Math 24, Ordinary Differential Equations UCSC Summer Session II, 2003 Instructor: José Agapito - PDF

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Math 4, Ordinary Differential Equations UCSC Summer Session II, 00 Instructor: José Agapito Phase Portrait, x = A x, case deta = 0 If deta = 0, then at least one eigenvalue is zero. There are basically

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Math 4, Ordinary Differential Equations UCSC Summer Session II, 00 Instructor: José Agapito Phase Portrait, x = A x, case deta = 0 If deta = 0, then at least one eigenvalue is zero. There are basically three cases to study. This does not mean that a matrix with determinant equal to zero is going to be one of these cases but its reduced echelon matrix will be similar to one of the matrices to be shown below or a combination of them. I) Real and distinct eigenvalues λ = α 0, λ = 0 A = ( α ) x = αx x = 0 Equilibrium points. Solve simultaneously αx = 0 and 0 = 0. We obtain that x = 0. This means that each point on the line x = 0 is an equilibrium point. At any other point (x, x ), there is no vertical direction (x = 0). Therefore, the phase portrait has a vertical line of equilibrium points and horizontal arrows pointing towards or away from that vertical line. This will depend on the sign of α. Positive sign means pointing outwards, negative sign means pointing inwards. To find the general solution to this system, we have a short way and a long way. Let s do it the long way. First, we compute the eigenvectors corresponding to these eigenvalues; i.e. you solve for (u, u ) in the system (( ) ( )) ( ) ( ) α 0 0 u 0 (A λi) u = 0 λ = u 0 ( ) ( ) ( ) 0 0 u 0 λ = α : = u 0 α u 0 = 0, u is free. We can then pick, for instance, v = (,0) as an eigenvector corresponding to λ. ( ) ( ) ( ) α 0 u 0 λ = α : = u 0 0 u 0 = 0, u is free. We can then pick, for instance, v = (0,) as an eigenvector corresponding to λ. Now, the general solution to the system above is ( ) ( ) ( x(t) = c e αt + c 0 e 0t 0 c e = λt c ). What is the short way? II) Both eigenvalues are zero λ = 0, λ = 0 A = ( ) x = 0 x = 0 Then, all points are equilibrium points. The general solution is (again, using the long way) ( ) ( ) ( ) x(t) = c e 0t + c 0 e 0t 0 c =. c What is the short way? III) Both eigenvalues are zero, but A is nilpotent λ = 0, λ = 0 A = ( ) x = x x = 0 Equilibrium points. Solve simultaneously x = 0 and 0 = 0. We obtain that x = 0. This means that each point on the line x = 0 is an equilibrium point. At any other point (x, x ), there is no vertical direction (x = 0). Therefore, the phase portrait has a horizontal line of equilibrium points and horizontal arrows pointing left or right depending whether they are above or below x = 0. To find the general solution to this system, we have a short way and a long way. Let s do it the long way. Again, we compute the eigenvectors corresponding to these eigenvalues; i.e. you solve for (u, u ) in the system ( ) ( ) ( ) 0 u 0 = u 0 0 u 0 = 0, u is free. We can then pick, for instance, v = (,0) as an eigenvector corresponding to λ = λ = 0. There is no way to compute a second non parallel eigenvector v. In this case, one computes what is called a generalized eigenvector. That is, one looks for a solution to the system (A λi) v = v. We find that v can be taken to be (0, ). Then, the general solution is ( ) [ ( ) ( )] ( ) x(t) = c e 0t + c 0 te 0t + e 0t 0 c + c = t 0 What is the short way? c Math 4 - Ordinary Differential Equations Instructor: José A. Agapito Ruiz Summer Session II - 00 (July 8 - August 9) Review Midterm The following problems have been taken from the textbook and the reference book. For your convenience, I have written them down explicitly. They are mostly homework, supplementary and miscellaneous problems, some of them have been slightly modified to fit better our purposes. You should also review your class notes to complement your preparation and be in better shape for the midterm. You will find the answers to these problems generally at the