# How to solve the differential equation $u_k(z)=-2\cfrac{\partial}{\partial z }(\cfrac{u_{k+1}(z)}{z})$?

$$e^{z\sqrt{1-t}}=\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!}$$

$$\frac{\partial}{\partial t }(e^{z\sqrt{1-t}})=\frac{\partial}{\partial t }(\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!})$$

$$\frac{-z}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=1}^\infty \frac{u_k(z)t^{k-1}}{{(k-1)}!}=\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)t^{k}}{{k}!}$$

$$\frac{-1}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)}{z}\frac{t^{k}}{{k}!}$$

$$\frac{\partial}{\partial z }(\frac{-1}{2\sqrt{1-t} }e^{z\sqrt{1-t}})=\frac{\partial}{\partial z }(\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)}{z}\frac{t^{k}}{{k}!})$$

$$\frac{-e^{z\sqrt{1-t}}}{2}=\sum \limits_{k=0}^\infty \frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})\frac{t^{k}}{{k}!}$$

$$\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!}=-2\sum \limits_{k=0}^\infty \frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})\frac{t^{k}}{{k}!}$$

$$u_k(z)=-2\frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})$$

for $t=0$,$u_0(z)=e^z$

How to find the general formula of $u_k(z)$ ? I would like to learn the methods to solve such differential equations.

EDIT:

We can find $u_1(z)$ as shown below

$$\frac{\partial}{\partial t }(e^{z\sqrt{1-t}})=\frac{-z}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=1}^\infty \frac{u_k(z)t^{k-1}}{{(k-1)}!}$$

for $t=0$, $u_1(z)=\cfrac{-ze^{z}}{2}$

If we continue to derivate in such way, we can find all $u_k(z)$ but it seems long method. I am looking for easier method.

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So $u_0(z)=e^z$, we get differential equation $$\operatorname{e} ^{z} = \frac{-2\frac{\partial u_1 (z)}{\partial z}}{z} + \frac{2 u_1 (z)}{z^{2}}$$ and solution $$u_1(z) =-\Biggl(\frac{\operatorname{e} ^{z}}{2} - C\Biggr)z$$ You'll need more initial conditions for unique answers to the DEs. – GEdgar Oct 16 '12 at 14:57
@GEdgar We have $e^{z\sqrt{1-t}}=\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!}$ – Mathlover Oct 16 '12 at 14:57

A typical approach for this kind of problems (Hermite, Legendre, etc) is to obtain recurrence relations.

You've proven that $$u_n(z) = - 2 \frac{d}{d z} \left(\frac{u_{n+1}(z)}{z}\right).$$ A more simple relation can be derived: $$\frac{d^2}{d z^2} e^{z\sqrt{1-t}} = (1-t) e^{z \sqrt{1-t}} = (1-t) \sum_0^\infty \frac{u_n(z)}{n!}t^n = \sum_0^\infty \frac{u_k''(z)}{n!}t^n$$ Arranging orders of $t$ and using your relation \begin{align} u_0'' - u_0 &= 0, \qquad k = 0,\\ u_n'' - u_n + n u_{n-1} &= 0, \qquad n \ge 1, \end{align} then $u_0^{(1)} = e^z$ and $u_0^{(2)} = e^{-z}$, and the ode $$u_n'' - \frac{2 n}{z} u_n' - \left(1 - \frac{2 n}{z^2}\right)u_n = 0, \qquad n \ge 1.$$ Taking the change of variables $$u_n(z) = z^{n + 1} v_n(z)$$ we have $$v_n'' + \frac{2}{z} v_n' - \big(n(n-1) + z^2\big) v_n = 0, \qquad n \ge 1.$$ hence $$u_n^{(1)}(z) = z^{n + 1} i_{-n}(z), \quad u_n^{(2)}(z) = z^{n + 1} k_{-n}(z),$$ where $i_n$ and $k_n$ are the Modified Spherical Bessel Functions of the first and second kind.

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Using the exponential and binomial series, $$e^{z\sqrt{1-t}} =\sum_{j=0}^\infty\frac{z^j(1-t)^{j/2}}{j!} =\sum_{k=0}^\infty\frac{t^k}{k!}\sum_{j=0}^\infty{j/2\choose k}\frac{(-1)^kk!z^j}{j!}$$

Performing the sum over $j$ gives the $u_j(z)$ in terms of modified Bessel functions of the first kind: $$u_j(z) =\sum_{j=0}^\infty{j/2\choose k}\frac{(-1)^kk!z^j}{j!} =\left(\pi\frac{z}{2}\right)^{\frac{1}{2}}\left(-\frac{z}{2}\right)^jI_{1/2-j}(z)$$

I'm not sure how to solve your differential equation, but I think trying to relate it to differential equations of Bessel functions might lead somewhere.

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