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

24.2 Leading Order Behavior of Differential EquationsIt is often useful to know the leading order behavior of the solutions to a differential equation.. Since the solutions at irregular

Differentiating this relation yields

x2 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 x2ex as x goes to infinity The controlling factor is ex For a fewexamples of this,

• x log x has the controlling factor x as x → ∞

• x−2e1/x has the controlling factor e1/x as x → 0

• x−1sin 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) ∼ a0(x) + a1(x) + a2(x) + · · · , as x → x0.where

a0(x)  a1(x)  a2(x)  · · · , as x → x0.The first term in the sum is the leading order behavior For a few examples,

• For sin x ∼ x − x3/6 + x5/120 − · · · as x → 0, the leading order behavior is x

• For f (x) ∼ ex(1 − 1/x + 1/x2− · · · ) as x → ∞, the leading order behavior is ex

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 theFrobenius 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 Frobenius method which always gives us the solution, we will use aheuristic approach that usually gives us the solution

Example 24.2.1 Consider the Airy equation

The Controlling Factor Since the solutions at irregular singular points often have exponential behavior, we makethe substitution y = es(x) into the differential equation for y.

us an approximate solution If one of the terms in the equation for s is much smaller than the others then we say thatthe remaining terms form a dominant balance in the limit as x → ∞

Assume that the s00 term is much smaller that the others, s00  (s0)2, x as x → ∞ This gives us

to the Airy equation are exp(±23x3/2) as x → ∞

The Leading Order Behavior of the Decaying Solution Let’s find the leading order behavior as x goes toinfinity of the solution with the controlling factor exp(−23x3/2) We substitute

s(x) = −2

3x

3/2

+ t(x), where t(x)  x3/2 as x → ∞into the differential equation for s

we see that the behavior of t is consistent with our assumptions.

Since u = c + o(1), eu = ec+o(1) The behavior of y is

Thus the full leading order behavior of the decaying solution is

The Controlling Factor Since the solutions at irregular singular points often have exponential behavior, we makethe substitution y = es(x) into the differential equation for y

We make the assumption that s00  (s0)2 as x → ∞ and we know that ν2/x2  1 as x → ∞ Thus we drop thesetwo terms from the equation to obtain an approximate equation for s.

(s0)2+ 1

xs

0− 1 ∼ 0This is a quadratic equation for s0, 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, therewill be a balance between the two dominant terms and we can neglect the third Let’s check the three possibilities.1

3

(s0)2 is small → 1

xs

0− 1 ∼ 0 → s0 ∼ xThis balance is not consistent as x2 6 1 as x → ∞

The only dominant balance that makes sense leads to s0 ∼ ±1 as x → ∞ Integrating this relationship,

Thus we see that the behavior we obtained for s is consistent with our initial assumption.

We have found two controlling factors, ex and e−x This is a good sign as we know that there must be two linearlyindependent 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 thedifferential equation for s

Checking our assumptions about the behavior of t,

we see that the solution is consistent with our assumptions

The leading order behavior to the solution with controlling factor e−x is

2x2 + u00+



− 12x + u

x and u00  1

x 2 as x → ∞ Thus we see that

we can neglect the u00 and (u0)2 terms

−2u0 + 1

4 − ν2

1

x2 ∼ 0

u0 ∼ 12

 1

4 − ν2

1

x2

u ∼ 12



ν2− 14

 1

x + c

u ∼ c as x → ∞

Since u = c + o(1), we can expand eu as ec+o(1) Thus we can write the leading order behavior as

Two linearly independent solutions to the modified Bessel equation are the modified Bessel functions, Iν(x) and

Kν(x) These functions have the asymptotic behavior

Iν(x) ∼ √1

2πxe

x, Kν(x) ∼

√π

√2xe

−x

as x → ∞

In Figure 24.1 K0(x) is plotted in a solid line and

√ π

√ 2xe−x is plotted in a dashed line We see that the leading orderbehavior of the solution as x goes to infinity gives a good approximation to the behavior even for fairly small values ofx

e−t2 dt

0 1 2 3 4 5 0.25

0.5 0.75 1 1.25 1.5 1.75

Figure 24.1: Plot of K0(x) and it’s leading order behavior

is used in statistics for its relation to the normal probability distribution We would like to find an approximation toerfc(x) for large x Using integration by parts,

erfc(x) = √2

π

Z ∞ x

 −12t





−2t e−t2 dt

= √2π

 −12t e

−t 2

∞

x

− √2π

Z ∞ x

t−2e−t2 dt

We examine the residual integral in this expression.

