Maths Compressed
Maths Compressed
Maths Compressed
b) Bessel’s eqn: 𝑥 2 𝑦 ′′ + 𝑥𝑦 ′ + 𝑥 2 − 𝑛2 𝑦 = 0
c) Hypergeometric equations
❑ Method for solving such ODE’s is:
b) Frobenius method
• Power series in (𝑥 − 𝑥0 ) is : σ∞
𝑚=0 𝑎𝑚 𝑥 − 𝑥0
𝑚
• Power series in 𝑥 is : σ∞ 𝑎
𝑚=0 𝑚 𝑥 𝑚
𝑥 𝑥2 𝑥3
𝑒 = 1+𝑥+ + 3! + ⋯ etc.
2
Technique of power series solution on ODE:
𝑦 ′ = σ∞
𝑚=1 𝑚 𝑎𝑚 𝑥
𝑚−1
= 𝑎1 + 2𝑎2 𝑥 + 3 𝑎3 𝑥 2 + ⋯
𝑎1 + 2𝑎2 𝑥 + 3 𝑎3 𝑥 2 + ⋯ − 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + ⋯ = 0
𝑎1 − 𝑎0 + 2 𝑎2 − 𝑎1 𝑥 + 3 𝑎3 − 𝑎2 𝑥 2 + ⋯ = 0
𝑥2 𝑥3
=𝑎0 1 + 𝑥 + 2! + 3!
+ ⋯ = 𝑎0 𝑒 𝑥
𝑄(𝑥) 𝑅(𝑥)
lim 𝑥 = finite and lim 𝑥 2 = finite
𝑥→𝑥0 𝑃(𝑥) 𝑥→𝑥0 𝑃(𝑥)
Note: For irregular singular point series solution does not apply.
Here 𝑃 𝑥 = 2𝑥 2 , 𝑄 𝑥 = −𝑥, 𝑅 𝑥 = 1 − 𝑥 2
−𝑥 1
lim 𝑥 = − 2 (finite)
𝑥→0 2𝑥 2
𝑥 2 1−𝑥 2 1
lim 2𝑥 2
= 2
(finite)
𝑥→0
❑ Here 𝑅 is called the radius of convergence. If the series converges for all 𝑥 we set 𝑅 = ∞
1
and =0
𝑅
We also know the radius of convergence formula.
Consider 𝑦 ′′ + 𝑝 𝑥 𝑦 ′ + 𝑞 𝑥 𝑦 = 𝑟(𝑥)…….(1)
If 𝑝, 𝑞, 𝑟 are analytic at 𝑥 = 𝑥0 then every solution of (1) is analytic at 𝑥 = 𝑥0 and can thus
be represented by a power series in powers of 𝑥 − 𝑥0 with radius of convergence 𝑅 > 0.
1 − 𝑥 2 𝑦 ′′ − 2𝑥𝑦 ′ + 𝑛 𝑛 + 1 𝑦 = 0 … … (1)
2𝑥 𝑛 𝑛+1
Dividing by (1 − 𝑥 2 ) we see that the coefficients − 1−𝑥2 and are analytic at 𝑥 = 0, so
1−𝑥 2
we can apply the power series method.
Put 𝑦 = σ∞ 𝑎
𝑚=0 𝑚 𝑥 𝑚
and its derivative in (1). Let 𝑛 𝑛 + 1 = 𝑘, we get
∞ ∞ ∞
1 − 𝑥 2 𝑚 𝑚 − 1 𝑎𝑚 𝑥 𝑚−2 − 2𝑥 𝑚 𝑎𝑚 𝑥 𝑚−1 + 𝑘 𝑎𝑚 𝑥 𝑚 = 0
𝑚=2 𝑚=1 𝑚=0
∞ ∞ ∞ ∞
𝑚 𝑚 − 1 𝑎𝑚 𝑥 𝑚−2 − 𝑚 𝑚 − 1 𝑎𝑚 𝑥 𝑚 − 2𝑚 𝑎𝑚 𝑥 𝑚 + 𝑘 𝑎𝑚 𝑥 𝑚 = 0
𝑚=2 𝑚=2 𝑚=1 𝑚=0
Put 𝑚 = 𝑠 + 2 in 1st series and 𝑚 = 𝑠 in the other series, we get
∞ ∞ ∞ ∞
𝑠 + 2 𝑠 + 1 𝑎𝑠+2 𝑥 𝑠 − 𝑠 𝑠 − 1 𝑎𝑠 𝑥 𝑠 − 2𝑠 𝑎𝑠 𝑥 𝑠 + 𝑘 𝑎𝑠 𝑥 𝑠 = 0
𝑠=0 𝑠=2 𝑠=1 𝑠=0
The sum of the coefficients of each powers of 𝑥 on the left side must be zero as right hand
side is zero.
