Ordinary Differential Equations/Frobenius Solution to the Hypergeometric Equation

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In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. This is a method that uses the series solution for a differential equation, where we assume the solution takes the form of a series. This is usually the method we use for complicated ordinary differential equations.

The solution of the hypergeometric differential equation is very important. For instance, Legendre's differential equation can be shown to be a special case of the hypergeometric differential equation. Hence, by solving the hypergeometric differential equation, one may directly compare its solutions to get the solutions of Legendre's differential equation, after making the necessary substitutions. For more details, please check the hypergeometric differential equation.

We shall prove that this equation has three singularities, namely at x = 0, x = 1 and around x = infinity. However, as these will turn out to be regular singular points, we will be able to assume a solution on the form of a series. Since this is a second-order differential equation, we must have two linearly independent solutions.

The problem however will be that our assumed solutions may or not be independent, or worse, may not even be defined (depending on the value of the parameters of the equation). This is why we shall study the different cases for the parameters and modify our assumed solution accordingly.

Contents

The equationEdit

Solve the hypergeometric equation around all singularities:

 

Solution around x = 0Edit

Let

 

Then

 

Hence, x = 0 and x = 1 are singular points. Let's start with x = 0. To see if it is regular, we study the following limits:

 

Hence, both limits exist and x = 0 is a regular singular point. Therefore, we assume the solution takes the form

 

with a0 ≠ 0. Hence,

 

Substituting these into the hypergeometric equation, we get

 

That is,

 

In order to simplify this equation, we need all powers to be the same, equal to r + c − 1, the smallest power. Hence, we switch the indices as follows:

 

Thus, isolating the first term of the sums starting from 0 we get

 

Now, from the linear independence of all powers of x, that is, of the functions 1, x, x2, etc., the coefficients of xk vanish for all k. Hence, from the first term, we have

 

which is the indicial equation. Since a0 ≠ 0, we have

 

Hence,

 

Also, from the rest of the terms, we have

 

Hence,

 

But

 

Hence, we get the recurrence relation

 

Let's now simplify this relation by giving ar in terms of a0 instead of ar−1. From the recurrence relation (note: below, expressions of the form (u)r refer to the Pochhammer symbol).

 

As we can see,

 

Hence, our assumed solution takes the form

 

We are now ready to study the solutions corresponding to the different cases for c1 − c2 = γ − 1 (this reduces to studying the nature of the parameter γ: whether it is an integer or not).

Analysis of the solution in terms of the difference γ − 1 of the two rootsEdit

γ not an integerEdit

Then y1 = y|c = 0 and y2 = y|c = 1 − γ. Since

 

we have

 

Hence,   Let A′ a0 = a and Ba0 = B. Then

 

γ = 1Edit

Then y1 = y|c = 0. Since γ = 1, we have

 

Hence,

 

To calculate this derivative, let

 

Then

 

But

 

Hence,

 

Differentiating both sides of the equation with respect to c, we get:

 

Hence,

 

Now,

 

Hence,

 

For c = 0, we get

 

Hence, y = Cy1 + Dy2. Let Ca0 = C and Da0 = D. Then

 

γ an integer and γ ≠ 1Edit

γ ≤ 0Edit

The value of   is  . To begin with, we shall simplify matters by concentrating a particular value of   and generalise the result at a later stage. We shall use the value  . The indicial equation has a root at  , and we see from the recurrence relation

 

that when   that that denominator has a factor   which vanishes when  . In this case, a solution can be obtained by putting   where   is a constant.

With this substitution, the coefficients of   vanish when   and  . The factor of   in the denominator of the recurrence relation cancels with that of the numerator when  . Hence, our solution takes the form

 

 

If we start the summation at   rather than   we see that

 

The result (as we have written it) generalises easily. For  , with   then

 

Obviously, if  , then  . The expression for   we have just given looks a little inelegant since we have a multiplicative constant apart from the usual arbitrary multiplicative constant  . Later, we shall see that we can recast things in such a way that this extra constant never appears

The other root to the indicial equation is  , but this gives us (apart from a multiplicative constant) the same result as found using  . This means we must take the partial derivative (w.r.t.  ) of the usual trial solution in order to find a second independent solution. If we define the linear operator   as

 

then since   in our case,

 

(We insist that  .) Taking the partial derivative w.r.t  ,

 

Note that we must evaluate the partial derivative at   (and not at the other root  ). Otherwise the right hand side is non-zero in the above, and we do not have a solution of  . The factor   is not cancelled for   and  . This part of the second independent solution is

   

Now we can turn our attention to the terms where the factor   cancels. First

 

After this, the recurrence relations give us

 

 

So, if   we have

 

We need the partial derivatives

   

Similarly, we can write

   

and

   

It becomes clear that for  

 

Here,   is the  th partial sum of the harmonic series, and by definition   and  .

