% problem notes Ch9, S2009 9.1-4: ====== Equilibria [==critical points] are constant solutions x=const, y=const. Because the derivative of a constant is zero, equilibria are determined from the linear algebraic equations 2x - 2y - 4 = 0, x - y + 3 = 0. This system has unique solution x=1, y=-1, found by frame sequence methods applied to the augmented matrix of the system. Other methods: elimination with equations, matrix inversion, Cramer's Rule. Only Figure 18 has a single equilibrium (1,-1), therefore the classification is a spiral. 9.1-8: ====== Equilibria [==critical points] are constant solutions x=const, y=const. Because the derivative of a constant is zero, equilibria are determined from the non-linear algebraic equations x - y - x^2 + xy = 0, - y - x^2 = 0. Solve the second for y in terms of x, then substitute into the first equation to get a cubic equation for x. The cubic x - x^3 has no constant term, therefore x factors out. The roots of x(1-x)(1+x) = 0 are x = 0, x = 1, x = -1. Then y = -x^2 gives the pairs (0,0), (1,-1), (-1,-1). It matches Figure 16 in the textbook [resp, saddle, node, spiral]. 9.1-18: ====== Maple 12 can be used to visualise the phase portrait and classify the equilibria as a center, spiral, node, saddle. To find the tool: TOOLS --> BROWSE --> Diff Equations --> ODE --> Phase Portraits Choose COPY TO CLIPBOARD and then paste the copy into a worksheet. You will then be able to change the data and click on the plot to add threaded curves. The matrix form u'=Au of this problem is [x(t)] [0 -1] u = [ ], A = [ ] [y(t)] [4 0] Then det(A-rI)=0 is r^2+4=0 with complex roots r = 2i, r = -2i and corresponding atoms cos(2t), sin(2t). The Cayley-Hamilton method implies that the general solution is u(t) = cos(2t) c_1 + sin(2t) c_2 where c_1, c_2 are constant vectors. Differentiate this relation and use u'=Au to obtain the two equations u(t) = cos(2t) c_1 + sin(2t) c_2 Au(t) = -2 sin(2t) c_1 + 2 cos(2t) c_2 then set t=0 to obtain equations for c_1, c_2 as follows: u0 = (1) c_1 + (0) c_2 Au0 = (0) c_1 + (2) c_2 Define vector u0 from initial conditions x(0)=a, y(0)=b, then c_1 = u0, c_2 = (1/2)Au0 implies u(t) = cos(2t) c_1 + sin(2t) c_2 = cos(2t) u0 + (1/2) sin(2t) Au0 [a] [0 -1][a] = cos(2t)[ ] + (1/2) sin(2t) [ ][0] [b] [4 0][b] [ a cos(2t) - (b/2) sin(2t)] = [ ] [ b cos(2t) + 2a sin(2t)] [ cos(2t) (-1/2) sin(2t)][a] = [ ][ ] [ cos(2t) 2 sin(2t)][b] Observation: The matrix W in the last display is the exponential matrix W=exp(At), a fundamental matrix which satisfies W=AW, W(0)=I. This process always works to find the exponential matrix, using the Cayley-Hamilton method. A more direct method is Putzer's formula for the exponential matrix. 9.2-2: ====== The characteristic equation is r^2-5r+6=0 with roots 2, 3 and then the atoms are exp(2t), exp(3t). The Cayley-Hamilton-Ziebur method solution is then u'=Au, u = exp(2t) d_1 + exp(3t) d_2 [d_1, d_2 are vectors] By looking at possiblities for d_1, d_2 we consider the example selected by taking d_1, d_2 to be the columns of the 2x2 identity matrix: x(t) = exp(2t), y(t) = exp(3t) We try to classify the equilibrium point x=y=0 as a spiral, center, node or saddle. Because the example has no trig terms, then u'=Au cannot be classified as a spiral or center. Remaining are the node and saddle. Because x --> infinity and y --> infinity as t --> infinity, then the example solution is unstable at t=infinity. But the limits are zero at t=-infinity. Therefore, the example is stable at t=-infinity. A saddle is always unstable at both extremes t=infinity and t=-infinity. So the classification must be a node. It can be further sub-classified as an improper node, in this case by matching the diagram in Figure 4, using a maple 12 phase portrait tool. From the atoms in the example, it is deduced that the improper node is unstable. 