Let us move on to the second discrete Fučík branch {C}_{2}^{d}. 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 m\in (1,N) and consider solutions which are nonnegative on {[0,m]}_{\mathbb{Z}},^{a} with values lying on the sine function (*cf.* Theorem 2 and Figure 2)

{x}_{P}(k)=sin\left(\frac{\pi}{m}k\right).

(9)

Since the solutions on the second branch change sign exactly once and lie on the sine function again, we seek constants C,\beta \in \mathbb{R} (*cf.* Theorem 2 and Figure 2) in the nonpositive part

{x}_{N}(k)=Csin\beta (k-N-1).

(10)

Then the vector (\lfloor m\rfloor and \lceil m\rceil denote floor and ceiling functions)

x=[{x}_{P},{x}_{N}]:={[0,{x}_{P}(1),\dots ,{x}_{P}(\lfloor m\rfloor ),{x}_{N}(\lceil m\rceil ),\dots ,{x}_{N}(N),0]}^{T}

is the solution of (1) coupled with the pair (\mu ,\nu ) given by

\mu =4{sin}^{2}\frac{\pi}{2m},\phantom{\rule{2em}{0ex}}\nu =4{sin}^{2}\frac{\beta}{2}.

(11)

Above, we considered m\in (1,N). This follows from the fact that if we had chosen m\le 1 or m\ge N, there would have been no positive/negative part and the solution would have laid on {C}_{1}^{d} 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 m=\lfloor m\rfloor =\lceil m\rceil, 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* m\in (1,N) *is an integer number*, *then*

\beta =\frac{\pi}{N+1-m},\phantom{\rule{2em}{0ex}}C=\frac{sin\frac{\pi}{m}}{sin\frac{\pi}{N+1-m}}.

(12)

*Moreover*, {x}_{P} *and* {x}_{N} *are equal in* m-1, *m* *and* m+1.

*Proof* If *m* is integer, then {x}_{P}(m)=0. Consequently, the difference equation (1) at k=m reduces to

{x}_{P}(m-1)+{x}_{N}(m+1)=0.

Exploiting the symmetry of sine functions, both {x}_{P} and {x}_{N} are equal in m-1, *m* and m+1. Since {x}_{N}(m)=0, we obtain that

\beta =\frac{\pi}{N+1-m}.

Finally, the equality of functions {x}_{P} and {x}_{N}, *e.g.*, at m+1, implies that

\begin{array}{c}sin\frac{\pi (m+1)}{m}=Csin\frac{\pi}{N+1-m}(m+1-N-1),\hfill \\ -sin\frac{\pi}{m}=-Csin\frac{\pi}{N+1-m},\hfill \\ C=\frac{sin\frac{\pi}{m}}{sin\frac{\pi}{N+1-m}}.\hfill \end{array}

□

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 \lfloor m\rfloor and \lceil m\rceil. Our first result states that the values of {x}_{P} and {x}_{N} coincide at \lfloor m\rfloor and \lceil m\rceil (*cf.* Figure 2).

**Lemma 4** (Necessary condition)

*Let* m\in (1,N) *be non*-*integer*. *Let* x=[{x}_{P},{x}_{N}] *be a nontrivial solution of* (1) *with* {x}_{P} *and* {x}_{N} *given by* (9) *and* (10). *Then the following equalities hold*:

{x}_{P}(\lfloor m\rfloor )={x}_{N}(\lfloor m\rfloor ),

(13)

{x}_{P}(\lceil m\rceil )={x}_{N}(\lceil m\rceil ).

(14)

*Proof* If {x}_{P} is given by (9), then the equation

{\mathrm{\Delta}}^{2}{x}_{P}(k-1)=\mu {x}_{P}(k)

holds for k=1,2,\dots ,\lfloor m\rfloor -1 if and only if (see the proof of Theorem 2)

\mu =4{sin}^{2}\frac{\pi}{2m}.

If we consider the difference equation (1) in \lfloor m\rfloor we get

{x}_{P}(\lfloor m\rfloor -1)-2{x}_{P}(\lfloor m\rfloor )+{x}_{N}(\lceil m\rceil )=4{sin}^{2}\frac{\pi}{2m}{x}_{P}(\lfloor m\rfloor ).

This equality holds if and only if (see (1))

{x}_{N}(\lceil m\rceil )={x}_{P}(\lceil m\rceil ),

which verifies (14).

Using the same argument at \lceil m\rceil for {x}_{N}, we obtain that (13) holds as well. □

This result enables us to get both peripheral parts of {C}_{2}^{d}.

**Corollary 5**

(\mu ,\nu )=(4{sin}^{2}\frac{\pi}{2m},2-\frac{sin\frac{(N-1)\pi}{m}}{sin\frac{N\pi}{m}})\in {C}_{2}^{d},\phantom{\rule{1em}{0ex}}m\in (N-1,N),

*and*

(\mu ,\nu )=(2-\frac{sin\frac{(N-1)\pi}{n}}{sin\frac{N\pi}{n}},4{sin}^{2}\frac{\pi}{2n})\in {C}_{2}^{d},\phantom{\rule{1em}{0ex}}n\in (N-1,N).

*Proof* Let us consider m\in (N-1,N). Then we have

{x}_{P}(N-1)=sin\left(\frac{(N-1)\pi}{m}\right),\phantom{\rule{2em}{0ex}}{x}_{P}(N)={x}_{N}(N)=sin\left(\frac{N\pi}{m}\right).

Then we can rewrite the equation (1) in k=N,

\begin{array}{c}-x(N-1)+2x(N)-x(N+1)=\nu x(N),\hfill \\ -sin\left(\frac{(N-1)\pi}{m}\right)+2sin\left(\frac{N\pi}{m}\right)=\nu sin\left(\frac{N\pi}{m}\right),\hfill \\ 2-\frac{sin\frac{(N-1)\pi}{m}}{sin\frac{N\pi}{m}}=\nu .\hfill \end{array}

This proves the former part of the statement. The latter follows from the mirror argument. □

Obviously, \nu >4 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 m\in (1,2)\cup (N-1,N), we could turn our attention to m\in (2,N-1) in the following. Applying (9) and (10), we can rewrite conditions (13) and (14) in the following way.

**Corollary 6** (Necessary condition II)

*Let* m\in (2,N-1) *be non*-*integer*. *If* {x}_{P} *and* {x}_{N} *have the form* (9) *and* (10), *then the following equality is satisfied*:

sin(\frac{\pi}{m}\lfloor m\rfloor )sin\beta (\lceil m\rceil -N-1)=sin(\frac{\pi}{m}\lceil m\rceil )sin\beta (\lfloor m\rfloor -N-1).

(15)

*Proof* One can rewrite equalities (13) and (14) into

\begin{array}{c}sin(\frac{\pi}{m}\lfloor m\rfloor )=Csin\beta (\lfloor m\rfloor -N-1),\hfill \\ sin(\frac{\pi}{m}\lceil m\rceil )=Csin\beta (\lceil m\rceil -N-1).\hfill \end{array}

Isolating *C* on the right-hand sides of both equations, we get

\frac{sin(\frac{\pi}{m}\lfloor m\rfloor )}{sin\beta (\lfloor m\rfloor -N-1)}=C=\frac{sin(\frac{\pi}{m}\lceil m\rceil )}{sin\beta (\lceil m\rceil -N-1)}.

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 \beta =\frac{\pi}{N+1-m} does not solve (15) for all *m*. In other words, if we define n:=\frac{\pi}{\beta}, we have m+n\ne N+1 for almost every m\in (1,N) (see Figure 2).

We have shown that

{C}_{2}^{d}{=}^{\mathbb{Z}}{C}_{2}^{d}{\cup}^{1}{C}_{2}^{d}{\cup}^{2}{C}_{2}^{d},

where (see Lemma 3 and (11))

{}^{\mathbb{Z}}C_{2}^{d}=\{(\mu ,\nu ):\mu =4{sin}^{2}\frac{\pi}{2m},\nu =4{sin}^{2}\frac{\pi}{2(N+1-m)},m\in {(1,N)}_{\mathbb{Z}}\},

and the parametrization of {}^{1}C_{2}^{d} is given by (5). Finally, we would like to get the parametrization of {}^{2}C_{2}^{d}, *i.e.*, consider m\in (2,N-1). Corollary 6 implies that one could get the parametrization of {}^{2}C_{2}^{d} if and only if one could solve (15). Equation (15) is equivalent to the problem of solving the transcendent equation

with A<0 and B\in \mathbb{Q}. Indeed, if we define

A=\frac{sin(\frac{\pi}{m}\lfloor m\rfloor )}{sin(\frac{\pi}{m}\lceil m\rceil )},\phantom{\rule{2em}{0ex}}C=\lceil m\rceil -N-1,\phantom{\rule{2em}{0ex}}D=\lfloor m\rfloor -N-1,

then the nonlinear equation (15) can be rewritten as

Asin(C\beta )=sin(D\beta ).

Considering m\in (\lfloor m\rfloor ,\lceil m\rceil ), we see that *D* and *C* are constant non-zero integers and A\in (-\mathrm{\infty},0). Consequently, if we substitute x=B\beta and D=BC, we get (16).

In order to get the parametrization of the complete second Fučík branch {C}_{2}^{d}, 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 m=\lfloor m\rfloor +\frac{1}{2}=\lceil m\rceil -\frac{1}{2}. Either the symmetry of sine functions or the direct computation yields that \beta =\frac{\pi}{N+1-m}. 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 {C}_{2}^{d} of the discrete spectrum slides monotonically along its continuous counterpart, coinciding in the ‘symmetric’ values of *m*, in which 2m\in \mathbb{N} (see Figure 3).

**Corollary 9** *The value of* *ν* *is decreasing in* *μ*.

*Proof* Putting \beta =\beta (m) and differentiating (15), we get

\begin{array}{c}{\beta}^{\prime}(m)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{cos(\frac{\pi}{m}\lfloor m\rfloor )\cdot \frac{\pi \lfloor m\rfloor}{{m}^{2}}\cdot sin\beta (\lceil m\rceil -N-1)-cos(\frac{\pi}{m}\lceil m\rceil )\cdot \frac{\pi \lceil m\rceil}{{m}^{2}}\cdot sin\beta (\lfloor m\rfloor -N-1)}{sin(\frac{\pi}{m}\lfloor m\rfloor )cos\beta (\lceil m\rceil -N-1)-sin(\frac{\pi}{m}\lceil m\rceil )cos\beta (\lfloor m\rfloor -N-1)}.\hfill \end{array}

Analyzing the signs of individual terms and observing that \frac{\pi}{m}\lfloor m\rfloor <\pi and that \frac{\pi}{m}\lceil m\rceil >\pi, we conclude that

and (11) yields the result. □