Differential Equations Project

Computer Lab Modules

 

Frobenius Series Solution of O.D.E.'s

 

 

Background. Frobenius series can be used to solve differential equations at a regular singular point. The series we substitute is [Graphics:e22.txtgr1.gif]. The parameter r must be chosen so that when the series is substituted into the D.E. the coefficient of the smallest power of x is zero. This is called the indicial equation. The recursive equation for the coefficients is obtained by setting the coefficient of [Graphics:e22.txtgr2.gif] equal to zero. We will attempt to have Mathematica solve these tasks.

 

Computer Lab Work.


Exercise 1. Use series to solve the D. E. [Graphics:e22.txtgr3.gif].

Solution. First find the indicial equation and it's roots.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr4.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr6.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr7.gif]

Form the set of equations to solve and do it.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr8.gif]

The first Frobenius solution is:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr9.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr10.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr11.gif]

Form the set of equations to solve and do it.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr12.gif]

The second Frobenius solution is:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr13.gif]

Observe that the coefficients that involve c[1] are that multiple of the first Frobenius solution. Hence we can set c[1] = 0.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr14.gif]

After you are done, use Mathematica's DSolve subroutine to get the answer and check out its series expansion.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr15.gif]

Now we plot the series approximations and the analytic solutions.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr16.gif]
[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr17.gif]

If we look at the series in more depth we will be able to obtain the analytic solutions as infinite sums. First find the recursive formula for the coefficients of [Graphics:e22.txtgr18.gif]. If you try this be sure to use the " := " replacement delayed structure to avoid an infinite recursion. Also, include the semicolon at the end of the lines.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr19.gif]

If you can't get the above computation to work, then just type in the recursive formula.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr20.gif]

Now look at each series individually. The first Frobenius series corresponds to [Graphics:e22.txtgr21.gif].

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr22.gif]

Now look at the second Frobenius series which corresponds to [Graphics:e22.txtgr23.gif].

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr24.gif]

When the explicit formulas for the coefficients of the first Frobenius are used we get:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr25.gif]

When the explicit formulas for the coefficients of the second Frobenius are used we get:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr26.gif]


Exercise 2. Use series to solve the D. E. [Graphics:e22.txtgr27.gif].

Solution. First find the indicial equation and it's roots.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr28.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr29.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr30.gif]

Form the set of equations to solve and do it.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr31.gif]

The first Frobenius solution is:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr32.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr33.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr34.gif]

Form the set of equations to solve and do it.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr35.gif]

The second Frobenius solution is:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr36.gif]

After you are done, use Mathematica's DSolve subroutine to get the answer and check out its series expansion.
This will require slight fussing around with the appropriate multiple of "Sinh".

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr37.gif]

At this time we could plot the series approximations and the analytic solutions. To see the difference in the graphs we will reduce the number of terms in the series.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr38.gif]
[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr39.gif]

If we look at the series in more depth we will be able to obtain the analytic solutions as infinite sums. First find the recursive formula for the coefficients of [Graphics:e22.txtgr40.gif]. If you try this be sure to use the " := " replacement delayed structure to avoid an infinite recursion. Also, include the semicolon at the end of the lines.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr41.gif]

If you can't get the above computation to work, then just type in the recursive formula.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr42.gif]

Now look at each series individually. The first Frobenius corresponds to [Graphics:e22.txtgr43.gif].

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr44.gif]

Now look at the second Frobenius series corresponds to r = 0.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr45.gif]

The explicit formulas for the coefficients for the first Frobenius series are:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr46.gif]
[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr47.gif]

Note. We cannot add up the infinite number of terms in this sequence because Mathematica does not know this formula yet.


Exercise 3. Use series to solve the D. E. [Graphics:e22.txtgr48.gif].

Solution. First find the indicial equation and it's roots.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr49.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr50.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr51.gif]

Form the set of equations to solve and do it.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr52.gif]

The first Frobenius solution is:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr53.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr54.gif]

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr55.gif]

Form the set of equations to solve and do it.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr56.gif]

The second Frobenius solution is:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr57.gif]

After you are done, use Mathematica's DSolve subroutine to get the answer and check out its series expansion.
This will require some fussing around with the appropriate multiple of the Bessel function.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr58.gif]

At this time we could plot the series approximations and the analytic solutions. To see the difference in the graphs we will reduce the number of terms in the series.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr59.gif]
[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr60.gif]

If we look at the series in more depth we will be able to obtain the analytic solutions as infinite sums. First find the recursive formula for the coefficients of [Graphics:e22.txtgr61.gif]. If you try this be sure to use the " := " replacement delayed structure to avoid an infinite recursion. Also, include the semicolon at the end of the lines.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr62.gif]

If you can't get the above computation to work, then just type in the recursive formula.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr63.gif]

Now look at each series individually. The first Frobenius corresponding to [Graphics:e22.txtgr64.gif] is:

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr65.gif]

Now look at the second Frobenius series corresponds to r = - 1.

[Graphics:e22.txtgr5.gif][Graphics:e22.txtgr66.gif]

Note. We cannot add up an infinite number of terms in this sequence because we do not have a closed formula for the coefficients c[k], it is a recursive formula and will exceed the finite recursion depth of Mathematica.

 

Solutions.

 

 

Return to the Differential Equations Project

Return to the Numerical Analysis Project

Return to the Complex Analysis Project

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews, 1998