%% 3. First-order scalar nonlinear ODEs %% % % \setcounter{page}{22} % %% % The first thing to do with nonlinear problems is enjoy them. % The chances are you can't solve a particular nonlinear % ODE analytically, but there % is a wonderful variety of effects to explore. % Incidentally, a change of notation applies % here: mathematicians usually write a linear differential operator % without parentheses, $y \mapsto Ly$, % but a nonlinear one with parentheses, $y \mapsto N(y)$. The % change of letters from $L$ to $N$ is also a reminder of nonlinearity. %% % For example, here is a basic linear IVP, really just an % integral since $y$ appears only in the term $y'$: % $$y' = 3\kern -.9pt\cos(t), \quad y(0) = 0. \eqno (3.1)$$ % Its solution is the sine wave $y(t) = 3\kern -.9pt\sin (t)$. ODEformats, N = chebop(0,20); N.lbc = 0; N.op = @(t,y) diff(y); rhs = chebfun('3*cos(t)',[0,20]); y = N\rhs; plot(y,CO,ivp), ylim([-4 4]) title('Fig.~3.1.~~Linear oscillation (3.1)',FS,11) %% % \vskip 1.02em %% % % \noindent % What happens if we add a nonlinear function of $y$ % to the operator? For example, let us construct an ODE % that behaves like (3.1) when $y$ has a small amplitude % but shuts off'' as $|y|$ increases. An example % of such an equation is % $$y' + |y|^2 y = 3\cos(t), \quad y(0) = 0. % \eqno (3.2)$$ % (We write $|y|^2 y$ instead of $y^3$ % to suggest the idea of an amplitude-dependent coefficient $|y|^2$ % multiplying the usual $y$ term.) % The next plot shows the new solution superimposed on the previous one. Note % that the curves are about the same at first, but diverge as the % amplitudes grow larger. % N.op = @(t,y) diff(y) + y^3; y2 = N\rhs; plot(y,CO,ivp), ylim([-4 4]) hold on, plot(y2,CO,ivpnl), hold off title('Fig.~3.2.~~Oscillation with nonlinear damping (3.2)',FS,11) %% % \vskip 1.02em %% % Here is a variant of the same nonlinear idea. Suppose that instead % of a penalty'' for large values of $|y|$ we impose a % barrier,'' preventing $|y|$ from reaching the value $1$. % Here is an equation with a logarithmic barrier: % $$y' - {1\over 2} \log(1-|y|) y = 3\cos(t). % \eqno (3.3)$$ % The solution is smooth, though it doesn't look it (see Exercise 3.1), % and confined to values $-1 < y < 1$. N.op = @(t,y) diff(y) - 0.5*log(1-abs(y))*y; y3 = N\rhs; plot(y,CO,ivp), ylim([-4 4]) hold on, plot(y3,CO,ivpnl), hold off title(['Fig.~3.3.~~Nonlinear damping by a ' ... 'logarithmic barrier (3.3)'],FS,11) %% % \vskip 1.02em %% % Examples like these, scalar problems of first order, are % rather limited. The variety will grow in later chapters % when we come to problems of higher order or with % multiple variables. %% % % Both (3.2) and (3.3) are written with nonlinearities % of the form $g(y) y$ for some function $g$ (equal to $|y|^2$ and % $-0.5 \log(1-|y|)$, respectively). This form suggests % the idea that locally, a nonlinear ODE should behave % approximately linearly. Near any time $t\approx t_0^{}$, it should % be possible to approximate the solution by the solution of a linear equation % derived, say, by series expansion. This is indeed the case, % at least if the coefficients are continuous, % and the idea will be made precise in Chapter~14. % %% % % For a few special classes of first-order nonlinear ODE\kern .5pt s, % analytical solutions are available. % The most important category of such problems are % the {\bf separable equations.} % In the last chapter we solved $y' = a(t)y$ by writing it as % $${dy\over y } = a(t) dt .