%% 2. First-order scalar linear ODEs %% % % \setcounter{page}{10} % %% % % Problem (1.1) of the last chapter % could be regarded as the prototype of all ODE\kern .5pt s. Slightly generalized, % it takes the form % $$y' - a y = 0, \quad y(0) = y_0^{},$$ % where $a$ and $y_0^{}$ are constants. This problem % has the solution % $$y(t) = y_0^{}e^{at},$$ % which is as good an argument as any for why the number % $e = 2.71828\dots$ is important. % Let us make it a theorem (the proof is in the paragraphs % following). % %% % % \smallskip % {\em % {\bf Theorem 2.1. Solution of first-order linear % autonomous scalar homogeneous IVP % (\textsf{\textbf{FLASHI}}).} % The problem % $$y'-a y = 0, \quad y(0) = y_0^{}, \eqno (2.1)$$ % where $a$ and $y_0^{}$ are constants, has the unique solution % $$y(t) = y_0^{} e^{at}. \eqno (2.2)$$ % } % %% % % \vskip -.5em % Here are images for eleven values of $a$ from $-10$ to $10$, showing % exponential growth for $a > 0$, exponential decay % for $a <0$, and a constant solution for $a=0$. % ODEformats, L = chebop(0,1); L.lbc = 1; for a = -10:2:10 L.op = @(t,y) diff(y) - a*y; y = L\0; plot(y,CO,ivp), hold on end ylim([0 3]), hold off title(['Fig.~2.1.~~Exponential growth or decay (2.2) ' ... ' depending on $a$'],FS,11) text(.33,0.67,'$a=-2$',IN,LT,FS,10) text(.66,1.17,'$a=0$',IN,LT,FS,10) text(.43,2.20,'$a=2$',IN,LT,FS,10) text(.23,2.30,'$a=4$',IN,LT,FS,10) %% % \vskip 1.02em %% % % We can derive (2.2) by writing (2.1) as % $${dy\over dt} = ay,$$ % or equivalently if $y \ne 0$ % $${dy\over y} = a\kern .5pt dt.$$ % (If $y(t)=0$ for some $t$ then $y(t)=0$ for all $t$, a case % readily handled separately.) % Note that the last step has separated the $y$ and $t$ terms % onto the two sides of the equation: we say that (2.1) is {\bf separable,} % and this technique is called {\bf separation of variables}. % We now integrate both sides of the equation to get % $$\log |y| = a\kern .3pt t + c$$ % for some constant $c$, % or after exponentiating both sides, % $$y(t) = C\kern -.2pt e^{a\kern .2pt t}$$ % with $C=e^c$ if $y>0$ and $C=-e^c$ if % $y<0$ (a similar adjustment works if $y$ is % complex). Taking $C=y_0^{}$ gives (2.2). % Moreoever, the solution is unique, since any solution $y(t)$ % must be of the form $C\kern -.2pt e^{a\kern .2pt t}$ by the reasoning just given, % and only $C=y_0^{}$ will match the initial condition. % %% % % If $a>0$, then $y(t)$ increases exponentially % with $t$, whereas if $a<0$ it decreases % exponentially. This simple % distinction between exponential growth and decay of solutions % to (2.1) is the starting point of the theory of stability % of dynamical systems, a recurring theme in the second % half of this book.\footnote{If $a$ is % a complex number $\alpha + i\beta$, % which makes perfectly good sense mathematically and changes % none of the formulas, then since $e^{at} = e^{\alpha t}(\cos \beta\kern .3pt t + % i\sin\beta\kern .3pt t),$ we have exponential increase for $\alpha >0$ and % exponential decrease for $\alpha <0$. In both cases the % solution oscillates as well as growing or decaying.} % %% % Equation (2.1) is autonomous (coefficients independent of $t$) % and homogeneous (zero right-hand side). % Nothing much changes if we make the problem % nonautonomous: we still obtain a % solution via the method of separation of variables. % What this means is that instead of a constant % coefficient $a$, we allow a variable coefficient function $a(t)$. % $$y'-a(t) y=0, \quad y(0) = y_0^{}. \eqno (2.3)$$ % We separate variables as before to get % $${dy\over y} = a(t)\kern .5pt dt,$$ % which implies % $$\log y(t) = \int_0^t a(s) ds + c,$$ % or equivalently % $$y(t) = C\exp\Bigl(\kern 1pt \int_0^t a(s) ds\Bigr) \eqno (2.4)$$ % for constants $c$ and $C = e^c$. Since the integral takes % the value $0$ for $t=0$, the right constant is $C=y_0^{}$. %% % % In the derivation just made we have tacitly assumed $a$ % is continuous. The integrals in (2.4) and the equation above % it make sense more generally, however, for example if % $a$ is just {\em piecewise continuous.