I ended up with a differential equation that looks like this: $$\frac{d^2y}{dx^2} + \frac 1 x \frac{dy}{dx} - \frac{ay}{x^2} + \left(b -\frac c x - e x \right )y = 0.$$ I tried with Mathematica. But could not get the sensible answer. May you help me out how to solve it or give me some references that I can go over please? Thanks.
Special Differential Equation
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0What is the domain of your problem? – 2012-10-14
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0Basically it is a radial component of the Schrodinger equation in Cylindrical co-ordinates. For this reason the domain has to be x>0. – 2012-10-16
5 Answers
Let $x=e^u$. I changed $e$ to $f$ in the equation to avoid confusions. Then, multiplying by $x^2$ gives $${x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} - ay + \left( {b{x^2} - cx - f{x^3}} \right)y = 0$$
Now, if $x=e^u$, then $$\eqalign{ & x\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cr & {x^2}\frac{{dy}}{{dx}} = \frac{{{d^2}y}}{{d{u^2}}} - \frac{{dy}}{{du}} \cr} $$ so the equation is
$$\frac{{{d^2}y}}{{d{u^2}}} - \frac{{dy}}{{du}} + \frac{{dy}}{{du}} - ay + \left( {b{e^{2u}} - c{e^u} - f{e^{3u}}} \right)y = 0$$
or $$\frac{{{d^2}y}}{{d{u^2}}} - ay + \left( {b{e^{2u}} - c{e^u} - f{e^{3u}}} \right)y = 0$$
$$\frac{{{d^2}y}}{{d{u^2}}} + \left( {b{e^{2u}} - c{e^u} - f{e^{3u}} - a} \right)y = 0$$
$$\frac{{{d^2}y}}{{d{u^2}}} + F\left( u \right)y = 0$$This is a $2^{\rm nd}$ degree DE. As Robert Israel points out, and shows in his answer, the solutions seem to be complicated. My inclination would be to aim for a series solution, finding the coefficients of $y$ recursively.
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0A similar approach would work for $x=-e^u$, wouldn't it? Or is there a reason we should assume $x>0$? – 2012-10-12
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0@CameronBuie I see no reason why it shouldn't, but I like to keep things positive. – 2012-10-12
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0How do you propose to solve $\dfrac{d^2y}{du^2} - F(u)y = 0$? First-order separable equations are easy, but not second-order. – 2012-10-12
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0@RobertIsrael You're right. I missed that. – 2012-10-12
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0Dear Peter, may you suggest the way to solve the last equation please? I tried with mathematica that gave a weired solution that I could not handle. – 2012-10-14
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0Elegant! As a mechanical engineer who dabbles in applied math, this is such an elegant answer! – 2012-10-14
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0@drN Well, I don't know how much it helped. It requires a lot of work from the last part to obtain a soultion. Pragabhava and R.Israel deserve more credit, I think. – 2012-10-14
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0@PeterTamaroff Yes, I just read the continued problem... of singularities. And I made a suggestion about changing x with x + delta where delta is a small number to avoid the singularity. – 2012-10-14
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0@PeterTamaroff, transfering the ODE of the form $p(x)\dfrac{d^2y}{dx^2}+q(x)\dfrac{dy}{dx}+r(x)y=0$ to the ODE of the form $\dfrac{d^2z}{dx^2}+f(x)z=0$ is ok, but the problem is that the reliability of eqworld.ipmnet.ru/en/methods/methods-ode/Khorasani2003.pdf is doubtful. – 2012-10-14
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0@nagendra, even you have questioned about the helpfulness of Peter Tamaroff's answer, you are impatient and still accept Peter Tamaroff's answer. This is your fault. In fact you still have room to redeem, as SE allow unaccept the accepted answer. – 2012-10-14
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0@Doraemonpaul, I do not mean to do that as I was asking for the complete solution and am still waiting for that. Thanks. – 2012-10-15
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0@PeterTamaroff I believe the sign on the last two equations is wrong. Shouldn't it be $+F(u)$? – 2012-10-15
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0@Pragabhava Yes. – 2012-10-15
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0@drN There is a known technique to transform the ode $$w'' + f(z) w'+ g(z) w= 0$$ to the _more qualitative_ $$ y'' + q(z) y = 0$$ using $$w = y e^{-\frac{1}{2}\int f(z) dz}$$ For a sound reference, see Olver's _Asymptotics and Special Functions_ (§7.1.1) – 2012-10-15
Series solution around zero:
The point $x= 0$ is a regular singular. Taking the anzats $$ y(x) = \sum_{n=0}^\infty q_n x^{n+s} $$ we have $$ \sum_{n=0}^\infty [(n+s)^2 - a]q_n x^{n+s-2} -\sum_{n=0}^\infty c q_n x^{n+s-1} + \sum_{n=0}^\infty b q_n x^{n+s} - \sum_{n=0}^\infty e q_n x^{n+s+1} = $$ \begin{multline} q_0(s^2 - a)x^{s-2} + \big[q_1\big((s+1)^2 - a\big) - c q_0\big] x^{s-1} + \\q_2 \big[\big((s+2)^2 - a\big) -c q_1 + b q_0\big]x^s + \sum_{n=0}^\infty \Big(...\Big) \end{multline}
The indicial equation is $s^2-a=0$, hence $s = \pm\sqrt{a}$.