$$ % If we generalize $y^{-1}$ to a continuous % function $b(y)$ that does not change sign, we get the equation % $$b(y) dy = a(t) dt ,$$ % and the two sides can be integrated as before to get a solution % at least locally. We record the result as a theorem. % %% % % \smallskip % {\em % {\bf Theorem 3.1. Solution of first-order % separable scalar homogeneous IVP % (\textsf{\textbf{FlaSHI}}).} % Let $a(t)$ be a continuous function of\/ $t$ and % let $b(y)$ be a continuous nonzero function of\/ $y$. % A solution $y(t)$ of the problem % $$b(y) dy = a(t) dt , \quad y(0) = y_0^{} \eqno (3.4)$$ % satisfies the equation % $$\int_{y_0^{}}^y b(x) dx = \int_0^t a(s) ds. \eqno (3.5)$$ % } % \smallskip % %% % % Equation (3.4) has the form $y' = f(t,y)$ (of type {\sf FlaSH}), % but not every equation $y' = f(t,y)$ can be separated in this way. % However, another case where separation is possible is if % $f$ is independent of $t$. % %% % % \smallskip % {\em % {\bf Theorem 3.2. Solution of first-order % autonomous separable scalar homogeneous IVP % (\textsf{\textbf{FlASHI}}).} % Let $b(y)$ be a continuous nonzero function of~$y$. % A solution $y(t)$ of the problem % $$b(y) dy = dt , \quad y(0) = y_0^{} \eqno (3.6)$$ % satisfies the equation % $$\int_{y_0}^y b(x) dx = t.$$ % } % \smallskip % %% % % \noindent % Similarly $y' = f(t,y)$ is separable if $f$ is independent % of $y$ --- it is just an integral. % %% % % A prototypical example of a problem of the form (3.6) is the ODE % $$y' = y^\alpha, \eqno (3.7)$$ % where $\alpha$ is a constant. Dividing by $y^\alpha$ gives % $dy/y^\alpha = dt$ and hence % $${y^{1-\alpha}\over 1-\alpha} = t - t_{\rm b}^{}$$ % for some constant $t_{\rm b}^{}$ (the letter b stands for % blowup''). This simplifies to % $$y = \big[(1-\alpha)(t - t_{\rm b}^{})\big]^{1/(1-\alpha)}. \eqno (3.8)$$ % For $\alpha>1$, this solution exhibits a phenomenon % of mathematical and physical interest, % {\em blowup in finite time}. % For a specific illustration, consider the IVP % $$y' = y^2, \quad y(0) = 1. \eqno (3.9)$$ % From (3.8) with $\alpha=2$, we see that the solution is % $$y = {1\over 1-t}, \eqno (3.10)$$ % which diverges to $\infty$ at $t_{\rm b}^{} = 1$. % Here is a plot of the solution up to the point where % it reaches the value $100$. % N = chebop(0,1); N.op = @(t,y) diff(y) - y^2; N.lbc = 1; N.maxnorm = 100; y = N\0; plot(y,CO,ivpnl), axis([0 1.12 0 120]) hold on, plot([1 1],[0 120],'--k'), hold off text(1.01,30,'$t_b = 1$','color','k',IN,LT) title('Fig.~3.4.~~Blowup in finite time (3.9)',FS,11) %% % \vskip 1.02em %% % % Behaviorally, (3.9) and (3.10) are rich in potential applications % and interpretations. % The ODE (3.9) describes a process where $y$ increases not just % in proportion to its current amplitude, but faster. For example, % one can imagine that $y(t)$ represents the temperature at time % $t$ of a smoldering haystack that smolders faster as it gets % hotter. The singularity at $t=1$ corresponds to the hay % catching fire --- spontaneous combustion or thermal runaway''. % (This simple idea is refined in the Application of Chapter~18.) % %% % Now let us vary the problem slightly and replace (3.9) by % $$y' = y + y^2, \quad y(0) = 1. \eqno (3.11)$$ % Since the values of $y'$ are now bigger, % we expect that blowup will occur % again and it will happen sooner. % Again a solution can be obtained by separation of variables. % We write % $${dy\over y + y^2} = dt,$$ % and integration gives % $$\log\left( {y\over 1+y}\right) = t-t_{\rm b}^{},$$ % that is, % $${y\over 1+y} = e^{t-t_{\rm b}},$$ % or equivalently % $$y = {1\over e^{t_{\rm b}-t} - 1}. \eqno (3.12)$$ % With the initial condition $y(0)=1$, we have $t_{\rm b}^{} = \log 2$, % confirming that as predicted, the blowup is earlier than in (3.10). N.op = @(t,y) diff(y) - y - y^2; y = N\0; plot(y,CO,ivpnl), axis([0 1.12 0 120]) hold on, plot(log(2)*[1 1],[0 120],'--k'), hold off text(.71,30,'$t_b = \log(2)$','color','k',IN,LT) title('Fig.