Solving a kind of differential equation

Is it possible to solve the following differential equation:

$g: \Bbb{R} \to \Bbb{R}$, $$g'(a)=a\cdot g(a-1),\ g(0)=\frac{1}{2}$$

I can't find any method for ordinary differential equations which works here.

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Wikipedia has a decent article on Delay differential equations, with an example very similar to yours under the heading "Solving DDEs". –  Yonatan N Mar 10 '12 at 10:13
Similar to (although also significantly different from) the Dickman rho-function, en.wikipedia.org/wiki/Dickman_function, beloved of Analytic Number Theorists. –  Gerry Myerson May 31 '12 at 4:38
Since books.google.com.hk/books?id=5n2sN8rBU28C is only trial-view version, some of the pages are blocked. Do the methods of determinating the number of linearly independent solutions in the general solutions of linear DDEs are in those blocked pages? –  doraemonpaul May 31 '12 at 23:59

Suppose a solution $g(x)$ is analytic in a neighbourhood of $0$, with radius of convergence $> 1$. Since $g(0)=1/2$, $g(x) = 1/2 + c_1 x + c_2 x^2 + c_3 x^3 + \ldots$, and $g'(x) = c_1 + 2 c_2 x + 3 c_3 x^2 + \ldots$. Now this is supposed to be $x g(x-1) = x \left(1/2 + c_1 (x-1) + c_2 (x-1)^2 + \ldots\right)$. Equating corresponding coefficients, \eqalign{c_1 &= 0\cr 2 c_2 &= 1/2 - c_1 + c_2 - c_3 + \ldots\cr 3 c_3 &= c_1 - 2 c_2 + 3 c_3 - \ldots\cr (n+1) c_{n+1} &= \sum_{k=n-1}^\infty (-1)^{k-n+1} {k \choose {n-1}} c_k } Truncating this infinite system of equations, here is an approximate solution involving $c_0$ to $c_{10}$:

$$g(x) = 0.5+ 0.2728329455\,{x}^{2}+ 0.07792814666\,{x}^{3}- 0.1526841461\,{x}^{4}+ 0.04393758550\,{x}^{5}+ 0.03918994575\,{x}^{6}- 0.01508119531\,{ x}^{7}- 0.005288135622\,{x}^{8}+ 0.002357480473\,{x}^{9}+ 0.0007572988537\,{x}^{10}$$

I tried this with up to $100$ coefficients, and it looks to me like the radius of convergence will be $\infty$: a plot of $|c_k|^{-1/k}$ looks like a quite convincing fit to a straight line with positive slope.

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I don't know many methods to solve a DDE equation analytically, but this particular one $$\dot g(a)=ag(a-1)$$ is linear, and the method of steps can be used. The only thing to note that the initial condition should be specified not at one point, but for an interval: $$g(a)=\frac 12,\quad -1\leq a\leq 0.$$ Using the initial condition, one has, for $0<a\leq 1$ $$\dot g(a)=\frac 12 a,\quad g(0)=\frac 12,$$ which is solved $$g(a)=\frac{a^2}{4}+\frac 12,\quad 0<a\leq 1.$$ Therefore, on the interval $1<a\leq 2$, one has $$\dot g(a)=\frac{a^3}{4}+\frac a2,\quad g(1)=\frac 34,$$ which can also be easily solved $$g(a)=\frac{a^4}{16}+\frac{a^2}{4}+\frac{7}{16},\quad 1<a\leq 2.$$ The process can be continued further.

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I think what's wanted is a smooth(er) solution. Yours is continuous, but doesn't have derivatives of all orders. –  Gerry Myerson Jun 1 '12 at 9:46
Yes, at the first point it is just continuous, at the second $C^1$ and so on. –  Artem Jun 1 '12 at 9:48

Even for example http://books.google.com.hk/books?id=5n2sN8rBU28C it do not mention the form of this DDE, so I think the only method of solving this DDE is the “integral kernel method”.

$g'(a)=a\cdot g(a-1)$

$g'(a)-a\cdot g(a-1)=0$

Let $g(a)=\int_Ce^{as}K(s)~ds$,

Then $(\int_Ce^{as}K(s)~ds)'-a\int_Ce^{(a-1)s}K(s)~ds=0$

$\int_Cse^{as}K(s)~ds-\int_Ce^{as}e^{-s}K(s)~d(as)=0$

$\int_Cse^{as}K(s)~ds-\int_Ce^{-s}K(s)~d(e^{as})=0$

$\int_Cse^{as}K(s)~ds-[e^{(a-1)s}K(s)]_C+\int_Ce^{as}~d(e^{-s}K(s))=0$

$\int_Cse^{as}K(s)~ds-[e^{(a-1)s}K(s)]_C+\int_C(e^{-s}K'(s)-e^{-s}K(s))e^{as}~ds=0$

