Proof. The statement $P^{\mathrm {fl}}(t)> 0$ for every $t \in (0, 1)$ is proven via a computer-assisted proof. The code can be found in the supplementary material, and we refer to Appendix B for details about the implementation.

With respect to $D_Z (b^{\mathrm {fl}}(t))> 0$ , note that this is a second-degree polynomial vanishing at $t=0$ . For $t> 0$ small enough, $D_Z (b^{\mathrm {fl}}(t))> 0$ since the slope of $b^{\mathrm {fl}}(t)$ coincides with the slope of the smooth solution (and $D_{Z, 1}> 0$ by Lemma A.11). Now, as (Lemma A.26), the second-degree polynomial $D_Z(b^{\mathrm {fl}}(t))$ cannot be non-positive at any $0 < t_0 < 1$ because otherwise $D_Z(b^{\mathrm {fl}}(t))$ would have three roots by continuity (one at $t=0$ , and two in $(0, 1)$ ).

Lastly, $D_W(b^{\mathrm {fl}}(t))$ is also a second-degree polynomial which is positive at $t=0$ and $t=1$ . It would need to have two roots in $[0, 1]$ in order to be negative for some $t \in [0, 1]$ . That is impossible since its derivative at $0$ is positive: $b^{\mathrm {fl}}(t)$ agrees up to first order with the smooth solution and $D_{W, 1}> 0$ by Lemma A.26.

Proof. We have that $D_W (b^{\mathrm {fl}}_{7/5}(t))$ is a third-degree polynomial. Calculating this polynomial at $r = r^\ast (7/5)$ , we obtain

$$ \begin{align*} D_W (b^{\mathrm{fl}}_{7/5}(t))\Big|_{r=r^\ast} = \frac{1}{396} \left(-52 \left(3 \sqrt{5}-10\right) t^3+\left(57 \sqrt{5}-91\right) t^2-33 \left(3 \sqrt{5}-5\right) t+198 \left(\sqrt{5}-1\right)\right). \end{align*} $$

As all the coefficients are positive, and we obtain that the coefficients of $D_W (b^{\mathrm {fl}}_{7/5}(t))$ are still positive for r sufficiently close to $r^\ast (7/5)$ , so that $D_W (b^{\mathrm {fl}}_{7/5}(t))> 0$ for $t \in (0, 1)$ .

With respect to $P^{\mathrm {fl}}_{7/5}(t)$ , we observe that this polynomial is a multiple of $t^3$ since the first three coefficients of $b^{\mathrm {fl}}_{7/5} (t)$ agree with those of the smooth solution passing through $P_s$ . We also have that $P_{7/5}^{\mathrm {fl}}(1) = 0$ because $N_W(P_\infty ) = N_Z (P_\infty ) = 0$ and $D_W(P_\infty ) = D_Z(P_\infty ) = 1$ . Moreover, we have that

$$ \begin{align*} P_{7/5}^{\mathrm{fl} \; \prime}(1) &= b^{\mathrm{fl} \; \prime}_{7/5, Z}(1) D_Z (P_\infty ) \nabla N_W (P_\infty) \cdot b^{\mathrm{fl} \; \prime}_{7/5} (1) - b^{\mathrm{fl} \; \prime}_{7/5, W} (1) D_W (P_\infty ) \nabla N_Z (P_\infty ) \cdot b^{\mathrm{fl} \; \prime}_{7/5} (1) \\ &= b^{\mathrm{fl} \; \prime}_{7/5, Z}(1) (-r, 0) \cdot b^{\mathrm{fl} \; \prime}_{7/5} (1) - b^{\mathrm{fl} \; \prime}_{7/5, W} (1) (0, -r) \cdot b^{\mathrm{fl} \; \prime}_{7/5} (1) = 0. \end{align*} $$

