%% 11. The fundamental existence theorem %% % % \setcounter{page}{129} % %% % % More than most mathematical fields, the subject of ODE\kern .5pt s is % founded on a key theorem: solutions to an IVP $y'= f(t,y)$, % $y(0) = y_0^{}$ % exist and are unique so long as $f$ is continuous % with respect to both variables and Lipschitz continuous with % respect to~$y$, as defined below. (If $\partial f/\partial y$ % exists and is uniformly bounded, that is enough to % imply Lipschitz continuity, since % a bound on $|\partial f / \partial y|$ will serve as % a Lipschitz constant.) % The theorem applies to % systems as well as scalars, which means that by the usual addition % of extra variables, it applies to higher-order as well as % first-order equations. We rely on this result throughout this % book. In this chapter we present one of the % standard proofs, due to Picard and % Lindel\"of.\footnote{We emphasize that this theorem and proof % represent just one particularly noteworthy % item from the theory of existence and % uniqueness for ODE\kern .5pt s. In fact, there is a whole book on the subject: % R. P. Agarwal and V. Lakshmikantham, % {\em Uniqueness and Nonuniqueness Criteria for Ordinary % Differential Equations,} World Scientific, 1993.} % %% % % Since an ODE prescribes the slope of a curve at each point, it % may seem obvious that a unique solution must always exist. However, % we saw examples in Chapter~3 showing this is not the % case. The problem $y'= y^2$, $y(0)=1$ of (3.9) has the solution % $y(t) = 1/(1-t)$ on $[\kern .3pt 0,d\kern 1pt]$ for $d<1$, but no % solution exists for $d\ge 1$; $y^2$ fails to be % Lipschitz continuous as $y\to\infty$. % The problem $y' = y^{1/2}$, $y(0) = 0$ of (3.16) % has the distinct solutions $y(t) = 0$ and $y(t) = t^2/4$, as well % as many others; $y^{1/2}$ is not Lipschitz continuous for $y\approx 0$. % So existence and uniqueness cannot be taken for % granted. % %% % % For simplicity, we will state and prove the theorem first in the scalar % case, and then indicate the modest changes needed for the % generalization to systems. % %% % % \smallskip % {\em % {\bf Theorem 11.1. Existence and uniqueness for a % first-order scalar ODE IVP (\textsf{\textbf{FlaShI}}).}\ \ If\/ $f$ is continuous % with respect to\/ $t$ and\/ $y$ % and Lipschitz continuous with respect to\/ $y$, then % the IVP % $$y'(t) = f(t,y), \quad % t\in [\kern .3pt 0,d\kern 1pt], ~y(0) = y_0^{} \eqno (11.1)$$ % has a solution, and it is unique.} % %% % % \smallskip % \noindent % By a solution to the IVP, we mean a continuously differentiable % function $y(t)$ satisfying (11.1). % The condition that $f$ is {\bf Lipschitz continuous} % with respect to $y$ means that there exists a constant % $K$ such that for all $t\in [\kern .3pt 0,d\kern 1pt]$ and % $y\in {\bf R}$, % $|f(t,y_2^{})-f(t,y_1^{})| \le K |y_2-y_1{}|$. % Although our discussion assumes $d>0$ for simplicity, % (11.1) is actually symmetric with respect to $t$, and the same result % holds for $t\in [\kern .2pt d,0\kern .3pt]$ with $d<0$. % %% % % The standard proof of Theorem 11.1 is based on the process % known as {\bf Picard iteration}. % We note that integration of (11.1) yields the % equation\footnote{This % is an example of a {\em Volterra integral equation.}} % $$y(t) = y_0^{} + \int_0^t f(s,y(s)) \kern .7pt ds . \eqno (11.2)$$ % In the Picard iteration, we consider % the sequence of functions defined by % $y^{(0)} = y_0^{}$ and then % $$y^{(k+1)}(t) = y_0^{}+\int_0^t\kern -1.5pt f(s,y^{(k)}(s)) \kern .7pt ds, % \quad k = 0,1,2,\dots , \eqno (11.3)$$ % or as we may write abstractly for an operator $N$, % $$y^{(k+1)} = N(y^{(k)}), \quad k = 0,1,2,\dots . \eqno (11.4)$$ % The proof consists of showing that under the given assumptions, % this successive substitution % process converges to a unique solution of (11.1), at least % on some smaller interval $[\kern .3pt 0,d/m]$. By a succession of % $m$ such steps we reach all of $[\kern .3pt 0,d\kern 1pt]$. % %% % Before presenting the mathematical argument, % let us see the construction in action. The problem % $$y' = \sin(y) + \sin(t), \quad t\in [0,8], ~~ y(0) = 1 \eqno (11.5)$$ % is an example of a nonlinear IVP whose defining function % $f(t,y)$ is continuous with respect to both variables and Lipschitz % continuous with respect to $y$; the Lipschitz constant % can be taken as $K=1$. % Here is a plot of iterates $k = 0,\dots,4$: ODEformats, d = 8; t = chebfun('t',[0 d]); y0 = 1; y = y0 + 0*t; ss = @(k) ['$k = ' int2str(k) '$']; for k = 0:4 plot(y,CO,ivpnl), hold on, ylim([-3 10]) text(1.015*d,y(end),ss(k),IN,LT) y = y0 + cumsum(sin(y)+sin(t)); end title(['Fig.~11.1.~~Picard iterates ' ... '$k = 0,\dots,4$ for (11.5)'],FS,11) xlabel('$t$',FS,10,IN,LT), ylabel('$y$',FS,10,IN,LT), hold off %% % \vskip 1.02em %% % % \noindent % A second plot shows $k = 5,\dots,9$. % for k = 5:9 plot(y,CO,ivpnl), hold on, ylim([0 7]) if k==8, text(1.015*d,y(end)-.25,ss(k),IN,LT) elseif k==5, text(1.015*d,y(end)+.25,ss(k),IN,LT) else text(1.015*d,y(end),ss(k),IN,LT), end y = y0 + cumsum(sin(y)+sin(t)); end title('Fig.~11.2.~~Picard iterates $k = 5,\dots,9$',FS,11) xlabel('$t$',FS,10,IN,LT), ylabel('$y$',FS,10,IN,LT), hold off %% % \vskip 1.02em %% % % \noindent % A third plot shows $k = 10,\dots ,14$, and this time, % we include the true solution (a dashed lined in red, mostly hidden % under the green curves) to confirm that % the iteration is converging successfully. % N = chebop(0,d); N.op = @(t,y) diff(y) - sin(y); N.lbc = y0; yexact = N\sin(t); plot(yexact,'--r'), ylim([1 6]), hold on for k = 10:14 plot(y,CO,ivpnl), text(1.015*d,y(end),ss(k),IN,LT) y = y0 + cumsum(sin(y)+sin(t)); end title(['Fig.~11.3.~~Picard iterates ' ... '$k = 10,\dots,14$ with exact solution'],FS,11) xlabel('$t$',FS,10,IN,LT), ylabel('$y$',FS,10,IN,LT), hold off %% % \vskip 1.02em %% % % These figures give a vivid impression of how % convergence works for the Picard iteration: % {\em from left to right.} % On a short interval like $[\kern .3pt 0,1]$, the convergence for % this example is rapid, whereas on the longer % interval $[\kern .5 pt 0,8\kern .5pt]$ the rate is not so good. % In fact, the successive iterates satisfy % $$y^{(0)}(t)-y(t) = O(t), ~~ y^{(1)}(t)-y(t) = O(t^2), ~~ % y^{(2)}(t)-y(t) = O(t^3), \dots . \eqno (11.6)$$ % These estimates apply even if $f$ is not smooth with % respect to $t$ or $y$ (Exercise~11.7). % For our example, the powers are readily confirmed on the computer % by plotting the errors of $y^{(0)},\dots, y^{(4)}$ % as functions of\/ $t$ on a log-log plot. (The vertical bars correspond % to points where the error happens to cross through zero.) % y = y0 + 0*t; ss = @(k) ['$k = ' int2str(k) '$']; tt = logspace(-2,log10(8),1200); for k = 0:4 err = abs(y(tt)-yexact(tt)); for j = 10:1000 if err(j)<=min(err(j-1:j+1)), err(j)=1e-20; end end loglog(tt,err,'k',LW,.7), hold on text(8.7,err(1),ss(k),IN,LT) y = y0 + cumsum(sin(y)+sin(t)); end xlabel('$t$',FS,10,IN,LT), ylabel('error',FS,9,IN,LT) title('Fig.~11.4.~~Errors of iterates $0,\dots,4$ for (11.5)',FS,11) axis([1e-2 8 1e-16 1e3]), hold off %% % \vskip 1.02em %% % % One way to prove Theorem 11.1 is to make (11.6) quantitative. % (See for example chap.