Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf

Differentiating this relation yields f  (x) ∼ − 1 x 2 as x → ∞. However, this is not true since f  (x) = − 1 x 2 − 2 x 3 sin(x 3 ) + 2 cos(x 3 ) ∼ − 1 x 2 as x → ∞. The Controlling Factor. The controlling factor is the most rapidly varying factor in an asymptotic relation. Consider a function f(x) that is asymptotic to x 2 e x as x goes to infinity. The controlling factor is e x . For a few examples of this, • x log x has the controlling factor x as x → ∞. • x −2 e 1/x has the controlling factor e 1/x as x → 0. • x −1 sin x has the controlling factor sin x as x → ∞. The Leading Behavior. Consider a function that is asymptotic to a sum of terms. f(x) ∼ a 0 (x) + a 1 (x) + a 2 (x) + ··· , as x → x 0 . where a 0 (x)  a 1 (x)  a 2 (x)  ··· , as x → x 0 . The first term in the sum is the leading order behavior. For a few examples, • For sin x ∼ x −x 3 /6 + x 5 /120 − ··· as x → 0, the leading order behavior is x. • For f(x) ∼ e x (1 − 1/x + 1/x 2 − ···) as x → ∞, the leading order behavior is e x . 1254 24.2 Leading Order Behavior of Differential Equations It is often useful to know the leading order behavior of the solutions to a differential equation. If we are considering a regular point or a regular singular point, the approach is straight forward. We simply use a Taylor expansion or the Frobenius method. However, if we are considering an irregular singular point, we will have to be a little more creative. Instead of an all encompassing theory like the Frobe niu s method which always gives us the solution, we will use a heuristic approach that usually gives us the solution. Example 24.2.1 Consider the Airy equation y  = xy. We 1 would like to know how the solutions of this equation b ehave as x → +∞. First we need to classify the point at infinity. The change of variables x = 1 t , y(x) = u(t), d dx = −t 2 d dt , d 2 dx 2 = t 4 d 2 dt 2 + 2t 3 d dt yields t 4 u  + 2t 3 u  = 1 t u u  + 2 t u  − 1 t 5 u = 0. Since the equation for u has an irregular singular point at zero, the equation for y has an irregular singular point at infinity. 1 Using ”We” may be a bit presumptuous on my part. Even if you don’t particularly want to know how the solutions behave, I urge you to just play along. This is an interesting section, I promise. 1255 The Controlling Factor. Since the solutions at irregular singular points often have exponential behavior, we make the substitution y = e s(x) into the differential equation for y. d 2 dx 2  e s  = x e s  s  + (s  ) 2  e s = x e s s  + (s  ) 2 = x The Dominant Balance. Now we have a differential equation for s that appears harder to solve than our equation for y. However, we did not introduce the substitution in order to obtain an equation that we could solve exactly. We are looking for an equation that we can solve approximately in the limit as x → ∞. If one of the terms in the equation for s is much smaller that the other two as x → ∞, then dropping that term and solving the simpler equation may give us an approximate solution. If one of the terms in the equation for s is m uch smaller than the others then we say that the remaining terms form a dominant balance in the limit as x → ∞. Assume that the s  term is much smaller that the others, s   (s  ) 2 , x as x → ∞. This gives us (s  ) 2 ∼ x s  ∼ ± √ x s ∼ ± 2 3 x 3/2 as x → ∞. Now let’s check our assumption that the s  term is small. Assuming that we can differentiate the asymptotic relation s  ∼ ± √ x, we obtain s  ∼ ± 1 2 x −1/2 as x → ∞. s   (s  ) 2 , x → x −1/2  x as x → ∞ Thus we see that the behavior we found for s is consistent with our assumption. The controlling factors for solutions to the Airy equation are exp(± 2 3 x 3/2 ) as x → ∞. 