Therefore, we see that $P_{7/5}^{\mathrm {fl}}(t) = (1-t)^2 t^3 Q_{7/5}^{\mathrm {fl}}(t)$ for some sixth-degree polynomial $Q_{7/5}^{\mathrm {fl}}(t)$ . We just need to show that $Q_{7/5}^{\mathrm {fl}}(t)> 0$ for $t \in (0, 1)$ . Calculating $Q_{7/5}^{\mathrm {fl}}(t)$ at $r=r^\ast $ , we obtain

$$ \begin{align*} Q_{7/5}^{\mathrm{fl}}(t)\Big|_{r=r^\ast} &= 375 \left(17845372 \sqrt{5}-22507109\right) t^6+25 \left(242253290 \sqrt{5}-88777981\right) t^5 \\&\quad+10 \left(1945028708 \sqrt{5}-2421950855\right) t^4+660 \left(12759056 \sqrt{5}-8257439\right) t^3 \\&\quad +7260 \left(1604200 \sqrt{5}-2790277\right) t^2-21780 \left(137 \sqrt{5}-38799\right) t+313632 \left(6133 \sqrt{5}-7995\right). \end{align*} $$

As all the coefficients are positive, they will be positive for r sufficiently close to $r^\ast $ , and therefore, for r sufficiently close to $r^\ast $ , we have that $Q_{7/5}^{\mathrm {fl}}(t)> 0$ for $t \in (0, 1)$ .

With respect to $D_Z$ , we have that $D_Z (b^{\mathrm {fl}}_{7/5}(t))$ is a multiple of t (as it vanishes at zero), and moreover,

(3.7) $$ \begin{align} \frac{1}{t} D_Z (b^{\mathrm{fl}}_{7/5}(t)) = D_{Z, 1} + \frac{D_{Z, 2}}{2} t + \frac16 D_{Z, 3}^{\mathrm{fl}}t^2 \end{align} $$

for some $D_{Z, 3}^{\mathrm {fl}}$ . We have that $D_{Z, 1}> 0$ from Lemma A.11, so the polynomial is initially positive. However, at $r = r^\ast $ , we have that $D_{Z, 1} = 0$ , $\frac {D_{Z, 2}}{2} = \frac {19-9\sqrt {5}}{264} < 0$ and $\frac {D_{Z, 3}^{\mathrm {fl}}}{6} = \frac {245 + 9\sqrt {5}}{264}> 0$ . Therefore, it is clear that for r sufficiently close to $r^\ast $ , we have two real roots of the second-degree polynomial (3.7), which we define to be $t^{\mathrm {in}}_{7/5}$ and $t^{\mathrm {out}}_{7/5}$ . Moreover, as $D_{Z, 3}^{\mathrm {fl}}> 0$ , the sign of (3.7) is positive except for $t \in [t^{\mathrm {in}}_{7/5}, t^{\mathrm {out}}_{7/5}]$ .

Lastly, we just need to show that $b^{\mathrm {fl}}_{7/5}(t^{\mathrm {in}}_{7/5})$ is located to the left of $\bar P_s$ ; that is, we need to check

(3.8)

This is checked in Lemma A.15.

Proof. Both curves $(W^{(r)}, Z^{(r)})(\xi )$ and $b^{\mathrm {nl}}_n(s)$ agree in their Taylor expansions around $P_s$ up to order n. If the $(n+1)$ -th coefficient is given by a term $(W_{n+1}, Z_{n+1}) \frac {t^{n+1}}{(n+1)!}$ , this will have a normal component over the tangent line of size $- (W_1, Z_1) \wedge (W_{n+1}, Z_{n+1}) \frac {t^{n+1}}{(n+1)!}$ (where the sign is positive for a deviation above the tangent line at $P_s$ and negative for a deviation below the tangent line at $P_s$ ).

Therefore, we just need to check

$$ \begin{align*} - (W_1, Z_1) \wedge (W_{n+1}, Z_{n+1})> 0 \Leftrightarrow W_1 Z_{n+1} - Z_1 W_{n+1} < 0. \end{align*} $$

This is done in Lemma A.26 for the case of $n = 3$ . For the case of $\gamma = 7/5$ and n odd sufficiently large, this will follow from Section 5, concretely from Corollary 5.9.

