Analysis and classic odes – In exponential polynomials

Suppose we have the following function $ f: mathbb {R} ^ {+} mapsto mathbb {R} $
$$ f (t) = sum_ {i = 1} ^ k P_i (t) exp ( alpha_i t), $$
where $ alpha_i $s are all algebraic numbers and $ P_i (t) $ are all polynomials with algebraic coefficients and a minor degree that $ m $.

There are several questions that interest me.

to. What is the maximum number? $ p $ such that there is a $ t_0 $ and for all $ r leq p $
$$ f ^ {(r)} (t_0) = 0. $$
second. Suppose $ t_1, cdots, t_q, cdots $ they are the true roots of $ f (t) = 0 $Is it possible to have a convergent sequence of $ t_q $? In other words, is it possible to have a Cauchy sequence? $ t_i $ such that $ f (t_i) = 0 $? If not, do we have a lower limit on the distance between different roots?

c. Is there an algorithm to calculate all the common real roots of $ f (t) $ Y $ f & # 39; (t) $?

Analysis and classic odes – Positivity and zeros of the Heun function

I am interested in understanding in which part of the complex a function of Heun can vanish or where (or its real part, for example) is positive. I have consulted Ronveaux's book, as well as the NIST Math Function Manual, and I could not find any information about this problem. Is there anything known in that direction? You could find information about where this happens for a hypergeometric function.

Analysis and classic odes – Real analytical function with a given set of values

We say that a strictly growing sequence. $ x_n $ of reals converges fast
to $ x $, if for each $ k in mathbb {N} $ sequence $ n ^ k cdot (x_n – x) $ is
locked. It is known that there is a $ C ^ infty $-function $ f $ such that $ f (1 / n) = x_n $ Y $ f (0) = x $.

In which case (sufficient condition in $ x_n $) exists real analytical function $ g $ such that $ g (1 / n_k) = x_ {n_k} $ Y $ g (0) = x $ by some subsequence $ x_ {n_k} $ ?

Analysis and classic odes – L1 linear approximation by parts of the convex function of a variable

Consider a convex function f (x) of a variable x in some interval [a,b].

Question:
What is known about the approximation L1 of f (x) by linear functions in parts?
How to find nodes?
Are there some algorithms and / or theoretical results?

Geometrically it means that we need to find a linear curve by sections that is below it, it is the closest area below f (x), in other words, the area of ​​secters is minimal.

If you need to find a single node, a clear geometry is easily obtained.
solution: place the node at the point x where the tanget is parallel to the chord f (a) -> f (b). (See figure 1).
If we have many points, this gives the condition of connection node with the neighboring nodes, so one could think of the nodes as in some system with peer interaction.

enter the description of the image here

I have a dynamic programming algorithm, but I think you should know a lot in s

Figure 2: Approximation with two nodes.

Analysis and classic odes – Product of values ​​of a function with matrix value over $ S ^ 1 $

Assume you have a function $ f: S ^ 1 rightarrow mathrm {GL} _d ( mathbb {C}) $ whose coefficients are Laurent polynomials $ f_ {i, j} (q) in mathbb {C}[q^{pm 1}]$

I am interested in obtaining the conditions for the spectral radius of the matrix.
$$ F_n = f (e ^ { frac {2i pi (n-1)} {n}}) ldots f (e ^ { frac {2i pi} {n}}) f (1) $ $
be growing exponentially with $ n. $

Yes $ d = 1, $ means that $ f (q) $ is a single polynomial of Laurent, you can easily find it $ F_n = e ^ {n m (f) + o (1)}, $ where $ m (f) $ it's Mahler's measure of $ f. $
If we also assume $ f (q) in mathbb {Z}[q^{pm 1}]$ so $ F_n $ is growing exponentially as long as $ f $ It is not a product of cyclotomic polynomials.

When $ d> 1, $ You can think about using the determinant, but that does not give you any information if $ f $ take values ​​in $ mathrm {SL} _d ( mathbb {C}). $

Of course there are many examples where the spectral radio of $ F_n $ it does not grow exponentially: yes $ f $ takes value in $ mathrm {SU} _d $ he also does $ F_n $ Y $ F_n $ has a spectral radius 1. The same would apply to the upper triangular matrices with diagonal $ 1. $

Ideally, these counterexamples would be general, which is the spectral radius of $ F_n $ It grows exponentially or the image of $ f $ live in a smaller lying group that $ mathrm {SL} _d ( mathbb {C}). $ But any condition sufficient for the spectral radius to grow exponentially would be interesting to me.