9.2-12: ======= The Jacobian matrix is matrix([[1,-2],[1,4]]) with characteristic equation r^2-5r+6=0, the same one as in the previous problem 9.2-2. By the same reasoning as used in 9.2-2, the equilibrium point of this exercise is an improper unstable node. 9.2-22: ======= In order to deal with the unusual equilibria, it is suggested that you do part of this problem in maple 12. STEP 1. Fire up the phase portrait tool in maple 12 and enter the differential equation. To find the tool: TOOLS --> BROWSE --> Diff Equations --> ODE --> Phase Portraits Choose COPY TO CLIPBOARD and then paste the copy into a worksheet. You will then be able to change the data to 9.2-22 data and click on the plot to add threaded curves. The equilibria are [0, 0], [-3.650281540, .5923591472], obtained from the maple code eq:=x+4*y-x*y^2=0,2*x-y+x^2*y=0; evalf(solve({eq},{x,y})); PLOT WINDOW. The window for the phase portrait can be something like -5 < x < 1, -1 < y < 1 in order to show enough detail. This window contains both equilibria. Try -10 < t < 10, which makes the threaded curves cover the white space in the window. STEP 2. Compute the Jacobian J(x,y). STEP 3. Find matrices A1=J(0,0) and A2=J(-3.650281540,0.5923591472). STEP 4. Find roots of the characteristic equation for 2x2 matrix A1. Use the roots to find the atoms. Classify the equilibrium point (0,0) of the linear problem u' = A1 u, by considering the example x=atom1, y=atom2 and its stability properties at t=infinity, t=-infinity. If the atoms have sines and cosines, then the classification is either a spiral or a center. Otherwise, separate the node from the saddle by the fact that a saddle is unstable at both infinity and -infinity. Repeat with A1 replaced by A2. The two classifications are unstable node and saddle [which order?]. STEP 5. Apply the basic theorem. It says the classification usually transfers directly, so we expect to see in the nonlinear phase portrait an unstable node and a saddle. Only the center and node might have to be replaced by a spiral in the nonlinear diagram. This doesn't happen here, as verified by a maple plot of the nonlinear system. 9.3-8: ====== The phrase "show" used in this problem means "compute." No proof is indicated or expected. STEP 1. Find the equilibria. The answer is reported in the problem header, equilibria = (0,0), (0,14), (20,0), (12,6). The equilibria are the constant solutions of (3), which implies x' = y' = 0 in textbook equation (3), giving the two equations (60 - 3x - 4y)x=0 (42 - 3y - 2x)y=0 which are solved by considering the four possible ways to make the factors in each equation zero. STEP 2. Compute J(x,y). Start with the vector function F represented by the right sides of textbook equation (3), that is, [(60 - 3x - 4y)x] F = [ ] [(42 - 3y - 2x)y] Then the partials F_x, F_y of the vector function F are [60 - 6x - 4y] [- 4x ] F_x = [ ] F_y = [ ] [ - 2y ] [42 - 6y - 2x] Copy the two column vectors into the columns of J(x,y). STEP 3. Compute J(0,14). Substitute x=0, y=14 into the matrix J(x,y) to get [ 4 0 ] J(0,14) = [ ] [-28 -42] STEP 4. Find the characteristic equation roots. The roots are r=-42, r=4 [triangular matrix]. STEP 5. Analysis of u'=J(0,14)u. The atoms for roots -42, 4 are exp(-42t), exp(4t). At both t=infinity and t=-infinity the limit is not zero, so the equilibrium (0,0) of this linear problem is unstable at both infinity and -infinity. The classification is a saddle. 9.3-9: ====== Steps 3,4,5 are repeated from 9.3-8, for the equilibrium (20,0) of textbook equation (3). The steps: STEP 3. Compute J(20,0). Substitute x=20, y=0 into the matrix J(x,y) to get [-60 -80] J(20,0) = [ ] [ 0 2] STEP 4. Find the characteristic equation roots. The roots are r=-60, r=2 [triangular matrix]. STEP 5. Analysis of u'=J(20,0)u. 9.3-10: ======= Steps 3,4,5 are repeated from 9.3-8, for the equilibrium (12,6) of textbook equation (3). The steps: STEP 3. Compute J(12,6). Substitute x=20, y=0 into the matrix J(x,y) to get [-36 -48] J(12,6) = [ ] [-12 -18] STEP 4. Find the