18 series solution and special functions
TRANSCRIPT
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18 Series Solution and Special Functions
18.1 INTRODUCTION
Generally the solutions of ordinary differential equations are obtainable in explicit form called a
closed form of the solution. However, many differential equations arising in physical problems are
linear but have variable coefficients and do not permit a general solution in terms of known functions.
For such equations, it is easier to find a solution in the form of an infinite convergent series called
power series solution. The series solution of certain differential equations give rise to special
functions such as Besselâs functions, Legendreâs polynomials, Lagurreâs polynomial, Hermiteâs
polynomial, Chebyshev polynomials. Strum-Liovelle problem based on orthogonality of functions is
also included which shows that Besselâs, Legendreâs and other equations can be determined from a
common point of view.
18.2 POWER SERIES SOLUTION OF DIFFERENTIAL EQUATIONS
Consider the differential equation
ð0 ð¥ ð2ðŠ
ðð¥2 + ð1 ð¥ ððŠ
ðð¥+ ð2 ð¥ ðŠ = 0 ⊠(1)
where ðð â²ð are polynomials in ð¥.
If ð0 ð â 0, then ð¥ = ð is called an ordinary point of (1), otherwise a singular point. Ordinary
point is also called a regular point of the equation.
A singular point ð¥ = ð of (1) is called regular singular point if, (1) can be put in the form
ð2ðŠ
ðð¥2 +ð1 ð¥
ð¥âð
ððŠ
ðð¥+
ð2 ð¥
ð¥âð 2 ðŠ = 0 ⊠(2)
provided ð1 ð¥ and ð2 ð¥ both possess derivatives of all orders in the neighborhood of ð.
A singular point which is not regular is called an irregular singular point.
Note: The power series method sometimes fails to yield a solution
e.g. ð¥2ðŠâ²â² + ð¥ðŠâ² + ðŠ = 0 âŠ(3)
dividing by ð¥2 throughout, ð¥2ðŠâ²â² + ð¥ðŠâ² + ðŠ = 0 âŠ(4)
Here neither of the terms ð1 ð¥ =1
ð¥ and ð2 ð¥ =
1
ð¥2 is defined at ð¥ = 0, so we cannot find a
power series representation for ð1 ð¥ or ð2 ð¥ that converges in an open interval containing
ð¥ = 0.
Theorem I: If ð¥ = ð is an ordinary point of the differential equation (1), i.e. ð0 ð â 0, then
series solution of (1) can be found as:
ðŠ = ð0 + ð1(ð¥ â ð) + ð2(ð¥ â ð)2 + ⯠⊠(5)
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Calculate the derivatives ððŠ
ðð¥,ð2ðŠ
ðð¥2 from (5), and substitute the values of y and its derivatives in
differential equation (1).
The values of the constants ð2 ,ð3 ,ð4 ,⊠are obtained by equating to zero the coefficients of
various powers of ð¥.
Putting the values of these constants in the solution (5), the desired power series solution of (1) is
obtained with ð0 ,ð1 as its arbitrary constants.
Theorem II: When ð¥ = ð is a regular singularity of (1) at least one of the solutions can be expressed
as,
ðŠ = (ð¥ â ð)ð [ð0 + ð1(ð¥ â ð) + ð2(ð¥ â ð)2 + ⯠] âŠ(6)
Theorem III: The series (5) and (6) are convergent at every point within the circle of convergence at
ð. A solution in series will be valid only if the series is convergent.
Example 1: Solve in series the equation ð ðð
ð ððâ ðð = ð.
Solution: Given differential equation is
ð2ðŠ
ðð¥2 â ð¥ðŠ = 0 ⊠(1)
Here ð0 ð¥ = 1, so ð0 0 = 1, i.e. ð¥ = 0 is the ordinary point of the differential equation (1).
Let the solution of differential equation (1) be
ðŠ = ð0 + ð1ð¥ + ð2ð¥2 + ð3ð¥
3 + ð4ð¥4 + ð5ð¥
5 + ⯠⊠(2)
Differentiating (2) w.r.t. ð¥,
ððŠ
ðð¥= ð1 + 2ð2ð¥ + 3ð3ð¥
2 + 4ð4ð¥3 + 5ð5ð¥
4 + ⯠⊠(3)
Again differentiating w.r.t ð¥
ð2ðŠ
ðð¥2 = 2ð2 + 6ð3ð¥ + 12ð4ð¥2 + 20ð5ð¥
3 + ⯠⊠(4)
Substitute values of y from (2) and its derivative from (4) in the differential equation (1), we get
2ð2 + 6ð3ð¥ + 12ð4ð¥2 + 20ð5ð¥
3 + â¯
âð¥ ð0 + ð1ð¥ + ð2ð¥2 + ð3ð¥
3 + ð4ð¥4 + ð5ð¥
5 + ⯠= 0
=> 2ð2 + 6ð3 â ð0 ð¥ + 12ð4 â ð1 ð¥2 + 20ð5 â ð2 ð¥
3 + ⯠= 0
Equating each of the coefficients to zero, we obtain the identities,
2ð2 = 0, 6ð3 â ð0 = 0, 12ð4 â ð1 = 0, 20ð5 â ð2 = 0
which further gives ð2 = 0, ð3 =1
6ð0 , ð4 =
1
12ð1 , ð5 =
1
20ð2 = 0
Generalizing the results, ðð+2 =ððâ1
ð+2 (ð+1) ⊠(5)
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Putting ð = 4, 5, 6⊠in (5), we get
ð6 =1
6 (5)ð3 =
1
6 6 (5)ð0 =
1
180ð0 ,
ð7 =1
7 (6)ð4 =
1
12 7 (6)ð1 =
1
504ð1 ,
ð8 = 0.
Using the values of the constants in (2), the general solution of differential equation (1) becomes
ðŠ = ð0 1 +1
6ð¥3 +
1
180ð¥6 + ⯠+ ð1 ð¥ +
1
12ð¥4 +
1
504ð¥7 + ⯠.
Example 2:
ASSIGNMENT 18.1
Solve the following differential equations in series
1. ð2ðŠ
ðð¥2 + ð¥ððŠ
ðð¥+ ðŠ = 0.
2. ð2ðŠ
ðð¥2 + ð¥ðŠ = 0.
3. 1 â ð¥2 ð2ðŠ
ðð¥2 â ð¥ððŠ
ðð¥+ 4ðŠ = 0.
4. ð2ðŠ
ðð¥2 + ðŠ = 0, given ðŠ 0 = 0.
5. 1 â ð¥2 ðŠâ²â² + 2ðŠ = 0,ððð£ðð ðŠ 0 = 4, ðŠâ² 0 = 5.
ANSWERS
1. ðŠ = ð0 1 âð¥2
2+
ð¥4
2.4â
ð¥6
2.4.6+ ⯠+ ð1 ð¥ â
ð¥3
3+
ð¥5
3.5â
ð¥7
3.5.7+ â¯
2. ðŠ = ð0 1 â1
3!ð¥3 +
1.4
6!ð¥6 â
1â4.7
9!ð¥9 âŠ
+ð1 ð¥ â1.2
4!ð¥4 +
2.7
7!ð¥7 + â¯
3. ðŠ = ð0 1 â 2ð¥2 + ð1ð¥ 1 âð¥2
2â
ð¥4
8â
3
6âð¥6
8â
5â3
8â6âð¥8
8ââ¯
4. ðŠ = ð0 ð¥ âð¥3
3!+
ð¥5
5!ââ¯
5. ðŠ = 4 + 5ð¥ â 4ð¥2 â5
3ð¥3 â
ð¥5
3â
ð¥7
7ââ¯
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18.3 FROBENIUS METHOD
This method is named after a German mathematician F.G. Frobenius (1849 â 1917) who is
known for his contributions to the theory of matrices and groups. This method is employed to find the
power series solution of the differential equation
ð0 ð¥ ð2ðŠ
ðð¥2 + ð1 ð¥ ððŠ
ðð¥+ ð2 ð¥ ðŠ = 0 ⊠(1)
when ð¥ = 0 is the regular singularity.
Working Procedure
(i) Let ðŠ = ð¥ð (ð0 + ð1ð¥ + ð2ð¥2 + ð3ð¥
3 + â¯+ ððð¥ð + ⯠) ⊠(2)
be the solution of the differential equation (1), where m is some real or complex number.
(ii) Substitute in (1) the values of ðŠ,ððŠ
ðð¥ ,
ð2ðŠ
ðð¥2 obtained by differentiating (2).
(iii) Find the indicial equation (a quadratic equation) by equating to zero the coefficient of the
lowest degree term in x.
(iv) Find the values of ð1 , ð2 , ð3 , ⯠in terms of ð0 by equating to zero the coefficients of
other powers of x.
(v) Find the roots ð1 , ð2 (say) of the indicial equation. The complete solution depends on
the nature of roots of the indicial equation.
Case I: Roots ðð, ðð are distinct and do not differ by an integer
In this case, the differential equation (1) has two linearly independent solutions of the following
forms:
ðŠ1 = ð¥ð1 (ð0 + ð1ð¥ + ð2ð¥2 + ð3ð¥
3 + ⯠)
ðŠ2 = ð¥ð1 (ð0 + ð1ð¥ + ð2ð¥2 + ð3ð¥
3 + ⯠)
The complete solution of the differential equation is given by
ðŠ = ð1ðŠ1 + ð2ðŠ2.
Example 3: Solve ððð ðð
ð ðð+ ð
ð ð
ð ð+ ð = ð
Solution: Given 4ð¥ð2ðŠ
ðð¥2 + 2ððŠ
ðð¥+ ðŠ = 0 ⊠(1)
Here ð¥ = 0 is a singular point, let its solution be
ðŠ = ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + ð3ð¥
ð+3 + ð4ð¥ð+4 + ⊠⊠(2)
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From equation (2)
ððŠ
ðð¥ = ðð0ð¥
ðâ1 + ð + 1 ð1ð¥ð + ð + 2 ð2ð¥
ð+1
+ ð + 3 ð3ð¥ð+2 + âŠâŠ ⊠(3)
ð2ðŠ
ðð¥2 = ð ð â 1 ð0ð¥ðâ2 + ð + 1 ð ð1ð¥
ðâ1
+ ð + 2 ð + 1 ð2ð¥ð + âŠâŠ ⊠(4)
Putting the above values in equation (1), we get
4ð¥ ð ð â 1 ð0ð¥ðâ2 + ð + 1 ð ð1ð¥
ðâ1 + ð + 2 ð + 1 ð2ð¥ð + âŠâŠ
+2 ðð0ð¥ðâ1 + ð + 1 ð1ð¥
ð + ð + 2 ð2ð¥ð+1 + ð + 3 ð3ð¥
ð+2 + âŠâŠ
+ ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + ð3ð¥
ð+3 + ð4ð¥ð+4 + âŠâŠ = 0 âŠ
(5)
Equating the coefficients of ð¥ðâ1 equal to zero
4ð ð â 1 ð0 + 2ðð0 = 0 â ð0 4ð2 â 4ð + 2ð = 0
Because ð0 â 0 â 4ð2 â 2ð = 0 i.e. ð = 0,1
2
⎠The solution of the indicial equation is ð1 = 0 and ð2 =1
2.
Here, the roots are real, distinct and do not differ by an integer.
⎠Its solution is ðŠ = ð1ðŠ1 + ð2ðŠ2 ⊠(6)
On equating coefficients of ð¥ð , we get
4 ð + 1 ðð1 + 2 ð + 1 ð1 + ð0 = 0 or
2 ð + 1 2ð + 1 ð1 = âð0
â ð1 =âð0
2 ð+1 (2ð+1) ⊠(7)
Likewise, 4 ð + 2 ð + 1 ð2 + 2 ð + 2 ð2 + ð1 = 0
ð + 2 4ð + 4 + 2 ð2 = âð1 or
2 ð + 2 2ð + 3 ð2 = âð1
â ð2 = âð1
2 ð+2 (2ð+3)=
ð0
22 ð+2 ð+1 2ð+1 (2ð+3) ⊠(8)
and 4 ð + 3 ð + 2 ð3 + 2 ð + 3 ð3 + ð2 = 0
ð + 3 4ð + 8 + 2 ð3 = âð2
2 ð + 3 2ð + 5 ð3 = âð0
22 ð+2 ð+1 2ð+1 (2ð+3)
â ð3 = âð0
23 ð+3 ð+2 ð+1 2ð+1 2ð+3 (2ð+5) and so on. ⊠(9)
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Thus, for ð = 0, we get
ðŠ(ð=0) = ðŠ1 = ð¥ð (ð0 + ð1ð¥ + ð2ð¥2 + âŠâŠ ) ð=0
= ð0 1 â1
2
ð¥
1.1+
1
22
ð¥2
2.1.1.3â
1
23
ð¥3
3.2.1.1.3.5+ âŠâŠ
= ð0 1 â ð¥
2
2!+
ð¥ 4
4!â
ð¥ 6
6!+ âŠâŠ = ð0 cos ð¥ ⊠(10)
Likewise for ð =1
2, we get
ðŠ(ð=
1
2)
= ðŠ2 = ð0ð¥1
2 1 â1
21
ð¥3
2.2
+1
22
ð¥2
5
2.3
2.2.4
â1
23
ð¥3
7
2.5
2.3
2.2.4.6
+ âŠâŠ
= ð0 ð¥ â ð¥
3
3!+
ð¥ 5
5!â
ð¥ 7
7!+ âŠâŠ = ð0 sin ð¥ ⊠(11)
Hence, on substituting the values of ðŠ1 and ðŠ2in equation (3), we get solution as:
ðŠ = ð1ðŠ1 + ð2ðŠ2 = ð¶1 cos ð¥ + ð¶2 sin ð¥ .
Example 4: Find the series solution of the equation
ðððð ðð
ð ððâ ð
ð ð
ð ð+ ð â ðð ð = ð
OR
Solve the equation ðððð ðð
ð ððâ ð
ð ð
ð ð+ ð â ðð ð = ð in power series.
Solution: Given 2ð¥2 ð2ðŠ
ðð¥2 â ð¥ððŠ
ðð¥+ 1 â ð¥2 ðŠ = 0 ⊠(1)
Let its solution be
ðŠ = ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + ð3ð¥
ð+3 + âŠâŠ ⊠(2)
So that ððŠ
ðð¥= ðð0ð¥
ðâ1 + ð + 1 ð1ð¥ð + ð + 2 ð2ð¥
ð+1
+ ð + 3 ð3ð¥ð+2 + âŠâŠ ⊠(3)
And ð2ðŠ
ðð¥2 = ð ð â 1 ð0ð¥ðâ2 + ð + 1 ð ð1ð¥
ðâ1
+ ð + 2 ð + 1 ð2ð¥ð + âŠâŠ ⊠(4)
On substituting the values of ðŠ, ððŠ
ðð¥,
ð2ðŠ
ðð¥2 in the given equation, we get
2ð¥2 ð ð â 1 ð0ð¥ðâ2 + ð + 1 ð ð1ð¥
ðâ1 + ð + 2 ð + 1 ð2ð¥ð + âŠâŠ
âð¥ ðð0ð¥ðâ1 + ð + 1 ð1ð¥
ð + ð + 2 ð2ð¥ð+1 + ð + 3 ð3ð¥
ð+2 + âŠâŠ
+ 1 â ð¥2 ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + ð3ð¥
ð+3 + âŠâŠ = 0
i.e. 2ð ð â 1 ð0ð¥ð + 2 ð + 1 ðð1ð¥
ð+1 + 2 ð + 2 ð + 1 ð2ð¥ð+1 + âŠâŠ
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â ðð0ð¥ð + ð + 1 ð1ð¥
ð+1 + ð + 2 ð2ð¥ð+2 + âŠâŠ
+ ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + âŠâŠ â ð0ð¥
ð+2 + ð1ð¥ð+3 + âŠâŠ ⊠(5)
On equating the coefficients of lowest power of ð¥ (i.e. ð¥ð ) equal to zero on both sides,
2ð ð â 1 ð0 âðð0 + ð0 = 0 ⊠(6)
â ð0 2ðâ 1 ð â 1 = 0
â Either ð0 = 0 or ð = 1,1
2 .
Now equating the coefficients of ð¥ð+1 equal to zero,
2 ð + 1 ð ð1 â ð + 1 ð1 + ð1 = 0
â ð1ð 2ðâ 1 = 0
Which implies either ð1 = 0 or ð = 0, but ð â 0,
⎠ð1 = 0 ⊠(7)
On comparing the coefficients of ð¥ð+2,
2 ð + 2 ð + 1 ð2 â ð + 2 ð2 + ð2 â ð0 = 0
â 2ð2 + 6ð + 4 â ð + 1 ð2 = ð0
â 2ð2 + 5ð + 3 ð2 = ð0
â ð2 = ð0
ð+1 2ð+3 . ⊠(8)
Likewise, on comparing the coefficients of ð¥ð+3,
2 ð + 3 ð + 2 ð3 â ð + 3 ð3 + ð3 â ð1 = 0
â 2 ð + 3 ð + 2 ð3 â ð + 3 + 1 ð3 = ð1
â ð3 = 0 (since ð1 = 0) ⊠(9)
Further, coefficients of ð¥ð+4,
2 ð + 4 ð + 3 ð4 â ð + 4 ð4 + ð4 â ð2 = 0
â 2 ð + 4 ð + 3 â ð + 4 + ð2 = ð2
2ð2 + 13ð + 21 ð4 = ð2
â ð4 = ð2
ð+3 2ð+7 and so on ⊠⊠(10)
Now for ð = 1, ð2 = ð0
1+1 2.1+3 =
ð0
2.5 from (8) ⊠(11)
ð4 = ð2
4.9=
ð0
2.5.4.9 from (10) ⊠(12)
âŠâŠ âŠâŠ âŠâŠ âŠâŠ âŠâŠ
For ð =1
2 , ð2 =
ð0
1
2+1 2.