back of the two books mentioned. This said, here is a beautiful way of explaining what I expect from you here, in the midterm, homework and in the final project : In working out your solutions, pretend you and I are having a discussion and that your job is to convince me, with an argument that is neither longer nor shorter than necessary, of the correctness of your position. If the right answer is it is never enough to say, but brevity will be rewarded.. Consider the initial value problem y y = t + et, y(0) = y o. Find the value of y o that separates solutions that grow positively as t from those that grow negatively. How does the solution that corresponds to this critical value of y o behave as t +?. Show that if a and λ are positive constants, and b is any real number, then every solution of the equation y + ay = be λt has the property that y 0 as t +. Hint: Consider the cases a = λ and a λ separately.. Solve the given differential equation by first, using the method of integrating factor and then, using variation of parameters. y y = t e t. 4. Consider the intial value problem y = ty (4 y), y(0) = y o. (a) Determine how the behavior of the solution as t increases depends on the initial value y o. Introduction to Bayesian Statistics, by David Draper, graduate course at UCSC, Winter 00. (b) Suppose that y o = 0.. Find the time T at which the solution first reaches the value.98.. Consider the equation (a) Show that the given equation is homogeneous. (b) Solve the differential equation. dy + y = 4x dx x + y. (c) Draw a direction field and some integral curves. Are they symmetric with respect to the origin? 6. Without solving the problem, determine an interval in which the solution of the given initial value problem is certain to exist: (4 t )y + ty = t, y( ) =. 7. (a) Verify that both y = t and y (t) = t /4 are solutions of the initial value problem y = t + (t + 4y) /, y() =. Where are these solutions valid? (b) Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of Theorem.4. in the textbook. (c) Show that y = ct + c, where c is an arbitrary constant, satisfies the differential equation in part (a) for t c. If c=-, the initial condition is also satisfied, and the solution y = y (t) is obtained. Show that there is no choice of c that gives the second solution y = y (t). 8. Sketch the graph of f(y) versus y. Determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable. dy dt = y( y ), y o Construct a direction field for the differential equations: (a) y = y x. (b) y = t + y. (c) v = u/. (d) x = x(x ). 0. Consider the equation u n+ = ρu n ( u n ), where ρ . (a) Draw a qualitatively correct stairstep diagram and thereby show that if u o 0, then u n as n +. (b) In a similar way determine what happens as n + if u o .. Solve the following differential equations and initial value problems: (a) dy dx = x y. x (b) (x + y)dx + (x + y)dy = 0, y() =. (c) dy dx = y, y(0) =. xy (d) ẍ 0ẋ + 64x = 0, x(0) =, ẋ(0) = 0. (e) ẍ 0ẋ + x = 0, x( ) = 0, ẋ( ) = 0. (f) ẍ + ẋ + x = 0, x() = 0, ẋ() =. Math 4 - Ordinary Differential Equations Instructor: José A. Agapito Ruiz Summer Session II - 00 (July 8 - August 9) Solution to some Review Midterm problems. Using variation of parameters, assume y = ϕe t. Then (ϕ e t + ϕe t ) ϕe t = t + e t. From here, which leads to ϕ = ϕ = te t + e t, te t dt + e t dt + C = te t 4 e t 4e t + C Therefore, Using the initial condition, which yields to y(t) = ( te t 4 e t 4e t + C)e t y 0 = C C = y 0 + 6, y(t) = t 4 4et + (y )e t ) = e ( t t 4/ 4et + (y e t e t e t ) As t +, the solution y(t) + if y o + 6 0; y(t) if y o + 6 y o = 6. 0 and y(t) if. Case (a = λ) y + λy = be λt. Using, for instance, variation of parameters, y = (bt + C)e λt. Case (a λ) In this case, y = y + ay = be λt. b a λ e λt + Ce at. Since λ and a are positive constants, the solution y(t) 0 as t + in both cases. 