$-[e^{(a-1)s}K(s)]_C+\int_C(e^{-s}K'(s)+(s-e^{-s})K(s))e^{as}~ds=0$

$\therefore e^{-s}K'(s)+(s-e^{-s})K(s)=0$

$e^{-s}K'(s)=(e^{-s}-s)K(s)$

$\dfrac{K'(s)}{K(s)}=1-se^s$

$\int\dfrac{K'(s)}{K(s)}~ds=\int(1-se^s)~ds$

$\ln K(s)=s-(s-1)e^s+c_1$

$K(s)=ce^{s-(s-1)e^s}$

$\therefore g(a)=\int_Cce^{(a+1)s-(s-1)e^s}~ds$

But since the above procedure in fact suitable for any complex number $s$,

$\therefore g_n(a)=\int_{a_n}^{b_n}c_ne^{(a+1)(p_n+q_ni)t-((p_n+q_ni)t-1)e^{(p_n+q_ni)t}}~d((p_n+q_ni)t)$

$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{(a+1)(p_n+q_ni)t-(p_nt-1+q_nti)e^{p_nt}e^{q_nti}}~dt$

$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{(a+1)(p_n+q_ni)t-(p_nt-1+q_nti)e^{p_nt}(\cos q_nt+i\sin q_nt)}~dt$

$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{(a+1)(p_n+q_ni)t-(p_nt-1+q_nti)(e^{p_nt}\cos q_nt+ie^{p_nt}\sin q_nt)}~dt$

$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{(a+1)(p_n+q_ni)t-(p_nt-1)e^{p_nt}\cos q_nt-i(p_nt-1)e^{p_nt}\sin q_nt-iq_nte^{p_nt}\cos q_nt+q_nte^{p_nt}\sin q_nt}~dt$

$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{p_n(a+1)t+q_nte^{p_nt}\sin q_nt-(p_nt-1)e^{p_nt}\cos q_nt}e^{(q_n(a+1)t-(p_nt-1)e^{p_nt}\sin q_nt-q_nte^{p_nt}\cos q_nt)i}~dt$

For some $a$-independent real number choices of $a_n$ , $b_n$ , $p_n$ and $q_n$ such that:

$\displaystyle\lim_{t\to a_n}e^{p_nat+q_nte^{p_nt}\sin q_nt-(p_nt-1)e^{p_nt}\cos q_nt}e^{(q_nat-(p_nt-1)e^{p_nt}\sin q_nt-q_nte^{p_nt}\cos q_nt)i}=\lim_{t\to b_n}e^{p_nat+q_nte^{p_nt}\sin q_nt-(p_nt-1)e^{p_nt}\cos q_nt}e^{(q_nat-(p_nt-1)e^{p_nt}\sin q_nt-q_nte^{p_nt}\cos q_nt)i}$

$\int_{a_n}^{b_n}e^{p_n(a+1)t+q_nte^{p_nt}\sin q_nt-(p_nt-1)e^{p_nt}\cos q_nt}e^{(q_n(a+1)t-(p_nt-1)e^{p_nt}\sin q_nt-q_nte^{p_nt}\cos q_nt)i}~dt$ converges

If only choose one group of linearly independent solution is enough, then $g_1(a)=c_1\int_{-\infty}^{\infty}e^{-\left|a+1\right|t+(t+1)e^{-t}}~dt$ is already a general solution. (Choose $a_1=-\infty$ , $b_1=\infty$ , $p_1=1$ , $q_1=0$ for $a+1\geq0$ ; $a_1=-\infty$ , $b_1=\infty$ , $p_1=-1$ , $q_1=0$ for $a+1\leq0$)

Although linear DDEs like linear ODEs and linear difference equations their general solutions are both the sum of linearly independent solutions, however, unlike linear ODEs and linear difference equations, it seems that there are no clear concepts about the number of linearly independent solutions in the general solutions of linear DDEs, so I don’t know whether $g_1(a)=c_1\int_{-\infty}^{\infty}e^{-\left|a+1\right|t+(t+1)e^{-t}}~dt$ is general enough or not.

If the particular solution $g_1(a)=c_1\int_{-\infty}^{\infty}e^{-\left|a+1\right|t+(t+1)e^{-t}}~dt$ is already general enough, then $g(a)=c\int_{-\infty}^{\infty}e^{-\left|a+1\right|t+(t+1)e^{-t}}~dt$.

Put $g(0)=\dfrac{1}{2}$:

$\dfrac{1}{2}=c\int_{-\infty}^{\infty}e^{-t+(t+1)e^{-t}}~dt$

$c=\dfrac{1}{2\int_{-\infty}^{\infty}e^{-t+(t+1)e^{-t}}~dt}$

$\therefore g(a)=\dfrac{\int_{-\infty}^{\infty}e^{-\left|a+1\right|t+(t+1)e^{-t}}~dt}{2\int_{-\infty}^{\infty}e^{-t+(t+1)e^{-t}}~dt}$

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That's why I further ask math.stackexchange.com/questions/152303/… for clarification. –  doraemonpaul Jun 1 '12 at 8:31