1
2+3
=ð0
3
2 . 4
=ð0
2.3 ⊠(13)
ð4 =ð2
ð+1 2ð+7 =
ð0
2.3
1
1
2+3
2
2+7
=ð0
2.3.4.7 ⊠(14)
âŠâŠ âŠâŠ âŠâŠ âŠâŠ âŠâŠ
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Thus ðŠ1 = ðŠ ð=1 = ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + ð3ð¥
ð+3 + âŠâŠ
= ð0ð¥ 1 +ð¥2
2.5+
ð¥4
2.4.5.9+
ð¥6
2.4.5.9.6.13+ âŠâŠ
ðŠ2 = ðŠ ð=
1
2
= ð0ð¥1
2 1 +ð¥2
2.3+
ð¥4
2.3.4.7+
ð¥6
2.3.4.5.7.11+ âŠâŠ
Hence ðŠ = ð¶1ðŠ1 + ð¶2ðŠ2.
Case II: Roots ðð, ðð are equal, i.e. ðð = ðð.
In this case, one of the linearly independent solutions ðŠ1 is obtained by substituting ð = ð1 and
the second solution is obtained as
ðŠ2 = ððŠ
ðð ð=ð1
.
Thus the complete solution is given by
ðŠ = ð1ðŠ1 + ð2 ððŠ
ðð ð=ð1
.
Example 5: Solve ðð ðð
ð ðð+
ð ð
ð ðâ ð = ð.
Solution: Given ð¥ð2ðŠ
ðð¥2 +ððŠ
ðð¥â ðŠ = 0 ⊠(1)
Let its solution be ðŠ = ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + âŠâŠ ⊠(2)
⎠ððŠ
ðð¥= ðð0ð¥
ðâ1 + ð + 1 ð1ð¥ð + ð + 2 ð2ð¥
ð+1 + âŠâŠ ⊠(3)
â ð2ðŠ
ðð¥2 = ð ð â 1 ð0ð¥ðâ2 + ð + 1 ð ð1ð¥
ðâ1
+ ð + 2 ð + 1 ð¥ð + âŠâŠ ⊠(4)
Putting the above values in equation (1), we have
ð¥ ð ð â 1 ð0ð¥ðâ2 + ð + 1 ð ð1ð¥
ðâ1 + ð + 2 ð + 1 ð2ð¥ð + âŠâŠ
+ ðð0ð¥ðâ1 + ð + 1 ð1ð¥
ð + ð + 2 ð2ð¥ð+1 + âŠâŠ
â ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + âŠâŠ ⊠(5)
Equating the coefficients of ð¥ðâ1 to zero,
ð ð0 + ð(ð â 1)ð0 = 0
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â Either ð0 = 0 or ð2 = 0
But ð0 â 0 ⎠ð = 0, 0.
Now equate the coefficients of ð¥ð on both sides,
ð + 1 ð1 + ð ð + 1 ð1 + ð0 = 0 or (ð + 1)2ð1 + ð0 = 0
â ð1 = âð0
(ð+1)2. ⊠(6)
Next equate the coefficients of ð¥ð+1 on both sides,
ð + 2 ð + 1 ð2 + ð + 2 ð2 â ð1 = 0
â ð + 2 ð2 ð + 1 + 1 â ð1 = 0 or ð + 2 2ð2 â ð1 = 0
â ð2 = ð1
ð+2 2 = ð0
ð+1 2 ð+2 2 and so on. ... (7)
Putting the values of ð1, ð2, âŠ.in the assumed series solution (2),
ðŠ = ð0ð¥ð 1 +
ð¥
ð+1 2 +ð¥2
ð+1 2 ð+2 2 +ð¥3
ð+1 2 ð+2 2 ð+3 2 + âŠâŠ ⊠(8)
Differentiating (8) partially with respect to ð
ððŠ
ðð = ð0ð¥
ð logð¥ 1 +ð¥
(ð+1)2 +ð¥2
ð+1 2 ð+2 2 + âŠâŠ
+ð0ð¥ð 0 â
2ð¥
ð+1 3 â2ð¥2
ð+1 2 ð+2 2 2ð+3
ð+1 (ð+2) + âŠâŠ
= ð0ð¥ð logð¥ 1 +
ð¥
ð+1 2 +ð¥2
ð+1 2 ð+2 2 +ð¥3
ð+1 2 ð+2 2 ð+3 2 + âŠâŠ
â2ð0ð¥ð
ð¥
ð+1 2 ð+1 +
ð¥2
ð+1 2 ð+2 2 1
ð+1 +
1
ð+2 +
ð¥3ð+12ð+22ð+32+ âŠâŠ ⊠(9)
Now ðŠ1 = ðŠ(ð=0) = ð0ð¥ 1 +ð¥
12 +ð¥2
12 .22 + âŠâŠ ⊠(10)
ðŠ2 = ððŠ
ðð ð=0
= ðŠ1 logð¥
â2ð0 ð¥
1!+
1
2!2 1 +1
2 ð¥2 +
1
3!2 1 +1
2+
1
3 ð¥3 + âŠâŠ ⊠(11)
Therefore, the complete solution is
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ðŠ = ð¶1 + ð¶2 logð¥ 1 +ð¥
1!2 +ð¥2
2!2 +ð¥3
2!3 + âŠâŠ
â2ð¶2 ð¥ +1
2!2 1 +1
2 ð¥2 +
1
3!2 1 +1
2+
1
3 ð¥3 + âŠâŠ .
Case III: Roots ðð, ðð are distinct and differ by an integer.
In this case, assume that ð1 < ð2 . If some of the coefficient of y series becomes infinite when
ð = ð1, we modify the form of y replacing ð0 by ð0 ð âð1 . Then the complete solution is given
by
ðŠ = ð1(ðŠ)ð2+ ð1
ððŠ
ðð ð1
Example 5: Solve the equation ð ð â ð ð ðð
ð ððâ ð
ð ð
ð ð+ ðð = ð
Solution: Given ð¥ 1 â ð¥ ððŠ
ðð¥â 3
ððŠ
ðð¥+ 2ðŠ = 0 ⊠(1)
Let its solution be ðŠ = ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + âŠâŠ ⊠(2)
⎠ððŠ
ðð¥= ðð0ð¥
ðâ1 + ð + 1 ð1ð¥ð + ð + 2 ð2ð¥
ð+1 + âŠâŠ ⊠(3)
and ð2ðŠ
ðð¥2 = ð ð â 1 ð0ð¥ðâ2 + ð + 1 ð ð1ð¥
ðâ1
+ ð + 2 ð + 1 ð2ð¥ð + âŠâŠ ⊠(4)
On substituting these values of ðŠ, ððŠ
ðð¥,
ð2ðŠ
ðð¥2 in the given differential equation,
ð¥ â ð¥2 ð ð â 1 ð0ð¥ðâ2 + ð + 1 ð ð1ð¥
ðâ1 + ð + 2 ð + 1 ð2ð¥ð + âŠâŠ
â3 ðð0ð¥ðâ1 + ð + 1 ð1ð¥
ð + ð + 2 ð2ð¥ð+1 + âŠâŠ +2 ð0ð¥
ð + ð1ð¥ð+1 + ð2ð¥
ð+2 +
âŠâŠ=0 ⊠(5)
On equating the coefficients of lowest power of ð¥ (i.e. ð¥ðâ1) on both sides,
ð0ð ð â 1 â 3ð0 = 0 or ð0 ð ð â 4 = 0
â Either ð0 = 0 or ð ð â 4 = 0
But as ð0 â 0 ⎠ð = 0, 4
Likewise, equate the coefficients of ð¥ð , ð¥ð+1, ð¥ð+2 equal to zero, and find out the values of
unknowns ð0, ð1, ð2 etc.
For the coefficients of ð¥ð ,
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11
âð ð â 1 ð0 â 3 ð + 1 ð1 + ð + 1 ðð1 + 2ð0 = 0
â ð â 3 ð + 1 ð1 = ð â 2 ð + 1 ð0
â ð1 = ðâ2
ðâ3 ð0 ⊠(6)
For the coefficient of ð¥ð+1,
â ð + 1 ðð1 + ð + 2 ð + 1 ð2 â 3 ð + 2 ð2 + 2ð1 = 0
â ð + 2 ð â 2 ð2 = ð â 1 ð + 2 ð1
â ð2 = ðâ1
ðâ2 ð1 =
ðâ1
ðâ3 ð0 ⊠(7)
Similarly,
ð3 =ð
ðâ1 ð2 =
ð
ðâ1
ðâ1
ðâ3 ð0 =
ð
ðâ3 ð0
ð4 = ð+1
ð ð3 =
ð+1
ð
ð
ðâ3 ð0 =
ð+1
ðâ3 ð0
ð5 = ð+2
ð+1 ð4 =
ð+2
ð+1
ð+1
ðâ3 ð0 =
ð+2
ðâ3 ð0 ⊠ð ð ðð
⊠(8)
⎠ðŠ = ð0ð¥ð 1 +
ðâ2
ðâ3 ð¥ +
ðâ1
ðâ3 ð¥2 +
ð
ðâ3 ð¥3 +
ð+1
ðâ3 ð¥4 + âŠâŠ ⊠(9)
Now, ðŠ1 = ðŠ ð=0 = ð0 1 +2
3ð¥ +
1
3ð¥2 â
1
3ð¥4 â âŠâŠ
and ðŠ2 = ðŠ ð=4 = ð0ð¥4 1 +
2
1ð¥ +
3
1ð¥2 +
4
1ð¥3 +
5
1ð¥4 + âŠâŠ
Hence the complete solution, ðŠ = ð1ðŠ1 + ð2ðŠ2.
ASSIGNMENT 18.2
Use Frobenius method to solve the following differential equations:
1. 9ð¥ 1 â ð¥ ð2ðŠ
ðð¥2 â 12ððŠ
ðð¥+ 4ðŠ = 0
2. 4ð¥ð2ðŠ
ðð¥2 + 2 1 â ð¥ ððŠ
ðð¥â ðŠ = 0
3. ð¥ð2ðŠ
ðð¥2 +ððŠ
ðð¥+ ð¥ðŠ = 0
4. ð¥ 1 â ð¥ ð2ðŠ
ðð¥2 â 1 + 3ð¥ ððŠ
ðð¥â ðŠ = 0
5. ð¥ð2ðŠ
ðð¥2 + 2ððŠ
ðð¥+ ð¥ðŠ = 0
6. 2ð¥2ðŠ â²â² + ð¥ðŠ â² â ð¥ + 1 ðŠ = 0
7. 2ð¥ 1 â ð¥ ð2ðŠ
ðð¥2 + 1 â ð¥ ððŠ
ðð¥+ 3ðŠ = 0
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ANSWERS
1. ðŠ = ð¶1 1 +1
3ð¥ +
1.4
3.6ð¥2 +
1.4.7
3.6.9ð¥3 + âŠâŠ
+ð¶2ð¥7/3 1 +
8
10ð¥ +
8.11
10.13ð¥2 +
8.11.14
10.13.16ð¥3 + âŠâŠ
2. ðŠ = ð¶1 1 +1
2.1!ð¥ +
1
22 . 2!ð¥2 +
1
23 . 3!ð¥3 + âŠâŠ
+ð¶2ð¥1
2 1 +1
1 .3ð¥ +
1
1.3.5ð¥2 +
1
1.3.5.7ð¥3 + âŠâŠ
3. ðŠ = (ð¶1 + ð¶2 logð¥) 1 â1
22 ð¥2 +
1
22 .42 ð¥4 â
1
22 .42 .62 ð¥6 + âŠâŠ
+ð¶2 1
22 ð¥2 â
1
22 .42 1 +1
2 ð¥4 +
1
22 .42 .62 1 +1
2+
1
3 ð¥6 + âŠâŠ
4. ðŠ = (ð¶1 + ð¶2 logð¥) 1.2ð¥2 + 2.3ð¥3 + 3.4ð¥4 + âŠâŠ
+ð¶2 â1 + ð¥ + 5ð¥2 + 11ð¥3 + âŠâŠ
5. ðŠ = ð¥â1(ð0 cosð¥ + ð1 sinð¥)
6. ðŠ = ð0ð¥ 1 +ð¥
5+
ð¥2
70+ ⯠+
ð1
ð¥ 1 â ð¥ â
ð¥2
2+ â¯
7. ðŠ = ð0 ð¥ 1 â ð¥ + ð1 1 â 3ð¥ +3ð¥2
1.3+
3ð¥3
3.5+
3ð¥4
5.7+ â¯
18.4 BESSELâS EQUATION
In applied mathematics, many physical problems involving vibrations or heat conduction in
cylindrical regions give rise the differential equation
ð¥2 ð2ðŠ
ðð¥2 + ð¥ððŠ
ðð¥+ ð¥2 â ð2 ðŠ = 0 ⊠(1)
which is known as the Besselâs differential equation of order n. The particular solutions of this
differential equation are called Besselâs functions of order n.
Let ðŠ = ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + âŠâŠ
⎠ððŠ
ðð¥= ðð0ð¥
ðâ1 + ð + 1 ð1ð¥ð + ð + 2 ð2ð¥
ð+1 + âŠâŠ
ð2ðŠ
ðð¥2 = ð ð â 1 ð0ð¥ðâ1 + ð + 1 ð ð1ð¥
ð
+ ð + 2 ð + 1 ð2ð¥ð + âŠâŠ
Putting these in the given differential equation, we get
âð¥2 ð ð â 1 ð0ð¥ðâ1 + ð + 1 ð ð1ð¥
ð + ð + 2 ð + 1 ð2ð¥ð + âŠâŠ + ð¥ ðð0ð¥
ðâ1 +
ð+1ð1ð¥ð+ð+2ð2ð¥ð+1+ âŠâŠ+[ð¥2âð2ð0ð¥ð+ð1ð¥ð+1+ð2ð¥ð+2+ âŠâŠ=0
Equating to zero, the coefficient of lowest degree term in ð¥, i.e. ð¥ð
ð ð â 1 ð0 + ðð0 â ð2ð0 = 0, ð0 â 0
⎠Indicial equation ð ð â 1 + ð â ð2 = 0
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13
â ð2 â ð2 = 0 ⎠ð = ±ð
Now coefficients of ð¥ð+1:
ð + 1 ðð1 + ð + 1 ð1 â ð2ð1 = 0
ð + 1 2 â ð2 ð1 = 0
â ð is fixed ð1 = 0
Coefficient of ð¥ð+2: ð + 2 ð + 1 ð2 + ð + 2 ð2 â ð2ð2 + ð0 = 0
ð + 2 2 â ð2 ð2 + ð0 = 0
⎠ð2 = âð0
ð+2 2âð2
Similarly, ð + 3 2 â ð2 ð3 + ð1 = 0
ð3 = âð1
ð+3 2âð2 = 0, as ð1 = 0
So ð1 = ð3 = ð5 = âŠâŠ = 0
ð4 = âð2
ð+4 2âð2 =ð0
ð+2 2âð2 ð+4 2âð2
So ðŠ = ð0ð¥ð 1 â
ð¥2
ð+2 2âð2 +ð¥4
ð+2 2âð2 ð+4 2âð2 â ⊠(2)
Case 1: For ð = 0, ð = 0 as ð = ±ð
ðŠðŒ = ð0 1 âð¥2
22 +ð¥4
22 .42 â âŠâŠ
ððŠ
ðð= ðŠ logð¥ + ð0ð¥
ð âð¥2
ð+2 2âð2 â2
ð+2 2âð2 +
ð¥4
ð+2 2âð2 ð+4 2âð2
â2
ð+2 2âð2 ââ2
ð+4 2âð2 + âŠâŠ
ðŠðŒðŒ = ððŠ
ðð ð=0
= ðŠðŒ logð¥ + ð0 âð¥2
2.â2
22 +ð¥4
22 .42 â2
22 ââ2
42 + âŠâŠ
So, the solution is ðŠ = ð¶1ðŠðŒ + ð¶2ðŠðŒðŒ
ðŠ = ð¶1ð0 + ð¶2 log ð¥ 1 âð¥2
22 +ð¥4
22 .42 â âŠâŠ
+ð¶2ð0 ð¥2
2.
2
22 â2ð¥4
22 .42 1
22 +2
42 + âŠâŠ
Case 2: For ð non integral and equal to ð (ð = ð) replace ð0 in equation (2) by 1
2ð ð+1
We get ðŠ0 =1
2ð ð+1ð¥ð 1 â
ð¥2
22 ð+1 +
ð¥4
22 ð+1 4.2 ð+2 + âŠâŠ
= ð¥
2 ð
1
ð+1â
1
ð+2 ð¥
2
2+
1
2! ð+3 ð¥
2
4+ âŠâŠ
= â1 ð1
ð ! ð+ð+1 ð¥
2 ð+2ð
= ðœð ð¥ âð=0
i.e. ðœð ð¥ = â1 ð1
ð ! ð+ð+1 ð¥
2 ð+2ð
âð=0 ⊠(3)
Similarly by putting ð = âð, we get the other solution
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ðœâð ð¥ = â1 ð1
ð! âð+ð+1 ð¥
2 âð+2ð
âð=0
The resulting solution is
ðŠ = ð¶1 ðœð ð¥ + ð¶2 ðœâð ð¥
ðœð ð¥ & ðœâð ð¥ as defined as above.
Case 3: If ð is integral
Let ðŠ = ð¢ ð¥ ðœð ð¥ ðŠâ² = ð¢â² ð¥ ðœð + ð¢ ðœðâ²
ðŠâ²â² = ð¢â²â² ðœð + 2ð¢â² ðœðâ² + ð¢ ðœð
â²â²
Putting these in ð¥2 ð2ðŠ
ðð¥2 + ð¥ððŠ
ðð¥+ ð¥2 â ð2 ðŠ = 0
â ð¥2 ð¢â²â² ðœð + 2ð¢â² ðœðâ² + ð¢ ðœð
â²â² + ð¥ ð¢â² ðœð + ð¢ ðœðâ² + ð¥2 â ð2 ð¢ ðœð = 0
â ð¢ ð¥2ðœðâ²â² + ð¥ðœð
â² + ð¥2 â ð2 ðœð + 2ð¢â²ð¥2ðœðâ² + ð¥2ð¢â²â² ðœð + ð¥ð¢â² ðœð = 0
Now ðœð ð¥ is a solution of ð¥2 ð2ðŠ
ðð¥2 + ð¥ððŠ
ðð¥+ ð¥2 â ð2 ðŠ = 0
⎠ð¥2ðœðâ²â² + ð¥ðœð
â² + ð¥2 â ð2 ðœð = 0
We get,
2ð¢â²ð¥2ðœðâ² + ð¥2ð¢â²â² ðœð + ð¥ð¢â² ðœð = 0
â 2ðœðâ²
ðœð+
ð¢ â²â²
ð¢ â² +1
ð¥= 0 (divide by ðœðð¢
â²ð¥2)
Integrating 2 logð ðœð + logð ð¢â² logð ð¥ = log ð
â ð¢â²ðœð2ð¥ = ðµ Where ðµ is constant of integration.
