4
$\begingroup$

Let $a,a_1,a_2,b \in {\mathbb R}$.

Being inspired by the answer to Solve $y''(x)=[a(x^2-1)^2+b]y(x)$ we found solutions of the following second order ODE : \begin{equation} \frac{d^2 y(x)}{d x^2} + \left( a x^4 + a_1 x^2 + a_2 x + b\right) y(x)=0 \end{equation} Indeed if we write: \begin{equation} y(x) = \exp\left( -\imath \frac{\sqrt{a}}{3} x^3 - \imath \frac{a_1}{2 \sqrt{a}} x\right) \cdot v(x) \end{equation} the function $v(x)$ satisfies the triconfluent Heun equation https://dlmf.nist.gov/31.12. We have: \begin{equation} \frac{d^2 v(u)}{d u^2} + u(u+\gamma) \frac{d v(u)}{d u} + (\alpha u - q) v(u)=0 \end{equation} where \begin{eqnarray} \gamma &=& \sqrt[3]{-1} 2^{5/6} \sqrt[6]{a} \sqrt{\frac{a_1}{a}}\\ \alpha &=& 1+\frac{\imath a_2}{2\sqrt{a}} \\ q &=& -\left( \frac{\sqrt[3]{-1} \left(4 \sqrt{2} a^{3/2} \sqrt{\frac{a_1}{a}}+2 i \sqrt{2} a a_2 \sqrt{\frac{a_1}{a}}+4 a b-a_1^2\right)}{4\ 2^{2/3} a^{4/3}}\right) \end{eqnarray} and \begin{equation} u:=\frac{(-1)^{1/6}}{2^{1/3} a^{1/6}}\left(x - \imath \sqrt{\frac{a_1}{(2 a)})}\right) \end{equation}

Here is a code snippet that verifies our claim:

a =.; a0 =.; a1 =.; a2 =.; b =.; m =.; n = -I Sqrt[a]/
   3; Clear[y]; Clear[u]; Clear[v];
y[x_] = Exp[n x^3] u[x];
myeqn = Collect[(D[
       y[x], {x, 2}] + (a x^4 + a1 x^2 + a2 x + b) y[
        x]) Exp[-n x^3], {u[x], u'[x], u''[x]}, Simplify];
u[x_] = Exp[m x] v[x]; m = -I a1/(2 Sqrt[a]);
myeqn1 = Collect[Simplify[myeqn Exp[-m x]], {v[x], v'[x], v''[x]}, 
   Simplify];
myeqn2 = Collect[
   myeqn1 /. x :> u + I Sqrt[a1/(2 a)] /. v[u + A_] :> v[u] /. 
     Derivative[1][v][u + A_] :> Derivative[1][v][u] /. 
    Derivative[2][v][u + A_] :> Derivative[2][v][u], {u[x], u'[x], 
    u''[x]}, Simplify];
Ab = (-1)^(1/6)/(2^(1/3) a^(1/6));
subst = {u :> Ab u, Derivative[1][v][u] :> 1/Ab Derivative[1][v][u], 
   Derivative[2][v][u] :> 1/(Ab)^2 Derivative[2][v][u]};
Collect[Expand[(Ab^2 myeqn2)] /. subst /. v[Ab u] :> v[u], {v[u], 
  v'[u], v''[u], u^_}, Simplify]

enter image description here

Update: Now let $a$,$a_0$,$a_1$,$a_2$ and $b$ be real numbers.

Likewise consider another second order ODE. We have: \begin{equation} \frac{d^2 y(x)}{d x^2} + \left( \frac{a}{x^4} + \frac{a_0}{x^3} + \frac{a_1}{x^2} + \frac{a_2}{x} +b\right) y(x)=0 \end{equation} Then by writing : \begin{equation} y(x)= x^{1+\frac{a_0}{2 \imath \sqrt{a}}} \exp\left[\imath \left(\frac{\sqrt{a}}{x} + \sqrt{b} x \right)\right] \cdot v(x) \end{equation} The function $v$ satisfies the doubly-confluent Heun equation. We have: \begin{equation} \frac{d^2 v(u)}{d u^2} + \left( \frac{\delta}{u^2} + \frac{\gamma}{u} + 1\right) \frac{d v(u)}{d u} + \frac{\alpha u-q}{u^2} v(u) = 0 \end{equation} where: \begin{eqnarray} \delta &=& 4 \sqrt{a b}\\ \gamma &=&2 - \frac{\imath a_0}{\sqrt{a}}\\ \alpha &=& 1-\frac{\imath a_0}{2 \sqrt{a}} - \frac{\imath a_2}{2 \sqrt{b}}\\ q &=& \frac{\imath a_0}{2 \sqrt{a}} + \frac{a_0^2}{4 a}-a_1-2 \sqrt{a b} \end{eqnarray} and $u:=x/(2 \imath \sqrt{b})$.