~3.5.~~Exponential blowup in finite time (3.11)',FS,11) %% % \vskip 1.02em %% % % We solved (3.11) by separation of variables. % As it happens, this is a special case of % a more general class of ODE\kern .5pt s that can also be solved analytically. % A {\bf Bernoulli equation}\footnote{Named after Jacob Bernoulli, the % brother of Johann and uncle of Daniel.} % is an ODE of the form % $$y' = a(t) y + b(t) y^p, \eqno (3.13)$$ % where $p$ is a constant and $a(t)$ and $b(t)$ are % given functions of $t$. % (If $p=0$, the problem is linear, and if $p=1$ it is % linear and homogeneous. We assume $p$ is not $0$ or $1$.) % If $a$ or $b$ is nonconstant, then separation of variables % will not work for (3.13), but there is a different approach that % still succeeds. % Let us multiply by $y^{-p}$ to get % $$y' y^{-p} = a(t) y^{1-p} + b(t).$$ % If we now make the substitution % $$u = y^{1-p}, \quad u' = (1-p) y^{-p} y',$$ % the equation becomes % $(1-p)^{-1} u' = a(t) u + b(t)$, that is, % $$u' = (1-p) a(t) u + (1-p) b(t).$$ % This is a linear ODE, which can accordingly be % solved by an integrating factor as described in % Theorem 2.3, or by the method of undetermined coefficients. % %% % For example, let us generalize the blowup problems (3.9) and (3.11) % a little further. Consider the ODE % $$y' = y + t\kern .5pt y^2, \quad y(0) = y_0^{}. \eqno (3.14)$$ % Comparing with (3.11), we see that for $y_0^{}=1$, the amplification % will be weaker here for $t\in [\kern .3pt 0,1)$, so we can expect blowup % at a time $t_{\rm b}^{} > \log 2$. % Dividing by $y^2$ converts the equation to % $${y'\over y^2} = {1\over y} + t, \quad y(0) = y_0^{},$$ % and the change of variables $u = y^{-1}$ converts this to % $$u' + u = - t, \quad u(0) = u_0^{} = {1\over y_0^{}}.$$ % Applying Theorem 2.3 or the method % of undetermined coefficients, we find that the % solution of this IVP is % $$u(t) = 1-t+ e^{-t} (u_0^{} - 1),$$ % that is, % $$y(t) = {1\over 1-t+ e^{-t} (y_0^{-1} - 1)}. \eqno (3.15)$$ %% % Examining (3.15), we see that $y_0^{}=1$, as in (3.9) and (3.11), is % a special case in which the exponential term does not appear. % For $y_0^{}=1$, therefore, the blowup occurs at exactly the same % time $t_{\rm b}^{} = 1$ as in (3.10). Larger $y_0^{}$ brings blowup sooner, % and smaller $y_0^{}$ defers it to later. Here are % solutions up to $t=1$ or $y=100$, whichever comes first, % for initial values $y_0^{} = 0.90,0.91, \dots, 1.00$. N.op = @(t,y) diff(y) - y - t*y^2; for y0 = 0.90:.01:1 N.lbc = y0; y = N\0; plot(y,LW,.7,CO,ivpnl), axis([0.95 1 0 120]), hold on title(['Fig.~3.6.~~Bernoulli eq.\ (3.14), ' ... '$y(0) = 0.90,0.91,\dots,1.00$'],FS,11) end hold off %% % \vskip 1.02em %% % % We have explored at some length the phenomenon of % blowup of solutions to an ODE.\ \ Physically, blowup % in finite time corresponds to an explosion or another % feedback process running out % of control. Mathematically, it illustrates % the phenomenon of {\bf nonexistence} of % solutions to certain nonlinear ODE problems. % Because of the blowup at $t_{\rm b}^{}=1$, for example, there % exists no solution to the problem (3.9) on the % interval $[0\kern .3pt, 2\kern .2pt ]$.\footnote{In % Chebfun, if the computed solution % hits the limit {\tt N.maxnorm}, then all further values are set to % {\tt NaN} --- not-a-number.} % There is a well-established general theory of existence and uniqueness % of solutions to ODE IVPs, presented in Chapter 11. % The fundamental result of this theory asserts that the IVP % $$y' = f(t,y), \quad t\in [\kern .3pt 0,\infty), \quad y(0) = y_0^{}$$ % is guaranteed to have a unique solution if $f$ is % continuous with respect to $t$ and Lipschitz continuous with % respect to $y$.\footnote{This means that there exists a constant % $C$ such that for all $t$ and $y_1^{}, y_2^{}$ in the range % of interest, $|f(t,y_2^{})-f(t,y_1^{})| \le C |y_2^{}-y_1{}|$. % Note that (3.9) loses Lipschitz continuity % as $|y|\to \infty,$ implying that existence of solutions % to this problem can fail only if $|y|$ diverges.} % %% % % From the ODE (3.7) with $\alpha<1$, we can develop % an example of {\bf nonuniqueness} that has a beautiful % physical interpretation. % Consider (3.7) with $\alpha = 1/2$, % $$y' = y^{1/2}, \quad y(0) = 0. \eqno (3.16)$$ % From (3.8) we obtain the solution % $$y(t) = {1\over 4} t^2.$$ % An equally valid solution, however, is $y(t) = 0$, and this is % the one that Chebfun will compute. Alternatively, a solution to % (3.16) might get going'' at any time $t_0^{}\ge 0$: % y(t) = \cases{ 0 & t\le t_0^{} % \cr\noalign{\vskip 4pt} {1\over 4}(t-t_0^{})^2 & t\ge t_0^{}} . % Thus there are not just two possible solutions % but an infinite family. Here is a plot of four of them. % t = chebfun('t',[0 6]); for t0 = 0:3 t = chebfun('t',[0, t0+2.6]); y = 0.25*(t-t0)^2*(t>t0); plot(y,CO,ivpnl), hold on end axis([0 6 -1 3]), hold off title('Fig.~3.7.~~Four solutions to (3.16)',FS,11) %% % \vskip 1.02em %% % % If $y\ge y_0^{}$ for some constant $y_0^{} > 0$, % then the right-hand side of (3.16) is % Lipschitz continuous with respect to $y$. This implies that uniqueness % can fail for this problem only at points with $y=0$, though for % any value of $t$. % In Chapter~11 we shall see examples where uniqueness fails % at isolated points such as $y=t=0$. % %% % % We mentioned a physical interpretation, and this involves % the {\em leaky bucket problem.} Suppose a bucket of water has a hole in % the bottom, so the water flows out. After a certain time, all the water % will be gone, and then the bucket remains empty for all time. % Here is the nonuniqueness effect % in the words of Corless and Jankowski:\footnote{See Variations on % a theme of Euler,'' {\em SIAM Review,} 2016. Nonuniqueness for the leaky % bucket is also discussed, among other places, in volume 1 of Hubbard and % West, {\em Differential Equations:\ A Dynamical Systems Approach.}} % \begin{quotation} % \noindent Given an empty bucket, there's no way to tell when it was full --- if % it ever was. % \end{quotation} % The connection with (3.16) is provided by {\em Torricelli's Law} of 1643 (Exercise % 3.12). % If $y>0$ is the height of water in a leaky bucket, then % $y$ decreases at a rate governed by the equation % $$y' = - C y^{1/2}$$ % for an appropriate constant $C$. This means that if we take $t$ to be % time measured {\em backward} from the present, the equation is % $$y' = C y^{1/2},$$ % which is essentially (3.16). % So Figure 3.7, with time reversed, can be interpreted % as a picture of the leaky bucket. % %% % % The last chapter was linear and this one is nonlinear, but both % have been restricted to scalar problems of first order. % Before turning to higher order problems and systems of equations, % we want to illustrate an invaluable trick for simplifying % computations involving particles in a plane: the use of % complex arithmetic. A particle moving in the $x$-$y$ plane % has two coordinates, $x(t)$ and $y(t)$. This suggests an % ODE with two dependent variables, but if we define % $z(t) = x(t) + iy(t)$, we have just a single, scalar variable % $z(t)$. This can be remarkably convenient. One reason is that % many particle interactions depend on distances % between points in the plane, and a distance in the $x$-$y$ plane % can be regarded as the absolute value of a complex vector. % %% % Our Application illustrates this use of a complex variable. %% % % \begin{center} % \hrulefill\\[1pt] % {\sc Application: classic pursuit problems}\\[-3pt] % \hrulefill % \end{center} % %% % % Suppose an antelope runs with speed 1 along the vertical line $x=1$, % starting at $(1,0)$ and going in the positive $y$-direction. A lioness % starts at $(0,0)$ and pursues the antelope, always moving directly toward % the antelope at a fixed speed $C$.\footnote{The cast % of characters is variable, as described in % P. J. Nahin, {\em Chases and Escapes: The Mathematics % of Pursuit and Evasion,} Princeton, 2006. % One may have a dog chasing a rabbit, a hawk chasing a sparrow, or, in % the original treatment by Pierre Bouguer in 1732, a pirate ship % chasing a merchant ship.}\ \ What path does the lioness follow, % and when and where does she catch the antelope? % %% % % The antelope's path is given as the function % $a(t) = 1 + i\kern .3pt t\kern .5pt$; the % unknown in this problem is the path of the lioness. % In complex arithmetic, we can regard this as a % function $z(t)$ given by the IVP % $$z' = C {a(t)-z(t)\over |a(t) - z(t)| }, \quad % z(0) = 0. \eqno (3.17)$$ % Here is a plot showing the lioness's track up to time % $t=4$ for $C = 0.5$, $1$, and~$1.1$. % tmax = 4; a = chebfun('1+1i*t',[0 tmax]); N = chebop(0,tmax); N.lbc = 0; CC = [.5 1 1.1]; for j = 1:3 C = CC(j); N.op = @(t,z) diff(z) - C*(a(t)-z)/abs(a(t)-z); subplot(1,3,j), plot(a,':r'), hold on, plot(a(end),'.r',MS,10) z = N\0; arrowplot(z,CO,ivpnl,YS,0.3), hold off text(0.75,4.6,['$C = ' num2str(C) '$'],FS,9,HA,CT) axis([0 1.5 0 5]), grid on end title('Fig.~3.8.~~The lion chases the antelope\kern .25in',FS,11,HA,RT) %% % \vskip 1.02em %% % % With $C=0.5$, the lioness will obviously never catch % the antelope, which soon sprints out of reach. % With $C=1$, she still never quite makes the catch. % The separation distance is % $0.50000015$ at $t=4$, converging exponentially % to $1/2$ as $t\to\infty$. For any $C>1$, on the other hand, % the catch will take place. We have stopped this experiment % before that point since there is a singularity involved. % %% % % Of course, the antelope may zig-zag. Here is another % run with $C=0.5$ and~$1$ % in which she makes a $90^\circ$ right turn at $t=2$. % As $t\to\infty$ in the latter case, the separation distance % approaches $0.2653\dots .$ % a = chebfun({'1+1i*t','-1+2i+t'},[0 2 4]); N = chebop(0,tmax); N.lbc = 0; CC = [.5 1]; clf for j = 1:2 C = CC(j); N.op = @(t,z) diff(z) - C*(a(t)-z)/abs(a(t)-z); subplot(1,2,j), plot(a,':r'), hold on, plot(a(end),'.r',MS,10) z = N\0; arrowplot(z,CO,ivpnl,YS,1.4), hold off text(1.85,2.7,['$C = ' num2str(C) '$'],FS,9,HA,CT) axis([0 3.5 0 3]), set(gca,'ytick',0:3), grid on end title('Fig.~3.9.~~The antelope takes evasive action\kern -.1in',FS,11,HA,RT) %% % \vskip 1.02em %% % This pursuit problem has only a single unknown trajectory, % making it a scalar problem in complex arithmetic and % thus fitting the theme of this chapter. % The main use of complex arithmetic, however, is for tracking systems % of multiple particles in the plane, and we shall use this method % to compute planar orbits of planets in Chapter 13 and electrons % in Exercises 13.5 and 19.4, % as well as looking at a multi-particle pursuit problem in % Exercise 10.1. %% % % \smallskip % {\sc History.