~12 of S\"uli and Mayers, {\em An % Introduction to Numerical Analysis,} Cambridge, 2003.) % We follow here the more abstract and elegant % approach of regarding the mapping $N$ of (11.4) % as a {\em contraction map} % in the Banach space $C([\kern .3pt 0,d\kern 1pt ])$ of continuous functions % on $[\kern .3pt 0,d\kern 1pt ]$ with the % supremum norm.\footnote{Do we really need Banach spaces? % Not really --- see p.~4 of Hastings and McLeod, % {\em Classical Methods in Ordinary Differential Equations with % Applications to Boundary Value Problems,} American Mathematical % Society, 2012.} % More precisely, it may be necessary to restrict the interval % to $[\kern .3pt 0,d/m]$ for some $m$ and take $m$ steps. % %% % % \smallskip % {\em Proof of Theorem\/ $11.1$.} % As described above, we set $y^{(0)} = y_0^{}$ and iterate with % the formula (11.3). % By induction it follows that % $y^{(k)}$ is well-defined and continuously differentiable % for each $k$. % From (11.2) we compute % $$y^{(k+1)}(t) - y^{(k)}(t) = \int_0^t\kern -1.5pt % \left[ f(s,y^{(k)}(s)) - f(s,y^{(k-1)}(s))\right ] ds.$$ % Since $t\le d$, this equation implies % $$\| y^{(k+1)}-y^{(k)}\|_\infty\le d K \|y^{(k)}-y^{(k-1)}\|_\infty, % \eqno (11.7)$$ % where $K$ is the Lipschitz constant. % Now suppose $dK < 1$. Then (11.7) asserts that the map % $N$ of (11.4) is a % {\em contraction} in the Banach space % $C([\kern .3pt 0,d\kern 1pt])$ of continuous functions % on $[\kern .3pt 0,d\kern 1pt]$ with norm $\|\cdot \|_\infty$. By a % standard result known as the {\em Banach fixed-point theorem\/} or % the {\em contraction mapping principle,} % a contraction map has a unique fixed point, i.e., a point $y$ with % $N(y) = y$, and the iteration converges to it. % In our context, this means that the iteration converges to a unique % function $y\in C([\kern .3pt 0, d\kern 1pt])$ that satisfies % $N(y) = y$, i.e., (11.2). From (11.2) it follows that $y$ is % in fact continuously differentiable and satisfies (11.1). % %% % % \def\qed{\vrule height4pt width3pt depth4pt} % This completes the proof if $dK < 1$. % What if $dK > 1$? In this case we may pick an integer % $m$ such that $dK/m < 1$ and use the same argument to establish % a unique solution in $C([\kern .3pt 0, d/m])$. % From here, a second % application of the fixed-point theorem gives a unique solution in % $C([\kern .2pt d/m,2d/m])$; and so on for $m$ steps. % \qed % %% % % \smallskip % Theorem 11.1 extends in an immediate manner to a first-order % system of $n$ equations. % \smallskip % %% % % {\em % {\bf Theorem 11.2. Existence and uniqueness % for a first-order system of ODE\kern .5pt s (\textsf{\textbf{FlashI}}).} % If\/ {\bf f} is continuous % with respect to\/ $t$ and\/ $\bf y$ and Lipschitz continuous % with respect to\/ $\bf y$, then the IVP % $${\bf y}'(t) = {\bf f}(t,{\bf y}), \quad % t\in [\kern .3pt 0,d\kern 1pt], ~{\bf y}(0) = {\bf y}_0 \eqno (11.8)$$ % has a solution, and it is unique.} % \smallskip % %% % % \noindent % The same proof works as before, with the Banach space % $C([\kern .3pt 0,d\kern 1pt])$ generalized to % $C^n([\kern .3pt 0,d\kern 1pt])$ and with $\|{\bf y}\|_\infty$ % now defined as the supremum for $t\in[\kern .3pt 0,d\kern 1pt]$ % of $\|{\bf y}(t)\|$, where $\|\cdot\|$ is any fixed norm on ${\bf R}^n$. % Lipschitz continuity is also defined with respect % to the latter norm. % We say that ${\bf f}(t,{\bf y})$ is Lipschitz continuous % with respect to $\bf y$ if there exists a constant % $K$ such that for all $t\in [\kern .3pt 0,d\kern 1pt]$ and % $y\in {\bf R}^n$, % $\|f(t,{\bf y}_2^{})-f(t,{\bf y}_1^{})\| \le K \|{\bf y}_2-{\bf y}_1{}\|$. % Since all norms are equivalent on a finite-dimensional