â ð¢â² = ðµ1
ð¥ðœð2
Integrating ð¢ = ðŽ + ðµ ðð¥
ð¥ ðœð ð¥ 2
So the solution of ðŠ in this case
ðŠ = ð¢ ð¥ ðœð ð¥
= ðŽðœð ð¥ + ðµðœð ð¥ ðð¥
ð¥ ðœð ð¥ 2
= ðŽðœð ð¥ + ðµðŠð ð¥
where ðŠð ð¥ = ðœð ð¥ ðð¥
ð¥ ðœð ð¥ 2 is the Besselâs function of the second kind and ðœð ð¥ is Besselâs
function of first kind.
18.5 RECURRENCE FORMULAE FOR ð±ð ð
The following relations are the recurrence formulae for Besselâs functions and are very useful in the
solution of Boundary value problems and in establishing various properties of Besselâs functions:
1. ð
ðð¥ ð¥ð ðœð(ð¥) = ð¥ð ðœðâ1(ð¥)
2. ð
ðð¥ ð¥âð ðœð(ð¥) = âð¥âð ðœð+1(ð¥)
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3. ðœð ð¥ =ð¥
2ð ðœðâ1 ð¥ + ðœð+1(ð¥)
4. ðœðâ² ð¥ =
1
2 ðœðâ1 ð¥ â ðœð+1(ð¥)
5. ðœðâ² ð¥ =
ð
ð¥ ðœð ð¥ â ðœð+1(ð¥)
6. ðœð+1 ð¥ =2ð
ð¥ ðœð ð¥ â ðœðâ1(ð¥)
18.6 EXPANSION FOR ð±ð AND ð±ð
We know that ðœð ð¥ = â1 ð1
ð! ð+ð+1 ð¥
2 ð+2ð
âð=0
Taking ð = 0 and 1 in above Besselâs function, we get
ðœ0 ð¥ = 1 â1
1! ð¥
2
2+
1
(2!)2 ð¥
2
4â
1
(3!)2 ð¥
2
6+ â¯
and ðœ1 ð¥ =ð¥
2 1 â
1
1! 2! ð¥
2
2+
1
2! 3! ð¥
2
4â
1
3! 4! ð¥
2
6+ â¯
18.7 VALUE OF ð±ðð
(ð)
In Besselâs functions, the function ðœ1/2 is the simplest one, as it can be expressed in finite form.
Taking ð = 1/2 in the value of ðœð ð¥ , we get
ðœ1/2 ð¥ = ð¥
2
1/2
1
Î 3
2 â
1
1! Î 5
2 ð¥
2
2+
1
2! Î 7
2 ð¥
2
4ââ¯
= ð¥
2
1/2
11
2Î
1
2 â
13
2â1
2 Î
1
2 ð¥
2
2+
1
2â5
2â3
2â1
2 Î
1
2 ð¥
2
4ââ¯
= ð¥
2Î 1
2
2
1!â
2 ð¥2
3!+
2 ð¥4
5!ââ¯
Now multiplying the series by ð¥
2 and outside by
2
ð¥ , we get
ðœ1/2 ð¥ = 2
ð¥ ð ð¥
1!â
ð¥3
3!+
ð¥5
5!â⯠=
2
ðð¥sinð¥
Similarly, taking ð = 1/2 in the value of ðœâð ð¥ , we get
ðœâ1/2 ð¥ = 2
ðð¥cosð¥
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18.8 GENERATING FUNCTION FOR ð±ð(ð)
To prove that ðð
ðð(ðâ
ð
ð) = ðð ð±ð(ð)â
ð=ââ .
We have ð1
2ð¥(ð¡â
1
ð¡) = ð
ð¥ð¡
2 à ðâð¥
2ð¡
= 1 +ð¥ð¡
2+
1
2 ! ð¥ð¡
2
2+
1
3 ! ð¥ð¡
2
3+ ⯠+ 1 â
ð¥
2ð¡+
1
2 ! ð¥
2ð¡
2â
1
3 ! ð¥
2ð¡
3+ â¯
The coefficient of ð¡ð in this product is
1
ð ! ð¥
2 ðâ
1
(ð+1) ! ð¥
2 ð+2
+1
2 ! (ð+1) ! ð¥
2 ð+4
â⯠= ðœð(ð¥)
As all the integral powers of t, both positive and negative occurs, we have
ð1
2ð¥ ð¡â
1
ð¡
= ðœ0 ð¥ + ð¡ðœ1 ð¥ + ð¡2ðœ2 ð¥ + ð¡3ðœ3 ð¥ + â¯
+ð¡â1ðœâ1 ð¥ + ð¡â2ðœâ2 ð¥ + ð¡â3ðœâ3 ð¥ + â¯
= ð¡ð ðœð(ð¥)âð=ââ
Thus the coefficients of different powers of t in the expansion of ð1
2ð¥ ð¡â
1
ð¡ give Besselâs functions of
various orders. Hence it is known as the generating function of Besselâs functions.
Example 6: Evaluate ðâððâ
ðð±ð ðð ð ð
Solution: We know that ðœð ð¥ =1
ð ððð ðð â ð¥ sinð ð
0ðð
For ð = 0,
ðœ0 ð¥ =1
ð cos ð¥ cosð ððð
0ðð
ðœ0 ð¥ =1
ð cos ð¥ sinð ððð
0
â ðœ0 ðð¥ =1
ð cos ðð¥ sinð ðð =
2
ð cos ðð¥ sinð ðð
ð
20
ð
0
So, ðâðð¥â
0ðœ0 ðð¥ ðð¥ = ðâðð¥
2
ð cos ðð¥ sinð ðð
ð
20
â
0ðð¥
=2
ð
ð âð+ð ð sin ð ð¥+ð âðâð ð sin ð ð¥
2
â
0
ð
20
dx dy
=1
ð
ð ð ð sin ðâð ð¥
ð ð sin ðâð +
ðâ ð ð sin ð+ð ð¥
â ð ð sin ð+ð
0
âð
20
ðð
=1
ð
â1
ð ð sin ðâð +
1
ð ð sin ð+ð
ð
20
ðð
=1
2ð
â2ð
ð2ð2ð ðð 2ðâð2 ððð
20
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=1
2ð
2ð
ð2+ð2ð ðð 2ððð
ð
20
=2ð
ð
ð ðð 2ð
ð2ð ðð 2ð+ð2ð¡ðð 2ððð
ð
20
=2ð
ð
ð ðð 2ð
ð2+ð2 ð¡ðð 2ð+ð2 ððð
20
Now take ð2 + ð2 tanð = ð¡
⎠ð2 + ð2 ð ðð2ð ðð = ðð¡
Further, if ð = 0 ðð¡ ððððððð ð¡ = 0
ð =ð
2 ðð¡ ððððððð ð¡ = â
⎠ðâðð¥ ðœ0 ðð¥ ðð¥ =2ð
ð
1
ð¡2+ð2
ðð¡
ð2+ð2
â
0
=2ð
ð ð2+ð2
1
ð2+ð¡2
â
0ðð¡
=2ð
ð ð2+ð2
1
ð tanâ1 ð¡
ð
0
â
=2
ð ð2+ð2 tanâ1
â
ð â tanâ1 0
=2
ð ð2+ð2 ð
2â 0 =
1
ð2+ð2
Example 7: Show that ð±ð ð = ðð
ððâ
ð
ð ð±ð ð + ð â
ðð
ðð ð±ð ð
Solution: We know
ðœð+1 ð¥ =2ð
ð¥ðœð ð¥ â ðœðâ1 ð¥
⎠for ð = 3
ðœ4 ð¥ =6
ð¥ðœ3 ð¥ â ðœ2 ð¥ (1)
For ð = 2
ðœ3 ð¥ =4
ð¥ðœ2 ð¥ â ðœ1 ð¥ (2)
For ð = 1
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ðœ2 ð¥ =2
ð¥ðœ1 ð¥ â ðœ0 ð¥ (3)
Substituting ðœ2 ð¥ from (3) in (2), we get
ðœ3 ð¥ =4
ð¥
2
ð¥ðœ1 ð¥ â ðœ0 ð¥ â ðœ1 ð¥
= 8
ð¥2 â 1 ðœ1 ð¥ â4
ð¥ðœ0 ð¥ (4)
Now substituting for ðœ2 ð¥ and ðœ3 ð¥ in (1), we will have
ðœ4 ð¥ =6
ð¥
8
ð¥2 â 1 ðœ1 ð¥ â4
ð¥ðœ0 ð¥ â
2
ð¥ðœ1 ð¥ â ðœ0 ð¥
= 48
ð¥3 â8
ð¥ ðœ1 ð¥ + 1 â
24
ð¥2 ðœ0 ð¥
Example 8: Show that
(i) ð±âð
ð
ð = ð±ðð
ð ððšð ð
(ii) ð±âð
ð
ð = â ð
ð ð ð¬ð¢ð§ ð +
ððšð¬ ð
ð
(iii) ð±âð
ð
ð = ð
ð ð ð
ðð¬ð¢ð§ð +
ðâðð
ððððšð¬ ð
Solution: (i) We know
ðœâ
1
2
ð¥ = 2
ðð¥ cosð¥ and ðœ1
2
ð¥ = 2
ðð¥ sinð¥
⎠ðœâ
12
ð¥
ðœ12
ð¥ =
2
ðð¥ cos ð¥
2
ðð¥ sin ð¥
= cotð¥
Hence ðœâ
1
2
ð¥ = ðœ1
2
ð¥ cotð¥
(ii) We know
ðœðâ1 ð¥ =2ð
ð¥ðœð ð¥ â ðœð+1 ð¥
⎠For ð = â1
2
ðœâ
3
2
ð¥ = â1
ð¥ðœâ
1
2
ð¥ â ðœ1
2
ð¥
(iii) We know
ðœðâ1 ð¥ =2ð
ð¥ðœð ð¥ â ðœð+1 ð¥
⎠For ð = â1
2
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ðœâ
3
2
ð¥ = â1
ð¥ðœâ
1
2
ð¥ â ðœ1
2
ð¥
= â1
ð¥
2
ðð¥cosð¥ â
2
ðð¥sinð¥
ðœâ
3
2
ð¥ = â 2
ðð¥
cos ð¥
ð¥+ sinð¥
(iv) We know
ðœðâ1 ð¥ =2ð
ð¥ðœð ð¥ â ðœð+1 ð¥ , ðœâ1
2
ð¥ = 2
ðð¥cosð¥, ðœ1
2
ð¥ =
2
ðð¥sinð¥
⎠for ð = â3
2
ðœâ
5
2
ð¥ = â3
ð¥ðœâ
3
2
ð¥ â ðœâ
1
2
ð¥
(1)
and for ð = â1
2
ðœâ
3
2
ð¥ = â1
ð¥ðœâ
1
2
ð¥ â ðœ1
2
ð¥ = â 2
ðð¥
cos ð¥
ð¥+ sinð¥
(2)
⎠from (1) and (2), we will have
ðœâ
5
2
ð¥ = â3
ð¥Ã â
2
ðð¥
cos ð¥
ð¥+ sinð¥ â
2
ðð¥ cosð¥
= 2
ðð¥
3
ð¥2 â 1 cosð¥ +3
ð¥sinð¥
Example 9: Prove that
(i) ð
ð ð ð±ð ð = âð±ð ð ,
(ii) ð
ð ð ð ð±ð ð = ðð±ð ð
(iii) ð
ð ð ððð±ð ðð = ð ððð±ðâð ð
(iv) ð
ð ð ðâðð±ð ð = âðâðð±ð+ð ð
Solutions: (i) We know that ð
ðð¥ ð¥âððœð ð¥ = âð¥âððœð+1 ð¥
For ð = 0, we will have
ð
ðð¥ ð¥0ðœ0 ð¥ = âð¥0ðœ1 ð¥
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ð
ðð¥ ðœ0 ð¥ = âðœ1 ð¥
(ii) We know
ð
ðð¥ ð¥ððœð ð¥ = ð¥ððœðâ1 ð¥
For ð = 1, it will give
ð
ðð¥ ð¥ ðœ1 ð¥ = ð¥ðœ0 ð¥
(iii) To prove ð
ðð¥ ð¥ððœð ðð¥ = ð ð¥ððœðâ1 ð¥
Let ðð¥ = ð¡ or ð¥ =ð¡
ð
⎠ð¥ððœð ðð¥ = ð¡
ð ððœð ð¡
Differentiating with respect to â²ð¥â², we get
ð
ðð¥ ð¥ððœð ðð¥ =
ð
ðð¡
ð¡
ð ððœð ð¡ .
ðð¡
ðð¥
=1
ðð .ð
ðð¡ ð¡ððœð ð¡ .ð,
=1
ððâ1 . ð¡ððœðâ1 ð¡ ,
=1
ððâ1 . ðð¥ ððœðâ1 ðð¥ ,
= ðð¥ððœðâ1 ðð¥
(iv) To prove
ð
ðð¥ ð¥âððœð ð¥ = âð¥âððœð+1 ð¥
We know
ðœð ð¥ = â1 ð1
ð ! ð+ð+1 ð¥
2 ð+2ð
âð=0
ð¥âððœð ð¥ = â1 ð1
ð! ð+ð+1 .
1
2ð+2ð . ð¥2ðâð=0
ð
ðð¥ ð¥âððœð ð¥ = â1 ð
1
ð! ð+ð+1
âð=1
1
2ð+2ð . 2ð ð¥2ðâ1
= âð¥âð â1 ðâ1 1
ðâ1 ! ð+1+ ðâ1 +1
âð=1 .
ð¥ð+1+2ð
2ðâ1+2ð
Taking ð â 1 = ð
= âð¥âð â1 ð .1
ð! ð+1+ð+1 .
ð¥
2 ð+1+2ð
âð=0
= âð¥âððœð+1 ð¥ .
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Example 10: Show by the use of recurrence formula, that
(i) ð±ðâ²â² ð =
ð
ð ð±ð ð â ð±ð ð
(ii) ð±ðâ²â² ð = âð±ð ð +
ð
ðð±ð ð
Solutions: (i) We know
ð
ðð¥ ð¥âððœð ð¥ = âð¥âððœð+1 ð¥
⎠for ð = 0
ð
ðð¥ ðœ0 ð¥ = âðœ1 ð¥
Differentiating with respect to â²ð¥â², we will have
ð2
ðð¥2 ðœ0 ð¥ = â
ð
ðð¥ ðœ1 ð¥
ðœ0â²â² ð¥ = âðœ1
â² ð¥
But ðœðâ² ð¥ =
1
2 ðœðâ1 ð¥ â ðœð+1 ð¥
⎠for ð = 1
ðœ1â² ð¥ =
1
2 ðœ0 ð¥ â ðœ2 ð¥
⎠ðœ0â²â² ð¥ = â
1
2 ðœ0 ð¥ â ðœ2 ð¥
=1
2 ðœ2 ð¥ â ðœ0 ð¥
(ii) We know
ðœðâ² ð¥ =
1
2 ðœðâ1 ð¥ â ðœð+1 ð¥
⎠ðœ1â² ð¥ =
1
2 ðœ0 ð¥ â ðœ2 ð¥
Differentiating with respect to â²ð¥â², we get
ðœ1â²â² ð¥ =
1
2 ðœ0
â² ð¥ â ðœ2â² ð¥
But ðœ0â² ð¥ = âðœ1 ð¥ and also
ðœðâ² ð¥ = ðœðâ1 ð¥ â
ð
ð¥ðœð ð¥
For ð = 2
ðœ2â² ð¥ = ðœ1 ð¥ â
2
ð¥ðœ2 ð¥
⎠ðœ1â²â² ð¥ =
1
2 âðœ1 ð¥ â ðœ1 ð¥ +
2
ð¥ðœ2 ð¥
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=1
ð¥ðœ2 ð¥ â ðœ1 ð¥
Example 11: Show that
(i) ð ð±ðâ²â²â² ð + ð ð±ð
â² ð + ð±ð ð = ð
(ii) ð ð±ðâ²â² ð = ð±ðâð ð â ð ð±ð ð + ð±ð+ð ð = ð
Solution: (i) We know ð
ðð¥ ð¥âððœð ð¥ = âð¥âððœð+1 ð¥
for ð = 0
ð
ðð¥ ðœ0 ð¥ = âðœ1 ð¥
Differentiating with respect to â²ð¥â², we get
ðœ0â²â² ð¥ = âðœ1
â² ð¥
ðœðâ² ð¥ =
1
2 ðœðâ1 ð¥ â ðœð+1 ð¥ (1)
⎠for ð = 1
ðœ1â² ð¥ =
1
2 ðœ0 ð¥ â ðœ2 ð¥
Differentiating again, it will give
ðœ0â²â²â² ð¥ =
1
2 âðœ0
â² ð¥ + ðœ2â² ð¥
=1
2 ðœ1 ð¥ + ðœ2
â² ð¥
From (1), for ð = ð
ðœ2â² ð¥ =
1
2 ðœ1 ð¥ â ðœ3 ð¥
⎠ðœ0â²â²â² ð¥ =
1
2 ðœ1 ð¥ +
1
2 ðœ1 ð¥ â ðœ3 ð¥
=1
4 3 ðœ1 ð¥ â ðœ3 ð¥
=1
4 â3 ðœ0
â² ð¥ â ðœ3 ð¥
⎠4 ðœ0â²â²â² ð¥ + 3 ðœ0
â² ð¥ + ðœ3 ð¥ = 0
(ii) We know
ðœðâ² ð¥ =
1
2 ðœðâ1 ð¥ â ðœð+1 ð¥ (1)
Differentiating with respect to â²ð¥â², we get
ðœðâ²â² ð¥ =
1
2 ðœðâ1
â² ð¥ â ðœð+1â² ð¥ (2)
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From (1) ðœðâ1â² ð¥ =
1
2 ðœðâ2 ð¥ â ðœð ð¥
and ðœð+1â² ð¥ =
1
2 ðœð ð¥ â ðœð+2 ð¥
⎠From (2), we get
ðœðâ²â² ð¥ =
1
2
1
2 ðœðâ2 ð¥ â ðœð ð¥ â
1
2 ðœð ð¥ â ðœð+2 ð¥
=1
4 ðœðâ2 ð¥ â 2 ðœð ð¥ + ðœð+2 ð¥
⎠4 ðœðâ²â² ð¥ = ðœðâ2 ð¥ â 2 ðœð ð¥ + ðœð+2 ð¥
Example 12: Prove that
(i) ð
ð ð ð±ð
ð ð =ð
ðð ð±ðâð
ð ð â ð±ð+ðð ð
(ii) ð
ð ð ð±ð
ð ð + ð±ð+ðð ð = ð
ð
ðð±ðð ð â
ð+ð
ðð±ð+ðð ð
Solutions: (i) LHS = 2 ðœð ð¥ ðœðâ² ð¥
But ðœð+1 ð¥ =2ð
ð¥ðœð ð¥ â ðœðâ1 ð¥
â ðœð ð¥ =ð¥
2ð ðœð+1 ð¥ + ðœðâ1 ð¥
and ðœðâ² ð¥ =
1
2 ðœðâ1 ð¥ â ðœð+1 ð¥
LHS= 2ðœð ð¥ ðœðâ² ð¥ = 2.