1256 The Leading Order Behavior of the Decaying Solution. Let’s find the leading order behavior as x goes to infinity of the solution with the controlling factor exp(− 2 3 x 3/2 ). We substitute s(x) = − 2 3 x 3/2 + t(x), where t(x)  x 3/2 as x → ∞ into the differential equation for s. s  + (s  ) 2 = x − 1 2 x −1/2 + t  + (−x 1/2 + t  ) 2 = x t  + (t  ) 2 − 2x 1/2 t  − 1 2 x −1/2 = 0 Assume that we can differentiate t  x 3/2 to obtain t   x 1/2 , t   x −1/2 as x → ∞. Since t   − 1 2 x −1/2 we drop the t  term. Also, t   x 1/2 implies that (t  ) 2  −2x 1/2 t  , so we drop the (t  ) 2 term. This gives us −2x 1/2 t  − 1 2 x −1/2 ∼ 0 t  ∼ − 1 4 x −1 t ∼ − 1 4 log x + c t ∼ − 1 4 log x as x → ∞. Checking our assumptions about t, t   x 1/2 → x −1  x 1/2 t   x −1/2 → x −2  x −1/2 1257 we see that the behavior of t is consistent with our assumptions. So far we have y(x) ∼ exp  − 2 3 x 3/2 − 1 4 log x + u(x)  as x → ∞, where u(x)  log x as x → ∞. To continue, we s ubs titute t(x) = − 1 4 log x + u(x) into the differential equation for t(x). t  + (t  ) 2 − 2x 1/2 t  − 1 2 x −1/2 = 0 1 4 x −2 + u  +  − 1 4 x −1 + u   2 − 2x 1/2  − 1 4 x −1 + u   − 1 2 x −1/2 = 0 u  + (u  ) 2 +  − 1 2 x −1 − 2x 1/2  u  + 5 16 x −2 = 0 Assume that we can differentiate the asymptotic relation for u to obtain u   x −1 , u   x −2 as x → ∞. We know that − 1 2 x −1 u   −2x 1/2 u  . Using our assumptions, u   x −2 → u   5 16 x −2 u   x −1 → (u  ) 2  5 16 x −2 . Thus we obtain −2x 1/2 u  + 5 16 x −2 ∼ 0 u  ∼ 5 32 x −5/2 u ∼ − 5 48 x −3/2 + c u ∼ c as x → ∞. 1258 Since u = c + o(1), e u = e c +o(1). The behavior of y is y ∼ x −1/4 exp  − 2 3 x 3/2  ( e c +o(1)) as x → ∞. Thus the full leading order behavior of the decaying solution is y ∼ (const)x −1/4 exp  − 2 3 x 3/2  as x → ∞. You can show that the leading behavior of the exponentially growing solution is y ∼ (const)x −1/4 exp  2 3 x 3/2  as x → ∞. Example 24.2.2 The Modified Bessel Equation. Consider the modified Bessel equation x 2 y  + xy  − (x 2 + ν 2 )y = 0. We would like to know how the solutions of this equation behave as x → +∞. First we need to classify the point at infinity. The change of variables x = 1 t , y(x) = u(t) yields 1 t 2 (t 4 u  + 2t 3 u  ) + 1 t (−t 2 u  ) −  1 t 2 + ν 2  u = 0 u  + 1 t u  −  1 t 4 + ν 2 t 2  u = 0 Since u(t) has an irregular singular point at t = 0, y(x) has an irregular singular point at infinity. The Controlling Factor. Since the solutions at irregular singular points often have exponential behavior, we make the substitution y = e s(x) into the differential equation for y. x 2 (s  + (s  ) 2 ) e s +xs  e s −(x 2 + ν 2 ) e s = 0 s  + (s  ) 2 + 1 x s  − (1 + ν 2 x 2 ) = 0 1259 We make the assumption that s   (s  ) 2 as x → ∞ and we know that ν 2 /x 2  1 as x → ∞. Thus we drop these two terms from the equation to obtain an approximate equation for s. (s  ) 2 + 1 x s  − 1 ∼ 0 This is a quadratic equation for s  , so we can solve it exactly. However, let us try to simplify the equation even further. Assume that as x goes to infinity one of the three terms is much smaller that the other two. If this is the case, there will be a balance between the two dominant terms and we can neglect the third. Let’s check the three possib ili ties. 1. 1 is small. → (s  ) 2 + 1 x s  ∼ 0 → s  ∼ − 1 x , 0 1  1 x 2 , 0 as x → ∞ so this balance is inconsistent. 2. 1 x s  is small. → (s  ) 2 − 1 ∼ 0 → s  ∼ ±1 This balance is consistent as 1 x  1 as x → ∞. 