%% 16. Multiple solutions of nonlinear BVPs % % \setcounter{page}{196} % %% % In the subject of linear algebra, existence and uniqueness % are straightforward. A scalar linear equation $ay = b$ has a unique % solution for any $b$ if $a\ne 0$, whereas if $a=0$, it either % has infinitely many solutions (if $b=0$) or % no solution at all (if $b\ne 0$). % For a system of $n$ linear equations ${\bf A}{\bf y}={\bf b}$, where ${\bf A}$ % is an $n\times n$ matrix, the situation is a generalization % of the same alternative. If ${\bf A}$ is % nonsingular, there is a unique solution for any $\bf b$, whereas % if ${\bf A}$ is singular, there are infinitely many solutions if % ${\bf b}\in \hbox{range}({\bf A})$ and % no solutions at all if ${\bf b}\not\in \hbox{range}({\bf A})$. %% % Nonlinear algebraic problems, by contrast, can do almost anything. % Consider first a nonlinear equation involving a scalar real variable $y$, % which without loss of generality we can write with a % zero right-hand side as % $f(y) = 0$. If $f(y) = \exp(y)$ there % is no solution; if $f(y) = y+\exp(y)$ there is one solution; and % if $f(y) = \hbox{round}(y)$ there % are infinitely many solutions, namely all the numbers % in the interval $(-0.5,0.5)$, together with perhaps $-0.5$ or $0.5$ depending % on exactly how you define the round'' function. % More important, it may happen that there are % multiple solutions that are separated from one another. % These may be finite in number, as with $f(y) = y^2-1$, whose solutions % are $y=\pm 1$, or they may be % infinite in number, as with $f(y) = \sin(\pi y),$ whose solutions % are all the integers. The function $f(y) = \sin(\pi y) + 0.01y^2$ % illustrates another possibility, that a problem % may have quite a few isolated solutions without having % infinitely many. ODEformats f = chebfun(@(y) sin(pi*y)+0.01*y.^2,[-15 15]); plot(f,'k') hold on, r = roots(f); plot(r,f(r),'.r',MS,12), hold off grid on, ylim([-2 4]) title(['Fig.~16.1.~~A nonlinear function $f(y)$ ' ... ' with many zeros'],FS,11) %% % \vskip 1.02em %% % % All this is for a scalar equation $f(y)=0$. For a nonlinear system % of equations ${\bf f}({\bf y}) = {\bf 0}$, further possibilities arise. % %% % Now let us look at differential as opposed to algebraic equations. % What can we say about the possibility of nonuniqueness, % that is, of problems with multiple solutions? %% % % For linear ODE\kern .5pt s, we have already made the main observations. % Theorems 2.1--2.4 asserted uniqueness of solutions % to first-order scalar linear IVPs, and these results carry over to % IVP linear systems and equations of % higher order. For linear BVPs, we saw in Chapter 6 % that the key matter is whether or not a problem has % an {\em eigenfunction}. % If not, there is a unique solution. If so, there are either % no solutions or a continuum of solutions.\footnote{For example, % $y''+y = 0$ with boundary condition $y(0)=0$ has the general solution % $y(x) = A\sin(x)$. On any interval $[\kern .3pt 0,L]$ where $L$ % is not an integer multiple of $\pi$, there is a unique solution for % any $b$ when % the second boundary condition $y(L) = b$ is specified. % If $L$ is an integer multiple of $\pi$, on the other hand, % there are infinitely many solutions if $b=0$ and no solutions % if $b\ne 0$.} % %% % % For nonlinear ODE IVPs, uniqueness is again usually not % a problem. Theorem~11.1 asserted existence and % uniqueness for any problem $y'(t) = f(t,y)$ % if~$f$ is continuous with respect to~$t$ and Lipschitz % continuous with respect to~$y$, and Theorem 11.2 made the % analogous statement for systems of IVPs.