Putting these together, for the case   we have a second solution

 

 

The two independent solutions for   (where   is a positive integer) are then

 

and

 

 

The general solution is as usual   where   and   are arbitrary constants. Now, if the reader consults a ``standard solution" for this case, such as given by Abramowitz and Stegun [1] in §15.5.21 (which we shall write down at the end of the next section) it shall be found that the   solution we have found looks somewhat different from the standard solution. In our solution for  , the first term in the infinite series part of   is a term in  . The first term in the corresponding infinite series in the standard solution is a term in  . The   term is missing from the standard solution. Nonetheless, the two solutions are entirely equivalent.

the standard form of the solution where γ ≤ 0Edit

The reason for the apparent discrepancy between the solution given above and the standard solution in Abramowitz and Stegun [1] §15.5.21 is that there are an infinite number of ways in which to represent the two independent solutions of the hypergeometric ODE. In the last section, for instance, we replaced   with  . Suppose though, we are given some function   which is continuous and finite everywhere in an arbitrarily small interval about  . Suppose we are also given

  and  

Then, if instead of replacing   with   we replace   with  , we still find we have a valid solution of the hypergeometric equation. Clearly, we have an infinity of possibilities for  . There is however a ``natural choice" for  . Suppose that   is the first non zero term in the first   solution with  . If we make   the reciprocal of  , then we won't have a multiplicative constant involved in   as we did in the previous section. From another point of veiw, we get the same result if we ``insist" that   is independent of  , and find   by using the recurrence relations backwards.


For the first   solution, the function   gives us (apart from multiplicative constant) the same   as we would have obtained using  . Suppose that using   gives rise to two independent solutions   and  . In the following we shall denote the solutions arrived at given some   as   and  .

The second solution requires us to take the partial derivative w.r.t  , and substituting the usual trial solution gives us

 

The operator   is the same linear operator discussed in the previous section. That is to say, the hypergeometric ODE is represented as  .

Evaluating the left hand side at   will give us a second independent solution. Note that this second solution   is in fact a linear combination of   and  . Any two independent linear combinations (  and  ) of   and   are independent solutions of  . The general solution can be written as a linear combination of   and   just as well as linear combinations of   and  . We shall review the special case where   that was considered in the last section. If we ``insist"  , then the recurrence relations yield

   

and

 

These three coefficients are all zero at   as expected. We have three terms involved in   by taking the partiial derivative w.r.t  , we denote the sum of the three terms involving these coefficients as   where

   

The reader may confirm that we can tidy this up and make it easy to generalise by putting

 

Next we can turn to the other coefficients, the recurrence relations yield

   

Setting   gives us

 

This is (apart from the multiplicative constant ) the same as  . Now, to find   we need partial derivatives

 

 

Then

 

we can re-write this as

 

The pattern soon becomes clear, and for  

 

Clearly, for  ,

 

The infinite series part of   is  , where

 

Now we can write (disregarding the arbitrary constant) for 

 

   

Some authors prefer to express the finite sums in this last result using the digamma function  . In particular, the following results are used

 

Here,   is the Euler-Mascheroni constant. Also

 

With these results we obtain the form given in Abramamowitz and Stegun §15.5.21, namely

     

The Standard" Form of the Solution γ > 1Edit

In this section, we shall concentrate on the ``standard solution", and we shall not replace   with  . We shall put   where  . For the root   of the indicial eqauation we had

 

where   in which case we are in trouble if  . For instance, if  , the denominator in the recurrence relations vanishes for  . We can use exactly the same methods that we have just used for the standard solution in the last section. We shall not (in the instance where  ) replace   with   as this will not give us the standard form of solution that we are after. Rather, we shall ``insist" that   as we did in the standard solution for   in the last section. (Recall that this defined the function   and that   will now be replaced with  .) Then we may work out the coefficients of   to   as functions of   using the recurrence relations backwards. There is nothing new to add here, and the reader may use the same methods as used in the last section to find the results of [1]§15.5.18 and §15.5.19, these are

 

and

   

Note that the powers of   in the finite sum part of   are now negative so that this sum diverges as  

Solution around x = 1Edit

Let us now study the singular point x = 1. To see if it is regular,

 

Hence, both limits exist and x = 1 is a regular singular point. Now, instead of assuming a solution on the form

 

we will try to express the solutions of this case in terms of the solutions for the point x = 0. We proceed as follows: we had the hypergeometric equation

 

Let z = 1 − x. Then

 

Hence, the equation takes the form

 

Since z = 1 − x, the solution of the hypergeometric equation at x = 1 is the same as the solution for this equation at z = 0. But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β − γ + 1. Hence, to get the solutions, we just make this substitution in the previous results. Note also that for x = 0, c1 = 0 and c2 = 1 − γ. Hence, in our case, c1 = 0 while c2 = γ − α − β. Let us now write the solutions. In the following we replaced each z by 1 - x.

Analysis of the solution in terms of the difference γ − α − β of the two rootsEdit

To simplify notation from now on denote γ − α − β by Δ, therefore γ = Δ + α + β.

Δ not an integerEdit

 

Δ = 0Edit

 

Δ is a non-zero integerEdit

Δ > 0Edit

 

Δ < 0Edit

 

Solution around infinityEdit

Finally, we study the singularity as x → ∞. Since we can't study this directly, we let x = s−1. Then the solution of the equation as x → ∞ is identical to the solution of the modified equation when s = 0. We had

 

Hence, the equation takes the new form

 

which reduces to

 

Let

 

As we said, we shall only study the solution when s = 0. As we can see, this is a singular point since P2(0) = 0. To see if it is regular,

 

Hence, both limits exist and s = 0 is a regular singular point. Therefore, we assume the solution takes the form

 

with a0 ≠ 0. Hence,

 

Substituting in the modified hypergeometric equation we get

 

And therefore:

 

i.e.,

 

In order to simplify this equation, we need all powers to be the same, equal to r + c, the smallest power. Hence, we switch the indices as follows

 

Thus, isolating the first term of the sums starting from 0 we get

 

Now, from the linear independence of all powers of s (i.e., of the functions 1, s, s2, ...), the coefficients of sk vanish for all k. Hence, from the first term we have

 

which is the indicial equation. Since a0 ≠ 0, we have

 

Hence, c1 = α and c2 = β.

Also, from the rest of the terms we have

 

Hence,

 

But

 

Hence, we get the recurrence relation

 

Let's now simplify this relation by giving ar in terms of a0 instead of ar−1. From the recurrence relation,

 

As we can see,

 

Hence, our assumed solution takes the form

 

We are now ready to study the solutions corresponding to the different cases for c1 − c2 = α − β.

Analysis of the solution in terms of the difference α − β of the two rootsEdit

α − β not an integerEdit

Then y1 = y|c = α and y2 = y|c = β. Since

 

we have

 

Hence, y = Ay1 + By2. Let Aa0 = A and Ba0 = B. Then, noting that s = x−1,

 

α − β = 0Edit

Then y1 = y|c = α. Since α = β, we have

 

Hence,

 

To calculate this derivative, let

 

Then using the method in the case γ = 1 above, we get

 

Now,

 

Hence,

 

Therefore:

 

Hence, y = C′y1 + D′y2. Let C′a0 = C and D′a0 = D. Noting that s = x−1,

 

α − β an integer and α − β ≠ 0Edit

α − β > 0Edit

From the recurrence relation

 

we see that when c = β (the smaller root), aα−β → ∞. Hence, we must make the substitution a0 = b0(cci), where ci is the root for which our solution is infinite. Hence, we take a0 = b0(c − β) and our assumed solution takes the new form

 

Then y1 = yb|c = β. As we can see, all terms before

 

vanish because of the c − β in the numerator.

But starting from this term, the c − β in the numerator vanishes. To see this, note that

 

Hence, our solution takes the form

 

Now,

 

To calculate this derivative, let

 

Then using the method in the case γ = 1 above we get

 

Now,

 

Hence,

 

Hence,

 

At c = α we get y2. Hence, y = Ey1 + Fy2. Let Eb0 = E and Fb0 = F. Noting that s = x−1 we get

 

α − β < 0Edit

From the symmetry of the situation here, we see that

 

ReferencesEdit

  • Ian Sneddon (1966). Special functions of mathematical physics and chemistry. OLIVER B. ISBN 978-0-05-001334-2. 

Abramowitz, Milton; Stegun, Irene A. (1964). Handbook of Mathematical Functions. New York: Dover. ISBN 978-0486612720. 

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