cF. Waleffe Math 221 supplementary notes, Aug. 2000The differential equation y0= y and the function exp(x)Motivation: Compound Interest [TF 6-11]Suppose you deposit an amount A0in a bank account paying an interest r (e.g. r = 6% peryear) with a compounding period of T , then the amount after one period will have grown toA(T ) = (1 + rT )A0. After the next compounding period it will be A(2T ) = (1 + rT )A(T) =(1 + rT )2A0and so on. After n periods it will beA(nT ) = (1 + rT )nA0. (1)The amount in deposit increases by a factor (1 + rT ) after each period: A(t + T) = (1 +rT )A(t). We can write this in the form of a difference equation:A(t + T ) − A(t)T= rA(t). (2)The solution of this difference equation is (1), check it! It should be clear that the com-pounding period is as important as the interest rate itself. Banks usually quote the interestrate (in units of yr−1) and the Annual Percentage Yield (APY) instead of the compoundingperiod. What is the connection between r, T and the APY? If the interest rate is 6% and Tis one month, what is the APY? If r = 6% and T = 1 week, what is the APY?• Continuous compounding: T → 0Continuous compounding corresponds to the limit T → 0, then the solution (1) becomesA(0) = A0. This is just the initial amount, what about A(t > 0)?! The difference equation(2) does not become so useless in the limit T → 0. The difference quotient becomes aderivative and the equation becomes the differential equationdAdt= rA. (3)We need to find the solution of this equation to determine A(t) and the proper limit of (1).• Non-dimensionalizationBefore trying to find the solution of (3) it is useful to reduce it to its bare essentials by usingnon-dimensional variables. This is an important step in mathematical modeling of scientificand engineering problems. The variables entering the problem are: A measured in dollars orcents, r an inverse time and t time. The units of r and t must be compatible, if r is takenper year, then t should be measured in years.Let x = rt and y(x) = A(t)/A0(assuming the original amount A0is not zero) then x and yare non-dimensional. By the chain rule and equation (3):dydx=dydtdtdx=1rA0dAdt= y.So the differential equation that we would like to solve isy0= y (4)with the initial condition y(0) = 1. This is the equation for which separation of variables[TF 4-2] y0= y ⇔ y−1dy = dx breaks down asRyndy = yn+1/(n + 1) but here n = −1.1• SolutionSketch solution using piecewise linear approximation over intervals of size ∆x (i.e. Euler’smethod). Algebraically, Euler’s method gives y(n∆x) ≈ (1 + ∆x)ny(0) (cf. (1) and Fig. 1).Solution by polynomial approximation (i.e. power series, i.e. Taylor’s formula [TF 3-10]):y(x) ≈ y(0) + y0(0) x +12y00(0) x2+13!y000(0) x3+ ··· (5)The differential equation y0= y implies y = y0= y00= ···, and y(0) = 1 thus y(0) = y0(0) =y00(0) = y000(0) = ··· = 1 andy(x) ≈ 1 + x + x2/2 + x3/6 + x4/4! + x5/5! + ··· (6)See Fig. 2. You can check using term-by-term differentiation that this is indeed the solutionof y0= y, if the sum does not end!. One can make sense of these infinite sums using limits[TF 16].The solution (6) is not a simple power law or polynomial (the sum does not end). It is anew function defined by the differential equation y0= y with y(0) = 1.But is the solution to that equation unique? After all y0=√y with y(0) = 0 has twosolutions y(x) = 0 and y(x) = x2/4 (check it).• Uniqueness:(1) by variation of parameters. Let y1(x) be a solution of y0= y with y(0) = 1. Lety(x) = u(x)y1(x) be another solution then u0y1= 0 and u = 1, so y(x) is identical to y1(x).[Tricky point: what if y1(x) = 0?](2) another proof: Assume u(x) and v(x) are two different solutions. Consider w(x) =u(x)v(a − x) where a is any constant. Then by the product and chain rules w0= 0 ∀a, xhence u(x)v(a − x) = v(a). Of course v(x)v(a − x) = v(a) by the same reasoning. Henceu(x)v(a − x) = v(x)v(a − x) for all a, x and u(x) = v(x).Either way, the solution of y0= y with y(0) = 1 is indeed unique, let’s call it y = exp(x).• Main property of exp(x):Note that exp(αx) is the unique solution of y0= αy with y(0) = 1, howeverddx[exp(x)]α= α[exp(x)]α−1[exp(x)]0= α[exp(x)]αso [exp(x)]αalso satisfies y0= αy with y(0) = 1, hence, by uniqueness,exp(αx) = [exp(x)]α.Nice application of the uniqueness proof. In particular, exp(x) = [exp(1)]x≡ exwheree = 1 + 1 + 1/2 + 1/3! + 1/4! + 1/5! + 1/6! + . . . = 2.718...Note that Euler’s method gives y(t) ≈ (1 + t/n)nand this suggests limn→∞(1 + t/n)n= et.The differential equation y00+ y = 0 can be treated similarly as well as in 4-4 (see 18-12 fora physical introduction). One could even go on to eix...20 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 111.21.41.61.822.22.42.62.8xyPiecewise linear approximations to y’=y with y(0)=1∆x=1, 1/2, 1/4, 1/8, 1/16, 1/32Figure 1: Piecewise linear approximations (Euler’s method)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 111.21.41.61.822.22.42.62.81+x+x2/2+x3/6 (dash)1+x+x2/21+x1+x+x2/2+x3/6+x4/24xyPower series approximations to y’=y with y(0)=1Figure 2: Polynomial approximations (Taylor
View Full Document
Unlocking...