\ \ For uniqueness to fail % for an IVP, $f$ must lack these continuity properties, % as in the example $y'= y^{1/2}$ of equation (3.16) or the % further examples discussed in Chapter~11. % %% % % When it comes to nonlinear ODE BVPs, however, anything % is possible. We saw illustrations of nonlinear nonuniqueness % with equation (5.10) and Exercise~9.2, and the BVPs associated % with the nonlinear pendulum of Chapter 9 also have % nonunique solutions, though we did not mention % that there. Phenomena of this kind % are the subject of this and the next two chapters. % %% % To make a start, consider the linear BVP % $$y'' = - y, \quad x\in [\kern.3pt 0,1], ~ y(0) = y(1) = 1. % \eqno (16.1)$$ % This problem has no eigenfunction, so there is a unique solution. N = chebop(0,1); N.lbc = 1; N.op = @(x,y) diff(y,2) + y; N.rbc = 1; y = N\0; plot(y,CO,bvp), ylim([0.9 1.3]) title(['Fig.~16.2.~~Linear BVP (16.1): maximum $' ... num2str(max(y)) '$'],FS,11) %% % \vskip 1.02em %% % % \noindent % Now consider the nonlinear variant in which the right-hand % side of (16.1) is replaced by its cube, % $$y'' = - y^3, \quad x\in [\kern.3pt 0,1], ~ y(0) = y(1) = 1. % \eqno (16.2)$$ % (A quintic as opposed to cubic nonlinear spring law'' appeared % in eq.~(4.8).) % If we give this problem to Chebfun, a solution is produced that % looks approximately like the last one, but about 10\% larger. % N.op = @(x,y) diff(y,2) + y^3; y1 = N\0; plot(y1,CO,bvpnl), ylim([0.9 1.3]) title(['Fig.~16.3.~~Nonlinear BVP (16.2): maximum $' ... num2str(max(y1)) '$'],FS,11) %% % \vskip 1.02em %% % % \noindent % However, this is not the only solution to (16.2). % First let us present another solution; then we shall explain % how we got it. % x = chebfun('x',[0 1]); N.init = 1-25*(x-x^2); y2 = N\0; plot(y2,CO,bvpnl), ylim([-6 4]) title('Fig.~16.4.~~Another solution to (16.2)',FS,11) %% % \vskip 1.02em %% % % The last two figures show that there exist % at least two distinct solutions % to (16.2). Now this is not a book of numerical analysis, but % a word must be said at this point about algorithms. In every area of % computational mathematics, to solve nonlinear problems, it is usually % necessary to use some kind of iteration, in which the % solution is approached via a sequence of linear problems. % The prototypical iteration % is Newton's method, and Chebfun uses a version % of Newton's method to solve nonlinear BVPs.\footnote{When % applied like this to find functions as opposed to just numbers, % Newton iteration is also called {\em Newton--Kantorovich iteration.}} % %% % % If a problem has more than one solution, which one will % an iteration like Newton's method converge to? The first % thing to be said is that sometimes, it may not converge % at all, and there is a large subject of % nonlinear numerical optimization that aims to improve matters % in this respect. When it does converge, however, the % solution it converges to is often one that is close to the % {\em initial guess} employed by the iteration. % Every Newton iteration starts from some initial guess or % other, even if it is the zero function. In the case of % Chebfun, if the user does not specify an initial guess % explicitly, then the iteration starts from a simple polynomial % in the variable $x$ constructed to match the boundary conditions, % and this is what Chebfun did to obtain the solution % plotted in Figure~16.3. % %% % % To get the second solution of (16.2), we overrode the % default by specifying a different initial % guess in the field \verb|N.init|. % This function, $1-25(x-x^2)$, was chosen because it has % approximately the right shape. % There are few guarantees in this business, but Chebfun % duly converged to the solution plotted in Figure~16.4. % Here we superimpose the initial guess on the plot. % hold on, plot(N.init,'--r'), hold off title('Fig.~16.5.