1

√π

Z ∞ x

and we expect that √1

πx−1e−x2 would be a good approximation to erfc(x) for large x In Figure 24.2 log(erfc(x)) isgraphed in a solid line and log√1

πx−1e−x2 is graphed in a dashed line We see that this first approximation to theerror function gives very good results even for moderate values of x Table24.1gives the error in this first approximationfor various values of x

If we continue integrating by parts, we might get a better approximation to the complementary error function

+√1π

Z ∞ x

Z ∞ x

− √1π

Z ∞ x

Z ∞ x

15

4 t

−6e−t2 dt

0.5 1 1.5 2 2.5 3

-10 -8 -6 -4 -2

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 threeterms give a much better approximation to erfc(x) than just the first term

At this point you might guess that you could continue this process indefinitely By repeated application of integration

by parts, you can obtain the series expansion

x erfc(x) One Term Relative Error Three Term Relative Error

an+1(x)

an(x)

< 1 → lim

n→∞

< 1

→ lim

n→∞

1x

= 0

Thus we see that our series diverges for all x Our conventional mathematical sense would tell us that this series isuseless, however we will see that this series is very useful as an asymptotic expansion of erfc(x)

Say we are working with a convergent series expansion of some function f (x).

For fixed N , erfc(x) −√2

-60 -40 -20

Figure 24.3: log(error in approximation)

In Figure 24.4we 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 x

A good rule of thumb for finding the optimal series is to find the smallest term in the series and take all of theterms up to but not including the smallest term as the optimal approximation This makes sense, because the nth term

is an approximation of the error incurred by using the first n − 1 terms In Figure 24.5 there is a plot of n versus thelogarithm of the nth term in the asymptotic expansion of erfc(3) We see that the tenth term is the smallest Thus, inthis case, our rule of thumb predicts the actual optimal series

5 10 15 20 25 -18

-16 -14

Figure 24.4: The logarithm of the error in using n terms

An asymptotic series may be convergent or divergent Most of the asymptotic series you encounter will be divergent

If the series is convergent, then we have that

5 10 15 20 25 -16

-14

Figure 24.5: The logarithm of the nth term in the expansion for x = 3

Let n(x) be some set of gauge functions The example that we are most familiar with is n(x) = xn If we say that

24.5 Asymptotic Expansions of Differential Equations

24.5.1 The Parabolic Cylinder Equation.

Controlling Factor Let us examine the behavior of the bounded solution of the parabolic cylinder equation as

Now let’s check if our assumption is consistent Substituting into s00  (s0)2 yields 1/2  x2/4 as x → +∞ which

is true Since the equation for y is second order, we would expect that there are two different behaviors as x → +∞.This is confirmed by the fact that we found two behaviors for s s ∼ −x2/4 corresponds to the solution that is bounded

at +∞ Thus the controlling factor of the leading behavior is e−x2/4

Leading Order Behavior Now we attempt to get a better approximation to s We make the substitution

s = −14x2+ t(x) into the equation for s where t  x2 as x → +∞

we see that they were consistent Now we wish to refine our approximation for t with the substitution t(x) =

ν log x + u(x) So far we have that

u0 ∼ ν

2− ν

x3 as x → +∞

Integrating this asymptotic relation,

Substituting this into the differential equation for y,

σ00(x) + (2νx−1− x)σ0(x) + ν(ν − 1)x−2σ = 0

x2σ00(x) + (2νx − x3)σ0(x) + ν(ν − 1)σ(x) = 0

Differentiating the expression for σ(x),

From the coefficient of x−n for n > 0,

n(n + 1)an− 2νnan+ (n + 2)an+2+ ν(ν − 1)an= 0(n + 2)an+2 = −[n(n + 1) − 2νn + ν(ν − 1)]an(n + 2)an+2 = −[n2+ n − 2νn + ν(ν − 1)]an(n + 2)an+2 = −(n − ν)(n − ν + 1)an

Thus the recursion formula for the an’s is

an+2x−n−2

anx−n

< 1 → lim

n→∞

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24. 5 Asymptotic Expansions of Differential Equations

24. 5.1 The Parabolic Cylinder Equation.

Controlling Factor Let us examine... elements u and v

d(u, v) ≡ ku − vkNote that d(u, v) = does not necessarily imply that u = v CONTINUE

25 .4 Linear Independence.

25.5 Orthogonality

Orthogonality... class="text_page_counter">Trang 34 < /span>

The cosine and this polynomial are plotted in the second graph in Figure 25.1 The least squares fit method usesinformation about