Applied Dynamical Systems MATH11140 Solutions and comments May 2021 1. 

Consider the partial differential equation, u = 6uu 3u, (1) t x x3 on the unbounded domain x (, ). This equation supports travelling wave solutions called solitons which are spatially localised, with u and its derivatives tending to zero as x . (a) By seeking solutions to equation (1) in the form u(x, t) = v(z), with z := x ct, and applying the boundary conditions that v and its derivatives vanish as z , show that travelling wave solutions are described by the planar dynamical system, where w := dv/dz. (b) Find and classify the fixed points of the planar dynamical system (2). dv w dz w = 3v2+cv , (2) [6 marks] [4 marks] (c) Sketch the phase portrait of (2). If there are hyperbolic fixed points, identify their stable and unstable subspaces and add them to your sketch. Label the orbit corresponding to the soliton. [10 marks] 2v + c for the soliton. [4 marks] c 2 cz v(z) = 2sech 2 . (3) You should assume that the wave is symmetric about z = 0 to eliminate an integration constant that appears. (Hint: Recall that 1tanh2z = sech2z, and d/(2b) = (1/b)tanh1(/b)+ constant.) [8 marks] Solution: (a) The ansatz u(x, t) = v(z) with z = x ct converts equation 1 into a third order ODE, d3v 6vdv cdv =0. dz3 dz dz Writing 6vdzv = dz(3v2) and integrating once we have d2v 3v2 cv = A, dz2 where A = 0 to satisfy the boundary conditions at z . Introducing the variable w := dv/dz we obtain the required planar dynamical system. (b) Fixed points are solutions of w 0= 3v2+cv , which yields two fixed points at (v, w) = (0, 0) and (v, w) = (c/3, 0). The Jacobian is 0 1 A= 6v+c 0 At v = 0 the eigenvalues are = c this is a hyperbolic saddle. At v = c/3 the eigenvalues are = ic a centre. (c) In addition to the fixed points, we note the v nullcline is w = 0, while the w nullclines are v=0andv=c/3. Thereisalsoasymmetryww,zz(reflectioninthev-axiswith (d) By forming an expression for dw/dv and integrating, show that w = v (e) Integrate this expression for w(v) to show that the solitons take the form 1 Applied Dynamical Systems MATH11140 Solutions and comments May 2021 Figure 1: Phase portrait for question 1(c) with c = 1 for illustration. Nullclines are dashed green, stable and unstable subspaces are blue and red lines respectively. Cyan curve shows the homoclinic orbit (v < 0). time reversed). Therefore we need only work out a single eigenvector of the hyperbolic saddle, the other can be added from symmetry, consider = + c, c 1 c cv=0, hence v = (1,c) (not normalised), and Eu = span{vc}. From symmetry we have vc = (1, c) and Es = span{vc}. With this and the nullclines, it is clear that there is a homoclinic orbit at the origin which encircles the centre at v = c/3 this is the soliton. An example phase portrait for c = 1 has been included in figure 1. (d) We use the chain rule and write dw = 3v2 +cv, dv w which can be immediately integrated to give w2/2 = v3 + cv2/2 + B. The integration constant B = 0 since the orbit of interest starts and ends at the fixed point (v,w) = (0,0). Take the square root to get the expression as it appears in the question. (e) Writing w = dv/dz and using the expression derived in (d) for the homoclinic orbit gives dv = z + C v 2v+c integrating dv =v2v+c dz The integration constant can be set to zero without loss of generality (as indicated in the question). Introducing := 2v + c the integral on the left hand side becomes 2 d/(2 c) = (2/c)tanh1(1+2v/c) (hint given in question). Rearranging we find v(z) = (c/2)sech2(cz/2) as required. Comment: Unseen equation (KdV) for the travelling waves (have seen Burgers, Fisher- Kolmogorov), but an almost identical two-dimensional system has been seen in the workshops. The final integration for the soliton will likely cause problems. 