Case $s = \sqrt{a}$
In this case \begin{align} q_1(2\sqrt{a} + 1) -c q_0 &=0,\\ q_2(4\sqrt{a} + 2) -c q_1 + b q_0 &=0, \end{align}
and the recurrence relation is $$ q_{m+3} = \frac{c q_{m+2} + b q_{m+1} - e q_m}{(m+3)(m + 3 + 2\sqrt{a})}. $$
This, assuming all my algebra is right.
What's important here is that $$ y(x) \sim x^\sqrt{a} z(x), $$
which, combined with the form of $F(u)$ obtained by Peter Tamaroff, might help to propose a solution of the type $$ y(x) = x^\sqrt{a} \exp[\sqrt{F(u)} x] v(x), $$ in a similiar way as it is done when solving the Hydrogen Atom or the Quantum Harmonic Oscillator as modern physics textbooks do (see Eisberg Fundamentals of Modern Physics, Hydrogen atom solution), that can lead to a relatively simple form for $v(x)$.
Edit 1
Taking the change of variables $$ y(x) = \frac{z(x)}{\sqrt{x}}, $$ you end up with the equation $$ z'' + \left\{\frac{1-4\alpha}{x^2} + \beta - \frac{\gamma}{x} - \epsilon x\right\}z = 0 $$ (where I've changed the constants to greek letters to avoid the $e$ confussion).
If $\epsilon = 0$ then, as Robert Israel points out, it reduces to the Whittaker Differential Equation (using the proper rescaling). Also, for $\alpha =1/4$ and $\gamma = 0$, you have the Airy Differential Equation.
Using the WKB approximation, $$ z(x) \sim A f^{-1/4} e^{\int f^{1/2} dx} + B f^{-1/4} e^{-\int f^{1/2} dx} $$ where $$ f(x) = \frac{1-4\alpha}{x^2} + \beta -\frac{\gamma}{x} -\epsilon x. $$ you can try to find the asymptotic behavior of $z$, which I believe is $$ z(x) \sim \frac{e^{-\frac{2}{3}(\epsilon^{1/3} x)^{3/2}}}{(\epsilon^{1/3} x)^{1/4}} $$ for $x > 0$, and play with the differential equation resulting from taking $$ z(x) = \frac{e^{-\frac{2}{3}(\epsilon^{1/3} x)^{3/2}}}{(\epsilon^{1/3} x)^{1/4}} v(x). $$
I don't know if there are closed form solutions in general. In the case $e=0$, Maple finds a solution using Whittaker M and W functions: $$y \left( x \right) =c_{{1}} {{\rm \bf M}\left({\frac {ic}{2\sqrt {b}}},\,\sqrt {a},\,2\,i\sqrt {b}x\right)} {\frac {1}{\sqrt {x}}}+c_{{2}} {{\rm \bf W}\left({\frac {ic}{2\sqrt {b}}},\,\sqrt {a},\,2\,i\sqrt {b}x\right)} {\frac {1}{\sqrt {x}}} $$ Another interesting special case is $a=1/4$, $c=0$, where Maple's solution involves Airy functions: $$ y \left( x \right) =c_{{1}} {\text{Ai}\left(-{\frac {b-ex}{ \left( -e \right) ^{2/3}}}\right)}{\frac {1}{\sqrt {x}}} +c_{{2}}{\text{Bi}\left(-{\frac {b-ex}{ \left( -e \right) ^{2/3}}}\right)}{\frac {1}{ \sqrt {x}}} $$
EDIT: Note that the scaling $x \to k x$ preserves the form of the differential equation with $(a,b,c,e) \to (a,k^2b, kc,k^3e)$. So if $e \ne 0$ we can assume WLOG that, say, $e=1$.
