ME 406 Assignment #9 Solutions

Transcription

1 ME 06 Assignment #9 Solutions sysid Mathematica 6.0., DynPac.0, ê8ê009 intreset; plotreset; imsize = 50; Problem The equations are x = -x, y = x +y, z = x + z. Let F = x + y +z. To show that the surface F = 0 is an invariant surface, we have show that points on the surface remain on the surface -- that is, df/dt evaluated on F = 0 is 0. We have df dt = x F x + y F y +z F z = -x(x) + (x +y) () + ( x + z) () = x + y +z, which is zero when evaluated on F = 0. Problem We define the system for DynPac. setstate@8x, y<d; setparm@8<d; slopevec = 9-x + y, x + y + x y=; sysname = "ProbNL"; eigval@80, 0<D 8-, < The eigenvalues are - and, hence neither has a zero real part, so the equilibrium is hyperbolic. It is in fact a saddle point, as DynPac tells us also.

2 Assignment#9sol.nb 0<D Abbreviations used in classifyd. L = linear, NL = nonlinear, R = repeated root. Z = one zero root, Z = two zero roots. This message printed once. unstable - saddle We now construct the linearized system. The matrix A of the linearized system is the derivative matrix evaluated at the equilibrium: A = dermatval@80, 0<D 88-, 0<, 8, << We save our nonlinear system and then define the linear system for DynPac. savesys@probnld; slopevec = A.statevec 8-x, x + y< sysname = "ProbLIN"; The manifolds of the linear system will be determined by the eigenvectors and eigenvalues. Eigensystem@AD 88-, <, 88-, <, 80, <<< Thus the stable manifold (associated with the eigenvalue -) is spanned by the vector {-,}. The unstable manifold (associated with the eigenvalue ) is spanned by {0,}. We plot these, using blue for stable, red for unstable. man = ParametricPlot@8 u, -u<, 8u, -, <, PlotStyle Ø RGBColor@0, 0, DD; man = ParametricPlot@80, u<, 8u, -, <, PlotStyle Ø RGBColor@, 0, 0DD;

3 Assignment#9sol.nb man<d ProbLIN It is actually just as easy to get these from DynPac by a numerical integration. We do that next. plrange = 88-.,.<, 8-.,.<<; asprat = ; axon = False; frameon = True; arrowflag = True; arrowvec = 8 ê <; rangeflag = True; ranger = plrange; t0 = 0.0; h = 0.0; nsteps = 800; eq = 80, 0<; eigstab = 8-, <; eigunstab = 80, <; eps = 0.0; initset = 8eq + eps * eigstab, eq - eps * eigstab, eq + eps * eigunstab, eq - eps * eigunstab< , 0.0<, 80.0, -0.0<, 80, 0.0<, 80, -0.0<< h = 8-0.0, -0.0, 0.0, 0.0<; setcolor@8blue, Blue, Red, Red<D;

4 Assignment#9sol.nb linport = portrait@initset, t0, h, nsteps,, D ProbLIN y Now we carry out the same construction for the full nonlinear system. restoresys@probnld; h = 8-0.0, -0.0, 0.0, 0.0<; x

5 Assignment#9sol.nb 5 nlport = portrait@initset, t0, h, nsteps,, D ProbNL y We combine these x labon = "Problem - Stable and Unstable Manifolds";

6 6 Assignment#9sol.nb nlportd Problem - Stable and Unstable Manifolds y x Problem We enter the matrix A. A = 880, ê, ê, ê, -7 ê <, 8 ê, ê, - ê, -5 ê, ê <, 8, - ê, - ê, - ê, 7 ê <, 85, 9 ê, - ê, -5 ê, 9 ê <, 85 ê, 5 ê,, -, 5 ê <<; MatrixForm@AD We get the eigenvalues and eigenvectors of A.

7 Assignment#9sol.nb 7 Eigensystem@AD :8-, -,,, <, ::-,,, 5, >, 80, 0, 0, 0, 0<, 80,, 0,, <, 80, 0, 0, 0, 0<, 80, 0, 0, 0, 0<>> We see that there is an eigenvalue - of multiplicity with one eigenvector, and an eigenvalue of multiplicity, also with one eigenvector. We will have to use generalized eigenvectors. We start with the eigenvalue -. Id = IdentityMatrix@5D 88, 0, 0, 0, 0<, 80,, 0, 0, 0<, 80, 0,, 0, 0<, 80, 0, 0,, 0<, 80, 0, 0, 0, << stabspan = NullSpace@MatrixPower@A + Id, DD 88, -, -, 0, <, 8-,,,, 0<< Thus the stable manifold is spanned by the two vectors stab = stabspan@@dd 8, -, -, 0, < stab = stabspan@@dd 8-,,,, 0< The original eigenvector must be in this space. We use this to verify our work so far. orig = 8- ê, ê, ê, 5 ê, <; Solve@C * stab + C * stab ã orig, 8C, C<D ::C Ø, C Ø 5 >> * stab + H5 ê L * stab :-,,, 5, > Everything checks. The most general initial condition which will decay to zero is the most general vector in this manifold, which is given by the expression below for arbitrary C and C. C * stab + C * stab 8 C - C, - C + C, - C + C, C, C< The unstable manifold is associated with the eigenvalue, and it is spanned by the generalized eigenvec-