2 Applied Dynamical Systems MATH11140 Solutions and comments May 2021 2. Consider the planar system x = y x x 2 , y = x y y2. (4) (a) Show that there is a non-hyperbolic fixed point at the origin, x = 0. Find the stable/unstable and centre subspaces of this point. Using your answer, find coordinates x,y, obtained from x,y by a rotation, such that the stable/unstable and centre subspaces are {y = 0} and {x = 0} respectively. (b) Consider now the modified system x = y x x 2 , y = ( + 1)x y y2. For which values of is x = 0 unstable? [12 marks] (5) [2 marks] (c) Now treat as an additional variable and append the additional equation = 0 to the planar dynamical system (5). Determine an equation for the two-dimensional centre manifold around (x, y, ) = (0, 0, 0) accurate to second order. (Hint: You will find it helpful to use your answer to (a), and you may give your answer in the new coordinate system.) [14 marks] (d) Derive an equation for the motion on the centre manifold accurate to second order. What type of bifurcation occurs at = 0? [10 marks] Solution: (a) The origin is clearly a fixed point for this system. Evaluating the Jacobian at the origin we have 1 1 A= 1 1. The eigenvalues are = 0 and = 2 this is a non-hyperbolic fixed point with a one-dimensional 1 1 centre subspace. Eigenvectors for = 0 solve Axc = 0, so select xc = 2 eigendirection we have 1 . For the stable 1 1 which yields xs = 2 1 . 1 1 1 1 x s = 0 , Clearly Ec = {y = x} and Es = {y = x}, which can be aligned with new coordinate axes by rotating the coordinate system through /4. Alternatively, the (orthonormal) matrix of eigenvectors 1/2 1/2 1/ 2 1/ 2 X := is a planar rotation through an angle /4. Therefore, we can define new coodinates x,y via x =XTx,suchthatEs ={y =0}andEc ={x =0}. (b) With the addition of the term involving the constant , the Jacobian at x = 0 is now 1 1 A= 1+ 1 . The trace is fixed at 2 but the determinant = the fixed point is therefore an unstable hyperbolic saddle when > 0, it is stable for < 0. (c) With as a variable and = 0, the Jacobian at the origin is now 1 1 0 A=1 1 0. 000 3 Applied Dynamical Systems MATH11140 Solutions and comments May 2021 By comparison with part (a) we note that the eigenvalues = 0 and = 2 are also eigenvalues in this three-dimensional problem with eigenvectors 1 1 2 2 x01=1 1 and x2=1 1 00 In addition, the = 0 equation yields an additional root = 0 with eigenvector x02 = (0,0,1). Therefore, the centre subspace Ec is now the plane {y = x}. To calculate a series expansion for the centre manifold, we first make use of a coordinate rotation (also see hint in question) in our dynamical system; the columns of the rotation matrix are eigenvectors in the problem 1/2 1/2 0 X:=1/ 2 1/ 2 0, 001 definingx =XTxandmakinguseoftheidentitiesx2+y2 =x2+y2 andx2y2 =2xy we find x = 2 x + 2 ( y x ) 2 x y 1 2 2 y = 2 ( y x ) 2 ( x + y ) = 0 . In this transformed system, Es is the x-axis and Ec is the plane {x = 0}. The primes on variables will be dropped from this point on for ease of notation. For the centre manifold we write x = Gc(y, ). With the condition that the manifold is tangent to the centre subspace at the origin we seek a series expansion x=ay2 +by+c2 +O(3) from the chain rule we have x = yGcy , and making use of the governing (rotated) equations and the fact that the motion is constrained to the manifold, ccc cc1c22 2G+2(yG) 2yG=yG 2(yG)2(G +y) Since we require the manifold only to second order, we can ignore many terms (e.g. Gc2, Gc); retaining terms to second order yields 2ay2 2by 2c2 + 21y = 0 + O(3) Comparing coefficients we find the only non-zero constant is b = 1/4, so an equation for the centre manifold accurate to second order is Gc(y,)= 41y+O(3) (d) For motion on the manifold we have = 0 and writing v in place of y, v = ( G c v ) 1 G c 2 + v 2 22 =v11 v2+O(3) 242 4 Applied Dynamical Systems MATH11140 Solutions and comments May 2021 v / = 0 (found to be an unstable saddle for > 0) and 2 + O(2) this is a transcritical bifurcation. With = constant there are equilibria at v Comment: A similar problem has been studied in the notes; extra work is required here to identify a nice coordinate system for deriving the centre manifold expansion which should be apparent from part (a). 