As @Pragabhava noted, the indicial roots are $\pm \sqrt{a}$, so unless $\sqrt{a}$ is an integer there will be two fundamental solutions of the form $$\eqalign{y_1(x) &= x^{\sqrt{a}} \left(1 + \sum_{j=1}^\infty u_j x^j\right)\cr y_2(x) &= x^{-\sqrt{a}} \left(1 + \sum_{j=1}^\infty v_j x^j\right)}$$ with coefficients satisfying the recurrences $(2 \sqrt{a} j+j^2) u_j - c u_{j-1} + b u_{j-2} - u_{j-3} = 0$ (with $u_0 = 1$, $u_j = 0$ for $j < 0$) and $(-2 \sqrt{a} j+j^2) v_j - c v_{j-1} + b v_{j-2} - v_{j-3} = 0$ (with $v_0 = 1$, $v_j = 0$ for $j < 0$). If $\sqrt{a}$ is an integer the second recurrence becomes singular at $j=2\sqrt{a}$, generally resulting in logarithmic terms. I don't think there are closed-form solutions for the recurrences.
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0But I can not make e = 0. – 2012-10-12
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0@nagendra, if you want to restrict $e\neq0$ , please state early in this question. – 2012-10-12
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0@Robert Israel, also have this special case: http://eqworld.ipmnet.ru/en/solutions/ode/ode0215.pdf – 2012-10-13
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0Sorry Doraemonpaul, for the inconvenience. I am looking for the solution with all the constants with finite value. The physics of my problem is contained in the two constants b and e. Thanks. – 2012-10-13
$\dfrac{d^2y}{dx^2}+\dfrac{1}{x}\dfrac{dy}{dx}-\dfrac{ay}{x^2}+\left(b-\dfrac{c}{x}-ex\right)y=0$
$\dfrac{d^2y}{dx^2}+\dfrac{1}{x}\dfrac{dy}{dx}-\left(ex-b+\dfrac{c}{x}+\dfrac{a}{x^2}\right)y=0$
Let $y=\dfrac{u}{\sqrt{x}}$ ,
Then $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x}}\dfrac{du}{dx}-\dfrac{u}{2x\sqrt{x}}$
$\dfrac{d^2y}{dx^2}=\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{2x\sqrt{x}}\dfrac{du}{dx}-\dfrac{1}{2x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}=\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}$
$\therefore\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}+\dfrac{3u}{4x^2\sqrt{x}}+\dfrac{1}{x\sqrt{x}}\dfrac{du}{dx}-\dfrac{u}{2x^2\sqrt{x}}-\left(ex-b+\dfrac{c}{x}+\dfrac{a}{x^2}\right)\dfrac{u}{\sqrt{x}}=0$
$\dfrac{1}{\sqrt{x}}\dfrac{d^2u}{dx^2}-\left(ex-b+\dfrac{c}{x}+\dfrac{4a-1}{4x^2}\right)\dfrac{u}{\sqrt{x}}=0$
$\dfrac{d^2u}{dx^2}-\left(ex-b+\dfrac{c}{x}+\dfrac{4a-1}{4x^2}\right)u=0$
The above ODE is hypergeometric only when the special cases below:
$1$. $e=0$
$2$. $b=0$ and $c=0$
$3$. $c=0$ and $a=\dfrac{1}{4}$
Other than the above special cases the above ODE is not hypergeometric.
Unfortunately it also not belongs to any confluent forms of Heun’s equation.
Therefore to solve the above ODE generally is extremely difficult.
One of the main reason is that the coefficient of $u$ has too many terms or contains too high power terms. The similar situation also appear in Titchmarsh's ODE.
