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I need help to find an analytical solution to:

$$p''(x)-k_1xp'(x)-k_2p(x)=0 \text{ where } k_1,k_2\in\mathbb R^+$$

with boundary conditions $p'(0)=0, p(r)=p(-r)=k_3$ where $r,k_3\in\mathbb R^+$. Mathematica gives as solution the ratio of two Hypergeometric $_1F_1$ (Kummer) functions but I'd like to know if Mathematica is right and a possible solution procedure.

Thanks a lot.

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I don't get a ratio... –  draks ... Apr 5 '12 at 10:57

1 Answer 1

Let $p(x)=\int_Ce^{xs}K(s)~ds$ ,

Then $(\int_Ce^{xs}K(s)~ds)''-k_1x(\int_Ce^{xs}K(s)~ds)'-k_2\int_Ce^{xs}K(s)~ds=0$

$\int_Cs^2e^{xs}K(s)~ds-k_1x\int_Cse^{xs}K(s)~ds-k_2\int_Ce^{xs}K(s)~ds=0$

$\int_C(s^2-k_2)e^{xs}K(s)~ds-\int_Ck_1se^{xs}K(s)~d(xs)=0$

$\int_C(s^2-k_2)e^{xs}K(s)~ds-\int_Ck_1sK(s)~d(e^{xs})=0$

$\int_C(s^2-k_2)e^{xs}K(s)~ds-[k_1se^{xs}K(s)]_C+\int_Ce^{xs}~d(k_1sK(s))=0$

$\int_C(s^2-k_2)e^{xs}K(s)~ds-[k_1se^{xs}K(s)]_C+\int_Ce^{xs}(k_1sK'(s)+k_1K(s))~ds=0$

$-~[k_1se^{xs}K(s)]_C+\int_C(k_1sK'(s)+(s^2+k_1-k_2)K(s))e^{xs}~ds=0$

$\therefore k_1sK'(s)+(s^2+k_1-k_2)K(s)=0$

$k_1sK'(s)=(k_2-k_1-s^2)K(s)$

$\dfrac{K'(s)}{K(s)}=\left(\dfrac{k_2}{k_1}-1\right)\dfrac{1}{s}-\dfrac{s}{k_1}$

$\int\dfrac{K'(s)}{K(s)}ds=\int\left(\left(\dfrac{k_2}{k_1}-1\right)\dfrac{1}{s}-\dfrac{s}{k_1}\right)ds$

$\ln K(s)=\left(\dfrac{k_2}{k_1}-1\right)\ln s-\dfrac{s^2}{2k_1}+c_1$

$K(s)=cs^{\frac{k_2}{k_1}-1}e^{-\frac{s^2}{2k_1}}$

$\therefore p(x)=\int_Ccs^{\frac{k_2}{k_1}-1}e^{-\frac{s^2}{2k_1}+xs}~ds$

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

$\therefore p_n(x)=\int_{a_n}^{b_n}c_n(m_nt)^{\frac{k_2}{k_1}-1}e^{-\frac{(m_nt)^2}{2k_1}+xm_nt}~d(m_nt)={m_n}^{\frac{k_2}{k_1}}c_n\int_{a_n}^{b_n}t^{\frac{k_2}{k_1}-1}e^{-\frac{{m_n}^2t^2}{2k_1}+m_nxt}~dt$

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

$\lim\limits_{t\to a_n}t^{\frac{k_2}{k_1}}e^{-\frac{{m_n}^2t^2}{2k_1}+m_nxt}=\lim\limits_{t\to b_n}t^{\frac{k_2}{k_1}}e^{-\frac{{m_n}^2t^2}{2k_1}+m_nxt}$

$\int_{a_n}^{b_n}t^{\frac{k_2}{k_1}-1}e^{-\frac{{m_n}^2t^2}{2k_1}+m_nxt}~dt$ converges

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

$\therefore p_1(x)=C_1\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\cosh xt~dt$ or $C_1\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\sinh xt~dt$

Hence $p(x)=C_1\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\sinh xt~dt+C_2\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\cosh xt~dt$

$p'(x)=C_1\int_0^\infty t^{\frac{k_2}{k_1}}e^{-\frac{t^2}{2k_1}}\cosh xt~dt+C_2\int_0^\infty t^{\frac{k_2}{k_1}}e^{-\frac{t^2}{2k_1}}\sinh xt~dt$

$p'(0)=0$ :

$C_1\int_0^\infty t^{\frac{k_2}{k_1}}e^{-\frac{t^2}{2k_1}}~dt=0$

$C_1=0$

$\therefore p(x)=C_2\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\cosh xt~dt$

$p(\pm r)=k_3$ :

$C_2\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\cosh rt~dt=k_3$

$C_2=\dfrac{k_3}{\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\cosh rt~dt}$

$\therefore p(x)=\dfrac{k_3\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\cosh xt~dt}{\int_0^\infty t^{\frac{k_2}{k_1}-1}e^{-\frac{t^2}{2k_1}}\cosh rt~dt}$

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