} Nonlinear equations have been % part of the study of ODE\kern .5pt s from the beginning. % For the first thirty years or so, until about % 1700, the equations were of first order. Then % higher order equations joined the discussion. % %% % % \smallskip % {\sc Our favorite reference.} Even if you don't read German, % the book {\em Differentialgleichungen:\ L\"osungsmethoden % und L\"osungen} by Erich Kamke % (Springer Fachmedien Wiesbaden, 1977) is extraordinary. % This book, which appeared in many editions starting in 1942, % features a collection of 1600 numbered % examples of ODE\kern .5pt s with their solutions --- the first 576 of % them, filling 103 pages, corresponding to first-order equations. % It is a monument to the % knowledge of ODE\kern .5pt s accumulated during their first 300 years % and a poignant indication of how different the world was % before computers. % %% % % \def\equals{\kern 1pt =\kern 1.5pt} % \begin{displaymath} % \framebox[4.7in][c]{\parbox{4.5in}{\vspace{2pt}\sl % {\sc Summary of Chapter 3.} % Some first-order nonlinear ODE\kern .5pt s can be solved by separation of % variables, and Bernoulli equations, of the form % $y' = a(t)y + b(t)y^p$, can be solved by the change of variables % $u = y^{1-p}$. Most other first-order % nonlinear problems cannot be solved % analytically. The ODE\/ $y' = f(t,y)$ has a unique solution % for all\/ $t$ if\/ $f$ is continuous with respect to\/ $t$ and Lipschitz % continuous with respect to\/ $y$. If these conditions do not hold, % existence and/or uniqueness may fail. \vspace{2pt}}} % \end{displaymath} % %% % % \small\parindent=0pt\parskip=1pt % {\em \underline{Exercise $3.1$}. Smoothness of the solution to the barrier problem.} % Despite appearances in Fig.~3.3, the solution $y(t)$ of (3.3) is smooth. % Confirm this by plotting $y,$ $y',$ and $y''$. % What are the minimum and maximum values of these three % functions over the interval % $[1,20\kern.3pt]$ (to exclude initial transients)? % (You can use a command like \verb|y{1,20}| to restrict a chebfun % to a subinterval.) % \par % {\em Exercise $3.2$. Analytical solutions via clever substitutions.} % One method used by experts in analytical solution of ODE\kern .5pt s is % to change variables. Find the general solutions of % the following problems % analytically using the substitutions indicated: % {\em (a)} $y' = e^{t-y}-e^{\kern .3pt t}$ ($u = e^y$), % {\em (\kern .7pt b)} $t\kern .3pt y' = y(\log(t\kern .3pt y)-1)$ ($u = t\kern .3pt y$), % {\em (c)} $2t\kern .3pt yy' = y^2-t$ ($u = y^2$). % \par % {\em Exercise $3.3$. A solution with compact support.} % Consider the IVP $y' = -t y/|y|^{1/2},$ $y(0) = 1$, where % $|y|^{1/2}$ represents the positive branch of the square root. % {\em (a)} Find analytically the (unique) solution for $t\in [\kern .3pt 0,1]$. % {\em (\kern .7pt b)} Find analytically the (unique) solution for $t\in [\kern .3pt 0,\infty)$. % \par % {\em \underline{Exercise $3.4$}. Spherical flame in microgravity.