ð¥
2ð ðœð+1 ð¥ + ðœðâ1 ð¥ Ã
1
2 ðœðâ1 ð¥ â ðœð+1 ð¥
=ð¥
2ð ðœðâ1
2 ð¥ â ðœð+12 ð¥ = RHS
Hence the result
(ii) LHS = 2ðœð ð¥ ðœðâ² ð¥ + 2ðœð+1 ð¥ ðœð+1
â² ð¥
But ðœðâ² ð¥ =
ð
ð¥ðœð ð¥ â ðœð+1 ð¥
and ðœðâ² ð¥ = ðœðâ1 ð¥ â
ð
ð¥ðœð ð¥
â ðœð+1â² ð¥ = ðœð ð¥ â
ð+1
ð¥ðœð+1 ð¥
⎠LHS = 2 ðœð ð¥ . ð
ð¥ðœð ð¥ â ðœð+1 ð¥ + 2 ðœð+1 ð¥ ðœð ð¥ â
ð+1
ð¥ðœð+1 ð¥
= 2 ð
ð¥ðœð
2 ð¥ â ðœð ð¥ . ðœð+1 ð¥ + ðœð+1 ð¥ . ðœð ð¥ âð+1
ð¥ðœð+1
2 ð¥
= 2 ð
ð¥ðœð
2 ð¥ âð+1
ð¥ðœð+1
2 ð¥ = RHS
Example 13: Prove that
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(i) ð±ð ð ð±ð ð = âð
ð ð±ð ð
ð
(ii) ð ð±ð ðð ð
ð=
ð
ðð±ð ðð
(iii) ðâððð±ð ðð â
ð=
ð
ðð+ðð
Solution: (i) We know ðœ0â² ð¥ = âðœ1 ð¥
⎠ðœ0 ð¥ ðœ1 ð¥ = â ðœ0 ð¥ ðœ0â² ð¥ ðð¥
= â1
2 ðœ0 ð¥
2
(ii) Let ðð¥ = ð¡, ⎠ððð¥ = ðð¡, 0 ð¡ð ð â 0 ð¡ð ðð
⎠ð¥ ðœ0 ðð¥ ð
0ðð¥ =
ð¡
ððœ0 ð¡ .
ðð¡
ð
ðð
0
=1
ð2 ð¡ðœ0 ð¡ ðð¡ðð
0=
1
ð2 ð
ðð¡
ðð
0 ð¡ ðœ1 ð¡ ðð¡
=1
ð2 ð¡ ðœ1 ð¡ 0
ðð =1
ð2 ðð ðœ1 ðð â 0. ð¥. ðœ1 0
=1
ðð ðœ1 ðð
(iii) ðâðð¥ ðœ0 ðð¥ ðð¥â
0
= ðâðð¥ .1
ð cos ðð¥ cosð ðð ðð¥ð
0
â
0
Integrating the order of integration, we get
=1
ð ðâðð¥
â
0cos ðð¥ cosð ðð¥ ðð
ð
0
= 1
ð
ðâðð¥
ð2+ð2ððð 2ð âð cos ðð¥ cosð + ð cosð sin ðð¥ cosð
0
â
ððð
0
=1
ð
ð
ð2+ð2ððð 2ð
ð
0ðð =
1
ð
ð ð ðð 2ð
ð2ð ðð 2ð+ð2 ð ð =2
ð
ð
0 ð ð ðð 2ð
ð2+ð2 +ð2ð¡ðð 2ð
ð
20
=2
ðð tanâ1
ð tan ð
ð2+ð2
0
ð
2Ã
ð
ð2+ð2
=2
ðð.
ð
ð2+ð2Ã
ð
2â 0 =
1
ð2+ð2
Example 14: Starting with series with generating functions, prove that
ðð ð±ð ð = ð ð±ðâð ð = ð ð±ðâð ð + ð±ð+ð ð and
ðð±ðâ² ð = ð ð±ð ð â ð ð±ð+ð ð (1)
Solutions: We know ð1
2ð¥ ð¡â
1
ð¡
= ð¡ððœð ð¥ âââ
Differentiating both sides with respect to â²ð¡â², we get
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1
2ð¥ 1 +
1
ð¡2 . ð1
2ð¥ ð¡â
1
ð¡
= ðð¡ðâ1ðœð ð¥ âââ
1
2ð¥ 1 +
1
ð¡2 ð¡ððœð ð¥ âââ = ð ð¡ðâ1ðœð ð¥
âââ
Equating the coefficients of ð¡ðâ1, we will have
1
2ð¥ ðœðâ1 ð¥ +
1
2ð¥ ðœð+1 ð¥ = ððœð ð¥
â 2ð ðœð ð¥ = ð¥ ðœðâ1 ð¥ + ðœð+1 ð¥ (2)
Now differentiating with respect to â²ð¥â², we get
1
2 ð¡ â
1
ð¡ ð
1
2ð¥ ð¡â
1
ð¡
= ð¡ððœðâ² ð¥ â
ââ
1
2 ð¡ â
1
ð¡ ð¡ððœð ð¥
âââ = ð¡ððœð
â² ð¥ âââ
Equating the coefficients of â²ð¡ð â², we will have
1
2ðœðâ1 ð¥ â
1
2ðœð+1 ð¥ = ðœð
â² ð¥
â ðœðâ² ð¥ =
1
2 ðœðâ1 ð¥ â ðœð+1 ð¥ (3)
From (2), substituting ðœðâ1 ð¥ in (3), we get
ðœðâ² ð¥ =
1
2
2ð
ð¥ðœð ð¥ â ðœð+1 ð¥ â ðœð+1 ð¥
â ðœðâ² ð¥ =
ð
ð¥ðœð ð¥ â ðœð+1 ð¥
Example 15: Establish the Jacobi series
ððšð¬ ð ððšð¬ ðœ = ð±ð â ðð±ð ððšð¬ððœ + ðð±ð ððšð¬ððœ ââŠâŠ
ð¬ð¢ð§ ð ððšð¬ ðœ = ð ð±ð ððšð¬ ðœ â ð±ð ððšð¬ððœ + ð±ð ððšð¬ ððœ ââŠâŠ
Solutions: We know ð1
2ð¥ ð¡â
1
ð¡
= ð¡ððœð ð¥ âââ
= ðœ0 ð¥ + ðœð ð¥ ð¡ð + â1 ð
1
ð¡ð â
ð=1 (1)
Now, let ð¡ = cosð + ð sinð and 1
ð¡= cosð â ð sinð
To get ð¡ð = cosðð + ð sinðð and ð¡âð = cosðð â ð sinðð
and thus ð¡ð + ð¡âð = 2 cosðð and ð¡ð â ð¡âð = 2 ð sinðð
⎠From (1)
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ððð¥ sin ð = ðœ0 ð¥ + 2ð ðœ1 ð¥ sinð + 2ðœ2 ð¥ cos 2ð
+2ððœ3 ð¥ sin 3ð + 2ðœ4 ð¥ cos 4ð + âŠâŠ
cos ð¥ sinð + ð sin ð¥ sinð = ðœ0 ð¥ + 2ðœ2 ð¥ cos 2ð + 2ðœ4 ð¥ cos 4ð + âŠâŠ
+ð 2ðœ1 ð¥ sinð + 2ðœ3 ð¥ sinð + âŠâŠ
Equating the real and imaginary parts, we get
cos ð¥ sinð = ðœ0 ð¥ + 2 ðœ2 ð¥ cos 2ð + ðœ4 ð¥ cos 4ð +âŠâŠ
and sin ð¥ sinð = 2 ðœ 1 ð¥ sinð + ðœ3 ð¥ sin 3ð + âŠâŠ
Replacing ð by ð
2â ð, we get
cos ð¥ cosð = ðœ0 ð¥ â 2 cos 2ð ðœ2 ð¥ + 2ðœ4 ð¥ cos 4ð + âŠâŠ
and sin ð¥ cosð = 2 ðœ1 ð¥ sinð â ðœ3 ð¥ sin 3ð + âŠâŠ
Example 16: Prove that
(i) ð¬ð¢ð§ ð = ð ð±ð ð â ð±ð ð + ð±ð ð â âŠâŠ
(ii) ððšð¬ ð = ð±ð ð â ðð±ð ð + ðð±ð ð â ðð±ð ð + âŠâŠ
(iii) ð = ð±ð + ðð±ð + ðð±ð + ðð±ð + âŠâŠ
Solution: We know
cos ð¥ sinð = ðœ0 ð¥ + 2 ðœ2 ð¥ cos 2ð + ðœ4 ð¥ cos 4ð +âŠâŠ
and sin ð¥ sinð = 2 ðœ1 ð¥ sinð + ðœ3 ð¥ sin 3ð + âŠâŠ
On taking ð =ð
2 , we will have
(ii) cosð¥ = ðœ0 ð¥ + 2 ðœ2 ð¥ cosð + ðœ4 ð¥ cos 2ð + ðœ6 ð¥ cos 3ð + âŠâŠ
= ðœ0 ð¥ + 2 âðœ2 ð¥ + ðœ4 ð¥ â ðœ6 ð¥ + âŠâŠ
= ðœ0 ð¥ â 2ðœ2 ð¥ + 2ðœ4 ð¥ â 2ðœ6 ð¥ + âŠâŠ
(i) sinð¥ = 2 ðœ1 ð¥ sinð
2+ ðœ3 ð¥ sin
3ð
2+ ðœ5 ð¥ sin
5ð
2+ âŠâŠ
= 2 ðœ1 ð¥ â ðœ3 ð¥ + ðœ5 ð¥ â âŠâŠ
(iii) Taking ð = 0, we get
cos 0 = 1 = ðœ 0 ð¥ + 2 ðœ2 ð¥ cos 2 à 0 + ðœ4 ð¥ cos 4 à 0 + âŠâŠ .
â 1 = ðœ0 ð¥ + 2ðœ2 ð¥ + 2ðœ4 ð¥ + âŠâŠ
ASSIGNMENT 18.3
1. Compute ðœ0 2 and ðœ1 1 correct to three decimal places.
2. Express ðœ5 ð¥ in terms of ðœ0 ð¥ and ðœ1 ð¥ .
3. Prove that
(a) ðœðâ²â² ð¥ =
1
4 ðœðâ2 ð¥ â 2ðœð ð¥ + ðœð+2 ð¥
(b) ð
ðð¥ ð¥ðœð ð¥ ðœð+1 ð¥ = ð¥ ðœð
2 ð¥ â ðœð+12 (ð¥) .
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4. Prove that ðœ5
2
ð¥ = 2
ðð¥
3âð¥2
ð¥2 sinð¥ â3
ð¥cosð¥ .
5. Prove that
(a) ðœ3 ð¥ ð ð¥ = ð â ðœ2 ð¥ â2
ð¥ ðœ1(ð¥)
(b) ð¥ðœð2 ð¥ ðð¥ =
1
2ð¥2 ðœ0
2 ð¥ + ðœ12(ð¥) .
6. Show that
a) ðœð ð¥ =1
ð cos ðð â ð¥ sinð ðð , ðð
0 being an integer.
b) ðœ0 ð¥ =1
ð cos ð¥ cosð ððð
0
c) ðœ02 + 2ðœ1
2 + 2ðœ22 + 2ðœ3
2 + ⯠= 1.
ANSWERS
1. 0.224, 0.44
2. ðœ5 ð¥ = 384
ð¥4 â72
ð¥2 â 1 ðœ1 ð¥ + 12
ð¥â
192
ð¥3 ðœ0 ð¥
18.9 EQUATIONS REDUCIBLE TO BESSELâS EQUATION
In differential calculus, we come across such differential equations which can be easily reduced to
Besselâs equation and thus can be solved by the means of Besselâs functions. The following are some
examples of such differential equations:
1. Reduce the differential equation ððð ðð
ð ðð+ ð
ð ð
ð ð+ ðððð â ðð ð = ð to the Besselâs Equation.
Putting ð¡ = ðð¥, so that ððŠ
ðð¥= ð
ððŠ
ðð¡ and
ð2ðŠ
ðð¥2 = ðð2ðŠ
ðð¡ 2 in the above differential equation, we get
ð¡2 ð2ðŠ
ðð¡2 + ð¡ððŠ
ðð¡+ ð¡2 â ð2 ðŠ = 0 , which is the Besselâs Form of Equation.
⎠Its solution is ðŠ = ð1ðœð ð¡ + ð2ðœâð ð¡ , ð is non-integral.
or ðŠ = ð1ðœð ð¡ + ð2ðð ð¡ , ð is integral.
Hence solution of the given differential equation is
ðŠ = ð1ðœð ðð¥ + ð2ðœâð ðð¥ , ð is non-integral.
or ðŠ = ð1ðœð ðð¥ + ð2ðð ðð¥ , ð is integral.
2. Reduce the differential equation ðð ðð
ð ðð+ ð
ð ð
ð ð+ ðððð = ð to the Besselâs Equation.
Putting ðŠ = ð¥ðð§, so that ððŠ
ðð¥= ð¥ð ðð§
ðð¥+ ðð¥ðâ1ð§
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and ð2ðŠ
ðð¥2 = ð¥ð ð2ð§
ðð¥ 2 + 2ðð¥ðâ1 ðð§
ðð¥+ ð ð â 1 ð¥ðâ2ð§ in the above differential equation, we get
ð¥ð+1 ð2ð§
ðð¥2 + (2ð + ð)ð¥ð ðð§
ðð¥+ ð2ð¥2 + ð2 + ð â 1 ð ð¥ðâ1ð§ = 0
Dividing throughout by ð¥ðâ1 and putting 2ð + ð = 1, we get
ð¥2 ð2ð§
ðð¥2 + ð¥ðð§
ðð¥+ ð2ð¥2 â ð2 ð§ = 0, which is the Besselâs Form of Equation.
And its solution is ð§ = ð1ðœð ðð¥ + ð2ðœâð ðð¥ , ð is non-integral.
or ðŠ = ð1ðœð ðð¥ + ð2ðð ðð¥ , ð is integral.
Hence solution of the given differential equation is
ðŠ = ð¥ð ð1ðœð ðð¥ + ð2ðœâð ðð¥ , ð is non-integral.
or ðŠ = ð¥ð ð1ðœð ðð¥ + ð2ðð ðð¥ , ð is integral.
3. Reduce the differential equation ðð ðð
ð ðð+ ð
ð ð
ð ð+ ððððð = ð to the Besselâs Equation.
Putting ðŠ = ð¡ð , so that ððŠ
ðð¥=
ððŠ
ðð¡.ðð¡
ðð¥=
1
ðð¡1âð ððŠ
ðð¡
and ð2ðŠ
ðð¥2 =ð
ðð¡
1
ðð¡1âð ððŠ
ðð¡ .
1
ðð¡1âð =
1
ð2 ð¡2â2ð ð2ðŠ
ðð¡ 2 +1âð
ð2 ð¡1â2ð ððŠ
ðð¡ in the above differential
equation, we get
1
ð2 ð¡2âð ð2ðŠ
ðð¡2 +1âð+ðð
ð2 ð¡1âð ððŠ
ðð¡+ ð2ð¡ðð ðŠ = 0
Multiplying throughout by ð2 ð¡1âð , we get
ð¡ ð2ðŠ
ðð¡2 + (1 âð + ðð)ððŠ
ðð¡+ (ðð)2ð¡ðð+ðâ1ðŠ = 0
To reduce this equation to the equation at point 2. above, we set ðð + ðâ 1 = 1
ð. ð.ð = 2/(ð + 1) and ð = 1 âð + ðð =ð+2ðâ1
ð+1. Thus we get the equation as
ð¡ ð2ðŠ
ðð¡2 + ðððŠ
ðð¡+ (ðð)2ð¡ðŠ = 0 which is similar to equation at point 2.
Hence its solution is
ðŠ = ð¥ð/ð ð1ðœð ðð ð¥1/ð + ð2ðœâð ðð ð¥1/ð , ð is non-integral.
or ðŠ = ð¥ð/ð ð1ðœð ðð ð¥1/ð + ð2ðð ðð ð¥1/ð , ð is integral.
18.10 ORTHOGONALITY OF BESSEL FUNCTIONS
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Prove that ðð±ð ð¶ð ð±ð ð·ð ð ðð
ð=
ð, ð¶ â ð·ð
ð ð±ð+ð ð¶
ð, ð¶ = ð·
ðððð ð ð¶,ð· ððð ððððð ðð ð±ð ð = ð.
