by the integrating factor µ(t = e t, e t y e t y = e t te t, recognize the left-hand side as the derivative of a product (recall that is the reason for the integrating factor! ( e t y = e t te t, integrate (integration by parts needed, e t y = ( e t te t dt = te t + c, where c is an arbitrary constant, and solve for y to get the general solution y = φ(t = t + cet, where c is an arbitrary constant. The initial condition φ(0 = gives c = and therefore φ(t = t + et. (d Use the formula for φ(t (the result of part (c to obtain a more accurate upper bound for the absolute value of the local truncation error that is valid for 0 t 0.. Solution: From part (c we have φ(t = t + et. 8

9 Differentiating twice or using the expression in part (a (they should give the same result! gives φ (t = 4e t, and therefore e n+ = e t n h, and we get another (presumably more accurate upper bound for the absolute value of the local truncation error, valid for 0 t 0., e n+ max 0 t 0. et h. (e For step size 0. use the results of parts (b and (d to compute (with a calculator the two upper bounds for e, the absolute value of the local truncation error in the first step, and compare them with the actual error e (accurate to 4 significant digits at t = 0. (see Questions (a, 5(c. (Your two upper bounds should be verified, but one is more accurate than the other. Solution: Because when the formula for the local truncation error is derived we assume the previous value from the Euler method is exact, y n = φ(t n, this can only be true if n = 0, so only e is realistic to calculate. To use the bound from part (b, we note that φ(t = t + et is an increasing, positive function on 0 t 0., so it attains its maximum absolute value at the right endpoint t = 0., and the first upper bound for e evaluates to ( max φ(t + 0. h = ( ( e (0. = t 0. To use the bound from part (d, we note that e t is also an increasing, positive function that attains its maximum absolute value at t = 0., and the second upper bound for e evaluates to max 0 t 0. et h = e 0. (0. = , which is slightly smaller than the previous upper bound. Both are upper bounds, the smaller one is more accurate. The actual error (using the results of (a for y and 5(c for φ(t e = φ(t y = (0. + e0..5 = = 0.040, is positive and its absolute value is indeed less than or equal to both of upper bounds calculated from parts (b and (d. Moreover, the upper bounds are reasonably accurate: close to the actual absolute value, but of course larger. (Note that e is positive, which indicates that the solution is greater than the approximation. The solution is concave upwards. 9

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