} We use this term in its % standard sense to refer to a function that is continuous apart % from at most a finite set of jump discontinuities. % With this in mind, the following % theorem and Theorem~2.3 are stated for piecewise continuous functions, % on the understanding that a solution is defined to be a continuous function % that is differentiable and satisfies the ODE everywhere % except at the points of discontinuity. % %% % % \smallskip % {\em % {\bf Theorem 2.2. Solution of first-order linear scalar homogeneous IVP % (\textsf{\textbf{FLaSHI}}).} % The problem % $$y'-a(t) y = 0, % \quad y(0) = y_0^{}, \eqno (2.5)$$ % where $y_0^{}$ is a constant and\/ $a(t)$ is a continuous or % piecewise continuous function, has the unique solution % $$y(t) = y_0^{} \exp\Bigl(\kern 1pt\int_0^t a(s) ds\Bigr). \eqno (2.6)$$ % } % %% % % \vskip -.5em % As an example, consider % $$y'=\sin(t^2) y, \quad t\in [\kern .3pt 0,8\kern .3pt ], \quad y(0) = 1. \eqno (2.7)$$ % A numerical solution gives an elegant oscillatory curve. % L = chebop(0,8); L.lbc = 1; L.op = @(t,y) diff(y) - sin(t^2)*y; y = L\0; plot(y,CO,ivp), ylim([0.5 3]) title('Fig.~2.2.~~Smooth homogeneous ODE (2.7)',FS,11) %% % \vskip 1.02em %% % For an example with a coefficient that is only piecewise continuous, % suppose we replace $\sin(t^2)$ by $\hbox{sign}(\sin(t^2))$, % $$y'=\hbox{sign}(\sin(t^2)) y, % \quad t\in [\kern .3pt 0,8\kern .3pt ], \quad y(0) = 1. \eqno (2.8)$$ % This corresponds to a bang-bang'' situation in which % the system is pushed one way and then the other, always % with amplitude 1. % Here is the solution. L.op = @(t,y) diff(y) - sign(sin(t^2))*y; y = L\0; plot(y,CO,ivp) title('Fig.~2.3.~~Nonsmooth homogeneous ODE (2.8)',FS,11) ylim([0 7]) %% % \vskip 1.02em %% % % \noindent % It is striking how this curve resembles the earlier one in form, % though the vertical scale has changed % considerably. Actually, this second example is mathematically % simpler than the first, since it consists of nothing but an % alternation of segments of exponential growth $C\kern -.2pt e^{\kern .3pt t}$ and % exponential decay $C\kern -.2pt e^{-t}$. % %% % % Our problem is still homogeneous. We are about to % take the next step and introduce a nonzero right-hand side. % Before doing this, however, let us examine the significance % of homogeneity from an abstract point of view. % Suppose we consider the ODE (2.5) {\em without\/} a boundary condition, % $$y'-a(t) y = 0, \quad t\in [\kern .3pt 0,d\kern .5pt]. \eqno (2.9)$$ % By the reasoning above, the solutions of this % equation consist precisely of all functions of the form % $C\kern -.5pt \exp(\int_0^t a(s) ds)$ for % any constant $C$. In other words, the set of solutions of (2.9) % is a {\em vector space of dimension $1$} spanned by the % basis function $\exp(\int_0^t a(s) ds)$.\footnote{We are using the idea of a vector % space in its standard abstract sense. A vector in this space % is a function $y(t)$.} % This conclusion applies to any first-order linear, scalar, homogeneous % ODE.