Proof: Let ð¢ = ðœð(ðŒð¥) and ð£ = ðœð(ðœð¥) are the solutions of the following differential equations
ð¥2ð¢â²â² + ð¥ð¢â² + ðŒ2ð¥2 â ð2 ð¢ = 0 (1)
ð¥2ð£ â²â² + ð¥ð£ â² + ðœ2ð¥2 â ð2 ð£ = 0 (2)
Multiplying equation (1) by ð£/ð¥ and equation (2) by ð¢/ð¥ and then on subtracting, we get
ð¥ ð¢â²â² ð£ â ð¢ð£ â²â² + ð¢â²ð£ â ð¢ð£ â² + ðŒ2 â ðœ2 ð¥ð¢ð£ = 0
â ð
ðð¥ ð¥(ð¢â²ð£ â ð¢ð£ â²) = (ðœ2 â ðŒ2)ð¥ð¢ð£ (3)
Now, integrating both sides of equation (3) within the limits 0 to 1, we get
ðœ2 â ðŒ2 ð¥ ð¢ð£ ðð¥1
0= ð¥(ð¢â²ð£ â ð¢ð£ â²) 0
1 = ð¢â²ð£ â ð¢ð£ â² (4)
Since ð¢ = ðœð(ðŒð¥) and ð£ = ðœð(ðœð¥)
⎠ð¢â² = ðŒ ðœðâ²(ðŒð¥) and ð£ â² = ðœ ðœð
â²(ðœð¥)
Substituting these values in equation (4), we get
ð¥ ðœð ðŒð¥ ðœð(ðœð¥) ðð¥1
0=
ðŒ ðœðâ² ðŒ ðœð ðœ âðœ ðœð
â² (ðœ)ðœð (ðŒ)
ðœ2âðŒ2 (5)
Case I: ð¶ â ð·
Since ðŒ,ðœ are roots of ðœð ð¥ = 0, so we have ðœð ðŒ = ðœð ðœ = 0. Thus equation (5) results in
ð¥ ðœð ðŒð¥ ðœð(ðœð¥) ðð¥1
0= 0 (6)
Case II: ð¶ = ð·
In this case RHS of (5) becomes 0/0 form. So to get its value, apply LâHospital Rule, by taking ðŒ as
constant and ðœ as variable approaching to ðŒ, we get
LimðœâðŒ ð¥ ðœð ðŒð¥ ðœð(ðœð¥) ðð¥1
0= LimðœâðŒ
ðŒ ðœðâ² ðŒ ðœð ðœ
ðœ2âðŒ2
0
0
or LimðœâðŒ ð¥ ðœð2 ðŒð¥ ðð¥
1
0= limðœâðŒ
ðŒ ðœðâ² ðŒ ðœð
â² ðœ
2ðœ
=1
2 ðœð
â²(ðŒ) 2
=1
2 ðœð+1(ðŒ) 2 ð¢ð ððð ðœð
â² = âðœð+1 (7)
The relations (6) and (7) are known as Orthogonality relations of Bessel functions.
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18.11 FOURIER BESSEL EXPANSION
If ð(ð) is a continuous function having finite number of oscillations in the interval ð, ð , then we
can write
ð ð = ðð ð±ð(ð¶ðð)âð=ð = ðð ð±ð ð¶ðð + ðð ð±ð ð¶ðð + â¯+ ðð ð±ð ð¶ðð + â¯(1)
where ð¶ð,ð¶ð,⊠are the positive roots of ð±ð ð = ð.
To determine the coefficients ðð, multiply both sides of (1) by ð¥ðœð(ðŒðð¥) and integrating within the
limits 0 to a, we get
ð¥ ð ð¥ ðœð ðŒðð¥ ðð¥ð
0= ðð ð¥ ðœð
2 ðŒðð¥ ðð¥ð
0= ðð
ð2
2 ðœð+1
2 ð ðŒð
â ðð =2
ð2ðœð+12 ð ðŒð
ð¥ ð ð¥ ðœð ðŒðð¥ ðð¥ð
0
The relation (1) is called Fourier Bessel Expansion of f(x).
18.12 BER AND BEI FUNCTIONS
The differential equation generally encountered in the field of electrical engineering for finding the
distribution of alternating currents in wires of circular cross section is as follows:
ð¥ð2ðŠ
ðð¥2 +ððŠ
ðð¥â ð ð¥ðŠ = 0 (1)
which is the special case of first form of differential equation reducible to Bessel equation with ð = 0
and ð2 = âð, so that ð = âð = ð ð = ð3
2 (Refer Art 18.9).
Thus, the general solution of differential equation (1) is given by
ðŠ = ð1ðœ0 ð3
2 ð¥ + ð2ð0 ð3
2 ð¥
Now ðœ0 ð3
2 ð¥ = 1 âð3ð¥2
22 +ð6ð¥4
(2!)2 24 âð9ð¥6
(3!)2 26 +ð12ð¥8
(4!)2 28 ââ¯
= 1 âð¥ 4
22 .42 +ð¥8
22 .42 .62 .82 ââ¯
+ð ð¥2
22 âð¥6
22 .42 .62 +ð¥10
22 .42 .62 .82 .102 â⯠(2)
which is complex for x is real.
The series in the brackets of (2) is defined as
ððð ð¥ = 1 âð¥4
22 .42 +ð¥8
22 .42 .62 .82 ââ¯
= 1 + (â1)ð .âð=1
ð¥4ð
22 .42 .62â¯(4ð)2
and ððð ð¥ =ð¥2
22 âð¥6
22 .42 .62 +ð¥10
22 .42 .62 .82 .102 ââ¯
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= â (â1)ð .âð=1
ð¥4ðâ2
22 .42 .62â¯(4ðâ2)2
where ber stands for Bessel real and bei for Bessel imaginary.
Thus we have ðœ0 ð3
2 ð¥ = ððð ð¥ + ð ððð(ð¥)
Similarly, decomposing ð0 ð3
2 ð¥ into real and imaginary parts, we obtain another two functions
known as ker (x) and kei (x).
Properties of ber and bei functions
1. ð
ðð¥ ð¥.
ð
ðð¥ ððð (ð¥) = âð¥ ððð (ð¥)
2. ð
ðð¥ ð¥.
ð
ðð¥ ððð (ð¥) = âð¥ ððð (ð¥)
Example 17: Solve ðâ²â² +ðâ²
ð+ ð â
ð
ððð ð = ð
Solution: ðŠâ²â² +ðŠ â²
ð¥+ 1 â
1
9ð¥2 ðŠ = 0
â ð¥2ðŠâ²â² + ð¥ðŠâ² + ð¥2 â1
9 ðŠ = 0
Comparing with Besselâs equation
ð¥2ðŠâ²â² + ð¥ðŠâ² + ð¥2 â ð2 ðŠ = 0
We find ð =1
3
⎠The solution of the given equation is ðŠ = ð1ðœ1
3
ð¥ + ð2ð1
3
ð¥
Example 18: Solve ðâ²â² +ðâ²
ð+ ð â
ð
ð.ðð ðð ð = ð
Solution: ðŠâ²â² +ðŠ â²
ð¥+ 1 â
1
6.25 ð¥ 2 ðŠ = 0
â ðŠâ²â² +ðŠ â²
ð¥+ 1 â
100
625 ð¥2 ðŠ = 0
Comparing with the Besselâs equation, we find ð =10
25=
2
5
⎠The solution of the given equation is ðŠ = ð1ðœ2
5
ð¥ + ð2ð2
5
ð¥
Example 19: Solve ððâ²â² + ðâ² +ð
ðð = ð
Solution: Let ð¡ = ð¥1
ð , so that
ððŠ
ðð¥=
ððŠ
ðð¡.ðð¡
ðð¥=
1
ðð¥
1
ðâ1 .
ððŠ
ðð¡=
1
ð ð¡ 1âð .
ððŠ
ðð¡
ð2ðŠ
ðð¥2 =ð
ðð¡
1
ðð¡ðâ1 .
ððŠ
ðð¡ ðð¡
ðð¥
= 1
ð. 1 âð ð¡âð .
ððŠ
ðð¡+
1
ðð¡1âð ð2ðŠ
ðð¡2 Ã1
ðð¡1âð
=1
ð2 1 âð ð¡1â2ð ððŠ
ðð¡+
1
ð2 ð¡2â2ð ð2ðŠ
ðð¡2
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⎠ð¥ð2ðŠ
ðð¥2 +ððŠ
ðð¥+
1
4ðŠ
=ð¡ð
ð2 1 âð ð¡1â2ð ððŠ
ðð¡+ ð¡ 2â2ð ð2ðŠ
ðð¡2 +1
ðð¡1âð ððŠ
ðð¡+
1
4ðŠ = 0
â 1
ð2 ð¡2âð ð2ðŠ
ðð¡2 +1âð
ð2 ð¡1âð ððŠ
ðð¡+
1
ðð¡1âð ððŠ
ðð¡+
1
4ðŠ = 0
â ð¡2 ð2ðŠ
ðð¡2 + 1 âð + ð ð¡ððŠ
ðð¡+
1
4ð2ð¡ððŠ = 0
â ð¡2 ð2ðŠ
ðð¡2 + ð¡ððŠ
ðð¡+
1
4ð2ð¡ððŠ = 0
Comparing with
ð¥ð2ðŠ
ðð¥2 + ðððŠ
ðð¥+ ð2ððŠ = 0
We get ð = 1, ð2 =ð2
4, ð â 1 = 1 it implies ð = 2
i.e. ð2 = 1 and ð =1âð
2= 0
⎠The solution of the given equation is
ðŠ = ð1ðœ0 ð¡ + ð2ð0 ð¡ = ð1ðœ0 ð¥ + ð2ð0 ð¥
Example 20: Solve ððâ²â² + ððâ² +ð
ððð = ð
Solution: Let ðŠ = ð¥ðð§ so that
ððŠ
ðð¥= ð¥ð ðð§
ðð¥+ ðð¥ðâ1ð§
ð2ðŠ
ðð¥2 = ð¥ð ð2ð§
ðð¥2 + 2ðð¥ðâ1 ðð§
ðð¥+ ð ð â 1 ð¥ðâ2ð§
⎠ð¥ðŠâ²â² + 2ðŠâ² +1
2ð¥ðŠ = 0
â ð¥ð+1 ð2ð§
ðð¥2 + 2ð + 2 ð¥ð ðð§
ðð¥+ ð ð â 1 + 2ð ð¥ðâ1 +
1
2ð¥ð+1ð§ = 0
â ð¥2 ð2ð§
ðð¥2 + 2 ð + 1 ð¥ðð§
ðð¥+ ð ð + 1 +
1
2ð¥2 ð§ = 0
Taking 2 ð + 1 = 1 i.e. ð = â1
2
â ð¥2 ð2ð§
ðð¥2 + ð¥ðð§
ðð¥+
1
2ð¥2 â
1
4 ð§ = 0
â ð§ = ð1ðœ1
2
1
2 ð¥ + ð2ð1
2
1
2 ð¥
â ðŠ = ð¥â1
2 ð1ðœ1
2
ð¥
2 + ð2ð1
2
ð¥
2
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Example 21: Solve ððâ²â² + ð = ð (1)
Solution: Let ð¡ = ð¥1
ð , so that
ððŠ
ðð¥=
ððŠ
ðð¡
ðð¡
ðð¥=
1
ðð¡1âð ððŠ
ðð¡
and ð2ðŠ
ðð¥2 =ð
ðð¡
1
ðð¡1âð ððŠ
ðð¡ Ã
ðð¡
ðð¥
= 1
ðð¡1âð ð2ðŠ
ðð¡2 +1
ð 1 âð ð¡âð .
ððŠ
ðð¡
1
ðð¡1âð
⎠ð¥ðŠâ²â² + ðŠ = 0
â ð¡ð 1
ð2 ð¡2â2ð ð2ðŠ
ðð¡2 +1
ð2 1 âð ð¡1â2ð ððŠ
ðð¡ + ðŠ = 0
â ð¡2âð ð2ðŠ
ðð¡2 + 1 âð ð¡1âð ððŠ
ðð¡+ ð2ðŠ = 0
â ð¡ð2ðŠ
ðð¡2 + 1 âð ððŠ
ðð¡+ ð2ð¡ðâ1ðŠ = 0 (2)
Comparing both
ð¥ðŠâ²â² + ððŠâ² + ð2ð¥ðŠ = 0
We will have
ð = 1 âð, ð = ð and ð â 1 = 1
i.e. ð = 2, ð = 2 and ð = 1 â 2 = â1
⎠ð =1âð
2=
1+1
2= 1
Hence the solution of the equation (2) will be
ðŠ = ð¡ ð1ðœ1 2ð¡ + ð2ð1 2ð¡
â ðŠ = ð¥1
2 ð1ðœ1 2 ð¥ + ð2ð1 2 ð¥
Example 22: Solve ðâ²â² + ðð âðð
ðð ð = ð (1)
Solution: Let ð¡ = ð¥1
ð or ð¥ = ð¡ð , so that
ððŠ
ðð¥=
ððŠ
ðð¡
ðð¡
ðð¥=
1
ðð¡1âð ððŠ
ðð¡
and ð2ðŠ
ðð¥2 =ð
ðð¡
1
ðð¡1âð ððŠ
ðð¡ Ã
ðð¡
ðð¥
= 1
ðð¡1âð ð2ðŠ
ðð¡2 +1
ð 1 âð ð¡âð .
ððŠ
ðð¡
1
ðð¡1âð
â 1
ð2 ð¡2â2ð ð2ðŠ
ðð¡2 +1
ð2 1 âð ð¡1â2ð ððŠ
ðð¡+ 9ð¡ð â
20
ð¡2ð ðŠ = 0
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â ð¡2 ð2ðŠ
ðð¡2 + 1 âð ð¡ððŠ
ðð¡+ ð2 9ð¡3ð â 20 ðŠ = 0
â ð¡2 ð2ðŠ
ðð¡2 + 1 âð ð¡ððŠ
ðð¡+ 9ð2ð¡3ð â 20ð2 ðŠ = 0
Taking 3ð = 2 i.e. ð =2
3, we will have
ð¡2 ð2ðŠ
ðð¡2 +1
3ð¡ððŠ
ðð¡+ 4ð¡2 â
80
9 ðŠ = 0 (2)
Now let ðŠ = ð¡ðð§ ð¡ , so that
ððŠ
ðð¡= ð¡ð
ðð§
ðð¡+ ðð¡ðâ1ð§ ,
ð2ðŠ
ðð¡2 = ð¡ðð2ð§
ðð¡2 + 2ðð¡ðâ1 ðð§
ðð¡+ ð ð â 1 ð¡ðâ2ð§
Substituting these in (2), we get
ð¡ð+2 ð2ð§
ðð¡2 + 2ð +1
3 ð¡ð+1 ðð§
ðð¡+ ð ð â 1 +
1
3ð â
80
9 ð¡ð + 4ð¡ð+2 ð§ = 0 (3)
Now for 2ð +1
3= 1, ð =
1
3
and ð ð â 1 +1
3ð â
80
9=
1
3Ã â
2
3+
1
3Ã
1
3â
80
9= â
81
9= â9
⎠Dividing (3) by ð¡ð and substituting for ð, we will have
ð¡2 ð2ð§
ðð¡2 + ð¡ðð§
ðð¡+ 4ð¡2 â 9 ð§ = 0 (4)
The solution of (4) is
ð§ = ð1ðœ3 2ð¡ + ð2ð3 2ð¡
â ðŠ = ð¡1
3 ð1ðœ3 2ð¡ + ð2ð3 2ð¡
â ðŠ = ð¥3
2
1
3 ð1ðœ3 2ð¥
3
2 + ð2ð3 2ð¥3
2
= ð¥1
2 ð1ðœ3 2ð¥3
2 + ð2ð3 2ð¥3
2
Example 23: Show that
(i) ððð±ð ð is a solution of the equation ððâ²â² + ð â ðð ðâ² + ðð = ð .
(ii) ðâðð±ð ð is the solution of the equation ððâ²â² + ð + ðð ðâ² + ðð = ð
Solution: Let ðŠ = ð¥ððœð ð¥
⎠ððŠ
ðð¥= ð¥ððœð
â² ð¥ + ðð¥ðâ1ðœð ð¥
and ð2ðŠ
ðð¥2 = ð¥ððœðâ²â² ð¥ + 2ðð¥ðâ1ðœð
â² ð¥ + ð ð â 1 ð¥ðâ2ðœð ð¥
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⎠ð¥ðŠâ²â² + 1 â 2ð ðŠâ² + ð¥ðŠ
= ð¥ð+1ðœðâ²â² ð¥ + 2ðð¥ððœð
â² ð¥ + ð ð â 1 ð¥ðâ1ðœð ð¥
+ 1 â 2ð ð¥ððœðâ² ð¥ + ðð¥ðâ1ðœð ð¥ + ð¥ð+1ðœð ð¥
= ð¥ð+1ðœðâ²â² ð¥ + ð¥ððœð
â² ð¥ 2ð + 1 â 2ð
+ ð ð â 1 + ð 1 â 2ð ð¥ðâ1 + ð¥ð+1 ðœð ð¥
= ð¥ðâ1 ð¥2ðœðâ²â² ð¥ + ðœð
â² ð¥ + ð¥2 â ð2 ðœð ð¥ = 0
As ðœð ð¥ is the Bessel function and is a solution of ð¥2ðŠâ²â² + ðŠâ² + ð¥2 â ð2 ðŠ = 0
Hence, ð¥ððœð ð¥ satisfy the given equation and therefore is a solution of it.
Example 24: Show under the transformations ð =ð
ð Besselâs equation becomes ðâ²â² +
ð +ðâððð
ððð ð = ð; Hence find the solution of this equation.
