When finding the solutions to the simple ODE

$$ y'- mxy= x^n \text{ ; } y(0) = 0$$

I found the following:

Let $P_n$ be the particular solution for each integer exponent $n$. Then if we define

$$P_0(x) = \exp\left(\displaystyle \frac{mx^2}{2}\right)\int_0^x \exp\left(-\displaystyle \frac{mt^2}{2}\right)dt$$

$$P_1(x) =\frac{1}{m}\left\{ \exp\left(\displaystyle \frac{mx^2}{2}\right)-1 \right\}$$

All other solutions are given by the following recursion

$$P(x)_{n+1} = \frac{n}{m}\left\{ P_{n-1}(x)-\frac{x^{n}}{n} \right\} $$

$$P'(x)_{n+1} = nxP(x)_{n-1}$$

Is there any theory on such functions? (I'm mostly interested in the last one, which is very similar to Bernoulli's:

$$B'_{n+1}(x) = (n+1)B_{n}(x)$$

ADD: The particular solutions are

$${P_{2n + 1}} = \frac{{\left( {2n} \right)!!}}{{{m^n}}}{P_1} - \sum\limits_{k = 1}^n {\frac{{\left( {2n} \right)!!}}{{\left( {2k} \right)!!}}} \frac{{{x^{2k}}}}{{{m^{n - k + 1}}}}$$

$${P_{2n}} = \frac{{\left( {2n - 1} \right)!!}}{{{m^n}}}{P_0} - \sum\limits_{k = 1}^n {\frac{{\left( {2n - 1} \right)!!}}{{\left( {2k - 1} \right)!!}}} \frac{{{x^{2k - 1}}}}{{{m^{n - k + 1}}}}$$

So maybe the importance will be in the polynomials (the sums).

  • $\begingroup$ Have you tried finding the power series and seeing if they match up some known functions? $\endgroup$
    – Aryabhata
    Feb 27, 2012 at 7:41
  • $\begingroup$ @Aryabhata I added the general expressions for the even and odd solutions. $\endgroup$
    – Pedro
    Feb 27, 2012 at 16:17
  • $\begingroup$ Polynomial sequences satisfying $p_n'(x) = np_{n-1}(x)$ are called Appell sequences, after Paul Appell (who also has a street in Paris named after him). Bernoulli polynomials, Hermite polynomials, and the most well known sequence of monomials are three examples. There's a fair-sized literature on them. But of course this doesn't answer the question. $\endgroup$ Feb 27, 2012 at 18:13
  • $\begingroup$ @MichaelHardy Thank you. I find out about that some time ago, but although the recurrence is similar it has an extra $x$, that's why I couldn't get much. $\endgroup$
    – Pedro
    Feb 27, 2012 at 18:23
  • $\begingroup$ I removed the orthogonal polynomials tag; the recursion for any orthogonal polynomial system necessarily has to be a three-term recursion, and the recursion of the polynomials being considered isn't three-term. $\endgroup$ May 1, 2012 at 17:52

1 Answer 1


Solution in terms of the incomplete gamma function

The behavior of the solutions to this ODE are somewhat obscured by its representation. Let $t = m x^2/2$ and $u_a = 2(m/2)^a P_{2a-1}$, where $a = (n+1)/2$. The ODE takes the form $$u_a'(t) - u_a(t) = t^{a-1},$$ with the boundary condition $u_a(0) = 0$. The Green's function for the operator $d/dt - 1$ satisfying the boundary condition is $G(t,t') = \Theta(t-t') e^{t-t'}$. Therefore, $$u_a(t) = e^t \int_0^t d z\, e^{-z} z^{a-1}.$$ The integral is the lower incomplete gamma function, so \begin{equation*} u_a(t) = e^t \gamma(a,t). \tag{1} \end{equation*} In terms of $P$ and $x$ we find $$\begin{eqnarray*} P_n(x) &=& \textstyle\frac{1}{2} \left(\frac{2}{m}\right)^{\frac{n+1}{2}} u_{\frac{n+1}{2}}\left(\frac{m x^2}{2}\right) \\ &=& \textstyle\frac{1}{2} \left(\frac{2}{m}\right)^{\frac{n+1}{2}} \exp\left(\frac{m x^2}{2}\right) \gamma\left(\frac{n+1}{2}, \frac{m x^2}{2}\right). \end{eqnarray*}$$

The interesting properties of $P_n$ are inherited from $\gamma$. For example, $$\begin{eqnarray*} u_a(t) &=& a^{-1} t^a M(1,a+1,t) \\ &=& \Gamma(a) t^a \sum_{k=0}^\infty \frac{t^k}{\Gamma(a+k+1)} \\ &=& \frac{t^a}{\displaystyle a-\frac{a t}{\displaystyle a+1+\frac{t}{\displaystyle a+2-\ldots}}}. \end{eqnarray*}$$ Above $M(a,b,z) = {}_1 F_1(a;b;z)$ is Kummer's confluent hypergeometric function. Have a look at Wikipedia and the DLMF on the incomplete gamma function and related functions.

In the question statement only integer and half-integer values of $a$ were discussed. Equation (1) above defines the function for all $\mathrm{Re}\, a >0$. In fact we can analytically continue to most of the complex plane, in which case $u_a(t)$ is meromorphic with simple poles at $a = 0, -1, -2, \ldots$. For convenience, $$\begin{eqnarray*} u_{1/2}(t) &=& \sqrt{\pi} \mathrm{erf}(\sqrt{t}) e^t \\ u_1(t) &=& -1 + e^t. \end{eqnarray*}$$ These are $P_0$ and $P_1$ in terms of $u$ and $t$.