\ \ The boundary % condition of an IVP selects one function out of the vector space. % Later we shall see that a second-order linear, scalar, homogeneous % ODE has a vector space of % solutions of dimension $2$, third-order gives dimension 3, and so on. % %% % % Now let us modify (2.9) to make the ODE inhomogeneous, % $$y'-a(t) y = g(t), \quad t\in [\kern .3pt 0,d\kern .5pt] \eqno (2.10)$$ % for some function $g(t)$. % Suppose that somehow or other, we find a function $y_{\rm p}^{}(t)$ % that satisfies (2.10). % The subscript \kern .7pt p \kern.7pt stands for particular:'' % a solution to a linear inhomogeneous % ODE is called a {\bf particular solution.} % Now let $y_{\rm h}^{}$, with h standing for % homogeneous,'' be any nonzero solution % to the homogeneous ODE (2.9), such as $\exp(\int_0^t a(s) ds)$ from Theorem 2.2. % Then for any constant $C$, the function % $$y_{\rm p}^{} + C y_{\rm h}^{} \eqno (2.11)$$ % is another solution to (2.10). This is % called the {\bf general solution} to (2.10). % In the language of vector spaces, we % can say that the set of solutions to the inhomogeneous ODE (2.10) is % an {\em affine space}, which means, a vector space shifted % by the addition of a constant vector (namely $y_{\rm p}$). % %% % % We have just presented the general framework for solving a first-order % linear scalar inhomogeneous IVP: % find a particular solution $y_{\rm p}$, then apply Theorem~2.2 to % find a nonzero solution $y_{\rm h}^{}$ to the homogeneous problem. % The solutions to the inhomogeneous problem are then all the % functions of the form $y_{\rm p} + Cy_{\rm h}^{}$ % for any constant $C$, and we pick $C$ to match the initial condition. % Beginning in Chapter~4 we shall extend the same idea % to higher order ODE\kern .5pt s and systems of linear ODE\kern .5pt s. % %% % % How do we find a particular solution $y_{\rm p}$? % There is a mechanical procedure that in principle % achieves this, multiplication by a function known as an % {\bf integrating factor}. (We shall see two pages along that in % practice, this approach may be more cumbersome % than necessary; the easier alternative is called the % method of undetermined coefficients.) Define % $$h(t) = \int_0^t a(s) ds .$$ % Then the product rule for differentiation gives % $$[ e^{-h(t)} y(t) ]' = e^{-h(t)} [ y'(t) - y(t) h'(t)] ,$$ % and since $h'(t) = a(t)$, (2.10) reduces this to % $$[ e^{-h(t)} y(t) ]' = e^{-h(t)} g(t).$$ % The function $e^{-h(t)}$ is the integrating factor. % We can integrate both sides to get % $$e^{-h(t)}y(t) - y(0) = \int_0^t e^{-h(s)} g(s) ds,$$ % that is, % $$y(t) = e^{h(t)} y(0) + e^{h(t)} \kern -2pt \int_0^t e^{-h(s)} g(s) ds.$$ % %% %% % % Let us formulate this conclusion as a theorem. % The derivation above can be regarded as a proof, at least apart % from the statement of uniqueness. % Alternatively, one could substitute (2.13) into (2.12) % and verify that it is a solution. % %% % % \smallskip % {\em % {\bf Theorem 2.3. Solution of first-order linear scalar inhomogeneous % IVP (\textsf{\textbf{FLaSh\kern .5pt I}}).} % The problem % $$y'-a(t) y = g(t), \quad y(0) = y_0^{}, \eqno (2.12)$$ % where $y_0^{}$ is a constant and % $a(t)$ and $g(t)$ are continuous or % piecewise continuous functions, has the unique solution % $$y(t) = y_0^{} \exp\Bigl(\kern 1pt\int_0^t a(s)ds\Bigr) % + \int_0^t g(s) \exp\Bigl(\kern 1pt\int_s^t a(r)dr\Bigr) ds. % \eqno (2.13)$$ % } % \smallskip % %% % % Equation (2.13) has an intuitive interpretation. We know % from Theorem 2.2 that the influence of an initial condition in % the homogeneous equation (2.5) is $y_0^{} \exp(\int_0^t a(s) ds).