~~Solution shown with initial guess',FS,11) %% % \vskip 1.02em %% % % As an autonomous scalar equation of second order, (16.2) can % be examined in the $y$-$y'$ phase plane. Here is an image showing the % phase plane trajectories of the two solutions just computed. % y1p = diff(y1); y2p = diff(y2); plot(y1([0 1]),y1p([0 1]),'.r',MS,10), hold on plot(y2([0 1]),y2p([0 1]),'.r',MS,10) arrowplot(y1,y1p,CO,bvpnl,MS,4) arrowplot(y2,y2p,CO,bvpnl,MS,4), hold off axis([-5 5 -20 20]), xlabel('$y$',FS,10), ylabel('$y''$',FS,10) text(1.2,-3,'Fig.~16.2',FS,9) text(-2.5,9.5,'Fig.~16.3',FS,9) title(['Fig.~16.6.~~Both solutions to (16.2) ' ... 'in the phase plane'],FS,11) %% % \vskip 1.02em %% % % \noindent % Each solution makes a half-circuit clockwise around the origin, % one on the right starting at $(y,y') = (1,0.7258)$, and % the other on the left starting at $(y,y') = (1,-13.4074)$. % By considering this picture we may be tempted to conjecture that in fact, % (16.2) has infinitely many additional solutions besides these: % at least two that wind around exactly once, % two that wind around 1 1/2 times, and so on (Exercise~16.3). % As the winding % numbers increase, so do the amplitudes. For example, here % is the next solution in the sequence, which we obtain by % starting with the initial guess $y(x) = 5\sin(2\pi x)$. % In the phase plane, this corresponds to a large oval winding % around one full revolution clockwise, beginning at % $(y,y') = (1,38.8855)$ (not shown). % N.init = 5*sin(2*pi*x); y3 = N\0; plot(y3,CO,bvpnl); y3p = diff(y3); hold on, plot(N.init,'--r'), hold off title(['Fig.~16.7.~~A third solution ' ... ' with the corresponding initial guess'],FS,11) %% % \vskip 1.02em %% % Whereas nonlinear BVPs can be rather delicate, nonlinear % IVPs are relatively straightforward. With this in mind, it is interesting % to solve (16.2) as an initial value problem, fixing the left-hand % value at $y(0) = 1$ and taking various choices $y'(0) = a$ in % the range $[-50,50\kern .3pt]$ for the left-hand derivative value, % $$y'' = - y^3, \quad x\in [\kern.3pt 0,1], ~ y(0) = 1, ~ y'(0) = a. % \eqno (16.3)$$ %% % \vskip -2em N.rbc = []; for a = -50:5:50 N.lbc = [1;a]; y = N\0; plot(y,LW,0.7,CO,ivpnl), hold on end title(['Fig.~16.8.~~Eq. (16.2) as an IVP ' ... 'with different initial slopes'],FS,11), hold off %% % \vskip 1.02em %% % % \noindent % If we study this image carefully, we can see that there are % five choices of initial slope $y'(0) = a$ that lead to the condition % $y(1)= 1$ being satisfied at the right. Regarding % $y(1)$ as a function of $a$, let us plot % this function over the given range $a\in [-50,50\kern .3pt]$. % fa = @(a) chebop(@(x,y) diff(y,2)+y^3,[0,1],[1;a],[])\0; f = chebfun(@(a) feval(fa(a),1),[-50,50],120); plot(f,'k') title(['Fig.~16.9.~~$y(1)$ as a function of ' ... 'the initial slope $a = y''(0)$'],FS,11) xlabel('initial slope $y''(0) = a$',FS,9,IN,LT) ylabel('$y(1)$',FS,10,IN,LT) %% % \vskip 1.02em %% % % \noindent % Solutions to the BVP (16.2) correspond to points where this curve % takes the value $1$. % r = roots(f-1); hold on, plot(r,f(r),'.r',MS,12), hold off title('Fig.~16.10.~~Five solutions of the BVP (16.2)',FS,11) %% % \vskip 1.02em %% % % \noindent % Five solutions appear in this range, % r' %% % % \noindent % and we have seen three of them already, % [y1p(0) y2p(0) y3p(0)] %% % % \noindent % In numerical analysis, the technique of varying a % slope at one end of an interval so as to satisfy a boundary condition at % the other end is known as the {\em shooting method.} % %% % % Let us turn to a more celebrated example of nonuniqueness, % one involving just two solutions rather than an infinite set, % the {\em Bratu equation:} % $$y'' + 3\kern -.7pt \exp(y) = 0 , % \quad x\in [\kern .3pt 0,1], ~ y(0) = y(1) = 0 . \eqno (16.4)$$ % Here we show two solutions on a single plot, one resulting % from Chebfun's default initial guess (the zero function) % and the other from the alternative initial guess $8(x-x^2)$. % N = chebop(0,1); N.op = @(x,y) diff(y,2) + 3*exp(y); N.lbc = 0; N.rbc = 0; y1 = N\0; plot(y1,CO,bvpnl) hold on, x = chebfun('x',[0 1]); N.init = 8*(x-x.^2); y2 = N\0; plot(y2,CO,bvpnl), ylim([-.5 2.5]), hold off title(['Fig.~16.11.~~Two solutions of the ' ... 'Bratu equation (16.4)'],FS,11) %% % \vskip 1.02em %% % % \noindent % These are the only two solutions to this problem, as we can % explain (though not quite rigorously prove without some % more work) by the same method of shooting as before. % N.rbc = []; for a = -8:2:16 N.lbc = [0;a]; y = N\0; plot(y,LW,0.7,CO,ivpnl), hold on end title(['Fig.~16.12.~~The Bratu eq.\ as an IVP ' ... ' with different initial slopes'],FS,11) ylim([-5 5]), hold off %% % \vskip 1.02em %% % % \noindent % In Chapter 18 we % shall consider the behavior of equation (16.4) as a parameter % is varied, namely the coefficient that here takes the value $3$. % %% % % We close this chapter with a final pair of examples % of nonlinear BVPs with multiple solutions. % In each case, shooting is a good method to % explore the variety of solutions, at least when the % coefficient multiplying the highest derivative % is not too small. % Another method of investigation for such problems is % {\em path-following,} to be considered in Chapter~18. % Besides these approaches, a further technique is the idea of % {\em deflation} (Exercise 16.4). % %% % First we consider the inhomogeneous equation % $$\varepsilon y'' + y + y^2 = 1 , \quad x\in [-1,1], ~ % y(\pm 1) = 0. \eqno (16.5)$$ % For $\varepsilon=0.2$ % this problem has four solutions, two symmetric and two asymmetric. % Higher values of $\varepsilon$ would have just two symmetric % solutions; the asymmetric ones emerge in a pitchfork % bifurcation (see next chapter) at a critical value of $\varepsilon$. N = chebop(-1,1); N.lbc = 0; N.rbc = 0; N.op = @(x,y) 0.2*diff(y,2) + y + y^2; x = chebfun('x'); N.init = x.^2-1; y1 = N\1; plot(y1,CO,bvpnl) ylim([-2.6 2.6]), hold on N.init = 1-x.^2; y2 = N\1; plot(y2,CO,bvpnl) N.init = sin(pi*x); y3 = N\1; plot(y3,CO,bvpnl) N.init = -sin(pi*x); y4 = N\1; plot(y4,CO,bvpnl), hold off title(['Fig.~16.13.~~Four solutions of (16.5) ' ... 'with $\varepsilon=0.2$'],FS,11,IN,LT) ylim([-2.5,3]) %% % \vskip 1.02em %% % % Equation (16.5) is a constant-coefficient variant of an % equation known as the {\em Carrier equation,} % $$\varepsilon y'' +2(1-x^2)y + y^2 = 1, \quad x\in [-1,1], ~ % y(\pm 1) = 0. \eqno (16.6)$$ % For $\varepsilon=0.2$ this also has four solutions, of similar structure. % N = chebop(-1,1); N.lbc = 0; N.rbc = 0; N.op = @(x,y) 0.2*diff(y,2) + 2*(1-x^2)*y + y^2; N.init = x.^2-1; y1 = N\1; plot(y1,CO,bvpnl) ylim([-2.5 2.5]), hold on N.init = 1-x.^2; y2 = N\1; plot(y2,CO,bvpnl) N.init = sin(pi*x); y3 = N\1; plot(y3,CO,bvpnl) N.init = -sin(pi*x); y4 = N\1; plot(y4,CO,bvpnl), hold off title(['Fig.~16.14.~~Four solutions of Carrier eq.\ (16.6) ' ... 'with $\varepsilon=0.2$'],FS,11,IN,LT) ylim([-2.5 3]) %% % \vskip 1.02em %% % % \begin{center} % \hrulefill\\[1pt] % {\sc Application: sending a spacecraft to a destination} \\[-3pt] % \hrulefill % \end{center} % %% % % There may be several ways to get a spacecraft from A to B, % even in a specified time interval. For example, suppose % the sun is fixed at the origin in the \hbox{$x$-$y$}~plane and a % spacecraft starts at position $(-2,1)$. Starting from the initial velocity % $(0.7,0.7)$, here is the orbit up to the time $T=6$. We have % solved this as a second-order IVP in two variables $x$ and $y$ defining % the position of the spacecraft. % T = 6; N = chebop(0,T); x0 = -2; y0 = 1; u0 = 0.7; v0 = 0.7; N.op = @(t,x,y) [diff(x,2) + x/(x^2+y^2)^1.5; ... diff(y,2) + y/(x^2+y^2)^1.5]; N.lbc = @(x,y) [x-x0; y-y0; diff(x)-u0; diff(y)-v0]; orange = [1 .7 0]; [x1,y1] = N\0; plot(0,0,'.',CO,orange,MS,30), hold on, grid on h1 = arrowplot(x1,y1,CO,ivpnl,LW,1.2,YS,2.5); axis([-5 5 -1.2 2.8]), set(gca,XT,-4:2:4,YT,0:2:2) title(['Fig.