space, this % definition is independent of the choice of norm. % %% % % \vskip .1em % \begin{center} % *~~~*~~~* % \end{center} % \vskip -1em % %% % The main business of this chapter is finished: the statement and % proof of Theorems 11.1 and 11.2, which assume that $f(t,y)$ is continuous % in $t$ and Lipschitz continuous in $y$. % Let us now explore around the edges a % little. Although these theorems mark a center point % of this field, there is more to be said about both % existence and uniqueness. %% % % The most important matter to note is % that continuity of $f$ with respect to~$t$ is a stronger assumption % than necessary for the Picard iteration argument. % The simplest next step, sufficient for the % examples explored in this book such as those of Figures~2.3 and~2.8, % is to suppose that $f$ is piecewise continuous as in Theorems~2.2 and~2.3. % When we say that the bivariate function $f(t,y)$ is % piecewise continuous with respect to $t$, we mean that % it is continuous with respect to $t$ except for at most a finite % set of jump discontinuities at points $t_1^{},\dots,t_k^{}$ that are % independent of $y$. % The proof by Picard iteration then goes through essentially as before, with % the Banach space $C([\kern .3 pt 0,d\kern .3pt])$ generalized to the % Banach space of piecewise continuous functions with jump discontinuities only % at the same points % $t_1^{},\dots,t_k^{}$, again with norm $\|\cdot\|_\infty^{}$. % %% % % \smallskip % {\em % {\bf Theorem 11.3. Existence and uniqueness % for discontinuous ODE\kern .5pt s (\textsf{\textbf{FlashI}}).} % The conclusions of Theorems~$11.1$ and\/~$11.2$ also hold % if\/ {\bf f} is piecewise continuous % with respect to\/ $t$ as defined above % and Lipschitz continuous with respect to\/ $\bf y$.} % \smallskip % %% % % For an extensive treatment of existence theorems for % discontinuous ODE\kern .5pt s, see % A. F. Filippov, {\em Differential Equations % with Discontinuous Righthand Sides,} Kluwer, 1988. % One theme in this theory is to extend results like % Theorem~11.3 to coefficients that are just integrable % with respect to $t$. More generally, one may also % weaken the condition that $f$ is Lipschitz continuous % with respect to $y$. % One line of such results takes $f$ to be continuous % but not necessarily Lipschitz continuous, which is % enough to guarantee existence of a solution at least locally, on some interval % $[\kern .3pt 0,\delta\kern .7pt ]$ with $\delta > 0\kern .5pt$; this is % the {\em Peano existence theorem.} % The solution need not be unique, however. % A standard example is equation (3.16), % $y' = y^{1/2}$, and here is another example, taken % from Ince's {\em Ordinary Differential Equations\/}: % y' = f(t,y), \quad f(t,y) = \cases{0 & t=y=0, % \cr\noalign{\vskip 4pt} % \displaystyle{4 t^3 y\over y^2 + t^4} & otherwise.} \eqno (11.9) % The function $f$ is continuous with respect % to $t$ and $y$ for all $t$ and $y$, and % away from the point $t=y=0$, it is (locally) Lipschitz continuous with % respect to $y$, ensuring existence and uniqueness of solutions % so long as they stay away from this point and $\pm \infty$. % Here is a display of some such solutions. % N = chebop(0,1); N.op = @(t,y) diff(y) - 4*t^3*y/(y^2+t^4); for y0 = [-1.4:.2:-.2 .2:.2:1.4] N.lbc = y0; y = N\0; plot(y,CO,ivpnl), hold on end ylim([-1.5 1.5]) title(['Fig.~11.5.~~Solutions of (11.9) ' ... 'that avoid the point $t=y=0$'],FS,11,IN,LT) %% % \vskip 1.02em %% % % \noindent % The interesting matter is what happens with solutions that % do touch the point $t=y=0$. % We can quickly spot three such solutions, % $y(t) = -t^2$, $y(t) = 0$, and $y(t)= t^2$. We add dashed black curves % for $-t^2$ and $t^2$ to the figure. % t = chebfun('t',[0,1]); plot(-t^2,'--k'), plot(t^2,'--k') title(['Fig.