$ % The idea behind (2.13) is that at each time $s$, the right-hand % side $g(s)$, as it were, injects a small amount of initial condition'' % at that point. Each such injection produces a contribution for % $t>s$, and the second integral in (2.13) adds up those contributions. % So inhomogeneous problems are like homogeneous ones, but with lots % of little initial conditions --- a continuum of initial % conditions --- along the way.\footnote{This idea can be made % precise by the use of the Dirac delta function, though we shall not % do that in this book. Another language used by engineers calls % the influence of a signal injected at a point the % associated {\em impulse response.}} This way of thinking leads to % the method known as {\em variation of constants} or % {\em variation of parameters,} which can be used % to give a different proof of Theorem 2.3 and also applies % to ODE\kern .5pt s of higher order. % %% % To illustrate these developments, let us consider an IVP that looks a bit % like the first nontrivial example of this chapter, (2.7). Instead of % taking $\sin(t^2)$ % as an oscillating coefficient on the $y$ term, we put it as an % inhomogeneous forcing function on the right-hand side, % $$y'+ y = \sin(t^2) , % \quad t\in [\kern .3pt 0,8\kern .3pt ], \quad y(0) = 0. \eqno (2.14)$$ % Here is the solution, a curve that decays toward zero % while being forced alternately up and down. L = chebop(0,8); L.lbc = 0; L.op = @(t,y) diff(y) + y; t = chebfun('t',[0 8]); g = sin(t^2); y = L\g; plot(y,CO,ivp) title('Fig.~2.4.~~Smooth inhomogeneous ODE (2.14)',FS,11) %% % \vskip 1.02em %% % As with the transition from (2.7) to (2.8), we can see the % same effect more cleanly if we replace $\sin(t^2)$ by % $\hbox{sign}(\sin(t^2))$, % $$y'+ y = \hbox{sign}(\sin(t^2)) , % \quad t\in [\kern .3pt 0,8\kern .3pt ], \quad y(0) = 0. \eqno (2.15)$$ %% % \vskip -1.02em L.op = @(t,y) diff(y) + y; g = sign(sin(t^2)); y = L\g; plot(y,CO,ivp) title('Fig.~2.5.~~Nonsmooth inhomogeneous ODE (2.15)',FS,11) %% % \vskip 1.02em %% % % \noindent % Everywhere on this Sydney Opera House'' % there is exponential decay towards~$0$, % though after the first few time units, the amplitude is small % enough that the curve is dominated by the right-hand side, % not the decay. % %% % Here is another example. If $g(t)=0$, the IVP % $$y' - \cos(t) = -10(y-\sin(t)) + g(t) , % \quad t\in [\kern .3pt 0,15], \quad y(0) = 0 \eqno (2.16)$$ % has solution $y(t) = \sin(t)$, as can be directly verified. % Suppose now we introduce a forcing function $g$ that consists of a % train of upward impulses located where $t$ is close to an % odd integer. Sydney Opera House turns into Batman. L = chebop(0,16); L.op = @(t,y) diff(y) - cos(t) + 10*(y-sin(t)); L.lbc = 0; t = chebfun('t',[0 16]); g = 10*(abs((t+1)/2-round((t+1)/2))<.05); y = L\g; plot(y,CO,ivp), ylim([-2 2]) title('Fig.~2.6.~~Forcing by discontinuous impulses (2.16)',FS,11) %% % \vskip 1.02em %% % % We mentioned before Theorem 2.3 that although the method % of integrating factors provides a general formula for % finding a particular solution of a % first-order linear scalar inhomogeneous IVP, % in many cases a simpler method is available. % This is the {\bf method of undetermined coefficients,} % and it consists of guessing the form of the solution, % then substituting to find coefficients.\footnote{Some % textbooks call it the % {\em method of judicious guessing.