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subject: Application Of Series [print this page]


In this page we are going to discuss about application of series concept .For solving general linear first order differential equations , constant coefficient second and higher order differential euations and a special case of a variable coefficient differential equation ( Euler-Cauchy type equation ). The solution of these equations were all closed form solutions in terms of standard functions. However, it is often not possible to express the solutions of vaiable coeficient equations in closed form using the standard functions. In such cases, we seek the solution as an infinite series in terms of the independent variable. Many of the important physical problems can be described by second order variable coefficient equations. Solutions of such equation can be obtained in terms of infinite series. The series solution methods can be classified into two categories: power series method and general series solution method( Frobenius method ).

Power Series Method

Power series soution: We now represent the results regarding the existence of a power series solution of a differential equation about an ordinary point x =x_0 .

Theorem 1:

Let x =x_0 be an ordinary point ( regular point ) of the equation a_0(x)y'' + a_1(x)y' + a_2(x)y =0 . Then, every solution of the equation is analytic at x =x_0 and has a power series expansion about the point x =x_0 , of the form

y( x ) = c_0 + c_1(x-x_0) + c_2(x-x_0)^2 + ........ where c_0,c_1,........ are constants.

Proof:

The proof is obvious. Since a_0(x_0) !=0 , we can write the given equation as y " + p(x) y ' +q(x)y = 0, where p(x) = (a_1(x))/(a_0(x)) , and q(x) = (a_2(x))/(a_0(x)) are analytic at x = x_0 . Hence, y''(x_0),y'''(x_0),....... exist and the taylor expansion of y(x), that is, power seies solution about x = x_0 exists. We note that every function which is analytic in the region [ x-x_0 ] less than R admits a converging power series representationsum_(m=0)^ooc_m(x-x_0)^m in the region.

Example 1:

Write a power series expansion of cos (alphax) , differentiate term by term and verify the derivative formula (d[cos(alphax)])/dx= - alpha sin(alphax) .

Solution :

The power series expansion of cos alphax is

cos alphax = 1- (alpha^2 x^2)/(2!) + (alpha^4 x^4)/(4!) - .....

Differentiating the right hand side term by term we obtain

d/(dx)[1- (alpha^2 x^2)/(2!) + (alpha^4 x^4)/(4!) - .....] = -alpha^2x + (alpha^4 x^3)/(3!) - (alpha^6 x^5)/(5!) +......

= -alpha[alphax - (alpha^3 x^3)/(3!) + (alpha^5 x^5)/(5!) - .......] = -alpha sin alphax .

General Series Soution(frobenius Method)

Series solution about a regular singular point:Frobenius method for obaining a series solution about a regular singular point of the equation: A_0(x) y'' + A_1(x) y' + A_2(x) y =0 .

Example 1:

Find the Frobenius series solution about x=0, of the equation (1-x^2)y'' - 2xy' + 6y =0.

Solution :

The point x = 0 is a regular point of the differential equation. substituting

y(x) = sum_(m=0)^ooc_mx^(m+r) , y'(x)= sum_(m=0)^oo(m+r)c_mx^(m+r-1),

y''(x) = sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2)

in the given equation, we obtain

sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo(m+r)(m+r-1)c_mx^(m+r) - 2 sum_(m=0)^oo(m+r)c_mx^(m+r) + 6 sum_(m=0)^ooc_mx^(m+r)=0 The owest degree term is the term containing x^(r-2) . Setting the coefficient of x^(r-2) to zero, we get

c_0r(r-1)=0 , c_0!=0 giving r=0,1.

Setting the coefficient of x^(r-1) to zero,we obtain c_1r(r+1) =0.

For r =1,c_1=0 and for r=0,c_1 is arbitrary. We shall now show that r=0 gives the complete solution.

combining the remaining terms , we get

sum_(m=2)^oo(m+r)(m+r-1)c_mx^(m+r-2) - sum_(m=0)^oo[(m+r)(m+r-1) + 2(m+r)-6]c_mx^(m+r)=0

Letting m-2=t in the first sum and changing the dummy variable t to m, we get

sum_(m=0)^oo[(m+r+2)(m+r+1)c_(m+2) - {(m+r)(m+r+1) - 6}c_m] x^(m+r)=0

Setting the coefficient of x^(m+r) to zero, we obtain

c_(m+2) = ((m+r)(m+r+1) - 6)/((m+r+1)(m+r+2))c_m , mgreater than or equal to 0.

We have for r = 0, c_(m+2)= (m(m+1) - 6)/((m+1)(m+2))c_m, mgreater than or equal to 0.

Therefore, c_2 = -3c_0, c_3 = -2/3c_1, c_4 = 0, c_5 = 3/10c_3 = -1/5c_1, c_6 = 0=c_8 = ......

The solution is given by y(x)=c_0(1-3x^2) + c_1(x -2/3x^3-1/5x^5-.....) .

For r=1 we have c_1 =0 and c_(m+2) = ((m+1)(m+2)-6)/((m+2)(m+3))c_m, mgreater than or equal to 0.

We have c_2 = -2/3c_0, c_4 = 3/10c_2 = -1/5c_0,....,c_3 = 0 = c_5 = ...... Therefore, we have

y_2(x) = c_0x[1-2/3x^2-1/5x^4-....] = c_0[x-2/3x^3 - 1/5x^5-.....]

But this solution is the constant multiple of the second solution in equation (1). The singular points of the equation are x = +- 1 and the series expansion is written about x=0. Therefore, the radius of convergence is R = 1.

by: nitinp




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