~11.6.~~Two solutions of (11.9) ' ... 'passing through $t=y=0$'],FS,11,IN,LT) %% % \vskip 1.02em %% % % \noindent % Between these two black curves, there is a continuum of % solutions, all passing through the point $t=y=0$. We compute % some numerically by marching backwards from $t=1$ to $t=0.01$, % plotting the results in orange. % N.domain = [.01 1]; N.lbc = []; for y0 = [-.85:.17:.85] N.rbc = y0; y = N\0; plot(y,CO,orange) end plot(-t^2,'--k'), plot(t^2,'--k') title(['Fig.~11.7.~~A continuum of solutions ' ... 'passing through $t=y=0$'],FS,11,IN,LT) hold off %% % \vskip 1.02em %% % % \noindent % Actually, this problem has an analytic solution. For any % $c$, the function % $$y(t) = c^2 - (t^4+c^4)^{1/2} \eqno (11.10)$$ % satisfies (11.9) and takes the value $y(0) = 0$. % The negative of (11.10) is also a solution. % %% % % In an image like Figure 11.7, nonuniqueness takes the form of % multiple solutions emanating from a single point % $t$ and $y(t)$. It is interesting to consider the same % effect in reverse time, solving the same ODE for example % from $t=1$ to $t=0$, as we discussed for the leaky bucket'' problem % of Chapter 3. (We also reversed time to generate Figure 11.7.) % Here, we see multiple trajectories that coalesce --- not just % approximately but exactly. In an ODE problem driven by % a Lipschitz continuous function $f$, two distinct solutions % can never coalesce, so in some sense information about the % initial condition can never be completely lost. Non-Lipschitz % problems like the leaky bucket viewed backward in time, % however, are different. % One might say that the adjoint of nonuniqueness is extinction. % %% % % \begin{center} % \hrulefill\\[1pt] % {\sc Application: designer nonuniqueness}\\[-3pt] % \hrulefill % \end{center} % %% % From examples like (3.16) and (11.9), one may get the impression % that nonuniqueness is elusive, to be found only if % one knows how to look in just the right hiding places. In fact, % generating nonunique solutions is as easy at sketching curves % on a sheet of paper, and Figure 11.7 shows us the sort % of curves that are needed. Take this diagram of functions % $a (\sin(\pi t))^2$ with $-1 \le a \le 1$, for example. t = chebfun('t',[-2,2]); for a = -1:.2:1 y = a*sin(pi*t)^2; plot(y,LW,.5,CO,ivpnl) axis([-1.7 1.7 -2.5 2.5]), hold on end title('Fig.~11.8.~~Sketching nonunique solution trajectories', ... FS,11,IN,LT) %% % \vskip 1.02em %% % % \noindent % Each of these curves $y(t)$ is smooth, with a well-defined derivative % at each point, which we may call $f(t,y)$. So the curves are solutions % of $y'(t) = f(t,y)$ for this choice of $f$. Yet % the trajectory emanating from $t=0$, $y= 0$ is obviously nonunique, as % is the trajectory emanating from $t=k$, $y=0$ for any integer $k$. % \noindent % %% % % In the figure, suppose we imagine that trajectories have been % specified as indicated for each value $a \in [-1, 1]$. If the ODE % is to be defined for all~$t$ and~$y$, we need to fill in the regions % above the top curve and below the bottom one. This can be % done in any number of ways, and here is one of them, based on % functions $\pm [b + (\sin(\pi t))^2]$ with $b>1$. % t = chebfun('t',[-2,2]); for b = .2:.2:1 y = b + sin(pi*t)^2; plot([y; -y],LW,.5,CO,ivpnl) end hold off title('Fig.~11.9.