3. (s  ) 2 is small. → 1 x s  − 1 ∼ 0 → s  ∼ x This balance is not consistent as x 2  1 as x → ∞. The only dominant balance that makes sense leads to s  ∼ ±1 as x → ∞. Integrating this relationship, s ∼ ±x + c ∼ ±x as x → ∞. Now let’s see if our assumption that we made to get the simplified equation for s is valid. Assuming that we can differentiate s  ∼ ±1, s   (s  ) 2 becomes d dx  ± 1 + o(1)    ± 1 + o(1)  2 0 + o(1/x)  1 1260 Thus we see that the behavior we obtained for s is consistent with our initial assumption. We have found two controlling factors, e x and e −x . This is a good sign as we know that there must be two line arly independent solutions to the equation. Leading Order Behavior. Now let’s find the full leading behavior of the solution with the controlling factor e −x . In order to find a better approximation for s, we substitute s(x) = −x + t(x), where t(x)  x as x → ∞, into the differential equation for s. s  + (s  ) 2 + 1 x s  −  1 + ν 2 x 2  = 0 t  + (−1 + t  ) 2 + 1 x (−1 + t  ) −  1 + ν 2 x 2  = 0 t  + (t  ) 2 +  1 x − 2  t  −  1 x + ν 2 x 2  = 0 We know that 1 x  2 and ν 2 x 2  1 x as x → ∞. Dropping these terms from the equation yields t  + (t  ) 2 − 2t  − 1 x ∼ 0. Assuming that we can differentiate the asymptotic relation for t, we obtain t   1 and t   1 x as x → ∞. We can drop t  . Since t  vanishes as x goes to infinity, (t  ) 2  t  . Thus we are left with −2t  − 1 x ∼ 0, as x → ∞. Integrating this relationship, t ∼ − 1 2 log x + c ∼ − 1 2 log x as x → ∞. 1261 Checking our assumptions about the behavior of t, t   1 → − 1 2x  1 t   1 x → 1 2x 2  1 x we see that the solution is consistent with our assumptions. The leading order behavior to the solution with controlling factor e −x is y(x) ∼ exp  −x − 1 2 log x + u(x)  = x −1/2 e −x+u(x) as x → ∞, where u(x)  log x. We substitute t = − 1 2 log x + u(x) into the differential equation for t in order to find the asymptotic behavior of u. t  + (t  ) 2 +  1 x − 2  t  −  1 x + ν 2 x 2  = 0 1 2x 2 + u  +  − 1 2x + u   2 +  1 x − 2  − 1 2x + u   −  1 x + ν 2 x 2  = 0 u  + (u  ) 2 − 2u  + 1 4x 2 − ν 2 x 2 = 0 Assuming that we can differentiate the asymptotic relation for u, u   1 x and u   1 x 2 as x → ∞. Thus we see that we can neglect the u  and (u  ) 2 terms. −2u  +  1 4 − ν 2  1 x 2 ∼ 0 u  ∼ 1 2  1 4 − ν 2  1 x 2 u ∼ 1 2  ν 2 − 1 4  1 x + c u ∼ c as x → ∞ 1262 [...]... (x) = 1 P1 (x) = x 3x2 − 1 P2 (x) = 2 3 5x − 3x P3 (x) = 2 35 x4 − 30 x2 + 3 P4 (x) = 8 Expanding cos(πx) in Legendre polynomials 4 cos(πx) ≈ cn Pn (x), n=0 and calculating the generalized Fourier coefficients with the formula cn = Pn | cos(πx) , Pn |Pn 1285 yields 15 45 (2π 2 − 21) P2 (x) + P4 (x) π2 4 105 = 4 [ (31 5 − 30 π 2 )x4 + ( 24 2 − 270)x2 + (27 − 2π 2 )] 8π cos(πx) ≈ − The cosine and this polynomial... by parts, you can obtain the series expansion 2 2 erfc(x) = √ e−x π ∞ n=0 1266 (−1)n (2n)! n!(2x)2n+1 x 1 2 3 4 5 6 7 8 9 10 erfc(x) 0.157 0.0 046 8 2.21 × 10−5 1. 54 × 10−8 1. 54 × 10−12 2.15 × 10−17 4. 18 × 10− 23 1.12 × 10−29 4. 14 × 10 37 2.09 × 10 45 One Term Relative Error 0 .32 03 0.1 044 0.0507 0.0296 0.0192 0.0 135 0.0100 0.0077 0.0061 0.0 049 Three Term Relative Error 0. 649 7 0.0182 0.0020 3. 9 · 10 4. .. expanded in a Taylor series about infinity Let us assume that we can expand the solution for y in the form x2 y(x) ∼ x exp − 4 x2 σ(x) = x exp − 4 ν ∞ an x−n as x → +∞ ν 2 where a0 = 1 Differentiating y = xν exp − x 4 n=0 σ(x), 1 2 2 y = νxν−1 − xν+1 e−x /4 σ(x) + xν e−x /4 σ (x) 2 1 1 1 1 2 2 y = ν(ν − 1)xν−2 − νxν − (ν + 1)xν + xν+2 e−x /4 σ(x) + 2 νxν−1 − xν+1 e−x /4 σ (x) 2 2 4 2 + xν e−x 2 /4 σ... π π x 2 ∞ ∞ 1 −x2 1 3 1 3 −5 −t2 1 15 −6 −t2 −1 =√ e x − x +√ − t e −√ t e dt 2 4 π π π x 4 x ∞ 1 1 3 1 15 −6 −t2 2 = √ e−x x−1 − x 3 + x−5 − √ t e dt 2 4 π π x 4 1265 2 0.5 1 2 1.5 2.5 3 -2 -4 -6 -8 -10 Figure 24. 