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 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.
Solve the hypergeometric equation around all singularities:
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 (it should be noted that this reduces to study the nature of the parameter γ: whether it is an an integer or not).
Analysis of the solution in terms of the difference γ − 1 of the two roots
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Then y1 = y|c = 0 and y2 = y|c = 1 − γ. Since
we have
Hence, Let A′ a0 = a and B′ a0 = B. Then
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 = C′ y1 + D′ y2. Let C′ a0 = C and D′ a0 = D. Then
From the recurrence relation
we see that when c = 0 (the smaller root), a1 − γ → ∞. Hence, we must make the substitution a0 = b0 (c - ci), where ci is the root for which our solution is infinite. Hence, we take a0 = b0 c and our assumed solution takes the new form
Then y1 = yb|c = 0. As we can see, all terms before
vanish because of the c in the numerator. Starting from this term however, the c in the numerator vanishes. To see this, note that
Hence, our solution takes the form
Now,
To calculate this derivative, let
Then following the method in the previous case, we get
Now,
Hence,
Hence,
At c = 1- γ, we get y2. Hence, y = E′ y1 + F′ y2. Let E′ b0 = E and F′ b0 = F. Then
From the recurrence relation
we see that when c = 1 - γ (the smaller root), aγ − 1 → ∞. Hence, we must make the substitution a0 = b0(c − ci), where ci is the root for which our solution is infinite. Hence, we take a0 = b0(c + γ - 1) and our assumed solution takes the new form:
Then y1 = yb|c = 1 - γ. All terms before
vanish because of the c + γ - 1 in the numerator. Starting from this term, however, the c + γ - 1 in the numerator vanishes. To see this, note that
Hence, our solution takes the form
Now,
To calculate this derivative, let
Then following the method in the second case above,
Now,
Hence,
At c = 0 we get y2. Hence, y = G′y1 + H′y2. Let G′b0 = E and H′b0 = F. Then
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. It should be noted 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 roots
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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's 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,
- and
Substituting in the modified hypergeometric equation we get
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 roots
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Then y1 = y|c = α and y2 = y|c = β. Since
- ,
we have
Hence, y = A′y1 + B′y2. Let A′a0 = A and B′a0 = B. Then, noting that s = x-1,
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
Hence,
For c = α we get
Hence, y = C′y1 + D′y2. Let C′a0 = C and D′a0 = D. Noting that s = x-1,
From the recurrence relation
we see that when c = β (the smaller root), aα - β → ∞. Hence, we must make the substitution a0 = b0(c − ci), 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 = E′y1 + F′y2. Let E′b0 = E and F′b0 = F. Noting that s = x-1 we get
From the symmetry of the situation here, we see that
- Ian Sneddon (1966). Special functions of mathematical physics and chemistry. OLIVER B. ISBN 978-0050013342.