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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    $\begingroup$ I'm familiar with $f_x$ as notation for the derivative of $f$, likewise $f'$, but $f_x'$ is a new one. $\endgroup$ – Gerry Myerson Dec 11 '12 at 5:38
  • $\begingroup$ 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. $\endgroup$ – Xiangyu Meng Dec 11 '12 at 5:40
  • $\begingroup$ 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. $\endgroup$ – doraemonpaul Dec 11 '12 at 8:15
  • $\begingroup$ 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. $\endgroup$ – doraemonpaul Dec 11 '12 at 9:00
  • $\begingroup$ @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. $\endgroup$ – Xiangyu Meng Dec 11 '12 at 16:42

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







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




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


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

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 $-\dfrac{s}{2a}>0$

  • $\begingroup$ 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. $\endgroup$ – Xiangyu Meng Mar 23 '13 at 16:54

Since this is a linear second order ODE of the type: \begin{equation} y^{''} + A_1(x) y^{'} +A_0(x) y = 0 \end{equation} it is known that it is reduced to a simpler form with the middle term missing by a substitution $y = \exp(-1/2 \int a_1(x) dx ) u(x)$ with the new function satisfying the following equation: \begin{equation} u^{''} - \underbrace{(\frac{A_1(x)^2}{4} + \frac{A_1^{'}(x)}{2} - A_0(x))}_{v(x)} \cdot u=0 \end{equation} The equation above is called the ${\bf input}$ equation.

Now the idea is to seek solutions of the above equation in terms of known specials functions whose arguments have been changed appropriately. In other words we make a following ansatz $u(x) = m(x) F_{1,2} (\xi(x))$ where \begin{equation} \left(\frac{d^2}{d x^2} + a_1(x) \frac{d}{d x} +a_0(x)\right) F_{1,2}(x)=0 \end{equation} is the ${\bf target}$ equation whose solutions are assumed to be known. Now by inserting the ansatz into the input equation and setting coefficients at $F$ and at $F^{'}$ to zero we get the follwoing equations for the changed variables: \begin{eqnarray} \frac{m^{'}}{m} &=& \frac{1}{2} \left( a_1 \cdot \xi^{'} - \frac{\xi^{''}} {\xi}\right) \quad (i)\\ v&=& \left(\frac{a_1^2}{4} + \frac{a_1^{'}}{2}-a_0\right)\cdot (\xi^{'})^2 + \left( \frac{3(\xi^{''})^2}{4(\xi^{'})^2} - \frac{\xi^{'''}}{2\xi^{'}}\right)\quad (ii) \end{eqnarray}

Now back to our example. We have $v(x) = a+2 s+a^2 x^2$. I suggest to try to map our equation to the confluent hypergeometric equation or to the Whittaker equation which is its symmetric form. Here we have \begin{eqnarray} a_1(x)&=&0 \\ a_0(x)&=&-1/4+\mu/x+(1/4-\nu^2)/x^2 \end{eqnarray} Now it turns out that if the function $v(x)$ is a rational function equation (ii) has solutions in form of rational functions as well. Theorem 1 on page 3 in [1] gives a recipe on constructing the such solutions. Since our function $v(x)$ is a polynomial of order two it turns out that the solution has to be a polynomial of order two, i.e. $\xi(x) = c_0 + c_1 x + c_2 x^2$. Inserting this into (ii) we get a system of non-linear equations for the parameters $c_0$,$c_1$,$c_2$ which gives a following solution $c_0=0$,$c_1=0$,$c_2=a$,$\mu=-(a+2 s)/(4 a)$,$\nu=1/4$. Therefore the solutions of the original equation read: \begin{eqnarray} y(x) &=& C_1 \cdot \exp(-\frac{a}{2} x^2) \frac{1}{\sqrt{x}} W_{-\frac{a+2 s}{4 a},\frac{1}{4}}(a x^2) + C_1 \cdot \exp(-\frac{a}{2} x^2) \frac{1}{\sqrt{x}} M_{-\frac{a+2 s}{4 a},\frac{1}{4}}(a x^2)\\ &=& C_1 \exp(-a x^2) x U(\frac{3}{4}+\frac{a+2 s}{4 a},\frac{3}{2},a x^2) + C_2 \exp(-a x^2) x F_{1,1}(\frac{3}{4}+\frac{a+2 s}{4 a},\frac{3}{2},a x^2) \end{eqnarray} Here $W$ and $M$ are the Whittaker functions and $U$ is the confluent hypergeometric function.

[1] M. Bronstein & S. Lafaille, ``Solutions of linear ordinary differential equations in terms of special functions'', Proceedings of ISSAC'2002, Lille, ACM Press, 23-28. https://www-sop.inria.fr/cafe/Manuel.Bronstein/


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