Special functions: evaluation of the asymptotic equations of parabolic cylinders that arise from coupled ODEs.

For the context, I'm studying Coulomb paper blocking in Averin's superconducting quantum dot contacts since 1998. Specifically, I'm trying to find out how he gets equation 11 from equation 10, which gives Landau Zener the probability of end up in a specific branch of Josephson's potential of a superconducting QPC.

Equation 10, which describes Schrodinger's equation of the problem at the specific limit that is considered, is given by a system of two coupled first-order ODEs:
begin {equation}
2 sqrt { frac {E_C} { Delta}} frac { partial psi_s} { partial x} = -s x psi_s / 2 + sqrt {R} psi _ {- s}
end {equation}

where $ s pm1 $.

If I (for convenience) I take now $ A = 2 sqrt { frac {E_C} { Delta}} $ Y $ B = sqrt {R} $Then, substituting the differential equations, there is a second order ODE that is resolved by parabolic cylinder functions. Specifically,
begin {equation}
psi _ {- 1} = c_1 D { frac {-B ^ 2-A} {A}} left ( frac {x} { sqrt {A}} right) + c_2
D _ { frac {B ^ 2} {A}} left ( frac {i x} { sqrt {A}} right)
end {equation}

Y

begin {equation} psi_1 =
– frac {1} {B} left (x c_2 D _ { frac {B ^ 2} {A}} left ( frac {ix} { sqrt {A}} right) + sqrt { TO }
left[c_1 D_{-frac{B^2}{A}}left(frac{x}{sqrt{A}}right)+i c_2
D_{frac{B^2+A}{A}}left(frac{i x}{sqrt{A}}right)right]straight) \
end {equation}

Now, in his article, he says that by evaluating the asymptotes of these functions, one can find the probability $ w $ for the state $ s = 1 $ (beloning to $ psi_1 $) from $ x rightarrow – infty $ finish in the state $ s = -1 $ to $ x rightarrow infty $. This leads to equation 11, which writes as
begin {equation}
w = frac {1} { Gamma ( lambda)} sqrt { frac {2 pi} { lambda}} left ( frac { lambda} {e} right) ^ lambda
end {equation}

where $ lambda = frac {R ^ 2} {2 sqrt {E_C / Delta}} $, and using our substitutions, we can identify $ lambda = B ^ 2 / A $, that we already saw happen in the solution for $ psi _ {- 1} $.

My question is how do you get this? The language in the document is (to me) a bit uncertain, but from the physics, I know that we must choose our boundary conditions (configuration $ c_1 $ Y $ c_2 $) such that $ | psi_1 (- infty) | ^ 2 = 1 $, $ | psi _ {- 1} (- infty) | ^ 2 = 0 $ and then evaluate $ | psi _ {- 1} ( infty) | ^ 2 $ Arrive $ w $. But it seems that I can not find a way to correctly find the values ​​of $ c_1 $ Y $ c_2 $, since it seems that I can not impose these boundary conditions. How is one of these asymptotics treated?

Maybe you already find a clue in the form of the answer itself, which clearly looks like a Stirling approach to $ lambda! $Therefore, it is likely that we are looking for some solution with a Gamma function, but I can not understand it. Also, I must point out that this problem is essentially a problem of Landau Zener tunnels for those who are familiar with that.

Some code to evaluate the ODE & # 39; s:

ff = f[x] /. Solve[A*g'[A*g'[A*g'[A*g'[x] == x * g[x]/ 2 + B * f[x], f[x]];
dff = D[ff, x];
gg = g[x] /.
DSolve[{A*f'[{A*f'[{A*f'[{A*f'[x] == -x * f[x]/ 2 + B * g[x] /. {F[x] -> ff,
F & # 39;[x] -> dff}}, g[x], X]dgg = D[gg, x];
ff = FullSimplify[ff/{sun[ff/{g[ff/{sol[ff/{g[x] -> gg, g & # 39;[x] -> dgg}]

Analysis and classic odes – Lower limit of the standard $ L ^ 2 $ of a polynomial

Leave $ sigma> 0 $ be arranged even for $ k in mathbb {N} cup {0 } $, we consider the polynomial
begin {equation}
varphi_k (x) = sum_ {j = 0} ^ {k} (-1) ^ j {k choose j} b_j , x ^ {2j} x en (-1,1),
end {equation}

where
begin {equation}
b_j = frac { Big (k + sigma + frac12 Big) _j} { Big ( frac12 Big) _j}
end {equation}

and to $ s in mathbb {R} $, $ (s) _j $ denotes the symbol of Pochhammer
begin {equation}
(s) j = { begin {cases} 1 & j = 0 \ s (s + 1) cdots (s + j-1) & j> 0. end {cases}}
end {equation}

In particular, $ varphi_0 (x) = 1 $.