Solution: We know that the Besselâs equation is
ð¥2ðŠâ²â² + ð¥ðŠâ² + ð¥2 â ð2 ðŠ = 0 (1)
Taking ðŠ =ð¢
ð¥
â ðŠâ² =1
ð¥ ð¢â² + â
1
2 ð¥â
3
2ð¢,
and ðŠâ²â² =1
ð¥ ð¢â²â² + 2 â
1
2 ð¥â
3
2 ð¢â² + â1
2 â
3
2 ð¥â
5
2 ð¢
Substituting these into (1), we get
ð¥2 1
ð¥ð¢â²â² â ð¥â
3
2ð¢â² +3
4ð¥â
5
2 ð¢ + ð¥ 1
ð¥ð¢â² + â
1
2 ð¥â
3
2 ð¢ + ð¥2 â ð2 ð¢
ð¥= 0
â ð¥3
2 ð¢â²â² + âð¥1
2 + ð¥1
2 ð¢â² + 3
4ð¥â
1
2 â1
2ð¥â
1
2 + ð¥3
2 âð2
ð¥ ð¢ = 0
â ð¥3
2 ð¢â²â² + ð¥3
2 +3â2â4ð2
4 ð¥ ð¢ = 0
â ð¢â²â² + 1 +1â4ð2
4ð¥2 ð¢ = 0 (2)
Hence the Besselâs equation (1) becomes (2) as desired.
Now the solution of (1) is ðŠ = ð1ðœð ð¥ + ð2ðð ð¥ (3)
â ð¢
ð¥= ð1ðœð ð¥ + ð2ðð ð¥
â ð¢ = ð¥ ð1ðœð ð¥ + ð2ðð ð¥
Example 25: By the use of the substitution ð =ð
ð so that the solution of the equation ðð
ð ðð
ð ðð+
ðð ð
ð ð+ ðð â
ð
ð ð = ð can be written in the form ð = ðð
ð¬ð¢ð§ð
ð+ ðð
ððšð¬ ð
ð.
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Solution: Taking ðŠ =ð¢
ð¥
â ððŠ
ðð¥=
1
ð¥
ðð¢
ðð¥â
1
2ð¥â
3
2 ð¢ and ð2ðŠ
ðð¥2 = ð¥â1
2 ð2ð¢
ðð¥2 â ð¥â3
2ððŠ
ðð¥+
3
4ð¥â
5
2 ð¢
Substituting these in the given equation, we get
ð¥2 ð¥â
1
2ð2ð¢
ðð¥2 â ð¥â3
2ðð¢
ðð¥+
3
4ð¥â
5
2 ð¢ + ð¥ ð¥â
1
2ðð¢
ðð¥â
1
2ð¥â
3
2 ð¢ + ð¥2 â1
4 ð¥â
1
2 ð¢ = 0
â ð¥3
2 ð2ð¢
ðð¥2 â ð¥1
2ðð¢
ðð¥+
3
4ð¥â
1
2ð¢ + ð¥1
2ðð¢
ðð¥â
1
2ð¥â
1
2ð¢ + ð¥3
2 â1
4ð¥â
1
2 ð¢ = 0
â ð¥3
2ð2ðŠ
ðð¥2 + ð¥3
2 ð¢ = 0 It implies ð2ðŠ
ðð¥2 + ð¢ = 0
Its Auxiliary equation is ð·2 + 1 = 0 it implies ð· = ±ð
⎠ð¢ ð¥ = ð1 cos ð¥ + ð2 sinð¥
Hence ðŠ =ð¢
ð¥= ð1
cos ð¥
ð¥+ ð2
sin ð¥
ð¥
Example 26: Show that
ð ððððð + ððððð ð ð = ð ððð ð.ðððâ²ðâ ððð ð.ðððâ² ð ð
ð
Solution: We know ððð ð¥ = 1 + â1 ðð¥4ð
22 .42 .62âŠ. 4ð 2âð=1
and ððð ð¥ = â â1 ðð¥4ðâ2
22 .42 .62âŠâŠ 4ðâ2 2âð=1
â ð
ðð¥ ð¥ ðððâ² ð¥ = ð¥ ððð ð¥
⎠ð¥ ððð2 ð¥ + ððð2 ð¥ ðð¥ð
0= ð¥ ððð ð¥ .ððð ð¥ + ð¥ ððð ð¥ ððð ð¥ ðð¥
ð
0
= ð
ðð¥ ð¥ ðððâ² ð¥ .ððð ð¥ â
ð
ðð¥ ð¥ ðððâ² ð¥ ððð ð¥ ðð¥
ð
0
= ððð ð¥. ð¥ ðððâ² ð¥ â ðððâ² ð¥ ð¥ ðððâ²ð¥ ðð¥ â ððð ð¥ ð¥ ðððâ² ð¥ +
ðððâ² ð¥ð¥ ðððâ²ð¥ðð¥0ð
= ð ððð ð. ðððâ²ð â ððð ð. ðððâ² ð hence proved
Example 27: If ðð,ðð,ðð,âŠâŠ ðð are the positive roots of ð±ð ð = ð, prove that
(i) ð
ð=
ð±ð ð¶ð ð
ð¶ðð±ð ð¶ð âð=ð (ii) ðð = ð
ð¶ððâð
ð¶ððð±ð ð¶ð
ð±ð ð¶ð ð âð=ð
Solutions: (i) Let the Fourier Bessel expression of 1
2 is
1
2= ðððœ0 ðŒð ð¥ â
ð=1 and integrating with
respect to â²ð¥â² from 0 to 1, we get
1
2ð¥ðœ0 ðŒð ð¥
1
0ðð¥ = ðð ð¥ðœ0
2 ðŒð ð¥ ðð¥ = ðð1
2 ðœ1 ðŒð
21
0
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37
â ðð1
2 ðœ1
2 ðŒð =1
2 ð¥ ðœ0 ðŒð ð¥
1
0
Let ðŒðð¥ = ð¡ it implies ðð¥ =ðð¡
ðŒð
ð¥ â 0, 1 It implies ð¡ â 0 ð¡ð ðŒð
=1
2
ð¡
ðŒð
ðŒð
0ðœ0 ð¡
ðð¡
ðŒð
=1
2ðŒð2 ð¡ ðœ0 ð¡ ðð¡
ðŒð
0=
1
2ðŒð2
ð
ðð¡ ð¡ ðœ1 ð¡ ðð¡
ðŒð
0
=1
2ðŒð2 ð¡ ðœ1 ð¡ 0
ðŒð =1
2ðŒð2 ðŒððœ1 ðŒð
⎠ðð1
2 ðœ1
2 ðŒð =1
2ðŒððœ1 ðŒð
It implies ðð =1
ðŒð ðœ1 ðŒð Hence
1
2=
1
ðŒð ðœ1 ðŒð ðœ0 ðŒð ð¥ â
ð=1
(ii) Let the Fourier-Bessel expansion of ð¥2 is ð¥2 = ðððœ0 ðŒð ð¥ âð=1 and integrating from 0 to 1,
we get ð¥3ðœ0 ðŒð ð¥ 1
0= ðð ð¥ ðœ0
2 ðŒð ð¥ ðð¥
â ðð1
2ðœ1
2 ðŒð = ð¡3
ðŒð3 ðœ0 ð¡
1
ðŒð
ðŒð
0ðð¡, if ðŒðð¥ = ð¡ it implies ðð¥ =
ðð¡
ðŒð
=1
ðŒð4 ð¡2 ð
ðð¡ ð¡ ðœ1 ð¡
ðŒð
0ðð¡
=1
ðŒð4 ð¡2 . ð¡ðœ1 ð¡ â 2ð¡ . ð¡ðœ1 ð¡ ðð¡ 0
ðŒð
=1
ðŒð4 ð¡
3ðœ1 ð¡ â 2 ð
ðð¡ ð¡2ðœ2 ð¡ ðð¡
0
ðŒð
=1
ðŒð4 ð¡
3ðœ1 ð¡ â 2 ð
ðð¡ ð¡2ðœ2 ð¡ ðð¡
0
ðŒð
=1
ðŒð4 ð¡3ðœ1 ð¡ â 2ð¡2ðœ2 ð¡ 0
ðŒð
=1
ðŒð4 ðŒð
3ðœ1 ðŒð â 2ðŒð2ðœ 2 ðŒð
=1
ðŒð2 ðŒððœ1 ðŒð â 2ðœ2 ðŒð
=1
ðŒð2 ðŒððœ1 ðŒð â 2
2
ðŒð ðœ1 ðŒð â ðœ0 ðŒð
=1
ðŒð2
ðŒð2â4
ðŒð ðœ1 ðŒð â ðœ0 ðŒð
= ðŒð
2â4
ðŒð3 ðœ1 ðŒð as ðœ0 ðŒð = 0
⎠ðð =2
ðœ1 ðŒð
ðŒð2â4
ðŒð3
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Hence ð¥2 = 2 ðŒð
2â4
ðŒð3ðœ1 ðŒð
ðœ0 ðŒð ð¥
Example 28: Expand ð ð = ðð in the interval ð < ð¥ < 3 in terms of function ð±ð ð¶ð ð where
ð¶ð are determined by ð±ð ðð¶ = ð.
Solution: Let the Fourier-Bessel expansion of ð ð¥ = ð¥2 is
ð¥2 = ðð ðœ1 ðŒð ð¥ âð=1 , multiplying both sides by ð¥ðœ 1 ðŒð ð¥ and integrating from 0
to 3,
we get
ð¥4 ðœ1 ðŒð ð¥ ðð¥ = ðð ð¥ðœ1 ðŒð ð¥ ðð¥3
0
3
0
Let ð¥ = 3ð¡ so that ðð¥ = 3ðð¡
8 ð¡4ðœ1 3ðŒð ð¡ ð¡
0 3ðð¡ = ðð 3ð¡ ðœ1
2 3ðŒð ð¡ 3ðð¡1
0
⎠ðð ð¡ðœ121
0 3ðŒð ð¡ ðð¡ = 27 ð¡4ðœ1 3ðŒð ð¡ ðð¡
1
0
ðð1
2ðœ2
2 3ðŒð = 27 ð§4
81ðŒð4 ðœ1 ð§
ðð§
3ðŒð
3ðŒð
0 (where 3ðŒðð¡ = ð§ and ðð¡ =
ðð§
3ðŒð)
=1
9ðŒð5 ð§4ðœ1 ð§ ðð§
3ðŒð
0
=1
9ðŒð5 ð§2 ð
ðð§ ð§2ðœ2 ð§ ðð§
3ðŒð
0
=1
9ðŒð5 ð§2 . ð§2ðœ2 ð§ â 2ð§. ð§2ðœ2 ð§ ðð§ 0
3ðŒð
=1
9ðŒð5 ð§
4ðœ2 ð§ â 2 ð
ðð§ ð§3ðœ 3 ð§ ðð§
0
3ðŒð
=1
9ðŒð5 ð§4ðœ2 ð§ â 2ð§3ðœ3 ð§ 0
3ðŒð
=1
9ðŒð5 81ðŒð
4ðœ2 3ðŒð â 2 à 27ðŒð3ðœ3 3ðŒð
=1
ðŒð2 9ðŒððœ2 3ðŒð â 2ðœ3 3ðŒð
⎠ðð =6
ðŒð2ðœ2
2 3ðŒð 3ðŒððœ2 3ðŒð â 2ðœ3 3ðŒð
Hence ð¥3 = 6 3ðŒð ðœ2 3ðŒð â2ðœ3 3ðŒð
ðŒð2ðœ2
2 3ðŒð ðœ1 ðŒð ð¥ â
ð=1
ASSIGNMENT 18.4
1. Solve the differential equations:
(i) ðŠâ²â² +ðŠ â²
ð¥+ 8 â
1
ð¥2 ðŠ = 0
(ii) 4ðŠâ²â² + 9ð¥ðŠ = 0
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39
(iii) ð¥2ðŠâ²â² â ð¥ðŠâ² + 4ð¥2ðŠ = 0
2. If ðŒ1 , ðŒ2 ,⊠, ðŒð are the positive roots of ðœ0 ð¥ = 0, show that
1
2=
ðœ0(ðŒð ð¥)
ðŒð ðœ1(ðŒð)
â
ð=1
3. Expand ð ð¥ = ð¥2 in the interval 0 < ð¥ < 2 in terms of ðœ2(ðŒð ð¥), where ðŒð are determined
by ðœ2 ðŒð = 0.
4. Prove that
(i) ð
ðð¥ ð¥.
ð
ðð¥ ððð (ð¥) = âð¥ ððð (ð¥)
(ii) ð
ðð¥ ð¥.
ð
ðð¥ ððð (ð¥) = âð¥ ððð (ð¥)
ANSWERS
1. (i) ðŠ = ð¶1ðœ1 2 2ð¥ + ð¶2ðâ1 2 2ð¥
(ii) ðŠ = ð¥ ð¶1ðœ1
3
ð¥3
2 + ð¶2ðâ1
3
ð¥3
2
(iii) ðŠ = ð¥ ð¶1ðœ1 2 ð¥ + ð¶2ð1 2ð¥
3. ð¥2 = 4 ðœ2(ðŒð ð¥)
ðŒð ðœ3(2 ðŒð )âð=1
18.13 LEGENDREâS EQUATION
Legendreâs equation is one of the important differential equations occurring in applied mathematics,
particularly in boundary value problems for spheres. It is given as
1 â ð¥2 ð2ðŠ
ðð¥2 â 2ð¥ððŠ
ðð¥+ ð ð + 1 ðŠ = 0 (1)
where n is given real number. In most applications, n takes integral values.
The singularities of this equation are ð¥ = ±1.
Substituting ðŠ = ð0ð¥ð + ð1ð¥
ð+1 + ð2ð¥ð+2 + ⯠(ð0 â 0) in (1), we get
ð0 ð ð â 1 ð¥ðâ2 + ð1 ð + 1 ðð¥ðâ1 + â¯
+ ðð+2 ð + ð + 2 ð + ð + 1 â ð + ð ð + ð + 1 â ð ð + 1 ðð ð¥ð+ð
+ ⯠= 0
Equating to zero the co-efficient lowest powers of x, i.e of ð¥ðâ2, we get
ð0 ð ð â 1 = 0 â ð = 0, 1 ð0 â 0
Equating to zero the co-efficient of ð¥ðâ1 and ð¥ð+ð , we get
ð1 ð + 1 ð = 0 (2)
ðð+2 ð + ð + 2 ð + ð + 1 â ð + ð ð + ð + 1 â ð ð + 1 ðð = 0 (3)
When ð = 0, (2) is satisfied and therefore ð1 â 0. Then (3) for ð = 0, 1,2, 3⊠gives
ð2 = âð ð+1
2!ð0; ð3 = â
ðâ1 ð+2
3!ð1;
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ð4 = â(ðâ2) ð+3
4 .3ð2 =
ð ðâ2 ð+1 (ð+3)
4!ð0;
ð5 = â(ðâ3) ð+4
5 . 4ð3 =
(ðâ1) ðâ3 ð+2 (ð+4)
5!ð1; ðð¡ð.
Therefore two independent solutions of (1) for ð = 0 are as follows:
ðŠ1 = ð0 1 âð ð+1
2!ð¥2 +
ð ðâ2 ð+1 (ð+3)
4!ð¥4 â⯠(4)
ðŠ2 = ð1 ð¥ â ðâ1 ð+2
3!ð¥3 +
(ðâ1) ðâ3 ð+2 (ð+4)
5!ð¥5 â⯠(5)
When ð = 1, 2 gives that ð1 = 0. Therefore (3) gives
ð3 = ð5 = ð7 = 0
ð2 = âð ð+1
2!ð0; ð4 =
ð ðâ2 ð+1 (ð+3)
4!ð0; ðð¡ð
Thus for ð = 1, we get the solution (5) again. Hence the general solution of (1) is given by ðŠ =
ðŠ1 + ðŠ2.
Further, it is worth to note that if n is positive even integer, then (4) terminates at the term
containing ð¥ð and ðŠ1 becomes a polynomial of degree n. Similarly, if n is positive odd integer, then
ðŠ2 becomes a polynomial of degree n. Thus, whenever n is a positive integer (even or odd), the
general solution of (1) always contains a polynomial of degree n and an infinite series.
These polynomial solutions, with ð0 and ð1 chosen properly so that the value of the polynomial
becomes one at ð¥ = 1, are called Legendreâs Polynomials of degree n and is denoted by ð·ð(ð). The
infinite series with ð0 and ð1 chosen properly is called Legendreâs Function of second kind and is
denoted by ðžð ð .
18.14 RODRIGUEâS FORMULA
Another presentation of Legendreâs Polynomials is given by
ð·ð ð =ð
ð! ðð ð ð
ð ðð ðð â ð
ð (1)
is known as Rodrigueâs Formula.
Proof: Let ð£ = ð¥2 â 1 ð , then ð£1 =ðð£
ðð¥= 2ðð¥ ð¥2 â 1 ðâ1
i.e. 1 â ð¥2 ð£1 + 2ðð¥ ð£ = 0 (2)
Differentiating (2), n+1 times by Leibnitzâ theorem,
1 â ð¥2 ð£ð+2 + ð + 1 â2ð¥ ð£ð+1 +1
2! ð + 1 ð â2 ð£ð
+2ð ð¥ð£ð+1 + (ð + 1)ð£ð = 0
or 1 â ð¥2 ð2(ð£ð )
ðð¥2 â 2ð¥ð(ð£ð )
ðð¥+ ð ð + 1 (ð£ð) = 0
which is Legendreâs Equation and ðð£ð is its solution. Also its finite series solution is ðð ð¥ .
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41
⎠ðð ð¥ = ðð£ð = ð ðð
ðð¥ ð ð¥2 â 1 ð (3)
Putting ð¥ = 1 in equation (3) for determining the value of the constant c, we get
1 = ð ðð
ðð¥ ð ð¥ â 1 ð ð¥ + 1 ð ð¥=1
= ð ð! ð¥ + 1 ð + ð¡ðððð ð€ðð¡ð ð¥ â 1 ððð ðð¡ð ððð€ððð ð¥=1
= ð.ð! 2 ð , ð. ð., ð =1
ð! 2ð
Substituting the value of c in (3), we get eqution (1) which is known as Rodrigueâs formula.
18.15 LEGENDREâS POLYNOMIALS
By Rodrigueâs formula we have
ð0 ð¥ = 1, ð1 ð¥ = ð¥,
ð2 ð¥ =1
2 3ð¥2 â 1 , ð3 ð¥ =
1
2 5ð¥3 â 3ð¥ ,
ð4 ð¥ =1
8 35ð¥4 â 30ð¥2 + 3 ,
ð5 ð¥ =1
8 63ð¥5 â 70ð¥3 + 15ð¥ .