The following Mathematica code snippet provides the "proof". We have:

a =.; a1 =.; a2 =.; b =.; a0 =.; m =.; n =.; p =.; Clear[y]; \
Clear[v]; Clear[m]; x =.;
m[x_] = x^(1 + a0/(2 I Sqrt[a])) Exp[I (Sqrt[a]/x + Sqrt[b] x)] ;
y[x_] = m[x] v[x];
myeqn = Collect[
   Simplify[(D[
        y[x], {x, 2}] + (a /x^4 + a0 /x^3 + a1 /x^2 + a2 /x + b) y[
         x])/m[x]], {v[x], v'[x], v''[x]}, Simplify];
myeqn = Collect[Simplify[myeqn ], {v[x], v'[x], v''[x], x^_}, 
   Simplify];
Ab = 1/(2 I Sqrt[b]);
subst = {x :> Ab x, Derivative[1][v][x] :> 1/Ab Derivative[1][v][x], 
   Derivative[2][v][x] :> 1/(Ab)^2 Derivative[2][v][x]};
Collect[Expand[(Ab^2 myeqn)] /. subst /. v[Ab x] :> v[x], {v[x], 
  v'[x], v''[x], x^_}, Simplify]

enter image description here

Finally let $a$,$a_0$,$a_1$,$a_2$ and $b$ be real numbers. Consider the following ODE. We have: \begin{equation} \frac{d^2 y(x)}{d x^2} + \left( a x^2 + a_0 x + a_1 + \frac{a_2}{x} +\frac{b}{x^2}\right) y(x)=0 \end{equation} Then by writing: \begin{equation} y(x)=\exp\left( -\frac{\imath}{2\sqrt{a}} x(a_0+a x)\right) \cdot x^{\frac{1}{2}(1+\sqrt{1-4 b})} \cdot v(x) \end{equation} the function $v$ satisfies the biconfluent Heun equation. We have: \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 &=& -\frac{1}{2}\left( 1-\imath \right) \frac{a_0}{a^{3/4}}\\ \gamma &=& - 1-\sqrt{1-4 b}\\ \alpha &=& \frac{4 a^{3/2} \left(\sqrt{1-4 b}+2\right)+4 \imath a a_1-\imath a_0^2}{8 a^{3/2}}\\ q &=& -\frac{(2+2 \imath) \sqrt{a} a_2+(1-i) a_0 \left(\sqrt{1-4 b}+1\right)}{4 a^{3/4}} \end{eqnarray} and $u:=(-1)^{1/4} x/(\sqrt{2} a^{1/4})$.

Again we used Mathematica to verify the result:

Clear[v]; Clear[y]; a =.; a0 =.; a1 =.; a2 =.; b =.; A =.; d =.; \
Clear[m]; Clear[y]; Clear[v];

m[x_] = E^(-((I x (a0 + a x))/(2 Sqrt[a]))) x^(
   1/2 (1 + Sqrt[1 - 4 b]));
y[x_] = m[x] v[x];
ll = Collect[
   Simplify[(D[
        y[x], {x, 2}] + (a x^2 + a0 x + a1 + a2/x + b/x^2) y[x])/
     m[x]], {v[x], v'[x], v''[x]}, Simplify];
ll = Collect[
   Simplify[ll/Coefficient[ll, v''[x]]], {v[x], v'[x], v''[x], x^_}, 
   Simplify];
Ab = (-1)^(1/4)/(Sqrt[2] a^(1/4));
subst = {x :> Ab x, Derivative[1][v][x] :> 1/Ab Derivative[1][v][x], 
   Derivative[2][v][x] :> 1/(Ab)^2 Derivative[2][v][x]};
ll1 = Collect[
  Ab^2 (ll /. subst /. v[Ab x] :> v[x]), {v[x], v'[x], v''[x], x^_}, 
  Simplify]

enter image description here Now my question would be twofold.

Firstly, is there any mathematical software which can handle confluent Heun functions (just as Mathematica handles hypergeometric functions for example). Secondly, can we actually find similar solutions (i.e. map our ODE onto hte Heun equation) in the case when the coefficient at the function $y(x)$ in the ODE is an arbitrary polynomial of order strictly bigger than two ?

$\endgroup$
1
$\begingroup$

I think there also be some more examples:

$1.$ ODE of the form $\dfrac{d^2y}{dx^2}+(a_4x^4+a_3x^3+a_2x^2+a_1x+a_0)y=0$ , $a_4\neq0$ can first convert to $\dfrac{d^2y}{dt^2}+(b_4t^4+b_2t^2+b_1t+b_0)y=0$ and then relates to Heun's Triconfluent Equation as above. The case of $a_4=0$ and $a_3\neq0$ is a big headache.

$2.$ ODE of the form $(x+a)^2(x+b)^2\dfrac{d^2y}{dx^2}+(c_3x^3+c_2x^2+c_1x+c_0)y=0$ , $c_3\neq0$ can convert to Heun's Confluent Equation by letting $y=(x+a)^p(x+b)^qu$ with choosing suitable values of $p$ and $q$ similar to Differential equation with nasty coefficients $ x^2(1-x)^2 y'' + (Ax + b)y = 0 $.

Anyway, I think the most difficulties appear in e.g. "slipped fingers from Heun-type ODEs" , i.e. for example in https://math.stackexchange.com/questions/2944492, Does Heun's differential equation have other known types confluent approach?, an odd question about solving ODE by MATLAB, Solutions in terms of the hypergeometric functions, differential equation nondevelopable, Solving differential equation, Why can't I solve this homogenous second order differential equation?, Special Differential Equation, solving second order differential equation, Solve the given initial value problem.I need your help., differential equation - solving a second-order ODE with variable coefficients, etc. Welcome to challenge! Good luck!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.