~16.15.~~A spacecraft trajectory starting ' ... 'at $(x,y) = (-2,1)$'],FS,11,IN,LT) %% % \vskip 1.02em %% % Of course, actual spacecraft are not simply launched into free % flight from an initial position and velocity; they have rockets % and make adjustments along the way. We are considering % a simplified problem. %% % % Now suppose our goal is to choose the initial conditions % so that the spacecraft will be at % position $(2,1)$ at time $T$. % This is a problem of the kind mentioned in the first footnote % of Chapter 5: a BVP in $t$ rather than % $x$. If we start from the initial guess % of a straight line orbit from $(-2,1)$ to $(2,1)$, % Chebfun finds a solution. % xT = 2; yT = 1; N.lbc = @(x,y) [x-x0; y-y0]; N.rbc = [xT; yT]; t = chebfun('t',[0 T]); N.init = [-2+2*t/3; 1+0*t]; [x2,y2] = N\0; delete(h1) arrowplot(x2,y2,CO,bvpnl,LW,1.2,YS,2.5) title('Fig.~16.16.~~Reaching $(2,1)$ at $T=6$',FS,11,IN,LT) %% % \vskip 1.02em %% % % \noindent % Here are the corresponding initial velocities. % u20 = deriv(x2,0); v20 = deriv(y2,0); disp([u20 v20]) %% % % \noindent % There is another solution, however, following a longer orbit the % other way around. Its average speed will be greater, % and it will pass closer to the sun. Here is this second solution plotted % as a dashed line, obtained by starting from another initial guess. % N.init = [-2+2*t/3; 1-2*sin(pi*t/T)]; [x3,y3] = N\0; arrowplot(x3,y3,LS,'--',CO,bvpnl,LW,1.2,YS,2.5) title('Fig.~16.17.~~Another solution',FS,11,IN,LT) %% % \vskip 1.02em %% % % \noindent % The vector of initial velocities now points down instead of up. % u30 = deriv(x3,0); v30 = deriv(y3,0); disp([u30 v30]) %% % % \noindent % Both the solutions we have just found are arcs of ellipses, as the reader % may confirm by extending the orbits to a later time. A good % choice is $T = 27.4$. % %% % This 2-body problem belongs to % Newtonian mechanics, and it would have given little difficulty to % scholars of earlier centuries. It is well known that any orbit % will be an ellipse or a hyperbola, or a parabola in the borderline % case. In the present example we got ellipses rather % than hyperbolas because the number $T$ was sufficiently large. % If the spacecraft needs to get from A to B faster, for example in % $T=2$ time units, then the simplest solution is a nearly-straight % trajectory that begins pointed almost at $B$, like firing a bullet. % This time the spacecraft has more than enough kinetic energy to % escape the sun's gravitational field, and the orbit is an arc of % a hyperbola rather than an ellipse. T = 2; N.domain = [0 T]; N.lbc = @(x,y) [x-x0; y-y0]; N.rbc = [xT; yT]; t = chebfun('t',[0 T]); N.init = [-2+2*t; 1+0*t]; hold off, plot(0,0,'.',CO,orange,MS,30), hold on, grid on [x4,y4] = N\0; arrowplot(x4,y4,CO,bvpnl,LW,1.2,YS,2.5) axis([-5 5 -1.2 2.8]), set(gca,XT,-5:5,YT,-1:3) title('Fig.~16.18.~~Faster hyperbolic trajectory: $T=2$',FS,11,IN,LT) %% % % \noindent % Again there is a second solution, % also a hyperbola, that passes the other way around. % N.init = [-2+2*t; 1-1.5*sin(pi*t/T)]; [x5,y5] = N\0; arrowplot(x5,y5,LS,'--',CO,bvpnl,LW,1.2,YS,2.5) title('Fig.~16.19.