Consider a slightly more general ODE. \begin{equation} \frac{d^2 y(x)}{d x^2} + \frac{1}{x} \frac{d y(x)}{d x} + \left(-\frac{a}{x^2} + b - \frac{c}{x} - e x + e_1 x^2 \right) y(x)=0 \end{equation} If we write: \begin{equation} y(x)=x^{\sqrt{a}}\cdot \exp\left( -\frac{\imath}{2 \sqrt{e_1}} x(-e+e_1 x)\right) \cdot v(\frac{(-1)^{3/4}}{\sqrt{2} e_1^{1/4}} x) \end{equation} Then the function $v(x)$ satisfies the biconfluent Heun equation:
\begin{equation} \frac{d^2 v(u)}{d u^2} -\left( \frac{\gamma}{u} + \delta + u\right)\frac{d v(u)}{d u} + \frac{\alpha u - q}{u} v(u) = 0 \end{equation} where \begin{eqnarray} \delta&=& -1-2\sqrt{a}\\ \gamma&=& \frac{\left(\frac{1}{2}+\frac{i}{2}\right) e}{e_1^{3/4}}\\ \alpha&=& \frac{\imath \left(8 \imath \left(\sqrt{a}+1\right) e_1^{3/2}-4 b e_1+e^2\right)}{8 e_1^{3/2}}\\ q&=&\frac{\left(\frac{1}{4}+\frac{i}{4}\right) \left(2 \sqrt{a} e+2 i c \sqrt{e_1}+e\right)}{e_1^{3/4}} \end{eqnarray}
In[1304]:= Clear[y]; Clear[v]; Clear[m]; Clear[w]; Clear[f];
m[x_] = Exp[-I/(2 Sqrt[e1]) x (-e + e1 x)] x^(Sqrt[ a]);
y[x_] = m[x] w[x];
myeqn = Collect[
Simplify[(y''[x] +
1/x y'[x] + (-a/x^2 + b - c/x - e x + e1 x^2) y[x])], {w[x],
w''[x]}, Expand];
myeqn1 = Collect[Simplify[myeqn/m[x]], {w[x], w'[x], w''[x], x^_},
Simplify];
T = (-1)^(3/4)/(Sqrt[2] e1^(1/4));
f[x_] = T x;
subst = {x :> f[x],
Derivative[1][w][x] :> 1/f'[x] Derivative[1][w][x],
Derivative[2][w][x] :> -f''[x]/(f'[x])^3 Derivative[1][w][x] +
1/(f'[x])^2 Derivative[2][w][x]};
Collect[T^2 (myeqn1 /. subst /. w[f[x]] :> w[x]), {w[x], w'[x],
w''[x], x^_}, Simplify]
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0The trick containing $0$ denominator should not work. – 2018-10-20
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0@ doraemonpaul I have a question if you do not mind. The confluent hypergeometric equation is obtained from the hypergeometric equation by replacing $x$ aby $x/a$ and taking the limit $a \rightarrow \infty$. How do we obtain the three confluent versions of the Heun equation from the Heun ODE itself? I guess we also have to take a limit with respect to some parameter? However with respect to which one? – 2018-10-22
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0At least [the known confluent forms of Heun’s equation](http://dlmf.nist.gov/31.12) have been well-developed by mathematicians, but the credibility for example in this answer when seriously taking $e_1=0$ is really doubtful. Obviously directly making $0$ to some terms in ODE to obtain the confluent/degenerate versions often have problems. That's why e.g. https://math.stackexchange.com/questions/2282856 and https://math.stackexchange.com/questions/1551913 is quite meaningful. – 2018-10-23
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0@ doraemonpaul I guess in order to answer your questions one needs to know how to obtain the confluent Heun ODEs from the Heun ODE. I know that the confluent one is obtained by replacing $(\epsilon,\beta,q)$ by $(a\epsilon,a\beta,a q)$ and taking $a\rightarrow \infty$. How are the other three ODEs obtained ? – 2018-10-23
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0https://www.maplesoft.com/support/help/Maple/view.aspx?path=Heun#CHEDerivation gives the details how the HeunC comes. – 2018-10-25
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0HeunD: https://www.maplesoft.com/support/help/Maple/view.aspx?path=HeunD – 2018-10-25
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0HeunB: https://www.maplesoft.com/support/help/Maple/view.aspx?path=HeunB – 2018-10-25
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0HeunT: https://www.maplesoft.com/support/help/Maple/view.aspx?path=HeunT – 2018-10-25