4. Consider the intial value problem y = ty (4 y), y(0) = y o. (a) First of all, right away we can see that there are two equilibrium solutions; namely y = 0 and y = 4. Now, assume y 0 and y 4. Using separable equations, we get t y(4 y) dy = dt ( 4 ) 4 y dy + y dy = t 6 + C ln y 4 y = t + C y 4 y = Ce t y = 4Ce t + Ce t Using the initial condition, y 0 = 4C + C Therefore, we can write the solution y(t) as C = y 0 4 y 0. 4 y(t) = ( ). 4 y0 y 0 e t + Conclusion. If y 0 = 4, we stay there forever because y(t) = 4 is an equilibrium. If y 0 = 0, we stay there forever again because y(t) = 0 is also an equilibrium. Assuming y 0 0 and y 0 4, regardless the value of y 0, the solution y(t) 4 as t. (b) Suppose that y o = 0.. After a simple evaluation in the formula above and solving for T in y(t ) =.98, we obtain T.96.. Consider the equation dy + y = 4x dx x + y. (a) This equation is homogeneous in the sense that if we divides the right hand side by x, we get dy dx = 4 + y x + y. x That is, the right hand side is a function depending on y x only. (b) Setting v = y x, we solve the corresponding differential equation derived from the original one, v + v + 4 v + = x dv dx, provided v and x 0. Furthermore, suppose v, 4. Then v + (v + 4)(v + ) dv = x dx and so ln v ln v + = ln x + C. Simplifying and writing at the end everything in terms on x and y, this can be written as where C, as usual, is an arbitrary constant. y + x y + 4x = C (c) Warning. If you use the mathematical visualization toolkit to get an sketch of the direction field and some integral curves for this differential equation, you will get a nonsense answer. You d better off sketching the graph by hand. Hint: Find the points where dy dy dx = 0 and where dx does not exist. You will have 4 regions to analize in the plane XY. You should get symmetric integral curves with respect to the origin. 6. First, we write (4 t )y + ty = t, y( ) = in standard form. This is y + t (4 t ) y = t (4 t, y( ) =. ) For this to make sense, t ±. Then, I have three intervals to consider: (, ), (, ) and (, + ). I take the one containing the initial point t 0 = ; namely (, ). 7. (a) y = t. = y = t + t + 4 4t = t + t = { t+t if t 0 t t+ if t 0 Clearly, the first branch of the expression above is the one that satisfies the differential equation with the initial condition y() =. This is solution is valid for t. y = t 4. t = y = t + t + 4( t 4 ) = t Here, the initial condition is easily verified. This solution is valid for any t R. (b) Notice that f(t, y) = t+ t +4y and so f y (t, y) = t +4y. This shows that f y is not continuous at (, ); therefore, it does not contradict the uniqueness part of Theorem.4. in the textbook. (c) y = ct + c. c = y = t + t + 4ct + 4c = = t + t + c { t+t+c if t + c 0 t t c if t + c 0 Clearly, we need to choose the first branch in the expression above to get the equality claimed. Then, t c. If c =, the initial condition is also satisfied and the solution y = y is obtained. If y = y, this yields to t = c. But c is meant to be a constant! Therefore, there is no choice of c that gives the second solution y = y. 8. Use the mathematical visualization toolkit (MVT) or any other graphing device to plot f(y) = y( y ) versus y. Using analisis of phase diagrams as done in class, you have that y = is asymptotically stable, y = 0 is unstable and y = is also asymptotically stable. 9. Use the MVT or any other graphing device to check your plots. The plots obtained with the MVT are correct.. (c) Forget about this differential equation. 4 Math 4, Ordinary Differential Equations UCSC Summer Session II, 00 Instructor: José Agapito Midterm Name: Solution All 6 questions should be answered on this exam using the backs of the sheets if necessary. Show all your work and justify your answers clearly. Be neat. Do your best!! / /0 / 4 /0 /0 6 / neat work / Total /00. ( points) For each of the questions below, indicate if the statement is true or false. Only here, you do not need to justify your answer. Each correct answer is worth point. a b c d e F F T F T (a) The differential equation y + y = y is linear. A first order linear differential equation is of the form y + p(t)y = g(t). (b) A second order linear differential equation is