Recurrence relation and derivative

The recurrence relation given in the statement of the question can be thought of as a consequence of that for $\gamma$, $$\gamma(a,t) = (a-1)\gamma(a-1,t) - t^{a-1}e^{-t}.$$ (This can be found by integration by parts.) Thus, we find $$u_{a+1} = a u_a - t^a.$$ The identity of interest takes the form $$\frac{d u_{a+1}}{d t} = a u_a.$$ Notice how the original representation obscures this relation. (Note that the formula for the Bernoulli polynomials should read $B_{n+1}' = (n+1)B_n$.) The boundary conditions for these two functions are completely different. The Bernoulli polynomials satisfy $B_n(0) = B_n$ whereas we have $u_a(0) = 0$. The sequence of functions $u_a$ is not an Appell sequence. We can't index the $u$ in any way such that $u_0$ is constant. Of course, the $u$ are not polynomials.

A related Appell sequence

However, if we change the boundary conditions and consider $v_a(t) = u_{a+1}(t) + c_a e^t$ (so that $v_a(0) = c_a$) and if $c_a = -a!$ the sequence is Appell, $$v'_{a+1} = (a+1)v_a, \hspace{5ex} v_0(t) = -1.$$ (Note that $v_{a-1}$ satisfies the differential equation $v' - v = t^{a-1}$.) These polynomials are not very interesting, $$\begin{eqnarray*} v_a &=& -e^t \Gamma(a+1,t) \\ &=& -a! e_a(t), \end{eqnarray*}$$ where $e_a(t) = \sum_{k=0}^a t^k/k!$ is the partial exponential sum. (Here we have assumed $a$ is integral. For half-integral $a$ this does not work.)

Polynomials associated with original boundary conditions

We now consider the polynomials associated with the original boundary condition. For integral $a$ we let $q_n = u_{n+1} - n!u_1$. Then $$q_n = -n! \sum_{k=1}^n \frac{t^k}{k!}.$$ These are just the polynomials $v$ found above without the constant term. Note that $\frac{1}{2} (\frac{2}{m})^{n+1} q_n(\frac{m x^2}{2}) = - \sum_{k=1}^n \frac{(2n)!!}{(2k)!!} \frac{x^{2k}}{m^{n-k+1}}$, since $n! = (2n)!!/2^n$.

For half-integral $a$ we let $r_n = u_{n+1/2} - \frac{\Gamma(n+1/2)}{\Gamma(1/2)} u_{1/2}$. Then $$r_n = -\Gamma(n+1/2) \sum_{k=1}^n \frac{t^{k-1/2}}{\Gamma(k+1/2)}.$$ Note that $\frac{1}{2} (\frac{2}{m})^{n+1/2} r_n(\frac{m x^2}{2}) = - \sum_{k=1}^n \frac{(2n-1)!!}{(2k-1)!!} \frac{x^{2k-1}}{m^{n-k+1}}$, since $\Gamma(n+1/2) = (2n-1)!!\sqrt{\pi}/2^n$.

The sums for $q$ and $r$, though finite, are fairly ordinary. Perhaps surprisingly so. Their coefficients aren't related to Bernoulli numbers or any other interesting sequences. It doesn't appear useful to study these further. In fact, it is somewhat unnatural to separate the solution into two parts that fail to solve the same differential equation.

Further exploration

To explore further the functions $P_n$ it is probably more fruitful to study the properties of the incomplete gamma function and related special functions in more detail. The DLMF will be very useful in this regard. Here are two examples of the sorts of relations one can find. (It is probably best to stick with $u$ and $t$, but one can always go to $P$ and $x$ using the dictionary given at the top of this document.)

Using the properties of $\gamma$ we find $$u_{a+n}(t) = (a)_n u_a(t) - t^a \sum_{k=0}^{n-1} \frac{\Gamma(a+n)}{\Gamma(a+k+1)} t^k,$$ where $(a)_n = \Gamma(a+n)/\Gamma(a)$ is Pochhammer's symbol. The recurrence relation is a special case of this formula when $n=1$. Lastly, a somewhat amazing formula, $$u_a(\lambda t) = \lambda^a \sum_{k=0}^\infty u_{a+k}(t) \frac{(1-\lambda)^k}{k!}.$$ There is much more to learn about your functions by exploiting the properties of $\gamma$. Cheers!

  • $\begingroup$ Thank you so much! I will definitely print this out and give it some time! Can you expand a little on the continued fraction? I can't get the pattern. $\endgroup$
    – Pedro
    Apr 25, 2012 at 1:28
  • $\begingroup$ With the work you just showed, I think they are more yours than mine now, ;). Can you give any small hints on the last two equations you provide? $\endgroup$
    – Pedro
    Apr 25, 2012 at 1:40
  • $\begingroup$ Also, I really know NOTHING about Green's function. Well, I only now it is defined as $$\eqalign{ & G\left( {x,t} \right) = t\left( {x - 1} \right);0 \leqslant t \leqslant x \cr & G\left( {x,t} \right) = x\left( {t - 1} \right);x \leqslant t \leqslant 1 \cr} $$ but that's all. $\endgroup$
    – Pedro
    Apr 25, 2012 at 1:55
  • $\begingroup$ @PeterTamaroff: You are quite welcome. Thank you for the interesting question! You can see the relevant formula for the continued fraction at Wikipedia, for example. The DLMF also has a continued fraction formula 8.9.2, but it has to be massaged somewhat since it is in terms of $\Gamma$. $\endgroup$
    – user26872
    Apr 25, 2012 at 2:14
  • $\begingroup$ @PeterTamaroff: The last two equations can be had by using relations found at the DLMF: see 8.8.7 and 8.15.1. $\endgroup$
    – user26872
    Apr 25, 2012 at 2:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .