Let us move on to the second discrete Fučík branch . It corresponds to the solutions of (1) that change the sign exactly once (see Figure 2 for illustration and basic notation). As in the continuous case, we try to paste together two sine functions. Without loss of generality, we assume that the solution is positive first. We choose and consider solutions which are nonnegative on ,a with values lying on the sine function (cf. Theorem 2 and Figure 2)
(9)
Since the solutions on the second branch change sign exactly once and lie on the sine function again, we seek constants (cf. Theorem 2 and Figure 2) in the nonpositive part
(10)
Then the vector ( and denote floor and ceiling functions)
is the solution of (1) coupled with the pair given by
(11)
Above, we considered . This follows from the fact that if we had chosen or , there would have been no positive/negative part and the solution would have laid on instead.
Our first observation is trivial and considers integer values of m. In this case, the transition between the positive and negative parts occurs exactly at , and we could easily compute β and C in (10). In other words, we could still talk about pasting in this case.
Lemma 3 If is an integer number, then
(12)
Moreover, and are equal in , m and .
Proof If m is integer, then . Consequently, the difference equation (1) at reduces to
Exploiting the symmetry of sine functions, both and are equal in , m and . Since , we obtain that
Finally, the equality of functions and , e.g., at , implies that
□
The analysis gets more complicated once we consider non-integer values of m. In this case, the transition between positive and negative parts occurs between and . Our first result states that the values of and coincide at and (cf. Figure 2).
Lemma 4 (Necessary condition)
Let be non-integer. Let be a nontrivial solution of (1) with and given by (9) and (10). Then the following equalities hold:
(13)
(14)
Proof If is given by (9), then the equation
holds for if and only if (see the proof of Theorem 2)
If we consider the difference equation (1) in we get
This equality holds if and only if (see (1))
which verifies (14).
Using the same argument at for , we obtain that (13) holds as well. □
This result enables us to get both peripheral parts of .
Corollary 5
and
Proof Let us consider . Then we have
Then we can rewrite the equation (1) in ,
This proves the former part of the statement. The latter follows from the mirror argument. □
Obviously, for m sufficiently close to N. This implies that (11) cannot hold for any β, i.e., not all the solutions on the second branch can be obtained as a composition of sine functions!
Since the problem is solved for , we could turn our attention to in the following. Applying (9) and (10), we can rewrite conditions (13) and (14) in the following way.
Corollary 6 (Necessary condition II)
Let be non-integer. If and have the form (9) and (10), then the following equality is satisfied:
(15)
Proof One can rewrite equalities (13) and (14) into
Isolating C on the right-hand sides of both equations, we get
Now, it suffices to multiply this equality by both denominators to get (15). □
Remark 7 (Anchoring)
Corollary 6 implies that the (continuous extensions of) sine functions (9) and (10) do not intersect at m in general. Indeed, we could see that does not solve (15) for all m. In other words, if we define , we have for almost every (see Figure 2).
We have shown that
where (see Lemma 3 and (11))
and the parametrization of is given by (5). Finally, we would like to get the parametrization of , i.e., consider . Corollary 6 implies that one could get the parametrization of if and only if one could solve (15). Equation (15) is equivalent to the problem of solving the transcendent equation
with and . Indeed, if we define
then the nonlinear equation (15) can be rewritten as
Considering , we see that D and C are constant non-zero integers and . Consequently, if we substitute and , we get (16).
In order to get the parametrization of the complete second Fučík branch , we need to solve (15) for β, or, equivalently, (16) for x. While this could be done pretty easily numerically, we state, in the final section, a conjecture that it is not possible to use a finite number of elementary functions to get such a parametrization. Meanwhile, we make two straightforward observations.
Remark 8 Considering non-integer values of m, we can solve equation (15) only in the symmetric case, in which . Either the symmetry of sine functions or the direct computation yields that . Similarly, as in the integer case (cf. Lemma 3), this value coincides with the value of the corresponding continuous problem as the positive and negative parts meet at m.
Combining this with the following straightforward corollary, we observe that the second branch of the discrete spectrum slides monotonically along its continuous counterpart, coinciding in the ‘symmetric’ values of m, in which (see Figure 3).
Corollary 9 The value of ν is decreasing in μ.
Proof Putting and differentiating (15), we get
Analyzing the signs of individual terms and observing that and that , we conclude that
and (11) yields the result. □