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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