# Dependence of a solution of a linear ODE on parameter

Is the following theorem known, or can be easily derived from known results?

Consider the differential equation $$w''-kz^{-1}w'=(\lambda+\phi(z))w,$$ where $$k>0$$ is fixed, $$\lambda$$ is a large (complex) parameter and $$\phi$$ is a complex valued function analytic on $$[0,1]$$. It is known that there exists a unique solution $$w_0(z,\lambda)$$ which is analytic at $$z=0$$ and $$w_0(0,\lambda)=1$$. It is also known that $$f(\lambda)=w_0(1,\lambda)$$ is an entire function.

Theorem (?). $$f(\lambda)=(1+o(1))\exp\sqrt{\lambda},$$ as $$\lambda\to\infty$$, $$|\arg\lambda|\leq\pi-\epsilon$$, where $$\sqrt{\lambda}$$ is the principal branch.

Remark. There is another solution, normalized by $$w_1(z,\lambda)\sim z^{k+1},z\to 0$$, and for this solution I know how to prove the result, and know the references, for example Olver, Asymptotics and special functions, Ch. 12.

More remarks: 1. When $$\phi=0$$ this reduces to Bessel's equation; the Theorem is true in this case but not trivial.

1. In the definition of $$w_0$$ the crucial word is ANALYTIC. There are infinitely many other solutions $$w$$ satisfying $$w(0)=1$$: adding to $$w_0$$ any multiple of $$w_1$$ does not change the value at $$0$$. And the conclusion of the Theorem does NOT hold for some of solutions satisfying $$w(0)=1$$. For example, when $$\phi=0$$ there is a solution with $$w(0)=1$$, for which $$w(1,\lambda)$$ decays exponentially for $$\lambda>0$$.

2. The most important case for me is when $$\phi$$ is even, which may help. But on my opinion, the problem is interesting in the general case as well.

Remark. Now I proved this, but the question remains whether this follows from some known, published results.

• Do you mean $w_1(0,\lambda)=0$ or something similar ? Mar 6 '19 at 19:20
• @Loic Teysser: Yes, I corrected. Mar 7 '19 at 0:49

Consider the analytic vector field $$X(z,w,W)=z \partial_z+zW\partial_w+(kW+z(\lambda+\phi(z))w)\partial_W$$ whose orbits project to the graphs of solutions $$z\mapsto w(z)$$ (it is simply the companion vector field given by $$W:=w'$$ and scaled by $$z$$ to remove the pole).

The set $$\{z=0\}\cap\{W=0\}$$ is a line of singularities, and the special solutions $$w_0,w_1$$ pass through that line. The linearization of $$X$$ at $$(0,0,0)$$ is the matrix $$M_0=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & k \end{array}\right]$$ and at $$(0,1,0)$$ it is $$M_1=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ \lambda & 0 & k \end{array}\right].$$ If $$k\neq 1$$ the matrix $$M_1$$ is diagonalizable: in the new linear basis of $$\mathbb C^3$$ the linear part of $$X$$ has now the form $$M_0$$. Observe that one can require that only $$W$$ be affected by the diagonalization transform. In particular, $$w_0$$ is left untouched and so is its value at $$1$$.

If I understood correctly the $$M_0$$ case has been dealt with already, although it could happen that this part gets broken by the diagonalization (which does not preserve the companion structure). But I do not know the specifics of the OP argument.

In the case $$k=1$$ the argument does not work since $$M_1$$ is not diagonalizable anymore. Maybe something similar can be cooked up.

• I mostly care about non-integer k. But need some time to digest what you wrote: in particular I don't understand what $R^3$ is doing in your answer (my functions $\phi$, $w$ and the number $\lambda$ are complex. Mar 7 '19 at 0:55
• Sorry, yes, the ambient space is complex of course, I was temporarily misled by the real interval and the term analytic instead of holomorphic. Mar 7 '19 at 6:52

Maybe, something to adjust in view of the following results of Maple.

Order := 3; dsolve(((D@@2)(w))(z)-(D(w))(z)/z = (lambda+phi(z))*w(z), w(z), series, z = 0);


$$w \left( z \right) ={\it \_C1}\,{z}^{2} \left( (1+ \left( {\frac { \lambda}{8}}+{\frac {\phi \left( 0 \right) }{8}} \right) {z}^{2}+O \left( {z}^{3} \right) ) \right) +{\it \_C2}\, \left( \ln \left( z \right) \left( ( \left( -\lambda-\phi \left( 0 \right) \right) {z}^ {2}+O \left( {z}^{3} \right) ) \right) +(-2+O \left( {z}^{3} \right) ) \right)$$

Order := 3; dsolve(((D@@2)(w))(z)-2*(D(w))(z)/z = (lambda+phi(z))*w(z), w(z), series, z = 0);


$$w \left( z \right) ={\it \_C1}\,{z}^{3} \left( (1+ \left( {\frac { \lambda}{10}}+{\frac {\phi \left( 0 \right) }{10}} \right) {z}^{2}+O \left( {z}^{3} \right) ) \right) +{\it \_C2}\, \left( (12+ \left( -6 \,\lambda-6\,\phi \left( 0 \right) \right) {z}^{2}+O \left( {z}^{3} \right) ) \right)$$