~~Filling in the whole $t$-$y$ plane',FS,11,IN,LT) %% % \vskip 1.02em %% % % \noindent % If $b$ is regarded as taking all values $1< b < \infty$, then we % have completed a specification of a function $f(t,y)$ that is % continuous with respect to both variables for all % $t$ and $y$. Since it has nonunique trajectories through certain % points, it cannot possibly be Lipschitz % continuous with respect to $y\kern .7pt.$ % And indeed it is not, at least when $t$ ranges over any interval % that includes one of the integers (Exercise~11.6). % %% % % Once we start sketching trajectories in the plane, we can see that % all kinds of effects are possible. Here is a function {\tt pinch} % that has the effect of pinching'' a square of diameter $2d$ centered % at $(t_0^{},y_0^{})$ so as to introduce nonuniqueness at this point. % pinch = @(t,y,t0,y0,d) y - (abs(y-y0) % \noindent % Rather than talk through the algebra we visualize the effect % of applying {\tt pinch(t,y,0,0,1)} to a set of horizontal trajectories: % t = linspace(-2,2,200)'; for ys = -2:.05:2 y = ys + 0*t; y = pinch(t,y,0,0,1); plot(t,y,LW,.4,CO,ivpnl), hold on end axis([-2 2 -2 2]), axis square title(['Fig.~11.10.~~Pinching trajectories gives nonuniqueness ' ... '\vrule height 16pt width 0pt depth 6pt'],FS,11,IN,LT), hold off set(gcf,'position',[380 220 540 300]), set(gca,YT,-2:2) %% % \vskip 1.02em %% % % \noindent % By pinching the plane at the middle point, we have introduced % nonuniqueness there. With a few more pinches we can introduce more % points of nonuniqueness. % for ys = -2:.05:2 y = ys + 0*t; y = pinch(t,y,0,0,1); y = pinch(t,y,1.3,1.3,.5); y = pinch(t,y,-1.3,-1.3,.5); y = pinch(t,y,1,-1,.3); y = pinch(t,y,-1,1,.3); plot(t,y,LW,.4,CO,ivpnl), hold on end axis([-2 2 -2 2]), axis square, hold off title(['Fig.~11.11.~~Multiple pinch points' ... '\vrule height 16pt width 0pt depth 6pt'],FS,11,IN,LT) set(gcf,'position',[380 220 540 300]), set(gca,YT,-2:2) %% % \vskip 1.02em %% % % \noindent % In a paper published in 1963, based on an earlier publication % by Lavrentieff in 1925, P. Hartman took this idea to a fascinating % limit with a fractal flavor (A differential % equation with non-unique solutions,'' % {\em American Mathematical Monthly}). % By introducing infinitely many pinches on successively smaller % scales, he showed that one can % construct a flow field with pinch points falling % arbitrarily close to every value $(t,y)$. This implies a % remarkable consequence: there is an ODE $y' = f(t,y)$ with the % property that for every choice of initial point $(t_0^{},y_0^{})$, % the equation has more than one solution on every % interval $[\kern .5pt t_0^{}, t_0^{}+\varepsilon\kern .5pt ]$. % %% % % \smallskip % {\sc History.} Theorem 11.2 is often called the % Picard--Lindel\"of theorem, % following publications by Picard in 1890 and Lindel\"of in 1894; % another landmark was Goursat's {\em Cours d'analyse % math\'ematique} of 1908. % It is also called the Cauchy--Lipschitz theorem, and an early % publication was by Liouville in 1838. % Peano's local existence theorem, without the Lip\-schitz requirement % or the assertion of uniqueness, appeared in 1890 and was % elaborated in his {\em Trait\'e d'Analyse,} though % the idea of successive substitution is much older.\ \ Banach's % fixed-point theorem appeared in 1922, and % Carath\'eodory in 1927 published one of the first % local existence theorems for ODE\kern .5pt s defined by % discontinuous coefficient functions~$\bf f$. % %% % % \smallskip % {\sc Our favorite reference.} % Discontinuous right-hand sides are ubiquitous in applications, % and their dynamical consequences are explored in % di Bernardo, Budd, Champneys, and Kowalczyk, % {\em Piecewise-smooth Dynamical Systems: Theory and Applications,} % Springer, 2008. % \smallskip % %% % % \begin{displaymath} % \framebox[4.7in][c]{\parbox{4.5in}{\vspace{2pt}\sl % {\sc Summary of Chapter 11.