} % (Adapted from section 7.9 of C. B. Moler, {\em Numerical Computing with % MATLAB,} SIAM, 2008. % A video of a growing flame sphere in microgravity can be found at % {\tt https://goo.gl/nQ5Vxd}.) % In the absence of gravity, a flame takes a nearly spherical shape. % Oxygen, which fuels the flame, enters the sphere at a rate proportional % to its surface area. Combustion consumes the oxygen at a rate % proportional to the volume. In appropriate units the radius % $r(t)$ is approximately governed by the ODE % $dr/dt = r^2-r^3$. We assume that $r(0)=r_0^{}$. % {\em (a)} Show that the solution satisfies the implicit equation % $\log(r/(1-r)) - 1/r = t + C$, % and give a formula for $C$ in terms of the initial condition. % {\em (\kern .7pt b)} Show that the time $t_h^{}$ at which % $r$ takes the value $1/2$ is approximately $1/r_0^{}$ as % $r_0^{}\to 0$. (Note: $1/x$ dominates $\log x$ as $x\to 0$.) % {\em (c)} On one graph, plot numerically obtained solutions for $0\le t\le % 1500$ and $r_0^{}=10^{-1},10^{-2},10^{-3}$. As in Exercise 3.1, plot % a zoom of the % figure for $r_0^{} = 10^{-3}$ to confirm that despite appearances, this % solution is smooth. % \par % {\em Exercise $3.5$. Multiple routes to the same solution.} % Consider the ODE $y' + \sin(y) = 0$. {\em (a)} Find the general % solution analytically using separation of variables. % {\em (\kern .7pt b)} Find it again by interchanging the independent and % dependent variables as in Exercise 2.3. % \par % {\em Exercise $3.6$. Some nonlinear problems.} % Suppose $y(0)=1$. Determine $y(1)$ analytically if % {\em (a)} $y' = y^{3/2}e^{\kern .3pt t}$, {\em (\kern .7pt b)} $(t+1)y' +3y = 0$, {\em (c)} $yy' = t$. % \par % {\em Exercise $3.7$. Fixed points and stability.} % A number $y_*^{}$ is a {\bf fixed point} of an autonomous ODE % $y' = f(y)$ if $f(y_*^{}) = 0\kern .7pt ;$ % it is {\bf stable} if $f'(y_*^{})<0$ % and {\bf unstable} if $f'(y_*^{})>0$. Find all fixed points and determine % their stability or instability for % {\em (a)} $y' = y + y^2$ (eq.~(3.11)), % {\em (\kern .7pt b)} $y' = y^2 - 1$, % {\em (c)} $y' = y - y^2$ (the logistic equation; see also Exercises % 3.15 and 3.16), and {\em (d)} $y' = \sin(y)$. % \par % {\em \underline{Exercise $3.8$}. Stable and unstable fixed points.} % {\em (a)} Now explore equation {\em (c)} of the last % exercise on the computer by making a plot of the trajectories emanating % from $y(0) = -1,-0.8, \dots, 1.8, 2$ on the interval % $t\in [\kern .3pt0,4]$. You will need to use the {\tt N.maxnorm} % feature since otherwise some curves will blow up. % {\em (\kern .7pt b)} Similarly, explore equation {\em (d)} of the last % exercise with a plot of the trajectories emanating % from $y(0) = -15,-14,\dots, 15$ on the interval % $t\in [\kern .3pt0,4]$. % \par % {\em \underline{Exercise $3.9$}. Ghost of a fixed point.} % Plot the solution of $y' = 1 - a \sin(y)$, $y(0)=0$ for % $t\in [\kern .3pt 0,200\kern.3pt]$ with $a=0.9$, $0.99$, and $0.999$. % In each case use {\tt roots(y-2*pi)} to determine the value % of $t$ for which $y(t) = 2\pi$. % \par % {\em \underline{Exercise $3.10$}. The Lambert W function.} % {\em (a)} % The {\em Lambert W function} is a function $W(t)$ defined % by the functional equation $W(t\kern .3pt e^{\kern .3pt t}\kern .2pt ) = t$. It satisfies the % differential equation $W'(t) = W(t)/(t+t\kern .3pt W(t))$ and takes the % value $W(e) = 1$, where $e = 2.718\dots.$ Use this information % to solve an ODE to compute the value $W(1)$. % {\em (\kern .7pt b)} The number just computed probably matches the % exact value of $W(1)$ (readily found on the Web) to about % 10 digits of accuracy, because that is Chebfun's default accuracy % for ODE solutions. To get more digits, execute % \verb|cheboppref.setDefaults('ivpAbsTol',1e-14)| before making % the calculation (try {\tt chebfunpref} and {\tt cheboppref} and % see Chapter 8 of the {\em Chebfun Guide} for more information). % How accurate is the new value? % Afterwards, return to the usual defaults by executing % \verb|cheboppref.setDefaults('factory')|. % \par % {\em \underline{Exercise $3.11$}. Antelope on a circle or square.