~~Another fast trajectory: $T=2$',FS,11,IN,LT) %% % \vskip 1.02em %% % % \noindent % The initial velocities are much greater than before. % In space travel, a speedy journey may cost a lot of energy! % u40 = deriv(x4,0); v40 = deriv(y4,0); disp([u40 v40]) u50 = deriv(x5,0); v50 = deriv(y5,0); disp([u50 v50]) %% % As we mentioned, the calculations above are classical. % When a third body is introduced, however, everything changes % and only numerical computations are available. For example, % suppose that instead of one % sun at $(0,0)$ we have two fixed stars, one at $(0,0)$ and the other at $(0,1)$. % All your experience with elliptical orbits is now irrelevant. % To see a little of the complexity, let us consider solutions up % to time $T=12$ beginning with $x'(0)=0.8$ with % $y'(0) = 0.549$ and $y'(0) = 0.4$. We see the % beginnings of two orbits that will remain bounded but % are certainly not ellipses. T = 12; N = chebop(0,T); u0 = .8; N.op = @(t,x,y) ... [diff(x,2) + x/(x^2+y^2)^1.5 + x/(x^2+(y-1)^2)^1.5; ... diff(y,2) + y/(x^2+y^2)^1.5 + (y-1)/(x^2+(y-1)^2)^1.5]; N.lbc = @(x,y) [x-x0; y-y0; diff(x)-u0; diff(y)-0.549]; hold off [x6,y6] = N\0; subplot(1,2,1) plot([0 0],[0 1],'.',CO,orange,MS,20), hold on, grid on arrowplot(x6,y6,CO,ivpnl,LW,1,YS,.8); axis([-4 4 -3.5 4.5]), set(gca,XT,-4:2:4,YT,-4:2:4) N.lbc = @(x,y) [x-x0; y-y0; diff(x)-u0; diff(y)-0.40]; hold off [x7,y7] = N\0; subplot(1,2,2) plot([0 0],[0 1],'.',CO,orange,MS,20), hold on, grid on arrowplot(x7,y7,CO,ivpnl,LW,1,YS,.8,MS,5); axis([-4 4 -3.5 4.5]), set(gca,XT,-4:2:4,YT,-4:2:4) title('Fig.~16.20.~~Orbits about two fixed suns',FS,11,HA,RT) %% % \vskip 1.02em %% % Here are the same orbits extended to time $T=140$. On the left we see % apparent periodicity. This is not typical; it results from the particular choice % $v'(0) = 0.549$. N.domain = [0 140]; clf N.lbc = @(x,y) [x-x0; y-y0; diff(x)-u0; diff(y)-0.549]; hold off [x6long,y6long] = N\0; subplot(1,2,1) arrowplot(x6long,y6long,CO,ivpnl,LW,.7,YS,.8,MS,5), hold on, grid on axis([-4 4 -3.5 4.5]), set(gca,XT,-4:2:4,YT,-4:2:4) plot([0 0],[0 1],'.',CO,orange,MS,20) N.lbc = @(x,y) [x-x0; y-y0; diff(x)-u0; diff(y)-0.40]; hold off [x7long,y7long] = N\0; subplot(1,2,2) arrowplot(x7long,y7long,CO,ivpnl,LW,.7,YS,.8), hold on, grid on axis([-4 4 -3.5 4.5]), set(gca,XT,-4:2:4,YT,-4:2:4) plot([0 0],[0 1],'.',CO,orange,MS,20) title('Fig.~16.21.~~More orbits about two suns',FS,11,HA,RT) %% % \vskip 1.02em %% % % \noindent % Still more complicated orbits can be found by exploring % values $v'(0)$ such as $0.456$, $0.455$, and $0.454$. % %% % % No solar system, of course, has two stars fixed motionless in space. % However, our two-star model still has relevance % to real problems. First, we could build a system just % like this involving marbles rolling on a surface with two % suitably shaped drains in a fixed position (there are demonstrations along these % lines at some science museums). Second, there are many binary star systems % that differ from this only in that the two stars are in orbit around % each other, which just adds a few more terms % to the equations. Third, a single star like our sun may have planets % orbiting around it, and spacecraft trajectories are often chosen to % swing close to some of the planets to get to their destination fast % with less cost in fuel. The plot of the 2016 movie % {\em The Martian} turns on such an % unexpected choice of orbit. % %% % % \smallskip % {\sc History.} We do not know who first focused on the % phenomenon that a nonlinear BVP can have multiple solutions, but % certainly one of earliest and most consequential examples of this kind % concerns the buckling of columns. Euler published his great % paper on this subject in 1759 (in French), On the strength of % columns.'' He writes, {\em And so we see that, however small the % force $F$ acting horizontally, it must always produce a certain % deflection, which is proportional to $F$ itself. % But it is not the same when the force % acts vertically, or if the column must sustain % a load from above. At first it seems that such a force, no matter how % great, could not bend the column: for there is no reason why % it should bend in one direction rather than another. But the % least inequality in the parts of the column, or the % least stress which it feels from any side, will soon furnish % a sufficient reason for it to bend in a particular direction.