characteristic equation r^2+54r+72=0. The roots are r=-27+3 sqrt(73), r=-27-3 sqrt(73). Please derive the quadratic equation and show details from the quadratic formula. STEP 5. Analysis of u'=J(12,6)u. 9.4-4: ====== The phrase "show" used in this problem means a proof is indicated or expected. You are expected to use the phase portrait tool in maple 12 to complete the graphical steps. See 9.2-22 above for details about how to use it. ALMOST LINEAR proof. Definition. A system x'=f(x,y), y'=g(x,y) is ALMOST LINEAR if and only if f(x,y), g(x,y) are continuously differentiable and the Jacobian matrix J(x,y) of f and g satisfies det(J(0,0)) not zero. The proof uses STEP 1 below, which computes J(x,y). Then | 1 -2 | det(J(0,0)) = | | = 5 is not zero. | 4 -3 | The remainder of the proof details are technical, requiring a statement about the continuous differentiability of the component functions f(x,y)= sin(x) cos(y) - 2y, g(x,y) = 4x - 3 cos(x) sin(y). STEP 2. Compute J(x,y). Start with the vector function F represented by the right sides of the textbook equation, that is, for functions f(x,y)= sin(x) cos(y) - 2y, g(x,y) = 4x - 3 cos(x) sin(y) define F by [f(x,y)] F = [ ] [g(x,y)] Then the partials F_x, F_y of the vector function F are [cos(x)cos(y) ] [-sin(x)cos(y)-2] F_x = [ ] F_y = [ ] [4 + 3sin(x)sin(y)] [-3cos(x)cos(y) ] Copy the two column vectors into the columns of J(x,y). STEP 3. Compute J(0,0). Substitute x=0, y=0 into the matrix J(x,y) to get [ 1 -2 ] J(0,14) = [ ] [ 4 -3 ] STEP 4. Find the characteristic equation roots. The roots are complex, r=-1+2i, r=-1-2i. Display the characteristic equation and quadratic formula details. STEP 5. Analysis of u'=J(0,0)u. The atoms for roots are exp(-t)cos(2t), exp(-t)sin(2t). At t=infinity the limit is zero, so the equilibrium (0,0) of this linear problem is asymptotically stable at infinity. Because the atoms represent origin rotations, The classification is a spiral. 9.4-8: ====== The phrase "classify" used in this problem means to decide on the classification of the equilibrium points of the nonlinear system, by applying textbook Theorem 2 in 9.2, paraphrased as follows: Node, saddle, spiral, center linear pictures transfer to nonlinear pictures, except the node and center might have to transfer a spiral. You are expected to use the phase portrait tool in maple 12 to complete the graphical steps. See 9.2-22 above for details about how to use it. STEP 1. Find the equilibria. Equilibria = (n Pi,0), n = any integer, positive, negative or zero. The equilibria are the constant solutions, which implies x' = y' = 0 in the textbook equation, giving the two equations 3 sin(x) + y = 0 sin(x) + 2y = 0 These are solved by isolating y in the first equation, substitute y into the second equation to get sin(x) + 2(-3 sin(x)) = 0 The conclusion is sin(x)=0 or x = n Pi and then from the first equation y = 0. STEP 2. Compute J(x,y). Start with the vector function F represented by the right sides of the textbook equation, that is, for functions f(x,y)= 3 sin(x) + y, g(x,y) = sin(x) + 2y, define F by [f(x,y)] F = [ ] [g(x,y)] Then the partials F_x, F_y of the vector function F are [3 cos(x)] [ 1 ] F_x = [ ] F_y = [ ] [ cos(x)] [ 2 ] Copy the two column vectors into the columns of J(x,y). STEP 3. Compute J(n Pi,0). Substitute x=0, y=0 into the matrix J(x,y) and use the identity cos(n Pi) = (-1)^n to get [ 3(-1)^n -2 ] J(n Pi,0) = [ ] [ (-1)^n -3 ] STEP 4. Find the characteristic equation roots. Consider the two cases n even and n odd separately. Display the characteristic equation and quadratic formula details. STEP 5. Analysis of u'=J(n Pi,0)u. Case n even. The atoms for the roots are exp(r1 t), exp(r2 t) where r1, r2 are both positive. Display details to verify an unstable node. Case n odd. The atoms for the roots are exp(r1 t), exp(r2 t) where r1>0, r2<0. Display details to verify a saddle [always unstable]. %%===end of Ch9 problem notes======