How to solve the Sturm-Liouville problem : $y'' + A(x) y = 0$?

Question

Consider the following Sturm-Liouville problem : $$ y'' + A(x) y = 0 \text{ on } [0, 2\pi] $$ where $A$ is a non-constant continuous function on $[0, 2\pi]$.

Are there analytical solutions to this problem ?

Answer 1

In general this does not have closed-form solutions (except the trivial solution $0$). One simple example where (as far as I am aware) no closed-form solution is known is $A(x) = x^3 - 1$.

Answer 2

Consider the general second order linear homogeneous ODE : $$\frac{d^2Y}{dx^2}+g(x)\frac{dY}{dx}+h(x)Y(x)=0 \tag 1$$ The change of function $$Y(x)=\exp\left(-\frac12\int g(x)dx \right)y(x)$$ transforms Eq.$(1)$ into Eq.$(2)$ $$\frac{d^2y}{dx^2}+A(x)y(x)=0 \tag 2$$ in which $\quad A(x)=h(x)-\frac14 g(x)^2-\frac12\frac{dg}{dx}$

Now in response to your question :

Insofar "analytically solving" means getting closed form of solutions, not infinite series.

Suppose that a genius be able to "analytically solve" any equation on the form $\quad y''+A(x)y(x)=0\quad$ this means that he could solve any equation on the form $(1)$. Certainly he would earn a Fields medal.

This means that your question is much too wide. A general method for "analytically solving" Eq.$(2)$ isn't known as well as Eq.$(1)$. One knows how to "analytically solve" only a few kind of equations on the form $(2)$ as well as Eq.$(1)$. This is possible if some special functions have been defined and standardized.

For example in the particular cases of polynomial $A(x)$ :

First degree polynomial : $\quad\frac{d^2y}{dx^2}-(x-a)y(x)=0$ $$y(x)=c_1\text{Ai}(x-a)+c_2\text{Bi}(x-a)$$ Ai and Bi are the Airy functions.

Second degree polynomial : $\quad\frac{d^2y}{dx^2}-(x-a)(x-b)y(x)=0$ $$y(x)=c_1\text{D}_{\frac{(a-b)^2-4}{8}}\left(\frac{2x-a-b}{\sqrt{2}} \right)+c_2\text{D}_{\frac{-(a-b)^2-4}{8}}\left(-\frac{2x-a-b}{\sqrt{2}} \right)$$ D$_\nu(z)$ is the parabolic cylinder function.

Third degree polynomial : $\quad\frac{d^2y}{dx^2}-(x-a)(x-b)(x-c)y(x)=0$

No closed form solution for any $a,b,c$. There is no convenient special function avalable.

But they are closed form solutions in case of particular values of $a,b,c$. For example if $a=b=c=0$ the solution is $\quad y(x)=c_1\sqrt{x}\text{ I}_{1/5}(\frac25 x^{5/2})+c_2\sqrt{x}\text{ I}_{1/5}(\frac25 x^{5/2})\quad$ with the modified Bessel function.

Fourth degree polynomial : Again no closed form solution in case of general fourth degree polynomial. But they are closed form solutions in case of some particular fourth degree polynomials.

Answer 3

Let us focus on the case when $A(x)$ is a "polynomial" meaning a linear combination of powers of $x$. Then the following cases (not listed above) are worth mentioning: \begin{eqnarray} \text{If} \quad A(x) &=& \frac{\frac{1}{4}-a^2 n^2}{x^2}+B^2 n^2 x^{2 n-2} \quad \text{then} \quad y(x) =\sqrt{x}\left( C_1 J_a[B x^n] + C_2 Y_a[B x^n] \right) \\ \text{If} \quad A(x) &=& \frac{1}{2} B n^2 (b-2 a) x^{n-2}+\frac{1-(b-1)^2 n^2}{4 x^2}-\frac{1}{4} B^2 n^2 x^{2 n-2} \quad \text{then} \quad y(x) = x^{\frac{1}{2} (b n-n+1)} e^{-\frac{B x^n}{2}} \left( C_1 F_{1,1}[a,b;B x^n] + C_2 U[a,b;B x^n] \right) \\ \text{If} \quad A(x) &=& -\frac{1}{4} B^2 n^2 x^{2 n-2}+B k n^2 x^{n-2}+\frac{\frac{1}{4}-\mu ^2 n^2}{x^2} \quad \text{then} \quad y(x) = x^{\frac{1-n}{2}} v(B x^n) \quad \text{where $v(x)$ satisfies the Whittaker equation} \\ \hline \\ \text{If} \quad A(x) &=& -\frac{1}{2} B n^2 (g-2 a) x^{n-2}-\frac{d^2 n^2 x^{-2 n-2}}{4 B^2}-\frac{1}{4} B^2 n^2 x^{2 n-2}-\frac{d (g-2) n^2 x^{-n-2}}{2 B}+\frac{1-n^2 \left(2 d+(g-1)^2+4 q\right)}{4 x^2} \quad \text{then} \quad y(x) = x^{\frac{1}{2} ((g-1) n+1)} e^{\frac{B^2 x^n-d x^{-n}}{2 B}} v(B x^n) \quad \text{where $v(x)$ satisfies the doubly-confluent Heun equation} \\ \text{If} \quad A(x) &=& \frac{1}{4} B^2 n^2 x^{2 n-2} \left(4 a-d^2-2 g+2\right)-\frac{1}{4} B^4 n^2 x^{4 n-2}-\frac{1}{2} B^3 d n^2 x^{3 n-2}-\frac{1}{2} B n^2 x^{n-2} (d g+2 q)+\frac{1-(g+1)^2 n^2}{4 x^2} \quad \text{then} \quad y(x) = x^{\frac{1}{2} (-g n-n+1)} e^{-\frac{1}{4} B x^n \left(B x^n+2 d\right)} v(B x^n) \quad \text{where $v(x)$ satisfies the bi-confluent Heun equation}\\ \text{If} \quad A(x) &=& (a-1) B^3 n^2 x^{3 n-2}-\frac{1}{4} B^6 n^2 x^{6 n-2}-\frac{1}{2} B^5 g n^2 x^{5 n-2}-\frac{1}{4} B^4 g^2 n^2 x^{4 n-2}-\frac{1}{2} B^2 n^2 (g+2 q) x^{2 n-2}+\frac{1-n^2}{4 x^2} \quad \text{then} \quad y(x) = x^{\frac{1-n}{2}} e^{\frac{1}{12} B^2 x^{2 n} \left(2 B x^n+3 g\right)} v(B x^n) \quad \text{where $v(x)$ satisfies the triconfluent Heun equation} \end{eqnarray}