homogeneous if it has constant coefficients. A second order linear differential equation is of the form y + p(t)y + q(t)y = g(t). For this to be homogeneous, g(t) 0. (c) For the differential equation y = t/, y(t) = 0 is not an equilibrium solution. An equilibrium solution to a differential equation is of the form y(t) = c, for any t R. (d) Given a first order linear differential equation with an initial point, the method of integrating factor and variation of parameters yield two different solutions. They both produce the same answer provided the hypotheses of the fundamental theorem of ordinary differential equations hold (Theorem on the existence and uniqueness of solutions to IVP). (e) The problem y = t(t ), y(t o ) = y o has a unique solution. Setting f(t, y) = t(t ), we can see that both f and f y are continuous on R. Therefore, for any initial condition, this problem has a unique solution. . Let α . Suppose y(t) is the solution to a given initial value problem (I.V.P) Where is this solution valid? Identify the second order linear differential equation that originated the solution y(t) and describe the behavior of y(t) as t +, t and t 0. a) (0 points) y(t) = e αt + e (α )t. First, there is no restriction to impose on t in the solution y(t) described above. Then, this solution is valid in R. Second, since α is a real number (α ) and y(t) is a linear combination of e αt and e (α )t, its corresponding characteristic equation is (r α)(r (α )) = 0, which yields to the homogeneous differential equation y (α )y + α(α )y = 0 Finally, notice that we can also write y(t) as y(t) = e αt ( + e t ). Then lim t + eαt ( + e t ) =, lim t eαt + e (α )t = 0 and lim e αt + e (α )t =. t 0 b) (0 points) y(t) = e αt e (α )t. First, there is no restriction to impose on t in the solution y(t) described above. Then, this solution is valid in R. Second, since α is a real number (α ) and y(t) is a linear combination of e αt and e (α )t, its corresponding characteristic equation is (r α)(r (α )) = 0, which yields to the homogeneous differential equation y (α )y + α(α )y = 0 Finally, notice that we can also write y(t) as y(t) = e αt ( e t ). Then lim t + eαt ( e t ) = +, lim t eαt e (α )t = 0 and lim e αt e (α )t =. t 0. ( points) Define a second order difference equation. Give an example. How do you find equilibrium solutions of a second order difference equation? Hint: Get a clue from first order difference equations. A second order difference equation is an expression of the form y n+ = f(n, y n, y n ), where n denotes a natural number. In other words, y n+ depends, in general, on n, y n and y n, and this dependence may be linear or not. An example is y n+ = y n y n. If there is an equilibrium, then there is a n such that for any n n, y n+ = y n holds. Using this last equation is how you find equilibrium solutions. 4. Consider the differential equation y + y = y. a) (0 points) Find all possible solutions to this equation. We can write y + y = y as dy dt = y y. Notice that if y = 0 or y =, the differential equation is satisfied. Therefore, they are equilibrium solutions. Now, suppose y 0 and y, then ( dy y y = dt y + ) dy = dt. y Integrating both sides, we get ln y ln y = t + C, where C is an arbitrary constant. Now, we exponentiate both sides and get rid of the absolute value in the left hand side by allowing our arbitrary constant to take negative values too. Then, y y = Ce t. () We solve for y(t) and get y(t) = Ce x () In summary, all possible solutions are y(t) = 0, y(t) = and y(t) as in (). b) ( points) Classify the equilibrium solutions as asymptotically stable, unstable, or semistable. The diagram below shows that y = 0 is asymptotically stable and y = is unstable Figure : Phase diagram of y + y = y. c) ( points) Suppose y(0) =. Determine y(t) explicitly. Using Formula (), we find C =. Therefore, the explicit solution y(t) with the given initial condition y(0) = is y = et. . Solve a) (0 points) dy dx = x + y xy y. We can write the given differential equation as (xy y)dy + (x + y )dx = 0. Setting M = x + y and N = xy y, we find that M y = y and N x = y. Therefore, our differential equation is exact. This means that there exists a function ψ defined on some open rectangle such that dψ = (xy y)dy + (x + y )dx. Therefore, we have Let us integrate with respect to x; we get ψ x = x + y and ψ y = xy y. ψ = x + y x + h(y). And now, let us differentiate ψ with respect to y and use the formula for ψ y above, to get xy y = ψ y = xy + d d h(y) = h(y) = y. dy dy Integrating with respect to y we obtain h(y) = y zero, there is no harm. In summary, we have + C. We can choose the constant C to be ψ = x + y x y = K, where K is an arbitrary constant. This is so because dψ = 0 on an open rectangle. You can also write y = k x 6x. b) (0 points) y + y + y = 0, y(0) =, y (0) = 0. The characteristic equation of this differential equation is r + r + = 0. Using the quadratic formula, we get r = ± i. Therefore, the general solution to this homogeneous equation is y(t) = C e t cos t + C e t sin t. Using the initial conditions we obtain the system = C and 0 = + C. In summary, the solution to this initial value problem is ( y(t) = e t cos t + sin ) t. 6. Direction fields. a) ( points) Does the plot below correspond to an autonomous or non-autonomous differential equation? Why? First, the vertical axis corresponds to the dependent variable, say y, and the horizontal to Figure : Plot y vs. t the independent one, say t, as usual. Thus, we have the plot of the direction field of a nonautonomous differential equation, because the directions (denoted by the arrows) for a given value of y change as t varies. b) (0 points) Construct a direction field for y = t y. Figure : Plot y vs. t Math 4 - Ordinary Differential Equations Instructor: José A. Agapito Ruiz Summer Session II - 00 (July 8 - August 9) Final You are free to use any combination of paper/pencil and/or packages like Mathematica, Maple or Matlab for calculations, graphs, reports, etc. If you know LaTeX, feel free to use it to present your final work. However you decide to present your work, please be always neat and clear. Remember, in preparing your solutions, keep in mind the style I mentioned last time as a guide: your job is to convince me, with an argument that is neither longer nor shorter than necessary, of the correctness of your position. If the right answer is it is never enough to say, but brevity will be rewarded.. (a) Find the general solution of the given system of equations and describe the behavior of the solution as t. Also draw a direction field and plot a few trajectories of the system. Remember to justify your work. (i) x = ( 4 ) ( x, (ii) x = ) ( x, (iii) x = (b) Solve the given value problem and describe the behavior of the solution as t. x = 0 x, x(0) = (c) Find the general solution of the given system of equations x = 4 x + 4 x + t x = 4 x 4 x + e t. (Taken from The Boston University Ordinary Differential Equations Project. This is the default project) Here you will study a nonlinear, first-order system known as the Predator-Prey model. The equations are dx dt = (9 αx y)x dy dt = ( + x)y where α 0 is a parameter. In other words, for different values of α we have different systems. The variable x is the population (in some scaled units) of prey (rabbits), and y is the population of predators (foxes). For a given value of α, we want to understand what happens to these populations as t. 0 ) x Your job is to investigate the phase portraits of these equations for various values of α in the interval 0 α. To get started, you might want to try α = 0,,,, 4 and. Think about what the phase portrait means in terms of the evolution of the x and y populations. Where are the equilibrium points? What types are they? What happens to a typical solution curve? Also, consider the behavior of the special solutions where either x = 0 or y = 0 (that is, solution curves lying on the x- or y-axes). Determine the bifurcation values of α. Namely, determine the values of α
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