A.5 Well-Posed Problems and Mathematical Models

In addition to uniqueness, another consequence of the inequality in (45) is the fact that solutions of the differential equation

dxdt=f(x,t)
(47)

depend continuously on the initial value x(a); that is, if x1(t) and x2(t) are two solutions of (47) on the interval atT such that the initial values x1(a) and x2(a) are sufficiently close to one another, then the values of x1(t) and x2(t) remain close to one another. In particular, if |x1(a)x2(a)|δ, then (45) implies that

|x1(t)x2(t)|δek(Ta)=ϵ
(48)

for all t with atT. Obviously, we can make ϵ as small as we wish by choosing 1 sufficiently close to zero.

This continuity of solutions of (47) with respect to initial values is important in practical applications where we are unlikely to know the initial value x0=x(a) with absolute precision. For example, suppose that the initial value problem

dxdt=f(x,t),x(a)=x0
(49)

models a population for which we know only that the initial population is within δ>0 of the assumed value x0. Then even if the function f(x, t) is accurate, the solution x(t) of (49) will be only an approximation to the actual population. But (45) implies that the actual population at time t will be within δek(Ta) of the approximate population x(t). Thus, on a given closed interval [a, T], x(t) will be a close approximation to the actual population provided that δ>0 is sufficiently small.

An initial value problem is usually considered well posed as a mathematical model for a real-world situation only if the differential equation has unique solutions that are continuous with respect to initial values. Otherwise it is unlikely that the initial value problem adequately mirrors the real-world situation.

An even stronger “continuous dependence” of solutions is often desirable. In addition to possible inaccuracy in the initial value, the function f(x, t) may not model precisely the physical situation. For instance, it may involve physical parameters (such as resistance coefficients) whose values cannot be measured with absolute precision. Birkhoff and Rota generalize the proof of Theorem 4 to establish the following result.

If μ>0 is small, then (51) implies that the functions f and g appearing in the two differential equations, though different, are “close” to each other. If ϵ>0 is given, then it is apparent from (52) that

|x(t)y(t)|ϵ
(53)

for all t in [a, T] if both |x(a)y(a)| and μ are sufficiently small. Thus Theorem 5 says (roughly) that if both the two initial values and the two differential equations in (50) are close to each other, then the two solutions remain close to each other for atT.

For example, suppose that a falling body is subject both to constant gravitational acceleration g and to resistance proportional to some power of its velocity, so (with the positive axis directed downward) its velocity v satisfies the differential equation

dvdt=gcvρ.
(54)

Assume, however, that only an approximation c¯ to the actual resistance c and an approximation ρ¯ to the actual exponent ρ are known. Then our mathematical model is based on the differential equation

dudt=gc¯uρ¯
(55)

instead of the actual equation in (54). Thus if we solve Eq. (55), we obtain only an approximation u(t) to the actual velocity v(t). But if the parameters c¯ and ρ¯ are sufficiently close to the actual values c and ρ, then the right-hand sides in (54) and (55) will be close to each other. If so, then Theorem 5 implies that the actual and approximate velocity functions v(t) and u(t) are close to each other. In this case, the approximation in (55) will be a good model of the actual physical situation.

Problems

In Problems 1 through 8, apply the successive approximation formula to compute yn(x) for n4. Then write the exponential series for which these approximations are partial sums (perhaps with the first term or two missing; for example,

ex1=x+12x2+16x3+124x4+).
  1. dydx=y, y(0)=3

     

  2. dydx=2y, y(0)=4

     

  3. dydx=2xy, y(0)=1

     

  4. dydx=3x2y, y(0)=2

     

  5. dydx=2y+2, y(0)=0

     

  6. dydx=x+y, y(0)=0

     

  7. dydx=2x(1+y), y(0)=0

     

  8. dydx=4x(y+2x2), y(0)=0

In Problems 9 through 12, compute the successive approximations yn(x) for n3; then compare them with the appropriate partial sums of the Taylor series of the exact solution.

  1. dydx=x+y, y(0)=1

     

  2. dydx=y+ex, y(0)=0

     

  3. dydx=y2, y(0)=1

     

  4. dydx=12y3, y(0)=1

     

  5. Apply the iterative formula in (16) to compute the first three successive approximations to the solution of the initial value problem

    dxdt=2xy,x(0)=1;dydt=3x2y,y(0)=1.
  6. Apply the matrix exponential series in (19) to solve (in closed form) the initial value problem

    x(t)=[1101]x,x(0)=[11].

    (Suggestion: Show first that

    [1101]n=[1n01]

    for each positive integer n.)

  7. For the initial value problem dy/dx=1+y3, y(1)=1, show that the second Picard approximation is

    y2(x)=1+2(x1)+3(x1)2+4(x1)3+2(x1)4.

    Then compute y2(1.1) and y2(1.2). The fourth-order Runge–Kutta method with step size h=0.005 yields y(1.1)1.2391 and y(1.2)1.6269.

  8. For the initial value problem dy/dx=x2+y2, y(0)=0, show that the third Picard approximation is

    y3(x)=13x3+163x7+22079x11+159535x15.

    Compute y3(1). The fourth-order Runge–Kutta method yields y(1)0.350232, both with step size h=0.05 and with step size h=0.025.

  9. Prove as follows the inequality |Ax|A|x|, where A is an m×m matrix with row vectors a1,a2,,am, and x is an m-dimensional vector. First note that the components of the vector Ax are a1x,a2x,,amx, so

    |Ax|=[n=1m(aix)2]1/2.

    Then use the Cauchy–Schwarz inequality (ax)2|a|2|x|2 for the dot product.

  10. Suppose that ϕ(t) is a differentiable function with

    ϕ(t)kϕ(t)(k>0)

    for ta. Multiply both sides by ekt, then transpose to show that

    ddt[ϕ(t)ekt]0

    for ta. Then apply the mean value theorem to conclude that

    ϕ(t)ϕ(a)ek(ta)

    for ta.

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