}} Often this % idea works easily when the right-hand side contains % exponentials, because exponentials are % preserved under differentiation. Sines, cosines, and % polynomials in the right-hand side are also often treatable % in this way. (The method is not restricted to first-order % equations.) % %% % For example, suppose we have the problem % $$y' + y = \exp(3\kern .2pt t) . \eqno (2.17)$$ % Knowing that differentiation preserves a % term involving $\exp(3\kern .2pt t)$, we consider the trial solution % $$y(t) = A\exp(3\kern .2pt t).$$ % Inserting this in (2.17) gives % $$(3A + A)\exp(3\kern .2pt t) = \exp(3\kern .2pt t),$$ % from which we see that a particular solution of (2.17) is % $$y_{\rm p}(t) = \textstyle{1\over 4}\exp(3\kern .2pt t).$$ % The general solution of (2.17) is accordingly % $$y(t) = \textstyle{1\over 4}\exp(3\kern .2pt t) + C \exp(-t). \eqno (2.18)$$ %% % As another example, consider the equation % $$y' + t y = t \exp(t^2). \eqno (2.19)$$ % The experienced eye will note that differentiating $w(t) = a \exp(t^2)$ % gives $w' = 2\kern .3pt t w$, so $w$ will satisfy (2.19) if % $2\kern .3pt tw + tw = (t/a)w$, that is, $a = 1/3$. Thus $\exp(t^2)/3$ is % a particular solution, and the general solution is % $$y(t) = {1\over 3} e^{t^2} + C\kern -.2pt e^{-t^2/2} . \eqno (2.20)$$ % Further examples of the method of undetermined coefficients % are explored in Exercise 2.2. %% % % For equations like these that are simple enough to solve exactly, % an alternative method may be to do it in an automated way on the computer. % If you type certain problems into % WolframAlpha{\small\textsuperscript{\textregistered}}, the solution appears % quickly. Such an approach might succeed for many of the exercises % to be found in elementary ODE textbooks, though it will be less % successful for the higher-order, nonlinear, and behavioral investigations % that are the bigger part of this field --- let alone for conceptual % understanding. % %% % % Theorem 2.2 concerns the special case of Theorem 2.3 in which the problem % is homogeneous, i.e., the right-hand side is zero. Another important % special case occurs when % the problem is autonomous, with no explicit dependence % on $t$. Here is a theorem recording the solution in that case. % %% % % \smallskip % {\em % {\bf Theorem 2.4. Solution of first-order linear scalar autonomous % IVP (\textsf{\textbf{FLAShI}}).} % The problem % $$y'-a y = g, \quad y(0) = y_0^{}, \eqno (2.21)$$ % where $y_0^{}$, % $a$, and $g$ are constants, has the unique solution % $$y(t) \,=\, y_0^{} e^{at} + {g\over a}\left[ e^{at}-1\right], % \eqno (2.22)$$ % or\/ $y(t) = y_0^{} + g\kern .3pt t$ if $a=0$. % } % \smallskip % %% % % \begin{center} % \hrulefill\\[1pt] % {\sc Application: elimination of caffeine from the bloodstream}\\[-3pt] % \hrulefill % \end{center} % %% % % If you drink a cup of coffee or a glass of wine, the body absorbs % a dose of caffeine or alcohol that takes some time to clear away. % The details of how this happens involve a complicated % interplay of many organs and processes, but a differential % equation model may capture the overall behavior. Here is % a model of the elimination of caffeine adapted from % R. Newton, et al., % Plasma and salivary pharmacokinetics of caffeine in man,'' % {\em European Journal of Clinical Pharmacology} 21 (1981), pp.~45--52. % It falls in the category of models known as {\em first-order % pharmacokinetics}. % %% % The essential point is that to a reasonable approximation, when % you're not drinking more coffee, % the caffeine level $c(t)$ in your blood is governed by the equation % $$c'=-k c,$$ % where $k$ is a positive rate constant. % This means that $c(t)$ will decay exponentially, and the decay % rate is often expressed as a half-life $t_{1/2}^{}$, defined by % $$e^{-k \kern .5pt t_{1/2}^{}} = {1\over 2}, \qquad % t_{1/2}^{} = {\log(2)\over k}.