Proof. We first formulate the barrier $b^{\mathrm {fl}}(t)$ in implicit form using the resultant

(3.11) $$ \begin{align} B^{\mathrm{fl}} (W, Z) = \left| \begin{matrix} \frac{B_2}{2} & B_1 W_1 & W_0 - W & 0 \\ 0 & \frac{B_2}{2} & B_1 W_1 & W_0 - W \\ \frac{B_3}{2} & B_1 Z_1 & Z_0 - Z & 0 \\ 0 & \frac{B_3}{2} & B_1 Z_1 & Z_0 - Z \\ \end{matrix} \right|, \end{align} $$

so that the equation $b^{\mathrm {fl}}(t_\star ) = b^{\mathrm {nl}}(s_\star )$ can be reformulated as $B^{\mathrm {fl}}(b^{\mathrm {nl}}_n (t_\star ) ) = 0$ . Let us fix $s_- = 35/100$ . We will divide the proof in five steps:

1. 1. For every $\gamma> 1$ , the polynomial $B^{\mathrm {fl}}(b^{\mathrm {nl}}_3 (s))$ is negative for $s>0$ sufficiently small for all $r\in (r_3, r_4)$ .

2. 2. For every $\gamma> 1$ , the polynomial $B^{\mathrm {fl}}(b^{\mathrm {nl}}_3 (s_-(k-3 ))$ is positive for all $r\in (r_3, r_4)$ .

3. 3. For every $\gamma> 1$ , we have $P^{\mathrm {nl}}_3(s)> 0$ for all $s\in (0, s_-(k-3))$ and $r\in (r_3, r_4)$ .

4. 4. For every $\gamma> 1$ , we have $D_W (b^{\mathrm {nl}}_3(s))> 0$ for all $s\in (0, s_-(k-3))$ and $r\in (r_3, r_4)$ .

5. 5. For every $\gamma> 1$ , we have $D_Z (b^{\mathrm {nl}}_3(s))> 0$ for all $s\in (0, s_- (k-3))$ and $r\in (r_3, r_4)$ .

From items 1 and 2, by continuity, there exists a value $s_\star \in (0, s_-(r-r_3)]$ such that $B^{\mathrm {fl}}(b^{\mathrm {nl}}_3(s_\star )) = 0$ , and therefore, there exists $t_\star $ such that $b^{\mathrm {fl}}(t_\star ) = b^{\mathrm {nl}}_3(s_\star )$ . Then, items 3, 4 and 5 give us the desired result.

Finally, we prove each of those steps with a computer-assisted proof. The code can be found in the supplementary material. We refer to Appendix B for details about the implementation.

Proof of Proposition 3.1.

For $\gamma = 7/5$ and n sufficiently large, we can consider the closed region $\mathcal {T}$ of the plane which has a corner at $P_s$ and is enclosed by $b^{\mathrm {nl}}_{7/5,n}(t)$ , $b^{\mathrm {fl}}_{7/5}(t)$ , $D_Z = 0$ , the diagonal $W = Z$ and the vertical line $W = W_0$ (starting at $(W_0, W_0)$ , ending at $P_s$ ). The intersection between $b^{\mathrm {nl}}_{7/5,n}(t)$ and $b^{\mathrm {fl}}_{7/5}(t)$ is proven in Proposition 3.6. There are two intersection points between $b^{\mathrm {fl}}_{7/5}(t)$ and $D_Z = 0$ as indicated in Proposition 3.3. Note that Proposition 3.3 gives us two cases. In the first case, we go from $P_s$ to $P_\infty $ by following $b^{\mathrm {nl}}_{7/5,n}(s)$ for $0\leq s \leq s_{7/5, \mathrm {int}}$ , then $b^{fl}_{7/5}$ up to $P^{\mathrm {in}}_{7/5}$ , then $D_Z = 0$ up to $P^{\mathrm { out}}_{7/5}$ and finally $b^{\mathrm {fl}}_{7/5}(t)$ up to $P_\infty $ . In the second case, the path is the same except that we connect directly $b^{\mathrm {nl}}_{3}$ with $D_Z = 0$ at $b^{\mathrm {nl}}_3(s_{7/5, \mathrm {int}}')$ . We should notice that in any case, the region of $D_Z = 0$ which forms part of $\mathcal {T}$ is always located to the left of $\bar P_s$ (by Proposition 3.6 or Proposition 3.3). For general $\gamma $ and $r \in (r_3, r_4)$ , the endpoint of $b^{\mathrm {fl}}(t)$ at $t=1$ is , so we consider the same region with the addition of the barrier $b^{\mathrm {extra}} (t)= (X_0 - t, Y_0 + t)$ for $t \in [0, \frac 12 (X_0 - Y_0)]$ . That is, we take $\mathcal {T}$ to be enclosed by $b^{\mathrm {nl}}_3(t)$ , $b^{\mathrm {fl}}(t)$ , $b^{\mathrm {extra}}(t)$ , $W+Z = 0$ and $W=W_0$ . The intersection between $b^{\mathrm {nl}}_3(t)$ and $b^{\mathrm {fl}}(t)$ is guaranteed by Proposition 3.5. We show a sketch of region $\mathcal {T}$ in Figure 4 and Figure 5.

