Second Assignment for Math 128B Due Fri. 1 Mar. 2002 Prof. W. Kahan page 1/6 March 15, 2002 12:58 pm For some positive number y 0 the solution y(x) of the initial value problemy(0) := y 0 and dy/dx = x – 1/y for all x ≥ 0remains positive and bounded; this means that some constant (bound) ß satisfies ß > y(x) ≥ 0 for all x ≥ 0 . Your task is to estimate y 0 as accurately as you can and justify your claims scientifically. You may use any numerical techniques and software available to you provided you understand what they do and describe them well enough that your results can be replicated.If y 0 is overestimated, the solution y(x) will grow too big; if y(X) is big enough for some X > 0 then y(x) will grow at least about as fast as (x–X) 2 /2 grows as x → + ∞ . Can you see why? On the other hand, if y 0 is underestimated then y(x) will plunge down to y(X) = 0 for some X > 0 and then no solution y(x) will exist for x > X ; can you see why the plunge must occur? Numerical experiments may help. It is possible to deduce that when the initial value y 0 is just right then 1/(x+1) < y(x) < 1/x for all x > 0 ; verifying this claim mathematically leads to a better understanding of the computational challenge involved in estimating the right y 0 . Differential Inequalities The simplest differential inequalities compare the solutions y(x) and Y(x) of two differential equations dy/dx = ƒ(x, y) and dY/dx = F(x, Y) when y(0) < Y(0) and ƒ(x, z) < F(x, z) for all z and all x ≥ 0 ; then also y(X) < Y(X) for all X > 0 provided y(x) and Y(x) and their first derivatives remain finite while 0 ≤ x < X . This is so because the difference Y–y is differentiable and therefore continuous and consequently cannot reverse sign without vanishing first. Let X be the infimum of positive arguments x , if any, for which Y(x) – y(x) ≤ 0 ; this would imply that Y(x) – y(x) > 0 while 0 ≤ x < X but Y(X) – y(X) = 0 , contradicting the sign of the derivative: Y ' (X) – y ' (X) = F(X, Y(X)) – ƒ(X, y(X)) > 0 wherever Y(X) = y(X) . Similarly, if y(0) > Y(0) and ƒ(x, z) > F(x, z) for all z and all x ≥ 0 , then y(X) > Y(X) for every X > 0 provided etc . In fact, for every x > 0 each solution y(x) of dy/dx = ƒ(x, y) is a strictly increasing function of y(0) and of ƒ(…) provided they determine the solution uniquely.The differential equation we wish to solve has dy/dx = ƒ(x, y) := x – 1/y . Wherever the graph of some solution y(x) cuts the graph of 1/x , the cut goes from left below to right above the latter graph because y(x) = 1/x at the cut implies y ' (x) = 0 > (1/x) ' = –1/x 2 there; this means that the graph of a solution y(x) can’t cut the graph of 1/x more than once. Consequently the graph of the desired bounded solution y(x) can’t cut the graph of 1/x at all; that this solution (if it exists) must satisfy y(x) < 1/x for all x > 0 follows from the following analysis: Suppose there were some X > 0 beyond which y(x) > 1/x for all x ≥ X . Then y ' (x) = x – 1/y(x) > 0 and so this y(x) > y(X) > 1/X while x ≥ X . Then y ' (x) = x – 1/y(x) > x – X , so y(x) > y(X) + ∫ Xx ( ξ –X)d ξ > 1/X + (x–X) 2 /2 , which grows unboundedly with x > X . Therefore only unbounded solutions y can ever violate y(x) < 1/x . If the graph of some solution y(x) cuts the graph of 1/(x+1) , the cut goes from left above to right below the latter graph since y(x) = 1/(x+1) implies y ' (x) = –1 < (1/(x+1)) ' = –1/(x+1) 2 there, so no solution’s graph can cut the graph of 1/(x+1) more than once. An analysis very like the foregoing shows that y(x) > 1/(1+x) if y(x) > 0 for all x ≥ 0 ; can you verify this? This document was created with FrameMaker404Second Assignment for Math 128B Due Fri. 1 Mar. 2002 Prof. W. Kahan page 2/6 March 15, 2002 12:58 pmIn fact, if 0 < y(X) ≤ 1/(X+1) for some X > 0 then y(Z) = 0 for some positive Z < X + 1/(X+1) beyond which this solution y(x) cannot be continued; further analysis reveals that 0 < y(x) ≤ √ 2Z–2x for 0 ≤ x ≤ Z , which reveals how abruptly such solutions y(x) plunge to zero. This revelation accords with experience of numerical solutions started from too small an initial value y 0 . To summarize: If any solution y(x) exists that stays positive and bounded for all x ≥ 0 , it lies strictly between 1/(x+1) and 1/x . Now, at most one such solution y(x) can exist because the differential equation dy/dx = x – 1/y is stable for decreasing x . This means that the difference between nearby solutions diminishes as x decreases; here is why: Suppose y(x) and y(x) + ∆ y(x) are two solutions with y(X) + ∆ y(X) > y(X) > 0 for some X > 0 . Provided both solutions stay positive throughout 0 ≤ x ≤ X , the general existence and uniqueness theory of differential equations ensures that their graphs do not cross (lest solutions through the crossing point not be determined uniquely). Therefore y(x) + ∆ y(x) > y(x) > 0 throughout 0 ≤ x ≤ X , and thend ∆ y/dx = ƒ(x, y+ ∆ y) – ƒ(x, y) = ∆ y(x)/(y(x)·(y(x)+ ∆ y(x))) > 0 ,and thus ∆ y(x) must be an increasing function of x . This implies 0 < ∆ y(0) < ∆ y(X) .If both solutions y(x) and y(x)+ ∆ y(x) stayed positive and bounded for all x ≥ 0 then, as we have seen above, 1/X > y(X) + ∆ y(X) > y(X) > 1/(X+1) . Consequently ∆ y(X) → 0 as X → + ∞ , forcing ∆ y(0) → 0 too. In other words, both positive bounded solutions would have to be the same solution, if it exists. To prove that the desired positive bounded solution exists, consider two solutions y(x) of the differential equation dy/dx = ƒ(x, y) := x – 1/y . One solution y(x) = Y(x; X) is determined by Y(X; X) := 1/X for some X > 0 ; the other solution y(x) = Y(x; X) has Y(X; X) := 1/(X+1) . Because solutions y(x) are monotonic functions of