8

I started with an ODE (first ODE) :

$-(1-x^2)y''(x) +x y'(x) - q(x) y(x) = \lambda y(x).$

Then I got a more sophisticated differential equation ( second one) and is given by $$-(1-x^2)y''(x) +x y'(x) - q(x)y(x) + \frac{1}{(1-x^2)}y(x) = \lambda y(x).$$ What I want is to understand the following:

Given a solution of equation (1), how can I generalize this one to a solution of equation (2).

3 Answers3

1

With convenient change of variable, the ODE can be transformed to another ODE on a simplified form : $\frac{y’’(t)}{y(t)}=P_2(\cosh(t))$ where $P_2$ is a polynomial of the second degree.

It is known that some particular ODE’s of this form can be solved in terms of MathieuS and MathieuC functions: http://mathworld.wolfram.com/MathieuDifferentialEquation.html For example, this is the case if one of the coefficient is null and the two others not null : in the case $\gamma=0$ or in the case $\alpha=0$. But not in the case $\lambda=0$. Other cases are solvable in case of particular relationship between the coefficients, in terms of Mathieu funcions or in terms of other special functions of lower level for particular values of the coefficients of the polynomial.

In the general case, i.e. $\alpha, \gamma, \lambda$ not null and not related one to another, a closed form for the solutions would certainly require special functions of higher level, which are not referenced as standard functions. So, I think that the ODE on its general form cannot be analytically solved in that sense.

enter image description here

The second EDO with one more term becomes on the reduced form :

enter image description here

A fortiori, in the general case, the solutions cannot be expressed with a finite number of standard functions. But closed forms of solutions could exist in some cases of particular values of the coefficients $\alpha, \gamma, \lambda, \beta$

Adding a term of different nature to an ODE generally changes a lot the form of the solution. A simplified example is shown below : The solution of the ODE which was expressed in terms of elementary functions becomes much more complicated because requiring special functions (associated Legendre) :

enter image description here

Another example which involves even more complicated functions :

enter image description here

A solution on the exponential form exists in case of particular relationships between the parameters $\gamma, \alpha, \lambda$ :

enter image description here

With the additional term, an approximative solution on exponential form can also be found in case of particular relationships between the parameters $\gamma, \alpha, \lambda, \beta$ :

enter image description here

But this is of low interest for practical application because it cannot be generalized to ODEs which parameters are not related on the same way.

All this draw to think that the analytic method is not convenient to tackel this problem in case of a so complicated ODE. Other approaches might be more convenient, especially systematic studies with numerical methods.

JJacquelin
  • 66,221
  • 3
  • 37
  • 87
  • One thing this representation doesn't do, unfortunately, is make clear that the first ODE has an solution of exponential form. And considering the objective is to see how that solution changes in the second ODE... – Semiclassical Jul 21 '14 at 06:27
  • I should note that, while this answer doesn't address the key issue at hand, I do consider it a useful answer insofar as it provides another representation of the ODE. – Semiclassical Jul 21 '14 at 12:55
  • I updated my first answer above, in order to be consistent with the new form of EDO in the wording of the question. In fact, my aim is not to propose a direct answer, but to answer indirectly. I hope that the simplified examples which are now added will clarify what I mean. – JJacquelin Jul 22 '14 at 14:03
  • Indeed. It's essentially a question of how the solution is 'perturbed' by $\beta != 0$. One can approximate that by an expansion in powers of $\beta$, but seeing the exact evolution of one solution into another isn't easy. – Semiclassical Jul 22 '14 at 14:09
  • "isn't easy" is to say the least ! My point is that expressing analytically the "perturbation" would require special functions of more higher level and probably not standard today. I should be less pessimistic if the range of variation of $x$ was limited in a specified small range (in order to allow to approximate the additional term $\frac{\beta}{x^2-1}$ with a more convenient function (polynomial for example). – JJacquelin Jul 22 '14 at 14:28
  • Agreed. I'm trying to recall whether one could at least convert the problem into an integral equation that represents that perturbation exactly ( not that it'll be any more soluble, but perhaps it'll make clearer precisely how high a threshold this is). – Semiclassical Jul 22 '14 at 14:34
  • If $\frac{\beta}{x^2-1}$ can be aproximate on the form $\beta(c_0+c_1x+c_2x^2)$ on a limited range of $x$, the exponential form of solution in the particular case of $\gamma=\alpha^2=-\lambda$ will remain on the same exponential form. The "perturbation" will appear by a change of exponent. $\sqrt\gamma$ which will be replaced by a coefficient where $c_0, c_1 , c_2$ are involved. – JJacquelin Jul 22 '14 at 14:44
  • This idea requires also some change into the relationionship between $\alpha , \gamma , \lambda$. – JJacquelin Jul 22 '14 at 14:53
  • I added to my previous answer some material about a polynomial approximation of the additional term. – JJacquelin Jul 23 '14 at 12:53
1