Figure 4

Region $\mathcal {T}$ for the case $\gamma = 7/5$ and r sufficiently close to $r^\ast $ .

Figure 5

Region $\mathcal {T}$ for the case $\gamma> 1$ and $r \in (r_3, r_4)$ .

We will now show that the region $\mathcal {T}$ does not intersect the line $D_W = 0$ . We first show that the line $D_W = 0$ does not intersect $b^{\mathrm {nl}}(t)$ and $b^{\mathrm {fl}}(t)$ . For the case $r \in (r_3, r_4)$ , this follows as a consequence of Proposition 3.5 and Proposition 3.2, and for the case $\gamma = 7/5$ , n odd sufficiently large, this is follows from Proposition 3.3 and Proposition 3.6. In the case $r \in (r_3, r_4)$ , the fact that the line $D_W = 0$ does not intersect $b^{\mathrm {extra}}(t)$ follows from Lemma A.19. Moreover, as $D_W$ is increasing with Z along $W = W_0$ and is also increasing with t along $(t, t)$ , we conclude that $D_W> 0$ in all $\partial \mathcal {T}$ . As $D_W = 0$ is a straight line and $\mathcal {T}$ is connected, we obtain that $D_W> 0$ in all $\mathcal {T}$ . The same reasoning allows us to say that $\partial \mathcal {T}$ only intersects $D_Z = 0$ at $P_s$ and (in the case where $\gamma = \frac 75$ and r sufficiently close to $r^\ast $ ) the points in the segment between $P^{\mathrm {in}}$ and $P^{\mathrm {out}}$ defined in Proposition 3.3.

The smooth solution $W^{(r)}(\xi ), Z^{(r)} (\xi )$ is in $\mathcal {T}$ for small enough $\xi $ due to Lemma 3.4 and the fact that $W_1 < 0$ (Lemmas A.26 and A.7). It cannot exit $\mathcal {T}$ through $b^{\mathrm {nl}}(t)$ , $b^{\mathrm {fl}}(t)$ or $b^{\mathrm {extra}}(t)$ due to Proposition 3.5 and Proposition 3.6 (for $b^{\mathrm {nl}}(t)$ ), Proposition 3.2 and Proposition 3.3 (for $b^{\mathrm {fl}}(t)$ ) and Lemma A.19 (for $b^{\mathrm {extra}}(t)$ , $r \in (r_3, r_4)$ ). In the case of $\gamma = \frac 75$ and r close enough to $r^\ast $ , it cannot exit through $D_Z = 0$ because that region is always located to the left of $\bar P_s$ , and the field points inwards there (Lemma A.22). It cannot exit through the line $W = Z$ because that line is an invariant of the field ( $N_W / D_W, N_Z / D_Z$ is proportional to $(1, 1)$ in that diagonal). Also, it cannot exit through $W = W_0$ since $N_W / D_W < 0$ ( $D_W> 0$ from last paragraph and $N_W < 0$ from Lemma A.20).