$$ %% % % Experiments show that a 300 mg oral dose of caffeine, such as % might be found in a large mug of drip-brewed coffee, % creates a concentration % of about 8~$\mu{\rm g}$/mL in the blood plasma. % This boost is followed by first-order % kinetics with a half-life $t_{1/2}$ of about 6 hours, % although the rate can vary a great deal from person to person. % Here we compute the rate constant $k$ from the half-life $t_{1/2}$ % and then define an ODE accordingly, with $t$ representing % time in hours over a 24-hour period: % t12 = 6; k = log(2)/t12; L = chebop(@(t,c) diff(c)+k*c,[-2,24]); %% % % \noindent % We will assume that 300 mg of caffeine will be ingested % over half an hour, increasing the caffeine concentration % at a rate of 16 $(\kern .7pt\mu{\rm g}/{\rm mL})/{\rm hr}$. % Suppose coffee is consumed at 7:00, 10:00, and 15:00. % Here is the intake schedule with % time measured from the first cup. % t = chebfun('t',[-2,24]); coffee = @(t0) 16*(t > t0)*(t < t0+0.5); intake = coffee(0) + coffee(3) + coffee(8); plot(intake,'k','jumpline','-'), axis([-2 24 -5 20]) xlabel('time since first cup (hours)',FS,9) title(['Fig.~2.7.~~Caffeine intake rate, ' ... '($\kern .7pt\mu{\rm g}$/mL)/hr'],IN,LT,FS,11) %% % \vskip 1.02em %% % % \noindent % Here is the caffeine level during the 24-hour period. % L.lbc = 0; c = L\intake; hold on, plot(c,CO,ivp), hold off title(['Fig.~2.8.~~Caffeine concentration in the blood, ' ... '$\mu{\rm g}$/mL'],IN,LT,FS,11) %% % \vskip 1.02em %% % % \noindent % As you can see, 24 hours is not long enough to % clear the caffeine from the bloodstream. % For the case of three cups per day repeated periodically, see % Exercise~19.7. % %% % % \smallskip % {\sc History.} Separation of variables goes % back to Leibniz and Johann Bernoulli in the 1690s. Integating factors % came a few years later with Euler and Lagrange. % \smallskip % %% % % {\sc Our favorite reference.} It is hard to find much to say about % first-order % scalar linear ODE\kern .5pt s. Noting, however, that a theme in this subject % is time scales of growth and decay processes, we draw % attention to the fascinating {\em Time in Powers of Ten: Natural % Phenomena and their Timescales} by 't Hooft and Vandoren, % World Scientific, 2014. % \smallskip % %% % % \def\equals{\kern 1pt =\kern 1.5pt} % \begin{displaymath} % \framebox[4.7in][c]{\parbox{4.5in}{\vspace{2pt}\sl % {\sc Summary of Chapter 2.} % A first-order linear scalar % ODE has a one-dimensional affine space of solutions of the % form $y_{\rm p} + Cy_{\rm h}^{}$, and % if an initial condition is specified, it has the unique solution % given by Theorem\/ $2.3$, which can be derived by % multiplying by an integrating factor. % In the special case where the % problem is homogeneous, we have $y_{\rm p}=0$, % and the solution to the IVP is given by Theorem\/ $2.2$, which % can be derived by separation of variables. % \vspace{2pt}}} % \end{displaymath} % %% % % \small\parindent=0pt\parskip=1pt % {\em Exercise $2.1$. Separation of variables.} % Use separation of variables to find general solutions % to {\em (a)} $y' = e^{y+t}$, {\em (\kern .7pt b)} $y' = t\kern .3pt y+y+t+1$, % {\em (c)} $y' = (t^2+2)/y$. % Confirm your answers with % WolframAlpha or some other computational tool. % Which of these equations are linear? % \par % {\em Exercise $2.2$. Method of undetermined coefficients.