In general, ðð ð¥ = â1 ð 2ðâ2ð !
2ð ð! ðâð ! ðâ2ð ! ð¥ðâ2ðð
ð=0 where ð =1
2ð ðð
1
2 (ð â 1) according as n is
even or odd.
This general expression for ðð ð¥ in terms of sum of finite number of terms can be derived easily
from Rodrigueâs formula.
Example 29: Show that ð·ð âð = âð ðð·ð ð
Solution: ðð ð¥ = â1 2 2ðâ2ð !
ð! ðâð ! ðâ2ð !ð¥ð±2ðð
ð=0
Where ð =ð
2 or
ðâ1
2
⎠Replacing ð¥ by â ð¥, we will get
ðð ð¥ = â1 ð 2ðâ2ð !
ð ! ðâ2ð ! ðâð ! â1 ðâ2ðð¥ðâ2ðð
ð=0
= â1 ð â1 ð 2ðâ2ð !
ð! ðâð ! ðâ2ð !ð¥ðâ2ðâ
ð=0 , as â1 2ð = 1
= â1 ððð ð¥
Example 30: Express the following in the Legendre Polynomials
(i) ððð + ð (ii) ðð + ððð â ð â ð (iii) ððð â ððð â ðð + ð
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Solution: We know ðð ð¥ =1
ð! 2ð ð·ð ð¥2 â 1 ð
⎠ð0 ð¥ = 1 ð1 ð¥ = ð¥ ð2 ð¥ =1
2 3x2 â 1 ð3 ð¥ =
1
2 5ð¥3 â 3ð¥
(i) 5ð¥3 + ð¥ = 2.1
2 5ð¥3 â 3ð¥ + 4ð¥ = 2ð3 ð¥ + 4ð1 ð¥
ð¥3 =1
5 2ð3 ð¥ + 3ð1 ð¥ , ð¥
2 =1
3 2ð2 ð¥ + ð0 ð¥ , ð¥ = ð1 ð¥ , 1 = ð0 ð¥
(ii) ð¥3 + 2ð¥2 â ð¥ â 3 =1
5 2ð3 ð¥ + 3ð1 ð¥ +
2
3 2ð2 ð¥ + ð0 ð¥ â ð1 ð¥ â ð0 ð¥
=2
5ð3 ð¥ +
4
3ð2 ð¥ â
2
5ð1 ð¥ â
7
3ð0 ð¥
(iii) 4ð¥3 â 2ð¥2 â 3ð¥ + 8 =4
5 2ð3 ð¥ + 3ð1 ð¥ â
2
3 2ð2 ð¥ + ð0 ð¥ â 3ð1 ð¥ + 8ð0 ð¥
=8
5ð3 ð¥ â
4
3ð2 ð¥ â
9
5ð1 ð¥ +
22
3ð0 ð¥
18.16 GENERATING FUCTION FOR ð·ð(ð)
To show that ð â ððð + ðð âð
ð = ððð·ð ð âð=ð
Proof: We know that
(1 â ð§)â1
2 = 1 +1
2ð§ +
1
2.3
2
2! ð§2 +
1
2.3
2.5
2
3! ð§3 + â¯
= 1 +2!
1! 2 22 ð§ +4!
2! 2 24 ð§2 +6!
3! 2 26 ð§3 + â¯
⎠1 â ð¡ 2ð¥ â ð¡ â
1
2 = 1 +2!
1! 2 22 ð¡ 2ð¥ â ð¡ +4!
2! 2 24 ð¡ 2ð¥ â ð¡ 2
+ â¯
+(2ðâ2ð)!
(ðâð)! 2 22ðâ2ð (ð¡ 2ð¥ â ð¡ )ðâð + â¯+(2ð)!
ð ! 2 22ð (ð¡ 2ð¥ â ð¡ )ð (1)
The term in ð¡ð from the term containing ð¡ðâð 2ð¥ â ð¡ ðâð
=(2ðâ2ð)!
(ðâð)! 2 22ðâ2ð ð¡ðâð .ð â ðð¶ð âð¡ ð 2ð¥ ðâ2ð
=(2ðâ2ð)!
(ðâð)! 2 22ðâ2ð Ã ðâð !
ð ! ðâ2ð ! â1 ðð¡ð 2ð¥ ðâ2ð =
â1 ð(2ðâ2ð)!
2ð ð! ðâð ! ðâ2ð !ð¡ðð¥ðâ2ð
Collecting all terms in ð¡ð which will occur in the term containing ð¡ð 2ð¥ â ð¡ ð and the proceeding
terms, we see that terms in ð¡ð
= â1 ð(2ðâ2ð)!
2ð ð! ðâð ! ðâ2ð !ð¡ðð¥ðâ2ðð
ð=0 = ðð ð¥ ð¡ð
where ð =1
2 ð ðð
1
2(ð â 1) according as n is even or odd.
Hence (1) can be written as 1 â 2ð¥ð¡ + ð¡2 â1
2 = ð¡ððð ð¥ âð=0 , which is known as generating
function of Legendreâs Polynomials.
18.17 RECURRENCE RELATION FOR ð·ð(ð)
I. ð + ð ð·ð+ð ð = ðð + ð ð·ð ð â ðð·ðâð ð .
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43
Proof: We have the generating functions
(1 â 2ð¥ð¡ + ð¡2)â1/2 = ðð(ð¥)âð=0 ð¡ð (1)
Differentiate partially w.r.t. t, we get
â1
2 1 â 2ð¥ð¡ + ð¡2 â
3
2(â2ð¥ + 2ð¡) = ðð(ð¥)âð=0 ðð¡ðâ1
1 â 2ð¥ð¡ + ð¡2 â3
2(ð¥ â ð¡) = ðð(ð¥)âð=0 ðð¡ðâ1 (2)
1 â 2ð¥ð¡ + ð¡2 â1
2(ð¥ â ð¡) = 1 â 2ð¥ð¡ + ð¡2 ðð(ð¥)âð=0 ðð¡ðâ1
(ð¥ â ð¡) ðð(ð¥)âð=0 ð¡ðâ1 = 1 â 2ð¥ð¡ + ð¡2 ðð(ð¥)â
ð=0 ðð¡ðâ1
Comparing the coefficients of ð¡ð from both sides, we get
ð¥ ðð ð¥ â ððâ1 ð¥ = ð + 1 ðð+1 ð¥ â 2ð¥ððð ð¥ + (ð â 1)ððâ1(ð¥)
ð + 1 ðð+1 ð¥ = 2ð + 1 ð¥ ðð ð¥ â ðððâ1(ð¥)
II. ð ð·ð ð = ðð·ðâ² ð â ð·â²
ðâð ð .
Proof: Differentiating (1) partially w.r.t x, we obtain
â1
2 1 â 2ð¥ð¡ + ð¡2 â
3
2(â2ð¡) = ððâ²(ð¥)â
ð=0 ð¡ð
ð¡ 1 â 2ð¥ð¡ + ð¡2 â3
2 = ððâ²(ð¥)â
ð=0 ð¡ð (3)
Dividing (2) by (3), we get
ð¥âð¡
ð¡=
ð ððâ² (ð¥)â
ð=0 ð¡ðâ1
ððâ² (ð¥)â
ð=0 ð¡ð
ð¥ â ð¡ ððâ² ð¥ â
ð=0 ð¡ð = ð¡. ð ððâ²(ð¥)â
ð=0 ð¡ðâ1 = ððâ²(ð¥)â
ð=0 ð¡ð
Comparing the coefficient of ð¡ð from both sides, we get
ð¥ððâ² ð¥ â ðâ²
ðâ1 ð¥ = ð ðð ð¥
III. (ðð + ð) ð·ð ð = ð·â²ð+ð ð â ð·â²
ðâð ð .
Proof: From relation I, we have
ð + 1 ðð+1 ð¥ = 2ð + 1 ðð ð¥ â ðððâ1 ð¥
Differentiating w.r.t x, we get
ð + 1 ðâ²ð+1 ð¥ = 2ð + 1 ðð ð¥ + 2ð + 1 ð¥ ðð
â² ð¥ â ððâ²ðâ1 ð¥ (4)
Using ð¥ððâ² ð¥ â ðâ²
ðâ1 ð¥ = ð ðð ð¥
Or ð¥ððâ² ð¥ = ð ðð ð¥ + ðâ²
ðâ1 ð¥ (5)
Now eliminating the term ð¥ððâ² ð¥ from (4) using (5), we get
ð + 1 ðâ²ð+1 ð¥ = 2ð + 1 ðð ð¥ + 2ð + 1 ð ðð ð¥ + ðâ²
ðâ1 ð¥ â ððâ²ðâ1 ð¥
ð + 1 ðâ²ð+1 ð¥ = ð + 1 2ð + 1 ðð ð¥ + ð + 1 ðâ²
ðâ1 ð¥
ðâ²ð+1 ð¥ = 2ð + 1 ðð ð¥ + ðâ²
ðâ1 ð¥
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44
2ð + 1 ðð ð¥ = ðâ²ð+1 ð¥ â ðâ²
ðâ1 ð¥
IV. ð·ðâ² ð = ð ð·â²
ðâð ð â ð ð·ðâð ð .
Proof: Rewriting (4) as
ð + 1 ðâ²ð+1 ð¥
= 2ð + 1 ðð ð¥ + ð + 1 ð¥ ððâ² ð¥ + ð ð¥ðâ²
ðâ1 ð¥ â ðâ²ðâ1 ð¥
= 2ð + 1 ðð ð¥ + ð + 1 ð¥ ððâ² ð¥ + ð2ðð ð¥
= ð + 1 ð¥ ððâ² ð¥ + ð2 + 2ð + 1 ðð ð¥
= ð + 1 ð¥ ððâ² ð¥ + (ð + 1)2ðð ð¥
ðâ²ð+1 ð¥ = ð¥ ðð
â² ð¥ + (ð + 1)ðð ð¥
V. (ð â ðð)ð·ðâ² ð = ð ð·ðâð ð â ð ð·ð ð .
Proof: From Relation II, we have
ð¥ððâ² ð¥ â ðâ²
ðâ1 ð¥ = ð ðð ð¥ (6)
Also from relation IV, we have
ðâ²ð ð¥ â ð¥ ððâ1
â² ð¥ = ð ððâ1 ð¥ (7)
Multiply equation (7) by x and subtracting form equation (6), we get
(1 â ð¥2)ððâ² ð¥ = ð ððâ1 ð¥ â ð¥ ðð ð¥
18.18 ORTHOGONALITY OF LEGENDRE`S POLYNOMIALS
The Legendre Polynomial ð·ð ð satisfy the following orthogonality property
ð·ð ð .ð·ð ð ð ð =
ð, ð â ð
ð
ðð + ð, ð = ð
ð
âð
Proof: Both of the cases are discussed as follows:
Case I: ð â ð
Let the Legendre polynomials ðð ð¥ and ðð ð¥ satisfy the differential equations
1 â ð¥2 ðâ²â²ð â 2ð¥ ðâ²
ð + ð ð + 1 ðð = 0 (1)
1 â ð¥2 ðâ²â²ð â 2ð¥ ðâ²
ð + ð ð + 1 ðð = 0 (2)
Multiplying (1) by ðð ð¥ and (2) ðð ð¥ and then subtracting we get
1 â ð¥2 ðâ²â²ð .ðð â ðâ²â²
ð .ðð â 2ð¥ ðâ²ð .ðð â ðâ²
ð .ðð
+ ð ð + 1 â ð(ð + 1) ðð .ðð = 0
ð
ðð¥ 1 â ð¥2 (ðâ²
ð .ðð â ðâ²ð .ðð ) + ð â ð (ð + ð + 1)ðððð = 0
ð â ð ð + ð + 1 ðððð = âð
ðð¥ 1 â ð¥2 (ðâ²
ð .ðð â ðâ²ð .ðð )
Integrating from -1 to 1 both sides
ð â ð ð + ð + 1 ðð ð¥ .ðð ð¥ ðð¥1
â1= â 1 â ð¥2 (ðâ²
ð .ðð â ðâ²ð .ðð ) â1
1 = 0
ðð ð¥ .ðð ð¥ ðð¥ = 01
â1
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Case II: ð = ð
We know from generating functions that
1 â 2ð¥ð¡ + ð¡2 â1
2 = ð¡ððð ð¥ âð=0 (3)
Squaring both sides and integrating w.r.t. x from -1 to 1, we get
1
1â2ð¥ð¡+ð¡2 ðð¥1
â1= ð¡ððð ð¥
âð=0 2ðð¥
1
â1 (4)
Now 1
1â2ð¥ð¡+ð¡2 ðð¥1
â1=
ln 1â2ð¥ð¡+ð¡2
â2ð¡ â1
1
= â1
2ð¡ ln 1 â 2ð¡ + ð¡2 â ln 1 + 2ð¡ + ð¡2
= â1
2ð¡ ln 1 â ð¡ 2 â ln 1 + ð¡ 2 = â
1
ð¡ ln(1 â ð¡) â ln(1 + ð¡ )
=1
ð¡ ln(1 + ð¡) â ln(1 â ð¡ )
=1
ð¡ ð¡ â
ð¡2
2+
ð¡3
3â⯠â âð¡ â
ð¡2
2â
ð¡3
3ââ¯
= 2 1 +ð¡2
3+
ð¡4
5+ â¯+
ð¡2ð
2ð+1+ ⯠(5)
Also ð¡ððð ð¥ âð=0 2ðð¥
1
â1= ð¡ððð ð¥
âð=0 . ð¡ððð ð¥
âð=0 ðð¥
1
â1
= ð¡2ððð2 ð¥
1
â1âð=0 ðð¥
= ð¡2ð ðð2 ð¥
1
â1âð=0 ðð¥ (6)
Using (5) and (6) in equation (4), we get
2 1 +ð¡2
3+
ð¡4
5+ â¯+
ð¡2ð
2ð+1+ ⯠= ð¡2ð ðð
2 ð¥ 1
â1âð=0 ðð¥
Comparing the coefficient of ð¡2ð on both sides we get
ðð2 ð¥
1
â1ðð¥ =
2
2ð+1 .
18.19 FOURIER LEGENDRE EXPANSION
If ð ð¥ be a continuous function and having continuous derivatives over the interval [-1, 1], then we
can write
ð ð¥ = ð¶ð ðð(ð¥)âð=0 (1)
To determine the coefficient ð¶ð , multiply both sides by ð ð(ð¥) and integrate form -1 to 1, we get
ð ð¥ .ðð ð¥ ðð¥1
â1= ð¶ð ðð
2(ð¥)1
â1ðð¥
(Remaining terms vanishes by the orthogonal property)
= ð¶ð .2
2ð+1
ð¶ð = ð +1
2 . ð ð¥ .ðð ð¥ ðð¥
1
â1 (2)
The series in (1) converges uniformly in interval [-1, 1], and is known as Fourier-Legendre Expansion
of ð ð¥ .
Example 31: Prove that (i) ð·ððâ² ð = ð and ð·ðð+ð
â² ð = âð ð ðð+ð !
ððð ð! ð
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46
Solution: We know ð¡ððð ð¥ âð=0 = 1 â 2ð¥ð¡ + ð¡2 â
1
2
Differentiating with respect to â²ð¥â², we get
ð¡ðððâ² ð¥ = â
1
2 1 â 2ð¥ð¡ + ð¡2 â
3
2 â2ð¡
= ð¡ 1 â 2ð¥ð¡ + ð¡2 â3
2
Putting ð¥ = 0, ððâ² 0 â
ð=0 = ð¡ 1 + ð¡2 â3
2
= ð¡ 1 â3
2ð¡2 +
â3
2Ãâ
5
2
2!ð¡4 + âŠâŠ+
â3
2Ãâ
5
2ÃâŠâŠ â
3
2âðâ1
ð !ð¡2ð + âŠ
Equating the coefficients of ð¡2ð and ð¡2ð+1, we get ð2ðâ² 0 = 0
ð2ð+1â² 0 = â1 ð
3Ã5ÃâŠâŠ 2ð+1
2ð ð!
= â1 ð 2ð+1 !
2ðð!22ð
ð2ð+1â² 0 = â1 ð
2ð+1 !