Therefore, as $\mathcal {T}$ is bounded, Proposition 1.6 yields that either $(W^{(r)}, Z^{(r)})$ converges to some point of $\mathcal {T}$ with $D_W = 0$ or $D_Z=0$ or it converges to some equilibrium point inside $\mathcal {T}$ as $\xi \rightarrow + \infty $ . There are no points with $D_W = 0$ due to the second paragraph. The solution cannot converge to the points of $D_Z = 0$ in the segment $[P^{\mathrm {in}}_{7/5}, P^{\mathrm {out}}_{7/5}]$ because the field points inwards to $\mathcal {T}$ in that segment (due to Proposition 3.3 and Lemma A.22). Now, we show that the situation where the solution converges to $P_s$ is also not possible. As $P_s$ is the point with minimum Z in $B_\varepsilon (P_s) \cap \mathcal {T}$ for $\varepsilon $ small enough,Footnote ³ we would need to have points in $\mathcal {T}$ arbitrarily close to $P_s$ with $\frac {N_Z}{D_Z}> 0$ . In order to show that this does not happen, we just need to check that $D_Z> 0$ and $N_Z> 0$ in $(\mathcal {T} \setminus \{ P_s \}) \cap B_\varepsilon (P_s)$ for $\varepsilon $ small enough. As the corner of $\partial \mathcal {T}$ at $P_s$ has less than $\pi $ radians (because $W_1 < 0$ ), we just need to check $\nabla D_Z (P_s) \cdot (W_1, Z_1)> 0$ , $\nabla D_Z (P_s) \cdot (0, 1)$ , $\nabla N_Z (P_s) \cdot (W_1, Z_1)> 0$ and $\nabla N_Z (P_s)\cdot (0,1)> 0$ . Clearly, $\partial _Z D_Z = (\gamma + 1)/4> 0$ and $D_{Z, 1}, N_{Z, 1}, \partial _Z N_Z (P_s)> 0$ due to Lemmas A.11, A.26 and A.7.

Thus, we conclude that $W^{(r)}, Z^{(r)}$ converge to some equilibrium point of $\mathcal {T}$ – that is, some point with $N_W = N_Z = 0$ . There are four solutions to that system, which are . The point clearly lies in the half-plane $Z> W$ , so it is not in $\mathcal {T}$ . The point $(-r, -r)$ is also never in $\mathcal {T}$ . For the case $\gamma = 7/5$ , n sufficiently large, this is trivial and for the $r \in (r_3, r_4)$ case, this is because $b^{\mathrm {extra}}(t)$ intersects $W=Z$ at some point above $(-r, -r)$ (Lemma A.26).

Finally, we show that the solution does not converge to . For the case $\gamma = 7/5$ with n sufficiently large, this point is discarded arguing that is not in $\mathcal {T}$ . We can parametrize $b^{\mathrm {fl}}_{7/5}(t)$ by W because $b^{\mathrm {fl}}_{7/5}(t)$ is decreasing by Lemma A.24, so all the points $(b^{\mathrm {fl}}_{7/5, W}(t), Z)$ with $Z < b^{\mathrm {fl}}_{7/5, Z}(t)$ are outside $\mathcal {T}$ . That is the case of due to Lemma A.24. Now we show that the solution does not converge to for the general $\gamma $ case with $r \in (r_3, r_4)$ . Note that is a saddle point of the field $(N_W D_Z, N_Z D_W)$ due to Lemma A.17. Therefore, there is only one direction along for trajectories converging to , and will be given by the eigenvector of negative eigenvalue, $v_-$ . We will see that there are not points in $\mathcal {T}$ approaching in direction $v_-$ (or $-v_-$ ), and that will conclude the proof.