} % Use the method of undetermined coefficients to find general % solutions to % {\em (a)} $y' = y + e^{\kern .3pt t}$ (try $y = a\kern .5pt t\kern .5pt e^{\kern .3pt t}$), % {\em (\kern .7pt b)} $y' = y + t\sin(t)$ (try $y = a\kern .5pt \sin(t) % + b\kern .7pt t\sin(t) + c\kern .3pt\cos(t) + d\kern .7pt t\cos(t)$), % {\em (c)} $y' = 2y + e^{\kern .3pt t} + 1$, % {\em (d)} $y' = 1 - 2\kern .3pt t\kern .3pt y$ {\em (Dawson's integral)}. % Confirm your answers with % WolframAlpha or some other computational tool. % \par % {\em Exercise $2.3$. Interchanging variables to make % a problem linear.} % {\em (a)} Although the differential equation $y' = y/(t+y)$ is nonlinear, % show that it becomes linear if it is rewritten as an % equation for $dt/dy$ rather than $dy/dt$. Solve this linear % equation analytically by determining an appropriate integrating % factor, and thereby also solve the original nonlinear equation. % {\em (\kern .7pt b)} If $y(0) = 1$, what is $y(1)$? Find the solution numerically or % analytically. % (An analytical solution involves a special function % known as the Lambert W-function.) % \par % \par % {\em Exercise $2.4$. Choosing a coefficient.} % Suppose $y_0^{}=1$ in (2.5). Give a function % $a(t)$ such that the solution $y(t)$ has $y(1)=2$. % \par % {\em Exercise $2.5$. No changes of sign.} % Consider an IVP (2.5) for which $y_0^{}$£ and % $a(t)$ are real (i.e., not complex). Show that the % solution $y(t)$ is positive for all $t$, negative % for all $t$, or zero for all $t$. What's the % strongest analogous result you can state in the % case where $a(t)$ is permitted to be complex? % \par % {\em \underline{Exercise $2.6$}. Local extrema of an oscillation.} % Let $t_{50}^{}$ denote the point where the solution of % $y' = -\cos(10/(1-t))$, $y(0) = 1$, achieves its 50th local % maximum (in Chebfun, {\tt [a,b] = max(y,'local')}). {\em (a)} Determine % $t_{50}^{}$ and $y(t_{50}^{})$ numerically, and plot $y(t)$ for % $0\le t \le t_{50}^{}$. {\em (\kern .7pt b)} Confirm the the exact value of % $t_{50}^{}$ analytically. % \par % {\em \underline{Exercise $2.7$}. Adjusting Batman's ears.} % {\em (a)} Use \verb|max(y{6,8})| to calculate the maximum value of $y(t)$ % in the interval $[6,8\kern .3pt ]$ for the problem of Fig.~2.6. % {\em (\kern .7pt b)} What happens to the plot, and to this maximum value, if the % impulse is made 3 times as wide with $1/3$ the amplitude? % {\em (c)} What if it is 1/3 as wide with 3 times the amplitude? % \par % {\em \underline{Exercise $2.8$}. Temperature of a dead body.} % It was observed in 1894 that a human body after death cools at a rate % that is nearly proportional to the difference between the body and the % surroundings.'' (This heat transfer principle is known as Newton's law of cooling. % The quote comes from de Saram, Webster, and Kathirgamatamby, Post-mortem % temperature and the time of death,'' {\em Journal of Criminal Law % Criminology and Police Science} 64 (1955), 562--577.) % Let's assume that the cooling rate is $-\beta\kern .3pt % \theta(t)$, where $\theta(t)$ is the difference between body and % surrounding temperature, and $\beta$ is an empirical constant. % De Saram et al.\ measured the temperatures of 41 executed prisoners % to test this model. % In one case the body temperature was $97.8^\circ$F at 11 AM and $96.2^\circ$F % at 1 PM.\ \ The room temperature was held at $86.4^\circ$F.\ \ {\em (a)} % Use the temperature readings to % find a numerical value for $\beta$.\ \ {\em (\kern .7pt b)} By assuming % that at the time of % death the body temperature was $98.6^\circ$F, determine an approximate % time of death. % \par % {\em \underline{Exercise $2.9$}. Heaviside function.