} The fundamental existence theorem % of ODE\kern .5pt s asserts that an ODE\/ ${\bf y}' = {\bf f}(t,{\bf y})$ subject to % initial data ${\bf y}(0) = {\bf y}_0$ has a unique solution for all\/ $t$\/ % if\/ $\bf f$ is continuous with respect to\/ $t$ and Lipschitz % continuous with respect to\/ $\bf y$. This result applies to % systems as well as scalars and thus covers higher-order % ODE\kern .5pt s too if appropriate conditions are specified on derivatives. % Both existence and uniqueness can fail if\/ $\bf f$ is merely continuous % with respect to\/ $\bf y$, though existence still holds locally. % \vspace{2pt}}} % \end{displaymath} % %% % \smallskip\small\parskip=1pt\parindent=0pt % {\em Exercise $11.1$. Cleve Moler's favorite ODE.}\ \ Consider % the IVP $(y')^2 + y^2 = 1$, $y(0) = 0$ with % the additional constraint $-1 \le y \le 1$. % {\em (a)} Show that there are exactly two solutions % for $t\in [\kern .3pt 0,1]$ and state formulas for them. % {\em (\kern .7pt b)} Show that there are infinitely many solutions % for $t\in [\kern .3pt 0,2\kern .3pt ]$. % \par % {\em Exercise $11.2$. From an integral equation to an IVP.}\ \ Convert % the integral equation $y(t) = e^{\kern .3pt t} + 4\int_0^t (t-s) y(s) ds$ % to an ODE IVP (including the initial condition). Determine % the solution analytically. % \par % {\em Exercise $11.3$. Differentiability and Lipschitz continuity.} % Prove that if a function $f(t,y)$ has a uniformly bounded derivative % $|\partial f/\partial y|$ for all $t$ and $y$, % then it is Lipschitz continuous. % \par % {\em Exercise $11.4$. Value of\/ $m$ for our example.} % For the example (11.5) of the opening pages of this chapter, % the Picard iteration converges over the whole interval % $[\kern .3pt 0,8]$. However, our proof of Theorem 11.1 % subdivides the interval into $m$ subintervals. For this example, % what is the smallest allowed number $m$? % \par % {\em Exercise $11.5$. A first-order two-point BVP.}\ \ {\em (a)} % Sketch the solution curves of $y' = |y|^{1/2}$ in % the $(t,y)$ plane. How do nonuniqueness and extinction effects % appear in this plot? % {\em (\kern .7pt b)} Suppose $y(0) = -1$ and $y(6) = 1$. There exists % a unique solution for $t\in [\kern .3pt 0,6\kern .3pt ]$ % for these data (an unusual situation in that this is % a first-order equation with two boundary conditions!). % Determine this solution analytically. % \par % {\em Exercise $11.6$. Figure\/ $11.9$.} % {\em (a)} Verify that the function $f$ described by Figure~11.9 is % not Lipschitz continuous. % {\em (\kern .7pt b)} Exactly % how many solutions are there on $[-2,2]$ to the ODE described by % this figure that satisfy $|y(t)|=1$ for $t = -1.5, -0.5, % 0.5, 1.5\kern .7pt ?$ % \par % {\em \underline{Exercise $11.7$}. Nonsmooth $f$.} % In the text it was mentioned that the estimates (11.6) apply % even if $f$ is not smooth. Verify this experimentally by % reproducing Figure 11.4 with $f$ changed to % {\em (a)} $\sin(y) + \hbox{sign}(\sin(50\kern .3pt t))$ and % {\em (\kern .7pt b)} $|\sin(10\kern .3pt y)| + \sin(t)$. % It is enough to work with the time interval $[\kern .3pt 0, 1]$. % \par % {\em \underline{Exercise $11.8$}. A growing exponent.} % A function $y(t)$ is continuously differentiable, nonnegative, and unbounded for % $t\in (0,4)$, where it satisfies % $y' = y^t$. For what subinterval of $(0,4)$ is % $y$ identically zero? % ({\em Hint.} Try integrating backward in time from $t=4$ to $t=1$ % with a boundary condition $y(4) = 4$ or $8$.) %