Let us fix the angles $-\pi \leq \theta _- < \theta _+ < \theta _- +\pi < \pi $ , so that $\theta _-, \theta _- +\pi $ indicate the angles of $v_-, -v_-$ and $\theta _+, \theta _+ + \pi $ indicate the angles of $v_+, -v_+$ . Locally around , the angular component of the field around a saddle point points counterclockwise in the region $\Theta _\circlearrowleft = (\theta _-, \theta _+)\cup (\theta _- + \pi , \theta _+ + \pi )$ , while it points clockwise in the region $\Theta _\circlearrowright = (\theta _+, \theta _- +\pi )\cup (\theta _- + \pi , \theta _+ + 2\pi )$ . Let $\theta _{\mathrm {fl}}, \theta _{\mathrm {extra}} \in [-\pi , \pi )$ be the angles at which those barriers arrive to . Note $\theta _{\mathrm {extra}} = 3\pi /4$ and note also $\theta _{\mathrm {fl}} < 3\pi /4$ by Lemma A.25. As the field points inwards to $\mathcal {T}$ on the barriers $b^{\mathrm {fl}}$ and $b^{\mathrm {extra}}$ , we get that $\theta _{\mathrm {fl}} \in \Theta _\circlearrowleft $ and $\theta _{\mathrm {extra}} \in \Theta _\circlearrowright $ . Therefore, the set $(\theta _{\mathrm {fl}}, \theta _{\mathrm {extra}}) \cap \{ \theta _+ - \pi , \theta _-, \theta _+, \theta _-+\pi , \theta _++\pi \}$ has an odd number of elements. If there are three elements, we get $\theta _{\mathrm {extra}} - \theta _{\mathrm {fl}}> \pi $ , which contradicts Lemma A.25, so there is only one element. If that element is $\theta _-$ or $\theta _- + \pi $ , we would get that $\theta _{\mathrm {fl}} \in \Theta _\circlearrowright $ and $\theta _{\mathrm {extra}} \in \Theta _\circlearrowleft $ , which is also a contradiction. Thus, $\theta _-, \theta _- + \pi \notin (\theta _{\mathrm {fl}}, \theta _{\mathrm {extra}})$ .

Proof. First, note that $P^{\mathrm {nr}}_n (t)$ is a multiple of $t^{n+1}$ since $b^{\mathrm {nr}}_n(t)$ matches the smooth solution up to n-th order. By Corollary 2.4, the only terms in $b^{\mathrm {nr}}_n (t)$ which are not $O(1)$ in a neighborhood of $r_n$ are $\frac {1}{n!} Z_n (-t)^n$ and $\frac {1}{(n+1)!} Z_n \beta _n (-t)^{n+1}$ , which are $O_{\beta _n}\left ( \frac {1}{k-n} \right )$ . Summing the terms, the asymptotics for $P^{\mathrm {nr}}_n (t)$ are given by

(4.2) $$ \begin{align} \begin{aligned} P^{\mathrm{nr}}_n (t) &= C_1 Z_n t^{n+1} + C_2 Z_n^2 t^{2n-1} \\ &\qquad + O_{\beta_n}(t^{n+1} ) + O_{\beta_n}\left( \frac{t^{n+2}}{k-n} \right) + O_{\beta_n}\left( \frac{t^{2n}}{(k-n)^2 } \right) + O_{\beta_n}\left( \frac{t^{3n-1}}{(k-n)^3 } \right) + O_{\beta_n}\left( \frac{t^{4n-1}}{(k-n)^4 } \right), \end{aligned} \end{align} $$

for small t and $|k-n|$ . In order to calculate $C_1$ , we take $n+1$ derivatives in $P^{\mathrm {nr}}_n (t)$ and look for the terms with a factor $Z_n$ (as the rest of terms will be $O(1)$ , thus contributing to $O(t^{n+1})$ in (4.2)). Note that $P^{\mathrm {nl}}_n(t)$ already involves one derivative, so we will have terms with $n+2$ derivatives in total. At $t=0$ , we have

$$ \begin{align*} (-1)^{n+2} \partial_t^{n+1} P^{\mathrm{nr}}_n &= (n+1)\beta_n Z_n (-\partial_t) D_Z (b^{\mathrm{nr}}_{n}) N_W (b^{\mathrm{nr}}_n) + Z_1 (-\partial_t)^{n+1} D_Z (b^{\mathrm{nr}}_n ) N_W (b^{\mathrm{nr}}_n )\\& \quad - W_1 (-\partial_t)^{n+1} N_Z (b^{\mathrm{nr}}_n ) D_W (b^{\mathrm{nr}}_n ) + O(|Z_n|) .\end{align*} $$