Coeff. Of 𝑥 0 : 2𝑎2 + 𝑛 𝑛 + 1 𝑎0 = 0
i.e. 𝑠 + 2 𝑠 + 1 𝑎𝑠+2 + 𝑛 − 𝑠 𝑛 + 𝑠 + 1 𝑎𝑠 = 0
𝑛−𝑠 𝑛+𝑠+1
Thereofore, 𝑎𝑠+2 = − 𝑎𝑠 , 𝑠 = 0,1,2, …
𝑠+2 𝑠+1
And so on.
Problem: Find the first four terms in each portion of the series solution around 𝑥0 = −2 for
𝑦 ′′ − 𝑥𝑦 = 0
Given 𝑥0 = −2
Then 𝑦 𝑥 = σ∞
𝑚=0 𝑎𝑚 𝑥 + 2
𝑚
, 𝑦 ′ 𝑥 = σ∞
𝑚=1 𝑚 𝑎𝑚 𝑥 + 2
𝑚−1
,
𝑦 ′′ 𝑥 = σ∞
𝑚=2 𝑚 𝑚 − 1 𝑎𝑚 (𝑥 + 2)
𝑚−2
Substituting in (1), we get
∞ ∞
𝑚−2 𝑚
𝑚 𝑚 − 1 𝑎𝑚 𝑥 + 2 − 𝑥 𝑎𝑚 𝑥 + 2 =0
𝑚=2 𝑚=0
∞ ∞
𝑚−2 𝑚
𝑚 𝑚 − 1 𝑎𝑚 𝑥 + 2 − (𝑥 + 2 − 2) 𝑎𝑚 𝑥 + 2 =0
𝑚=2 𝑚=0
∞ ∞ ∞
𝑚−2 𝑚+1 𝑚
𝑚 𝑚 − 1 𝑎𝑚 𝑥 + 2 − 𝑎𝑚 𝑥 + 2 + 2 𝑎𝑚 𝑥 + 2 =0
𝑚=2 𝑚=0 𝑚=0
Put 𝑚 = 𝑠 + 2 in 1st series, 𝑚 = 𝑠 − 1 in the 2nd series and 𝑚 = 𝑠 in the 3rd series, we get
∞ ∞ ∞
𝑠 𝑠 𝑠
𝑠 + 2 𝑠 + 1 𝑎𝑠+2 𝑥 + 2 − 𝑎𝑠−1 𝑥 + 2 + 2 𝑎𝑠 𝑥 + 2 =0
𝑠=0 𝑠=1 𝑠=0
∞
𝑠
⟹ 2𝑎2 + 2𝑎0 + 𝑠 + 2 𝑠 + 1 𝑎𝑠+2 − 𝑎𝑠−1 + 2 𝑎𝑠 𝑥 + 2 =0
𝑠=1
𝑠 = 0: 2𝑎2 + 2𝑎0 = 0
𝑎𝑠−1 −2 𝑎𝑠
𝑎𝑠+2 = (𝑠+2)(𝑠+1)
𝑎0 −2𝑎1 𝑎0 𝑎1
𝑠 = 1: 𝑎3 = 3.2
= 6
− 3
𝑎1 − 2𝑎2 𝑎1 − 2(−𝑎0 ) 𝑎0 𝑎1
𝑠 = 2: 𝑎4 = = = +
4.3 4.3 6 12
𝑎2 −2𝑎3 𝑎 𝑎
𝑠 = 3: 𝑎5 = = − 150 + 301
5.4
2 3 4 5
= 𝑎0 + 𝑎1 𝑥 + 2 + 𝑎2 𝑥 + 2 + 𝑎3 𝑥 + 2 + 𝑎4 𝑥 + 2 + 𝑎5 𝑥 + 2 + ⋯.
2 𝑎0 𝑎1 3 𝑎0 𝑎1 4
=𝑎0 + 𝑎1 𝑥 + 2 − 𝑎0 𝑥 + 2 + − 𝑥+2 + + 12 𝑥+2
6 3 6
𝑎0 𝑎1 5
+ − + 𝑥+2 +⋯
15 30
2 1 3 1 4 1 5
=𝑎0 1 − 𝑥 + 2 +6 𝑥+2 + 𝑥+2 − 𝑥+2 +⋯
6 15
1 3 1 4 1 5
+𝑎1 [ 𝑥 + 2 − 𝑥+2 + 𝑥+2 + 𝑥+2 + ⋯]
3 12 30
𝑥 − 𝑥0 𝑟 σ∞
𝑛=0 𝑐𝑛 𝑥 − 𝑥0
𝑛
……… (1)
∞
𝑟 𝑛
𝑦 = 𝑥 − 𝑥0 𝑐𝑛 𝑥 − 𝑥0 … … … . (3)
𝑛=0
for the DE (2), with 𝑥0 as a regular singular point. Here 𝑟 is chosen so that 𝑐0 ≠ 0.