} % Many of the examples of this chapter involve ODE\kern .5pt s forced by % discontinuous right-hand sides. Another way to formulate such % problems is with the {\em step function} or {\bf Heaviside function} % $H(t)$, which takes the value $0$ for $t<0$ and $1$ for $t>0\kern .5pt;$ % the value at $t=0$ is $0.5$. % Execute the instructions \verb|L = chebop(0,2),| % \verb|L.op = @(t,y) diff(y)+y,| \verb|L.lbc = 1,| % \verb|t = chebfun('t',[0,2]),| \verb|f = heaviside(t-1)|, % \verb|y = L\f|.\ \ Write down the IVP that is being solved here, % plot the solution $y$ just computed, and derive an exact formula % for this function. How accurate is the computed value $y(2)\kern .5pt ?$ % \par % {\em \underline{Exercise $2.10$}. An ODE from Newton (\/$1671$) and Taylor % series approximations.} % One of the first IVPs ever considered was % the equation $y'=1-3\kern .15pt t+y+t^2+t\kern .3pt y$ % with initial condition $y(0)=0$, presented by Isaac Newton in 1671. % Newton solved this by using what we now call a Taylor series, % obtaining the representation % $y(t)=t-t^2+t^3/3-t^4/6+t^5/30-t^6/45+\cdots.$ % What is the maximum difference between this 6-term approximation and % the true solution on the intervals {\em (a)} $[\kern .5pt 0,0.5],$ % {\em (\kern .7pt b)} $[\kern .5pt 0,1],$ % {\em (c)} $[\kern .5pt 0,2\kern .3pt],$ and % {\em (d)} $[\kern .5pt 0,4]\kern.7pt?$\ \ Give the numbers % and also plot the absolute value of % the error as a function of $\kern .8pt t$ on a log-log scale. % Comment on the plot. % \par % {\em \underline{Exercise $2.11$}. Equation with sensitive solutions.} % Consider the equation $y' + t\kern .3pt y = f(t)$ for $t\in [-4,4]$ with $y(-4)=0$. % {\em (a)} What is $\max_{t\in[-4,4]}^{} y(t)$ if $f(t) = \sin(t)\kern .5pt?$ % What does this change to if $f(t) = \sin(0.85 t)\kern .5pt?$ % {\em (\kern .7pt b)} What term in the formulas of Theorem~2.3 makes it possible % for these numbers to vary so greatly? % \par % {\em \underline{Exercise $2.12$}. Thiele's equation for life insurance.} % Simple life insurance is purchased for a length of time $T$ and % costs the insured party a fixed premium $P$ per year. % If the insured dies during the term of the policy, the insurer must pay a % benefit of $S$. Otherwise, the insured receives nothing. The insurer % is required to keep money in reserve % for this policy in order to be able % to pay out all the claims likely to result from a pooled group of % individuals. This is done via the {\em Thiele differential equation,} % first derived in 1875 and published in 1910. % Let $V(t)$ for $0\le t \le T$ be the amount of reserve % needed for a policy purchased at time $t=0$. Thiele's equation is % $V'= P + \delta V(t) - \mu(t) (S-V(t)),$ % where $\delta$ is the {\em force of interest} (the interest % rate) and $\mu(t)$ is the {\em force of mortality} (the % probability of a person dying per unit time). When the policy % expires, no reserve is needed, so $V(T)=0$.\footnote{The % property of a time-dependent equation being governed % by a {\em final condition} rather than an initial condition % arises commonly in financial modelling. The famous example % involving a PDE is the Black--Scholes equation.} % {\em (a)} % Plot the solution $V(t)$ % for $\mu(t) = 0.007(t+5)$, $\delta = 0.03$, $P = 800,$ % $S = 1e6,$ and $T = 25.$ % {\em (\kern .7pt b)} What is the maximal value of $V(t)\kern .5pt$? Is it % less than or equal to $S$\kern .5pt? Interpret this inequality % financially. {\em (c)} Assuming $\delta \ge 0$, % prove that any solution of the % ODE problem will satisfy this inequality. % \par % {\em Exercise $2.13$. Without the method of undetermined coefficients.} % Solve (2.17) using Theorem~2.3. %