Collecting the terms involving $Z_n\beta _n$ , we define

(4.3) $$ \begin{align} \begin{aligned} C_3 &= -(n+1)D_{Z, 1} N_W (P_s) - Z_1 \partial_Z D_Z (P_s) N_W (P_s) + W_1 \partial_Z N_Z (P_s) D_W (P_s) \\ &= N_W(P_s) \left( -(n+1)D_{Z, 1} - Z_1 \partial_Z D_Z (P_s) + \partial_Z N_Z (P_s)\right), \end{aligned}\end{align} $$

so that we obtain

$$\begin{align*}(-\partial_t)^{n+1} P^{\mathrm{nr}}_n (0) = C_3 Z_n \beta_n + O(|Z_n|). \end{align*}$$

As $\left |r_n-r\right |\leq \varepsilon $ and we are taking $\varepsilon $ small, we may evaluate the sign of $C_3$ by looking at its sign at $r=r_n$ , which by continuity will remain the same sign in a neighborhood of $k=n$ . Let us recall from Lemma 2.1 that $k = \frac {\check D_{Z, 1}}{D_{Z, 1}}$ , where $\check D_{Z, 1} = \nabla D_Z (P_s) \cdot (W_1, \check Z_1)$ and $\check Z_1$ was defined in (2.1). Thus, using (4.3) at $k = n = \frac {\check D_{Z, 1}}{D_{Z, 1}}$ yields

(4.4) $$ \begin{align} \frac{C_3}{N_{W, 0}} &= - \check D_{Z, 1} - D_{Z, 1} - Z_1 \partial_Z D_Z(P_s) + \partial_Z N_Z (P_s)\nonumber \\ &= -D_{Z, 1} - \partial_W D_Z (P_s) W_1 - \partial_Z D_Z (P_s) \left( \check Z_1 + Z_1 \right) + \partial_Z N_Z(P_s). \end{align} $$

Note $Z_1, \check Z_1$ are the two solutions of the second-degree equation

$$ \begin{align*} \partial_Z D_Z z_1^2 + \left( \partial_W D_Z (P_s) W_1 - \partial_W N_Z (P_s) \right)z_1 - \partial_W N_Z (P_s) W_1 = 0, \end{align*} $$

and hence,

$$\begin{align*}Z_1+ \check Z_1=-\frac{\partial_W D_Z (P_s) W_1 - \partial_W N_Z (P_s)}{\partial_Z D_Z (P_s)}. \end{align*}$$

Substituting this expression into (4.4) and canceling terms, we deduce

$$\begin{align*}C_3 = - N_{W, 0} D_{Z, 1}. \end{align*}$$

Using Lemmas A.26 and A.11, we conclude that $C_3> 0$ for $k=n$ , and thus, it is positive for r in a sufficiently small neighborhood of $r_n$ .

We have that

$$ \begin{align*} C_1 = \frac{1}{(n+1)!Z_n}(-\partial_t)^{n+1} P^{\mathrm{nr}}_n (0) \geq \frac{(-1)^{n+1}}{(n+1)!} \left( \beta_n C_3 + O(1)\right). \end{align*} $$

Therefore, choosing $\beta _n$ sufficiently large, r close enough to $r_n$ and using that $C_3> 0$ ,

(4.5) $$ \begin{align} \text{sign}(C_1) = (-1)^{n+1} , \quad \quad \left| C_1 \right|> \frac{\beta_n |C_3|}{2(n+1)!}. \end{align} $$

We calculate the term $C_2$ in (4.2). Taking $2n-1$ derivatives in $P^{\mathrm {nr}}_n (t)$ and looking for terms with $Z_n^2$ , we obtain

$$ \begin{align*} (-1)^{2n} \partial_t ^{2n-1} P^{\mathrm{nr}}_n (t) = \binom{2n-1}{n} Z_n \partial_t^n D_Z (b^{\mathrm{nr}}_n ) N_W (P_s) + O_{\beta_n}\left( Z_n \right) . \end{align*} $$

Therefore,

(4.6) $$ \begin{align} C_2 = \frac{1}{(2n-1)!} \binom{2n-1}{n} \partial_Z D_Z (P_s) N_W (P_s) = \frac{1+\alpha}{2} \frac{1}{n!(n-1)!} N_W(P_s) < 0, \end{align} $$

where the sign is due to Lemma A.26.