My question is this.

by $ -1 <a <b <1 $exists? $ c> 0 $ such that
begin {equation}
int_a ^ b | varphi_k (x) | ^ 2 dx geq c
end {equation}

even for any $ k in mathbb {N} cup {0 } $?

Unless I'm wrong, a direct calculation yields
begin {equation}
int_a ^ b | varphi_k (x) | ^ 2 dx = sum_ {j = 0} ^ k sum _ { ell = 0} ^ k (-1) ^ {j + ell} {k choose j} {k choose ell} frac {b_j b _ { ell}} {2 (j + ell) +1} , (b ^ {2 (j + ell) +1} -a ^ {2 (j + ell) +1} ).
end {equation}

But I do not see how I can join this double sum from below.

Observation: I do not know if it is of any use, one can notice that $ varphi_k $ It is a hypergeometric function of the form. $ {} _ {2} F_ {1} (- k, k + sigma + frac12; frac12; x ^ 2) $ (See https://en.wikipedia.org/wiki/Hypergeometricometric_function).

Edit: The same question is open when $ k $ It's strange, but this time you consider it.
begin {equation}
varphi_k (x) = sum_ {j = 0} ^ {k} (-1) ^ j {k choose j} c_j , x ^ {2j + 1},
end {equation}

with
begin {equation}
c_j = frac { Big (k + sigma + frac32 Big) _j} { Big ( frac32 Big) _j}.
end {equation}

I suspect that the methodology is similar to that of the even case.

Realize: I have asked the question in MSE, where it is subject to a reward: https://math.stackexchange.com/questions/3154394/bounding-a-polynomial-from-below.

plotting – How to plot the portiere phase and the ODES solutions

Dear all I know how to plot the portiere phase for the non-linear odes system

But I did not know why the result of Show Parametric solutions with portiere phase are not good.

I wrote the code in the following way:

Sun[{N0_, I0_}?NumericQ] : =
First @ NDSolve[{N1'
r N1
I1 & # 39;
m + N1
I1[0] == I0}, {N1, I1}, {t, 0, 365}];
P1 = ParametricPlot

The
Evaluate[{N1[{N1[{N1[{N1
30}, PlotRange -> All, AspectRatio -> Complete, PlotRange -> Complete,
Frame -> True, MaxRecursion -> 8] 

where

r = 0.431201; [Beta] = 2.99 * 10 ^ -6;
[Eta] = 0.2; [Sigma] = 0.7; [Rho] = 0.003; m = 0.427; [Delta] = 
0.57; [Mu] = .82;

and I used StreamPlot as follows

F[N1_, I1_] = r N1 (1 - [Beta] N1) - [Eta] N1 I1;
Sun[N1_, I1_] = [Sigma] + ([Rho] N1 I1) /
m + N1) - [Delta] I1 - [Mu] N1 I1;
Sun[{N1_, I1_}] = {f[N1, I1]g[N1, I1]};

so

StreamPlot[{F[{F[{F[{f[N1, I1]g[N1, I1]}, {N1, 0, 30}, {I1, 0, 30},
StreamStyle -> Blue, AspectRatio -> Automatic, Frame -> True,
Axes -> False, AxesLabel -> {"N1", "I1"}]Show[StreamPlot[{F[StreamPlot[{F[StreamPlot[{F[StreamPlot[{f[N1, I1]g[N1, I1]}, {N1, 0, 30}, {I1, 0, 30},
StreamPoints -> Fine, StreamStyle -> Blue, AspectRatio -> 1/2,
Frame -> True, AxesLabel -> {"N1", "I1"}, StreamPoints -> Fine,
PlotRange -> All], P1]

the final result was,

enter the description of the image here

Can someone help me improve my result?