22ð ð!2
Example 32: Prove that
(i) ð â ðð ð·ðâ² ð = ð + ð ðð·ð ð â ð·ð+ð ð
(ii) ðð + ð ð â ðð ð·ðâ² ð = ð ð + ð ð·ðâð ð â ð·ð+ð ð
(iii) ð·ð ð = ð·ð+ðâ² ð â ððð·ð
â² ð + ð·ðâðâ² ð
Solution: We know ð¡ððð ð¥ = 1 â 2ð¥ð¡ + ð¡2 â1
2
(i) Differentiating with respect to â²ð¡â² and equating the coefficients of ð¡ð , we will get
ð + 1 ðð+1 ð¥ = 2ð + 1 ð¥ðð ð¥ â ðððâ1 ð¥ (1)
Now differentiating with respect to â²ð¥â² and using the derivative with respect â²ð¡â², we get
ððð ð¥ = ð¥ððâ² ð¥ â ððâ1
â² ð¥ (2)
From (1) & (2), we can derive
2ð + 1 ðð ð¥ = ðð+1â² ð¥ â ððâ1
â² ð¥ (3)
ððâ² ð¥ = ð¥ððâ1
â² ð¥ + ðððâ1 ð¥ (4)
From (1) & (4) eliminate ððâ1 ð¥
ð + 1 ðð+1 ð¥ + ððâ² ð¥ = 2ð + 1 ð¥ðð ð¥ + ð¥ððâ1
â² ð¥ (5)
= 2ð + 1 ð¥ðð ð¥ + ð¥ ð¥ððâ² ð¥ â ððð ð¥ , From (4)
1 â ð¥2 ððâ² ð¥ = ð + 1 ð¥ðð ð¥ â ð + 1 ðð+1 ð¥
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= ð + 1 ð¥ðð ð¥ â ðð+1 ð¥
(i) Eliminating ððâ1â² ð¥ from (2) & (4), we get
1 â ð¥2 ððâ² ð¥ = ð ððâ1 ð¥ â ð¥ðð ð¥
= ð ððâ1 ð¥ â1
2ð+1 ð + 1 ðð+1 ð¥ + ðððâ1 ð¥
=ð
2ð+1 2ð + 1 â ð ððâ1 ð¥ â ð + 1 ðð+1 ð¥
2ð + 1 1 â ð¥2 ððâ² ð¥ = ð ð + 1 ððâ1 ð¥ â ðð+1 ð¥
(ii) (3)â2 Ã(2) gives
ðð ð¥ = ðð+1â² ð¥ â 2ð¥ðð
â² ð¥ + ððâ1â² ð¥
Example 33: Using the Rodrigueâs formula, show that
ð
ð ð ð â ðð
ð
ð ð ð·ð ð + ð ð + ð ð·ð ð = ð
Solution: We know that ðð ð¥ =1
2ð ð! ð·ð ð¥2 â 1 ð =
1
2ðð!ð·ðð, ð = ð¥2 â 1 ð
Now differentiating â²ðâ² with respect to â²ð¥â² , we get
ð1 = 2ðð¥ ð¥2 â 1 ðâ1 or ð¥2 â 1 ð1 = 2ðð¥ð or 1 â ð¥2 ð1 + 2ðð¥ð = 0
Differentiating ð + 1 times, we get
1 â ð¥2 ðð+2 + ð + 1 ðð+1 â2ð¥ + ð + 1 ð
2!ðð â2 + 2ðð¥ðð+1
+ ð + 1 2ððð = 0
1 â ð¥2 ðð+2 â 2ð¥ ð + 1 â ð ðð+1 + ðð âð ð + 1 + 2 ð + 1 ð = 0
1 â ð¥2 ðð+2 â 2ð¥ðð+1 + ð ð + 1 ðð = 0
1 â ð¥2 ð2
ðð¥2 ðð â 2ð¥ð
ðð¥ðð + ð ð + 1 ðð = 0
But ðð = ð·ðð = 2ðð!ðð ð¥
1 â ð¥2 ð2
ðð¥2 2ðð!ðð ð¥ â 2ð¥
ð
ðð¥ 2ðð!ðð ð¥ + ð ð + 1 2ðð!ðð ð¥ = 0
1 â ð¥2 ð2
ðð¥2 ðð ð¥ â 2ð¥ð
ðð¥ðð ð¥ + ð ð + 1 ðð ð¥ = 0
Example 34: Prove that
(i) ð·ðð ð ð ð = ðð
ð
(ii) ððð·ð ð ð ð = ðð
âð ð < ð
Solution: (i) we know 2ð + 1 ðð ð¥ = ðð+1â² ð¥ â ððâ1
â² ð¥
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⎠4ð + 1 ð2ð ð¥ = ð2ð+1â² ð¥ â ð2ðâ1
â² ð¥
Integrating both sides
4ð + 1 ð2ð ð¥ ðð¥1
0= ð2ð+1 ð¥ â ð2ðâ1 ð¥ 0
1
= ð2ð+1 1 â ð2ðâ1 1 â ð2ð+1 0 â ð2ðâ1 0
= 1 â 1 â 0 â 0 = 0
(ii) ð¥ððð ð¥ ðð¥ = ð¥ð 1
2ðð!ð·ð ð¥2 â 1 ð
1
â1
1
â1
=1
2ðð! ð¥ðð·ðâ1 ð¥2 â 1 ð â1
1 â ðð¥ðâ1ð·ðâ1 ð¥2 â 1 ððð¥1
â1
=1
2ðð! 0 âð ð¥ðâ1ð·ðâ1 ð¥2 â 1 ððð¥
1
â1
=1
2ðð!à â1 ð ð·ðâð ð¥2 â 1 ððð¥
1
â1
=1
2ðð! â1 ð ð·ðâðâ1 ð¥2 â 1 ð â1
1 = 0
As ð·ðâðâ1 ð¥ â 1 ð ð¥ + 1 ð = 0 will contain terms in ð¥ â 1 and ð¥ + 1 both and hence
when ð¥ = ±, the value is zero.
Example 35: Prove that ð·ð ð ð â ððð + ðð âð
ðð ð =ððð
ðð+ð
ð
âð.
Solution: We know 1 â 2ð¥ð + ð2 â1
2 = ðððð ð¥
⎠ðð ð¥ 1
â1 ðððð ð¥ ðð¥ = ðð ðð ð¥ ðð ð¥ ðð¥
1
â1
= ðð 0, ð â ð 2
2ð+1, ð = ð
=2ðð
2ð+1
Example 36: Show that
(i) ðð·ð ð ð·ðâð ð ð ð =ðð
ðððâð
ð
âð
(ii) ððð
âðð·ð+ð ð ð·ðâð ð ð ð =
ðð ð+ð
ððâð ðð+ð ðð+ð
(iii) ð â ðð ð
âð ð·ð
â² ð ðð ð =ðð ð+ð
ðð+ð
(iv) ð â ðð ð·ðâ² ð ð·ð
â² ð ð ð = ðð
âð
Solution: (i) We know ð + 1 ðð+1 ð¥ = 2ð + 1 ð¥ðð ð¥ â ðððâ1 ð¥
ð¥ðð ð¥ =1
2ð+1 ð + 1 ðð+1 ð¥ + ðððâ1 ð¥
⎠ð¥ðð ð¥ ððâ1 ð¥ ðð¥1
â1
= 1
2ð+1 ð + 1 ðð+1 ð¥ + ðððâ1 ð¥ ððâ1 ð¥ ðð¥
1
â1
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49
=ð+1
2ð+1 ðð+1 ð¥ ððâ1 ð¥ ðð¥ +
ð
2ð+1
1
â1 ððâ12 ð¥
1
â1ðð¥
=ð+1
2ð+1Ã 0 +
ð
2ð+1Ã
2
2ðâ1=
2ð
4ð2â1
(ii) We know ð¥ðð ð¥ =1
2ð+1 ð + 1 ðð+1 + ðððâ1 ð¥
Changing ð â ð + 1
ð¥ðð+1 ð¥ =1
2ð+1 ð + 2 ðð+2 ð¥ + ð + 1 ðð ð¥
and changing ð â ð â 1
ð¥ððâ1 ð¥ =1
2ðâ1 ððð ð¥ + ð â 1 ððâ2 ð¥
⎠ð¥2ðð+1 ð¥ ððâ1 ð¥ ðð¥1
â1
= 1
2ð+3 ð + 2 ðð+2 ð¥ + ð + 1 ðð ð¥ à ððð ð¥ + ð â 1 ððâ2 ð¥ ðð¥
1
â1
=1
2ðâ1 2ð+3 0 + 0 + ð ð + 1 Ã
2
2ð+1+ 0
=2ð ð+1
2ðâ1 2ð+1 2ð+3
(iii) 1 â ð¥2 ððâ² ð¥ 2ðð¥
1
â1= 1 â ð¥2 ðð
â² ð¥ .ððâ² ð¥
1
â1
= 1 â ð¥2 ððâ² ð¥ ðð ð¥ â1
1 â ð
ðð¥ 1 â ð¥2 ðð
â² ð¥ 1
â1ðð ð¥ ðð¥
= 0 â âð ð + 1 ðð ð¥ ðð¥1
â1= ð ð + 1 ðð
2 ð¥ 1
â1ðð¥ =
2ð ð+1
2ð+1
(iv) 1 â ð¥2 ððâ² ð¥ ðð
â² ð¥ ðð¥1
â1
= 1 â ð¥2 ððâ² ð¥ ðð ð¥ â1
1 â ð
ðð¥ 1 â ð¥2 ðð
â² ð¥ 1
â1ðð ð¥ ðð¥
= 0 â 0 + ð ð + 1 ðð ð¥ ðð ð¥ 1
â1ðð¥ = ð ð + 1 à 0 = 0
Example 37: Expand the following functions in terms of Legendreâs polynomials in the interval
ð,âð
(i) ð ð = ðð + ððð â ð â ð (ii) ð ð = ðð + ðð + ððð â ð â ð
Solution: (i) We know ð ð¥ = ðððð ð¥ âð=0
Where ðð = ð +1
2 ð ð¥ ðð ð¥ ðð¥
1
â1
⎠ð0 = 0 +1
2 ð¥3 + 2ð¥2 â ð¥ â 3 à 1
1
â1ðð¥
=1
2
ð¥4
4+ 2
ð¥3
3â
ð¥2
2â 3ð¥
â1
1
=1
2 0 +
4
3â 6 =
2
3â 3 = â
7
3
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ð1 = 1 +1
2 ð¥3 + 2ð¥2 â ð¥ â 3 ð¥ ðð¥
1
â1=
3
2 ð¥4 + 2ð¥3 â ð¥2 â 3ð¥
1
â1ðð¥
= â3
2
2
5â
2
3 = 3
3â5
15 = â
6
15= â
2
5
ð2 = 2 +1
2 ð¥3 + 2ð¥2 â ð¥ â 3
1
2 3ð¥2 â 1 ðð¥
1
â1
=5
4 3ð¥5 â ð¥3 + 6ð¥4 â 2ð¥2 â 3ð¥3 + ð¥ â 9ð¥2 + 3 ðð¥
1
â1
=5
4 6 Ã
2
5â
4
3â
9Ã2
3+ 6 =
4
3
ð ð¥ = â7
3ð0 ð¥ â
2
5ð1 ð¥ +
4
3ð2 ð¥ + âŠâŠ
(i) ð ð¥ = ð¥4 + ð¥3 + 2ð¥2 â ð¥ â 3 = ðððð ð¥
ðð = ð +1
2 ð ð¥ ðð ð¥ ðð¥
1
â1
ð0 = 0 +1
2 ð¥4 + ð¥3 + 2ð¥2 â ð¥ â 3 à 1 ðð¥
1
â1
=1
2
2
5+
4
3â 6 =
1
2Ã
6+20â90
15= â
32
15
ð1 = 1 +1
2 ð¥4 + ð¥3 + 2ð¥2 â ð¥ â 3 à ð¥ ðð¥
1
â1
ð1 =3
2
2
5â
2
3 =
3
2Ã
6â10
15= â
2
5
ð2 = 2 +1
2 ð¥4 + ð¥3 + 2ð¥2 â ð¥ â 3 Ã
1
2 3ð¥2 â 1
1
â1ðð¥
=5
4 3ð¥6 â ð¥4 + 3ð¥5 â ð¥3 + 6ð¥4 â 2ð¥2 â 3ð¥3 + ð¥ â 9ð¥2 + 3 ðð¥
1
â1
=5
4
6
7â
2
5+
12
5â
4
3â 6 + 6 =
40
21
ð ð¥ = â32
15ð0 ð¥ â
2
5ð1 ð¥ +
40
21ð2 ð¥ + âŠâŠ
ASSIGNMENT 18.5
1. Show that ðâ²ð âð¥ = â1 ð+1ðâ²
ð ð¥ .
2. Evaluate the following:
(i) ð23ð ð¥ ðð¥
1
0
(ii) ð¥ð .ðð ð¥ ðð¥1
â1,ð€ðððð ð ðð ðð ððð¡ðððð ððð ð ð¡ððð ð.
3. Express 8 ð5 ð¥ â 8 ð4 ð¥ â 2 ð2 ð¥ + 5ð0(ð¥) in terms of polynomial of x.
4. Use Rodrigues formulae to obtain ð3(ð¥) and ð4 ð¥ .
5. Find the value of cos ð¡ .ð3(sin ð¡) ðð¡ð/2
0.
6. Prove that
(i) ðð (ð¥)
1â2ð¥ð§+ð§2 ðð¥ =
2ð§ð
2ð+1
1
â1
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(ii) ðð(ð¥) ðð¥ = 01
â1 except when ð = 0 in which case the value of the integral is 2.
ANSWERS
2. (i) 1
6ð+1
(ii) 0
3. 63 ð¥5 â 35 ð¥4 â 70ð¥3 + 27ð¥2 + 15ð¥ + 3
4. ð3 ð¥ =1
2 5ð¥3 â 3ð¥ ,ð4 ð¥ =
1
8 35ð¥4 â 30ð¥2 + 3
5. -1/8
18.20 STRUM â LIOUVILLE PROBLEMS
A differential equation of the form
ð ð¥ ðŠâ² â² + ð ð¥ + ð ð(ð¥) ðŠ = 0 (1)
is called Strum-Liouville Equation where ð is a real number.
Instead of initial conditions, this equation is usually subjected to the boundary conditions on the
interval ð, ð as
ðŒ1 ðŠ ð + ðŒ2 ðŠâ² ð = 0, ðœ1 ðŠ ð + ðœ2 ðŠâ² ð = 0 (2)
where ðŒ1 ,ðŒ2 ,ðœ1 ,ðœ2 are real constants such that either ðŒ1 ðð ðŒ2 are not zero and ðœ1 ðð ðœ2 are not
zero.
The non trivial solutions of the differential equation (1) subjected to the conditions (2) exists only for
specific values of ð, which values are termed as Eigen values or Characteristic values of the equation
(1). And the non trivial solution of (1) corresponding to these Eigen values are termed as Eigen
functions or Characteristic functions.
18.21 ORTHOGONALITY OF EIGEN FUNCTIONS
Two functions ðŠð (ð¥) and ðŠð(ð¥) defined on some interval ð, ð are said to be orthogonal on this
interval with respect to the weight function ð ð¥ > 0, if
ð ð¥ ð
ððŠð ð¥ ðŠð ð¥ ðð¥ = 0 ððð ð â ð
Also the norm ðŠð of the function ðŠð (ð¥) is defined to be non negative square root of
ð ð¥ ð
ð ðŠð ð¥
2ðð¥. Thus ðŠð = ð ð¥
ð
ð ðŠð ð¥
2ðð¥.
The functions which are orthogonal and having the norm unity are said to be orthonormal functions.
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52
Theorem: If ðð(ð) and ðð(ð) are two eigen functions of the Strum-Liouville problem
corresponding to eigen values ðð and ðð respectively (where ð â ð), then the eigen functions are
orthogonal w.r.t. the weight function ð(ð) over the interval ð,ð .
Proof: Since distinct eigen values and their corresponding eigen functions are the solutions of the
Stum Liouville equation (1), so we can write it as
ð ð¥ ðŠðâ² â² + ð ð¥ + ðð ð(ð¥) ðŠð = 0
ð ð¥ ðŠðâ² â² + ð ð¥ + ðð ð(ð¥) ðŠð = 0
Multiplying first equation by ðŠð and the second equation by ðŠð , and then subtracting, we get
ðð â ðð ð ð¥ ðŠððŠð = ðŠð ð ð¥ ðŠðâ² â² â ðŠð ð ð¥ ðŠð
â² â²
=ð
ðð¥ ð ð¥ ðŠð
â² ðŠð â ð ð¥ ðŠðâ² ðŠð
Now integrating both sides w.r.t. x from a to b, we get
ðð â ðð ð ðŠððŠððð¥ = ð ð¥ ðŠðâ² ðŠð â ð ð¥ ðŠð
â² ðŠð ððð
ð
= ð ð ðŠðâ² ð ðŠð ð â ðŠð
â² ð ðŠð ð â ð ð ðŠðâ² ð ðŠð ð â ðŠð
â² ð ðŠð ð
The R.H.S. will vanish if the boundary conditions are of one of the followings forms:
I. ðŠ ð = ðŠ ð = 0
II. ðŠâ² ð = ðŠâ² ð = 0
III. ðŒ1 ðŠ ð + ðŒ2 ðŠâ² ð = 0, ðœ1 ðŠ ð + ðœ 2 ðŠâ² ð = 0
where ðŒ1 ,ðŒ2 ,ðœ1 ,ðœ2 are real constants such that either ðŒ1 ðð ðŒ2 are not zero and ðœ1 ðð ðœ2 are not
zero. Thus in each of the three cases we get
ð ðŠððŠððð¥ = 0, (ð â ð)ð
ð
which shows that the eigen functions ðŠð (ð¥) and ðŠð(ð¥) are orthogonal w.r.t. the weight function ð(ð¥)
over the interval ð, ð .
Example 38: For Strum-Liouville problem ðâ²â² + ðð = ð, ð ð = ð, ð ð = ð find the eigen
functions.
Solution: For ð = âðŸ2, the general solution of the equation is given by
ðŠ ð¥ = ð¶1 ððŸð¥ + ð¶2 ðâðŸð¥
Using the above mentioned boundary conditions we get ð¶1 = ð¶2 = 0. Hence ðŠ ð¥ = 0 is not an
eigen function.
Also for ð = ðŸ2, the general solution of the equation is given by
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53
ðŠ ð¥ = ð¶1 cos ðŸð¥ + ð¶2 sinðŸð¥
Using ðŠ 0 = 0, we get ð¶1 = 0
Using ðŠ ð = 0, we get ð¶2 sin ðŸð = 0 => sinðŸð = 0
⎠ðŸ ð = ðð => ðŸ = ð, ð = ±1, ±2, ±3,âŠ.
Thus the eigen values are ð = 0, 1, 4, 9,⊠and taking ð¶2 = 1, we obtain the eigen functions as
ðŠð ð¥ = sinðð¥ , ð = 0, 1, 2,âŠ
ASSIGNMENT 18.6
1. Find the eigen values of each of the following Stum Liouville problems and prove their
orthogonality:
i) ðŠâ²â² + ððŠ = 0, ðŠ 0 = 0, ðŠ ð = 0
ii) ðŠâ²â² + ððŠ = 0, ðŠâ² 0 = 0, ðŠâ² ð = 0
iii) ðŠâ²â² + ððŠ = 0, ðŠ ð = ðŠ(âð), ðŠâ² ð = ðŠâ²(âð)
2. Show that the eigen values of the boundary value problem ðŠâ²â² + ððŠ = 0, ðŠ 0 = 0, ðŠ ð +
ðŠâ² ð = 0 satisfies ð + tan ð ð = 0.
ANSWERS
1. (i) sinððð¥
ð, ð = 0, 1, 2,âŠ
(ii) cosððð¥
ð, ð = 0, 1, 2,âŠ
(iii) 1, sinð¥ , cos ð¥ , sin 2ð¥ , cos 2ð¥ ,âŠ