You can transform your final equation to Sturm-Liouville form by the substitution $$ y = (1-x^{2})^{1/4}f. $$ I believe that the new equation has the Sturm-Liouville form $$ -\frac{d}{dx}(1-x^{2})\frac{df}{dx}+\left[\frac{u^{2}}{1-x^{2}}+\frac{1}{4}-\alpha x-\gamma x^{2}\right]f= \lambda f. $$ This form makes the equation look a little more classical. And it seems to suggest you'll have an orthogonality for the original solutions in the form $$ \int_{-1}^{1}\frac{1}{\sqrt{1-x^{2}}}y_{1}y_{2}\,dx = 0. $$ Of course you'll need to find the right endpoint conditions.

Classical Case: If you start with the equation $$ (1-x^{2})y''-2xy'+n(n+1)y = 0, $$ and you differentiate all terms $m$ times, you get $$ (1-x^{2})y^{(m+2)}+{{m}\choose{1}}(-2x)y^{(m+1)}+{{m}\choose{2}}(-2)y^{(m)}\\ -2xy^{(m+1)}+{{m}\choose{1}}(-2)y^{(m)}\\+n(n+1)y^{(m)}=0. $$ That gives you a different form of the Associated Legendre Equation $$ (1-x^{2})y^{(m+2)}-2(m+1)xy^{(m+1)}+[n(n+1)-m(m-1)-2m]y^{(m)}=0. (?) $$ This is a related second-order differentiation equation for $y^{(m)}$ which, after a change of variable $y=(1-x^{2})^{-m/2}f$ leads to the standard form of the Associated Legendre equation.

Even if you try clearing denominators in your posted DiffEQ and trying the same trick, the order of the derived equation will be too high. The classical trick works only because the coefficients are polynomial with orders that match the order of the derivative for that term, including the lowest order term which is classically constant.

You need some other type of ladder operator, and I'm not good at constructing those devices. BTW: I think it was Schrodinger who came up with ladder operators. I ran across one of his early publications on that. Amazing genius.

Disintegrating By Parts
  • 87,459
  • 5
  • 65
  • 149
  • 1
    Classically, those are related for $\alpha=\gamma=0$ only for $u=1,2,3,\cdots$. Optimistically, the best you could hope for is that the above would be related for certain values of $u$ only. However, I doubt you'll get much because the reason these equations are related in the classical case is that, if you differentiation the first equation $m$ times you arrive at a new non-selfadjoint equation whose selfadjoint normal form is the second equation. So, the solutions are related by a multiplicative factor (fractional power of $(1-x^{2})$) that was needed to put the derived eqn. into normal form – Disintegrating By Parts Aug 04 '14 at 14:02
  • 1
    @user159356 I posted a new problem with an outline of solution which has to do with endpoint conditions for the Legendre operator. I wanted to expose the classical way of dealing with endpoint conditions, and how to find them. I think you'll be surprised at what is forced onto a function just by knowing that $f \in L^{2}[-1,1]$ and $-((1-x^{2})f')'\in L^{2}[-1,1]$. The endpoint conditions become asymptotics. This arose out of our discussions. http://math.stackexchange.com/questions/886775/selfadjoint-restrictions-of-legendre-operator-fracddx1-x2-fracddx – Disintegrating By Parts Aug 04 '14 at 14:41
  • Operator algebras are studied extensively in Mathematics, starting from John vonNeumann's work. Unfortunately, that buries the unbounded nature of observables, avoid issues of endpoint/boundary conditions, but it does allow you to play with groups. I think the ladder operators are related to group invariants, but I'm not sure. – Disintegrating By Parts Aug 04 '14 at 15:03
-1

This isn't an answer but it's too long for a comment. Let $t=x^2$ and note that $$ x \frac{d}{dx}=x \frac{dt}{dx}\frac{d}{dt}=x\cdot 2x\frac{d}{dx}=2t\frac{d}{dt},\\ x^2\frac{d^2}{dx^2} = x \frac{d}{dx}\left(x \frac{d}{dx}-1\right)=2 t \frac{d}{dt}\left(2t\frac{d}{dt}-1\right)=4t^2 \frac{d^2}{dt^2}+2t \frac{d}{dt}.$$

If we then multiply your differential equation by $x^2$, we can thus use these identities to rewrite as $$-4t^2(1-t)y''(t)+2t(2t-1)y'(t) \\- 4 (2t-1) \left( \alpha + \gamma (2t-1) \right) y(t)+\frac{4u^2-1}{4t(1-t)}y(t) = \lambda \, t \, y(t)$$

This may be easier to approach, though it's still quite busy.

Semiclassical
  • 15,842