2: Logarithm of the Approximation to the Complementary Error Function The error in approximating erfc(x) with the first three terms is given in Table 24. 1 We see that for x ≥ 2 the three terms... Figure 24. 1 K0 (x) is plotted in a solid line and √2x e−x is plotted in a dashed line We see that the leading order behavior of the solution as x goes to infinity gives a good approximation to the behavior even for fairly small values of x 24. 3 Integration by Parts Example 24. 3. 1 The complementary error function 2 erfc(x) = √ π 12 63 ∞ 2 e−t dt x 2 1.75 1.5 1.25 1 0.75 0.5 0.25 0 1 2 3 4 5 Figure 24. 1:... the least squares fit of a fourth degree polynomial to cos(πx) The set of functions {1, x, x2 , x3 , x4 } is independent, but not orthogonal in the interval [−1, 1] Using Gramm-Schmidt orthogo12 84 nalization, φ0 = 1 1|x =x 1|1 x|x2 1 1|x2 − x = x2 − φ2 = x2 − 1|1 x|x 3 3 3 = x3 − x 5 3 6 2 4 4 = x − x − 7 35 A widely used set of functions in mathematics is the set of Legendre polynomials {P0 (x), P1... n!(2x)2n+1 as x → ∞ In Figure 24. 3 the logarithm of the difference between the one term, ten term and twenty term approximations and the complementary error function are graphed in coarse, medium, and fine dashed lines, respectively *Optimal Asymptotic Series Of the three approximations, the one term is best for x 2, the ten term is best for 2 x 4, and the twenty term is best for 4 x This leads us to the concept... 1268 1 2 3 4 5 6 -20 -40 -60 Figure 24. 3: log(error in approximation) In Figure 24. 4 we see a plot of the number of terms in the approximation versus the logarithm of the error for x = 3 Thus we see that the optimal asymptotic approximation is the first nine terms After nine terms the error gets larger It was inevitable that the error would start to grow after some point as the series diverges for all... very good results even for moderate values of x Table 24. 1 gives the error in this first approximation for various values of x If we continue integrating by parts, we might get a better approximation to the complementary error function ∞ 1 1 2 2 erfc(x) = √ x−1 e−x − √ t−2 e−t dt π π x ∞ ∞ 1 1 1 3 4 −t2 1 2 2 = √ x−1 e−x − √ − t 3 e−t +√ t e dt 2 π π π x 2 x ∞ 1 −x2 1 3 4 −t2 1 3 −1 =√ e +√ t e dt x... our asymptotic expansion for y y ∼ xν e−x 2 /4 1− ν(ν − 1) −2 ν(ν − 1)(ν − 2)(ν − 3) 4 x + x − ··· 21 1! 22 2! diverges for all x However this solution is still very useful If we only use a finite number of terms, we will get a very good numerical approximation for large x In Figure 24. 6 the one term, two term, and three term asymptotic approximations are shown in rough, medium, and fine dashing, respectively . 0. 649 7 2 0.0 046 8 0.1 044 0.0182 3 2.21 × 10 −5 0.0507 0.0020 4 1. 54 × 10 −8 0.0296 3. 9 · 10 4 5 1. 54 × 10 −12 0.0192 1.1 · 10 4 6 2.15 × 10 −17 0.0 135 3. 7 · 10 −5 7 4. 18 × 10 − 23 0.0100 1.5. = 1 √ π x −1 e −x 2 − 1 √ π  ∞ x t −2 e −t 2 dt = 1 √ π x −1 e −x 2 − 1 √ π  − 1 2 t 3 e −t 2  ∞ x + 1 √ π  ∞ x 3 2 t 4 e −t 2 dt = 1 √ π e −x 2  x −1 − 1 2 x 3  + 1 √ π  ∞ x 3 2 t 4 e −t 2 dt = 1 √ π e −x 2  x −1 − 1 2 x 3  + 1 √ π  − 3 4 t −5 e −t 2  ∞ x − 1 √ π  ∞ x 15 4 t −6 e −t 2 dt = 1 √ π e −x 2  x −1 − 1 2 x 3 + 3 4 x −5  − 1 √ π  ∞ x 15 4 t −6 e −t 2 dt 1265 0.5 1 1.5 2 2.5. to the behavior even for fairly small values of x. 24. 3 Integration by Parts Example 24. 3. 1 The complementary error function erfc(x) = 2 √ π  ∞ x e −t 2 dt 12 63 0 1 2 3 4 5 0.25 0.5 0.75 1 1.25 1.5 1.75 2 Figure