𝑦 ′ 𝑥 = 𝑛 + 𝑟 𝑐𝑛 𝑥 − 𝑥0 𝑛+𝑟−1
… . . (5)
𝑛=0
and 𝑦 ′′ 𝑥 = σ∞
𝑛=0 𝑛 + 𝑟 𝑛 + 𝑟 − 1 𝑐𝑛 𝑥 − 𝑥0
𝑛+𝑟−2
… … . . (6)
Here the coefficients 𝐾𝑖 (𝑖 = 0,1,2,3. . ) are functions of the exponent 𝑟 and the constant
coefficients 𝑐𝑛′ 𝑠. Also 𝑘 is an integer.
Step III: Since (4) is the solution of DE, then we must have
𝐾0 = 𝐾1 = 𝐾2 = ⋯ = 0
Step V: Let 𝑦1 (𝑥) and 𝑦2 (𝑥) be two non trivial linearly independent solutions of the DE (2).
Then the general solution of (2) is 𝑦 𝑥 = 𝐴 𝑦1 𝑥 + 𝐵𝑦2 (𝑥), where 𝐴, 𝐵 are arbitrary
constants.
Then with the known exponent 𝑟 and the known coefficients 𝑐𝑛′ 𝑠 , 𝑦1 (𝑥) , one of the two
solutions of (2) is of the form (4). The form of the second (other) solution 𝑦2 (𝑥) may be
similar to (4) (with different 𝑟 and different coefficient) or may contain a logarithmic term.
The form of 𝑦2 𝑥 will be indicated by the indicial equation. There are three cases:
Case 1: Distinct roots not differing by an integer 1,2,3,…
∞
𝑟1 𝑛
𝑦1 𝑥 = 𝑥 − 𝑥0 𝑐𝑛 𝑥 − 𝑥0 … … . . (7)
𝑛=0
and 𝑦2 𝑥 = 𝑥 − 𝑥0 𝑟2 σ∞ 𝑛
𝑛=0 𝑏𝑛 𝑥 − 𝑥0 , 𝑐0 ≠ 0, 𝑏0 ≠ 0 … … … (8)
and 𝑦2 𝑥 = 𝑥 − 𝑥0 𝑟 σ∞
𝑛=0 𝑏𝑛 𝑥 − 𝑥0
𝑛
+ 𝐴∗ 𝑦1 𝑥 ln 𝑥 − 𝑥0 … … . . (9)
Suppose 𝑟1 − 𝑟2 = 0. Then
∞
𝑟1
𝑦1 𝑥 = 𝑥 − 𝑥0 𝑐𝑛 𝑥 − 𝑥0 𝑛 , 𝑐0 ≠ 0 … . . (7)
𝑛=0
𝑟1 +1 σ∞ 𝑛
and 𝑦2 𝑥 = 𝑥 − 𝑥0 𝑛=0 𝑏𝑛 𝑥 − 𝑥0 + 𝑦1 𝑥 ln 𝑥 − 𝑥0 … … … . . (10)
Note 2: In Case 2, 𝑟1 − 𝑟2 = 𝑁, where 𝑁 = 1,2, … the second solution 𝑦2 (𝑥) may not contain
the logarithm term 𝐴∗ 𝑦1 𝑥 ln(𝑥 − 𝑥0 ) . In some cases 𝐴∗ is zero, so 𝑦2 is same as (8).
Note 3: In Case 3, 𝑟1 − 𝑟2 = 0 (double root), 𝑦2 (𝑥) always contain the logarithm term
𝑦1 𝑥 ln 𝑥 − 𝑥0 and is never of the simple form (8).
Note 4: In Case 2, roots differing by an integer in logarithmic case, take the first solution
𝜕 𝑦1 (𝑥,𝑟)
𝑦1 𝑥, 𝑟 , then the second solution is i.e. partial derivative of 𝑦1 (𝑥, 𝑟) w.r.t. r. Then
𝜕𝑟
taking 𝑟 as the smaller root 𝑟2 , we get the two linearly independent solutions 𝑦1 𝑥, 𝑟2 and
𝜕 𝑦1 (𝑥,𝑟2 )
. Here the second solution 𝑦2 contains the logarithm term.
𝜕𝑟
Note 5: In both Cases 2 and 3, the second solution containing logarithmic term can be
obtained by reduction of order.