} % As in Figs.~3.8 and 3.9, suppose a lion begins % at $z=0$ at $t=0$ and chases the antelope with % equal speeds, i.e., $C=1$ in (3.17). % {\em (a)} Suppose the antelope runs around the unit circle with % position $a(t) = e^{it}$. Plot the lion's trajectory over the % interval $t\in [\kern .3pt 0, 2\pi]$. % Plot the distance between the two animals as a function of $t$. At roughly % what time $t$ will the distance fall below $0.01\kern .5pt?$ % {\em (\kern .7pt b)} Suppose the antelope runs around the unit square, which % you can construct with % \verb|a = 1 + cumsum(round(exp(pi*.25i*(t+2))/sqrt(2)))| if % {\tt t} is a chebfun for $t$ defined over the time interval of interest. % Plot the lion's trajectory over the % interval $t\in [\kern .3pt 0, 8]$. % Again plot the distance as a function of $t$ and estimate when this % will fall below $0.01$. % \par % {\em Exercise $3.12$. The leaky bucket problem --- % Torricelli's law (1643).} % A cylindrical tank has a hole at the bottom. Water flows % out, making the height $y(t)$ of the water decrease from its % initial value $y_0^{}$ to $0$ after a certain time. Derive the ODE for % this process by considering energy, as follows. The water in % the tank has a certain {\em potential energy} determined by $y(t)$ % and hence decreasing at a rate determined by $y$ and $y'$. % As water leaves the % hole, this is converted to an equal amount of {\em kinetic energy} % determined by $y'$ and hence increasing at a rate determined by % $y'$. By balancing these two, explain why the ODE has the % form $y' = -Cy^{1/2}$. Solve the equation analytically % and show that $y(t) = 0$ is reached at a finite time $t$. % If $t_0^{}$ is doubled, what effect will this have on the drainage time? % \par % {\em \underline{Exercise $3.13$}. Blowup equation with a % complex coefficient.} % Consider equation (3.9) except with a complex coefficient: % $y' = Cy^2$, $y(0) = 1$, where $C$ is a constant with a nonzero % imaginary part. % {\em (a)} Write down the analytical solution, valid for all $t$. % {\em (\kern .7pt b)} Plot the solutions corresponding to $C=1+1i$ and $C = 1+0.1i$. % \par % {\em \underline{Exercise $3.14$}. Complex nonlinear oscillator.} % The equation % $y' = iy + 0.1(1-|y|^2) y$ might be regarded as a kind % of complex, first-order analogue of the van der Pol equation. % {\em (a)} Compute and plot the solution for $y(0)=0.1$ and % $t\in [\kern .3pt 0,100\kern .5pt]$. % {\em (\kern .7pt b)} Use $\tt roots(real(y))$ to determine the period % of the oscillation, and explain why this is the value that appears. % \par % {\em \underline{Exercise $3.15$}. Logistic equation.} % Positive solutions of $y' = y$ grow exponentially, but % positive solutions of % $y' = (1-y/Y)y$, where $Y>0$ is a constant known as % the {\em carrying capacity}, asymptote to $y = Y$ as $t\to\infty$. % {\em (a)} Solve the equation analytically for $y(0) = y_0^{}$. % {\em (\kern .7pt b)} Solve it numerically with $Y=5$ for $t\in [\kern .3pt 0,5]$ % and make a plot of trajectories with % $y_0^{} = -5,-4,\dots, 10$. % (This equation was derived by Pierre-Fran\c cois Verhulst in the % 1830s as a model of population growth after he read Malthus's % {\em Essay on the Principle of Population.}) % \par % {\em \underline{Exercise $3.16$}. Logistic equation with harvesting.} % Suppose a population is naturally governed by logistic growth but is regularly % harvested, obeying $y'=y(1-y/Y)-H$ for positive constants % $Y$ and $H$. % {\em (a)} Show theoretically that the existence of a steady real solution % satisfying $y'=0$ requires \$H