} % \smallskip % %% % % {\sc Our favorite reference.} Many features of % our world, from the fundamental laws of physics to the pattern % of stripes on a tiger, can be seen as having arisen by the % process called {\em symmetry breaking}, in which a choice is made % among a multiplicity of potential solutions. % For a popular account of such effects see % Stewart and Golubitsky, {\em Fearful Symmetry: Is God % a Geometer?,} first published in 1993. % \smallskip % %% % % \begin{displaymath} % \framebox[4.7in][c]{\parbox{4.5in}{\vspace{2pt}\sl % {\sc Summary of Chapter 16.} Nonlinear ODE BVPs can % have no solutions, one solution, finitely many solutions, % or infinitely many solutions. Some strategies for % investigating multiple solutions include (a) phase plane % analysis, (b) shooting, (c) path-following, and % (d) deflation. In general, finding all % solutions of a nonlinear BVP is difficult. % \vspace{2pt}}} % \end{displaymath} % %% % \smallskip\small\parskip=1pt\parindent=0pt % {\em \underline{Exercise $16.1$}. Fisher equation.} % The function $y(x)$ satisfies $y''+y-y^2 = 0$ % for $x\in [-1,1]$ with $y(-1) = 1$, $y(1) = 0$. % {\em (a)} If $y(0)\approx 0.6$, what is $y(0.5)\kern .5pt?$\ \ Plot % the solution. % {\em (\kern .7pt b)} If $y(0)\approx -2.5$, what is % $y(0.5)\kern .5pt?$\ \ Plot the solution. % {\em (c)} Sketch both of the orbits just described in the phase plane. % \par % {\em Exercise $16.2$. Bounce pass.} % A ball is thrown from player A to player B, 5 meters away, % starting and finishing at height 1 meter. This is an idealized % ball that travels as a point mass with no air resistance or rotation % and bounces perfectly with equal angles and speeds of % impact and rebound. The pass is a % slow one: it takes a full 3 seconds to get from A to B. % {\em (a)} Assuming the ball does not bounce, sketch its % trajectory. You do not need to write any differential equations. % {\em (\kern .7pt b)} Assuming the ball bounces once, sketch all of its % possible trajectories. % Again you do not need to write any differential equations. % {\em (c)} Now consider all possible solutions to this BVP, % with any number of bounces. % Assume it takes 0.45 seconds for a point mass to fall % from a height of 1 meter. Exactly how % many solutions are there all together? Sketch them. % \par % {\em \underline{Exercise $16.3$}. Multiple solutions of cubic % oscillator.} % Figure 16.6 shows phase plane plots of two of the five solutions of (16.2) % indicated in Figure 16.10. % Expand the plot to include all five solutions. How % do you think the conjectures stated after Figure 16.6 about % solutions with various winding numbers should be corrected? % \par % {\em \underline{Exercise $16.4$}. Deflation.} % {\em (a)} Equation (16.2) can be written $N(y) = 0$, where $N$ is a nonlinear % operator applying to functions on $[\kern .3pt 0,1]$ with % boundary values equal to~$1$. Compute the solution % to (16.2) shown in Figure 16.3, and % call this function~$Y$. Now consider the new nonlinear operator % $M(y) = N(y)(1 + \|y - Y\|^{-1})$, where $\|\cdot\|$ is the % 2-norm. Note that $M(y) = 0$ for % $y\ne Y$ if and only if $N(y) = 0$. Use Chebfun to solve % $M(y) = 0$ numerically. Which % solution from Figure 16.10 do you get? % {\em (\kern .7pt b)} This process is automated by % the Chebfun {\tt deflate} command. Show that you get the % same solution with \verb|deflate(N,Y,1,1)|. Which solution % do you get with \verb|deflate(N,Y,2,0)|? % (For information on deflation, see % Farrell, Birkisson, and Funke, Deflation techniques for % finding distinct solutions of nonlinear partial differential % equations,'' {\em SIAM Journal on Scientific Computing,} 2015.) %