8 8 Assignment#9sol.nb The unstable manifold is associated with the eigenvalue, and it is spanned by the generalized eigenvectors associated with that eigenvalue. unstabspan = NullSpace@MatrixPower@A - Id, DD 88-,, 0, 0, <, 8, 0, 0,, 0<, 8, -,, 0, 0<< Once again we check our work by verifying that the original eigenvector is in the space spanned by these. unstab = unstabspan@@dd 8-,, 0, 0, < unstab = unstabspan@@dd 8, 0, 0,, 0< unstab = unstabspan@@dd 8, -,, 0, 0< orig = 80,, 0,, <; Solve@C * unstab + C * unstab + C * unstab ã orig, 8C, C, C<D 88C Ø, C Ø, C Ø 0<< * unstab + * unstab 80,, 0,, < Problem We define the system for DynPac. setstate@8x, y, z<d; setparm@8<d; sysname = "Problem "; slopevec = 9-y + x z - x, x + y z + x y z, -z - x - y + z + SinAx E=; We find the eigenvalues at the origin. eigsys@80, 0, 0<D 88-, Â, -Â<, 880, 0, <, 8Â,, 0<, 8-Â,, 0<<< We have a complex conjugate pair, so the origin is a non-hyperbolic equilibrium, and its stability is not determined from the linearization. We see from the eigenvalues that there is a stable manifold of dimension and a center manifold of dimension. We now obtain a local approximation for the center manifold, which in turn will lead to a stability determination.

9 Assignment#9sol.nb 9 We see from the eigenvalues that the linear stable manifold is the z-axis, and the linear center manifold is the x-y plane. The nonlinear center manifold will have the form z = h(x, y) with h(0,0) = 0, h x (0,0) = 0, and h y (0,0) = 0. We differentiate this equation with respect to time to get () z = h x x + h y y. When the expressions for the time derivatives of x, y, and z are entered into (), with z replaced by h, we get the basic partial differential equation satisfied by h. We enter this equation into Mathematica. We define first a modification of the slope vector in which z is replaced by h. ClearAll@hD slopevec 9-x - y + x z, x + y z + x y z, -x - y - z + z + SinAx E= () dotvec = slopevec êê. z Ø h 9h x - x - y, x + h y + h x y, -h + h - x - y + SinAx E= Because we are going to keep only through cubic terms, we replace Sin[x ] with x. dotvec = dotvec ê. SinAx E Ø x 9h x - x - y, x + h y + h x y, -h + h - x + x - y = Now we expand h in a Taylor series, keeping terms through order. Note that the constant and linear terms are absent because of the conditions in (). h = a x + b x y + c y + d x + f x y + g x y + k y a x + d x + b x y + f x y + c y + g x y + k y Now we write z - h x x - h y y for Mathematica. hequat = dotvec@@dd - D@h, xd dotvec@@dd - D@h, yd dotvec@@dd -x - a x + x - d x - b x y - f x y - y - c y - g x y - k y + Ia x + d x + b x y + f x y + c y + g x y + k y M - I a x + d x + b y + 6 f x y + g y M I-x - y + x Ia x + d x + b x y + f x y + c y + g x y + k y MM - I b x + f x + c y + 6 g x y + k y M Ix + y Ia x + d x + b x y + f x y + c y + g x y + k y M + x y Ia x + d x + b x y + f x y + c y + g x y + k y MM This quantity must vanish. It is a power series in x and y, so terms of any given order must vanish separately.

10 0 Assignment#9sol.nb This quantity must vanish. It is a power series in x and y, so terms of any given order must vanish separately. We impose these conditions, starting with the quadratic terms. coeffxx = H ê L D@hequat, 8x, <D ê. Thread@8x, y< Ø 80, 0<D H- - a - bl coeffyy = H ê L D@hequat, 8y, <D ê. Thread@8x, y< Ø 80, 0<D H- + b - cl coeffxy = D@D@hequat, xd, yd ê. Thread@8x, y< Ø 80, 0<D a - b - c We solve these for a,b,c. quadans = Solve@8coeffxx ã 0, coeffxy ã 0, coeffyy ã 0<, 8a, b, c<d 88a Ø -, b Ø 0, c Ø -<< Thus the quadratic approximation to the center manifold is hquad = -x - y -x - y Now consider the orbits on the center manifold. We have xdot = slopevec@@dd ê. z Ø hquad -x - y + x I-x - y M ydot = slopevec@@dd ê. z Ø hquad x + y I-x - y M + x y I-x - y M Correct to third order, we have the following differential equations on the center manifold: dx dt = -y - x(x + y ) + higher order, dy dt = x -y(x + y ) + higher order. We convert to polar coordinates to get rr = -r + higher order. Equation () shows the stability of the equilibrium at the origin. () () Had there been no conclusion from this analysis, we would have had to return to the calculation of the center manifold and keep more terms.