I have generated the examples above by going to the respective ODE and then changing the abscissa as $x \rightarrow B x^n$ followed by a change in the ordinate so that the coefficient at the first derivative is anihilated.

The proofs are given in the following Mathematica code snippet:

In[2263]:= (*Polynomial case.*)
n =.; Clear[f]; Clear[y]; Clear[v]; a =.; b =.; c =.; B =.; g =.; d \
=.; q =.; x =.; k =.; mu =.;
FullSimplify[(((1/4 - a^2 n^2)/x^2 + B^2 n^2 x^(-2 + 2 n)) # + 
     D[#, {x, 2}]) &@{Sqrt[x] BesselJ[a, B x^n], 
   Sqrt[x] BesselY[a, B x^n]}]
FullSimplify[(((1 - (-1 + b)^2 n^2)/(4 x^2) + 
        1/2 B (-2 a + b) n^2 x^(-2 + n) - 
        1/4 B^2 n^2 x^(-2 + 2 n)) # + D[#, {x, 2}]) &@{E^(-((B x^n)/
     2)) x^(-(1/2) (-1 + n - b n)) Hypergeometric1F1[a, b, B x^n], 
   E^(-((B x^n)/2)) x^(-(1/2) (-1 + n - b n))
     HypergeometricU[a, b, B x^n]}]
FullSimplify[(((1/4 - mu^2 n^2)/x^2 + B k n^2 x^(-2 + n) - 
         1/4 B^2 n^2 x^(-2 + 2 n)) # + 
      D[#, {x, 2}]) &@{x^(-(1/2) (-1 + n)) v[B x^n]} /. 
  Derivative[2][v][x_] :> -(-1/4 + k/x + (1/4 - mu^2)/x^2) v[x]]

FullSimplify[(((1 - n^2 (2 d + (-1 + g)^2 + 4 q))/(4 x^2) - (
         d^2 n^2 x^(-2 - 2 n))/(4 B^2) - (d (-2 + g) n^2 x^(-2 - n))/(
         2 B) - 1/2 B (-2 a + g) n^2 x^(-2 + n) - 
         1/4 B^2 n^2 x^(-2 + 2 n)) # + D[#, {x, 2}]) &@{x^(
     1/2 (1 + (-1 + g) n)) E^((-d x^-n + B^2 x^n)/(2 B)) v[B x^n]} /. 
  Derivative[2][v][
    x_] :> -(1 + g/x + d/x^2) v'[x] - (a x - q)/x^2 v[x] ]
FullSimplify[(((1 - (1 + g)^2 n^2)/(4 x^2) - 
         1/2 B n^2 (d g + 2 q) x^(-2 + n) + 
         1/4 B^2 (2 + 4 a - d^2 - 2 g) n^2 x^(-2 + 2 n) - 
         1/2 B^3 d n^2 x^(-2 + 3 n) - 1/4 B^4 n^2 x^(-2 + 4 n)) # + 
      D[#, {x, 2}]) &@{x^(-(1/2) (-1 + n + g n))
      E^(-(1/4) B x^n (2 d + B x^n)) v[B x^n]} /. 
  Derivative[2][v][x_] :>  (g/x + d + x) v'[x] - (a x - q)/x v[x] ]
FullSimplify[(((1 - n^2)/(4 x^2) - 
         1/2 B^2 n^2 (g + 2 q) x^(-2 + 
           2 n) + (-1 + a) B^3 n^2 x^(-2 + 3 n) - 
         1/4 B^4 g^2 n^2 x^(-2 + 4 n) - 1/2 B^5 g n^2 x^(-2 + 5 n) - 
         1/4 B^6 n^2 x^(-2 + 6 n)) # + 
      D[#, {x, 2}]) &@{x^(-(1/2) (-1 + n)) E^(
     1/12 B^2 x^(2 n) (3 g + 2 B x^n)) v[B x^n]} /. 
  Derivative[2][v][x_] :> - (g + x) x v'[x] - (a x - q) v[x] ]


Out[2264]= {0, 0}

Out[2265]= {0, 0}

Out[2266]= {0}

Out[2267]= {0}

Out[2268]= {0}

Out[2269]= {0}

Those examples clearly do not exhaust the problem. One can also construct exact solutions to 2nd order ODEs using more sophisticated methods like a symmetric product or a gauge transformation Gauge transformation of differential equations. .