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How to solve the following ordinary differential equation \begin{equation} f_{xx}^{^{\prime \prime }}+2axf_{x}^{^{\prime }}-2sf=0 \end{equation} with $a$ and $s$ are arbitrary real numbers and the boundary conditions \begin{equation} f\left( \delta \right) =f\left( -\delta \right) =1 \end{equation}

After checking some books, I got the general solution \begin{equation} f\left( x\right) =C_{1}\Phi \left( -\frac{s}{2a},\frac{1}{2},-ax^{2}\right) +C_{2}\Psi \left( -\frac{s}{2a},\frac{1}{2},-ax^{2}\right) \end{equation} where $\Phi$ and $\Psi$ are degenerate hypergeometric functions. Are there any other forms of the solution, such as double integral or serious?

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I'm familiar with $f_x$ as notation for the derivative of $f$, likewise $f'$, but $f_x'$ is a new one. –  Gerry Myerson Dec 11 '12 at 5:38
    
It means the derivative of $f$ with respect to $x$. It is used in some books, such as "Handbook of Exact Solutions for Ordinary Differential Equations" by Andrei D. Polyanin and Valentin F. Zaitsev. –  Xiangyu Meng Dec 11 '12 at 5:40
    
Of course there are other forms of the solution, you should find those by another methods, see math.stackexchange.com/questions/97586 for details. Even $\Phi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)$ and $\Psi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)$ themselves are in fact have other forms, as they are only symbols. –  doraemonpaul Dec 11 '12 at 8:15
    
Don't limit yourself on the view on only known special functions, tw.knowledge.yahoo.com/question/question?qid=1011072501482 is a very good example that the ODE whose the solution have quite nice forms of non-known special functions rather than known special functions. –  doraemonpaul Dec 11 '12 at 9:00
    
@doraemonpaul: I checked your method math.stackexchange.com/questions/97586. It may not apply here. Since in this problem, the parameter $a$ may be zero. –  Xiangyu Meng Dec 11 '12 at 16:42

1 Answer 1

To maintain the meaning of this question, $a\neq0$ and $s\neq0$ should be restricted

Let $f(x)=\int_Ce^{xu}K(u)~du$ ,

Then $(\int_Ce^{xu}K(u)~du)''+2ax(\int_Ce^{xu}K(u)~du)'-2s\int_Ce^{xu}K(u)~du=0$

$\int_Cu^2e^{xu}K(u)~du+2ax\int_Cue^{xu}K(u)~du-2s\int_Ce^{xu}K(u)~du=0$

$\int_C(u^2-2s)e^{xu}K(u)~du+\int_C2aue^{xu}K(u)~d(xu)=0$

$\int_C(u^2-2s)e^{xu}K(u)~du+\int_C2auK(u)~d(e^{xu})=0$

$\int_C(u^2-2s)e^{xu}K(u)~du+[2aue^{xu}K(u)]_C-\int_Ce^{xu}~d(2auK(u))=0$

$\int_C(u^2-2s)e^{xu}K(u)~du+[2aue^{xu}K(u)]_C-\int_Ce^{xu}(2auK'(u)+2aK(u))~du=0$

$[2aue^{xu}K(u)]_C-\int_Ce^{xu}(2auK'(u)-(u^2-2a-2s)K(u))~du=0$

$\therefore 2auK'(u)-(u^2-2a-2s)K(u)=0$

$2auK'(u)=(u^2-2a-2s)K(u)$

$\dfrac{K'(u)}{K(u)}=\dfrac{u}{2a}+\left(-\dfrac{s}{a}-1\right)\dfrac{1}{u}$

$\int\dfrac{K'(u)}{K(u)}du=\int\left(\dfrac{u}{2a}+\left(-\dfrac{s}{a}-1\right)\dfrac{1}{u}\right)du$

$\ln K(u)=\dfrac{u^2}{4a}+\left(-\dfrac{s}{a}-1\right)\ln u+c_1$

$K(u)=cu^{-\frac{s}{a}-1}e^{\frac{u^2}{4a}}$

$\therefore f(x)=\int_Ccu^{-\frac{s}{a}-1}e^{\frac{u^2}{4a}+xu}~du$

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

$\therefore f_n(x)=\int_{a_n}^{b_n}c_n(m_nt)^{-\frac{s}{a}-1}e^{\frac{(m_nt)^2}{4a}+xm_nt}~d(m_nt)={m_n}^{-\frac{s}{a}}c_n\int_{a_n}^{b_n}t^{-\frac{s}{a}-1}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}~dt$

For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:

$\lim\limits_{t\to a_n}t^{-\frac{s}{a}}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}=\lim\limits_{t\to b_n}t^{-\frac{s}{a}}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}$

$\int_{a_n}^{b_n}t^{-\frac{s}{a}-1}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}~dt$ converges

Case $1$ : $a>0$ and $s<0$

For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm i$

$\therefore f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\cos xt~dt$ or $C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\sin xt~dt$

Hence $f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\sin xt~dt+C_2\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\cos xt~dt$

Case $2$ : $a<0$ and $s>0$

For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm1$

$\therefore f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\cosh xt~dt$ or $C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\sinh xt~dt$

Hence $f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\sinh xt~dt+C_2\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\cosh xt~dt$

Compare with $\Phi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)$ and $\Psi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)$ , according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_functions#Integral_representations,

$\Phi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)\propto\int_0^1t^{-\frac{s}{2a}-1}(1-t)^{\frac{s}{2a}-\frac{1}{2}}e^{-ax^2t}~dt$ when $\dfrac{1}{2}>-\dfrac{s}{2a}>0$

$\Psi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)\propto\int_0^\infty t^{-\frac{s}{2a}-1}(1+t)^{\frac{s}{2a}-\frac{1}{2}}e^{ax^2t}~dt$ when $-\dfrac{s}{2a}>0$

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Thanks for your solution. I think that $s=0$ and $a=0$ should not be excluded. Since when $a=0$, the problem becomes very easy. Similar for $s=0$. We should have a uniform expression. –  Xiangyu Meng Mar 23 '13 at 16:54

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