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 usually the method we use for complicated ordinary differential equations.
The solution of the hypergeometric differential equation is very important.
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.
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.
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.
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).
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.
in the denominator of the recurrence relation cancels with that of the numerator when
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
th partial sum of the harmonic series, and by definition
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
, and substituting the usual trial solution gives us
, we denote the sum of the three terms involving these coefficients as
The reader may confirm that we can tidy this up and make it easy to generalise by putting
Some authors prefer to express the finite sums in this last result using the digamma function
With these results we obtain the form given in Abramamowitz and Stegun §15.5.21, namely
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
Hence, to get the solutions, we just make this substitution in the previous results.
To simplify notation from now on denote γ − α − β by Δ, therefore γ = Δ + α + β.
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.
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 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, From the recurrence relation we see that when c = β (the smaller root), aα−β → ∞.
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.
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 Abramowitz, Milton; Stegun, Irene A.