3. Consider the following three-dimensional system in cylindrical polar coordinates r = r(1 r2) = (6) z = z(1 z2) where r [0,), S1, z R and > 0 is a constant. (a) Determine the flow, t(x), for (6) and use your answer to discuss the stability of any equilibria. [8 marks] (b) Characterise the -limit sets of all points x R3. [8 marks] (c) Now consider the section : {(r,,z) : r > 0, = 0}. Use your solution to (a) to define the Poincar e map P 0 : . [6 marks] (d) Find the fixed points of the Poincar e map P0 and determine their stability. How is this result connected to the -limit sets you identified in part (b)? [8 marks] Solution: (a) We first integrate the second equation to find = 0 + t. For the r- and z- dynamics we note that the equations are the same, though z can be negative; there is symmetry z z. We have to integrate = t. This is straightforward after using partial fractions and yields after some rearrangement r0 r(t) = r02 + (1 r02)e2t , with an identical equation for z(t). The flow is therefore r/r2 + (1 r2)e2t t(r, , z) = + t z/z2 + (1 z2)e2t Equilibria satisfy x = t(x) t. Any equilibria must be at r = 0 on the account of the azimuthal dynamics. The z-component of the flow indicates that z is steady for z = 0 and z = 1 (see symmetry noted earlier). The equilibria are therefore at (r, z) = (0, 0) and (r, z) = (0, 1). Note that limt t(x)z = 1 for z > 0, and limt t(x)r = 1 for all r > 0. Therefore all three equilibria are unstable; the origin is a repeller while z = 1 are saddles. (b) In addition to the equilibria described above, the limiting behaviour limtr(t;r0) = 1 indicates the presence of three periodic orbits, = {t(r = 1, = 0,z = 1) : 0 t < 2/} and0 ={t(r=1,=0,z=0):0t<2/}. Thesurface{z=0}isinvariantandformsa separatrix between orbits attracted to one of the two stable limit cycles , initial conditions on this surface are attracted to the unstable periodic orbit 0. This information now allows us to define the -limit sets. We will here describe only z 0, there is an identical copy below the z-axis by symmetry. First consider the equilbrium at the origin, x {0}, L(x) = {0}. Now consider initial conditions on the z-axis, x {(r,z) : r = 0,z 0}, L(x) = {(0,+1)}. On the other hand, for intial conditions on the plane x {z = 0 : r > 0}, L = 0. Finally, for all other initial conditions x {(r,,z) : r > 0,z > 0}, the limit set is the attracting limit cycle L(x) = +. r d r0 (1)(1+) 5 Applied Dynamical Systems MATH11140 Solutions and comments May 2021 (c) We write r = 1 and z = 2 at the first crossing of . Due to the constant angular velocity, repeated cuts of are spaced 2/ apart in time. Therefore, using the solutions for r and z we have (d) We look for fixed points = P 0 (), which yields (1, 2) = (1, 0) and the pair of solutions (1,2) = (1,1). The fixed points of the map correspond to the periodic orbits found for the continuous system in (b). To determine stability, we compute the Jacobian of the map DP 0 : e4/(2 + (1 2)e4/)3/2 DP0()= 1 1 0 0 2 4/ 3/2 . (8) /2 + (1 2)e4/ P0()= 1 1 1 (7) 2/2 + (1 2)e4/ The eigenvalues at the fixed points are simply the diagonal elements. All fixed points are at 1 = 1; at 2 = 0 we find 1 = e4/ and 2 = e2/, i.e. |1| < 1 and |2| > 1, indicating an unstable saddle. This is in agreement with the limit sets from (b), the periodic orbit 0 is attracting only for initial conditions in the plane {z = 0}. At 2 = 1 both eigenvalues are e4/ asymptotically stable fixed points to match the attracting limit cycles . Comment: A fairly standard problem: the r and equations have appeared several times in the course, though here they are embedded in a three dimensional system. Poincar e maps have caused some pain on assignments. e 4/ 2 /(2 +(12)e ) 6