Now that we have analyzed $C_1$ and $C_2$ in (4.5) and (4.6); let us go back to (4.2) and consider the cases of n odd and even separately.

Odd n case. We have that

$$\begin{align*}C_1> \frac{\beta_n C_3}{2(n+1)!} > 0\quad\mbox{and}\quad C_2 < 0\,\end{align*}$$

by (4.5) and (4.6). Moreover, we have that for odd n and r sufficiently close to $r_n$ from above, $Z_n> 0$ (Corollary 5.8 and Lemma A.31); moreover, assuming in addition that $\beta _n$ is sufficiently large, and $t^{n-2} \leq c^{n-2} (k-n) \beta _n$ , the equation (4.2) yields

$$ \begin{align*} P^{\mathrm{nr}}_n(t)> \frac{1}{2} C_1 Z_n t^{n+1} + C_2 Z_n^2 t^{2n-1} > t^{n+1} Z_n \beta_n \left( \frac{C_3}{4(n+1)!} - c^{n-2} |C_2| Z_n (k-n) \right). \end{align*} $$

Hence, since $Z_n (k-n) = O(1)$ , and we may choose c sufficiently small, we obtain that $ P^{\mathrm {nr}}_n(t)>0$ .

Even n case. For this case, we obtain that for $\beta _n$ sufficiently large,

$$\begin{align*}C_1 < -\frac{\beta_n C_3}{2(n+1)!} < 0\quad\mbox{and}\quad C_2 < 0\,\end{align*}$$

by (4.5) and (4.6). Moreover, for n even and r close to $r_n$ from below, we have that $Z_n> 0$ (Lemma A.31 and Corollary 5.8). Therefore, taking $\beta _n$ sufficiently large, r close enough from below to $r_n$ and $t \leq c(n-k)^{1/n}$ , (4.2) yields

$$ \begin{align*} P^{\mathrm{nr}}_n(t) &< \frac{1}{2}C_2 Z_n^2 t^{2n-1} + O_{\beta_n}\left( \frac{t^{2n}}{(n-k)^2} + \frac{t^{3n-1}}{(n-k)^3} + \frac{t^{4n-1}}{(n-k)^4} \right) \\ &= Z_n^2 t^{2n-1} \left( \frac{C_2}{2} + O_{\beta_n} \left( c(n-k)^{1/n} +c^n+c^{2n}\right) \right), \end{align*} $$

which is negative as long as c is chosen small enough since $C_2 < 0$ .

Proof. We have that $D_Z (b^{\mathrm {nr}}_{n+1} (t))$ is a $(n+2)$ -th degree polynomial, which is multiple of t since

$$\begin{align*}D_Z (b^{\mathrm{nr}}_{n+1} (0)) = D_Z (P_s) = 0.\end{align*}$$

Moreover, we have that at $t=0$ ,

(4.7) $$ \begin{align} \begin{cases} \partial_t^i D_Z (b^{\mathrm{nr}}_{n+1} (t)) = (-1)^i \nabla D_Z (P_s) (W_{n+1}, Z_{n+1}), \qquad \text{ for } 1\leq i \leq n+1, \\ \partial_t^{n+2}D_Z (b^{\mathrm{nr}}_{n+1} (t)) = (-1) \nabla D_Z (P_s) (0, Z_{n+1} \beta_{n+1} ). \end{cases} \end{align} $$

Corollary 2.4 tells us that all the coefficients of $D_Z(b^{\mathrm {nr}}_{n+1}(t))$ are $O(1)$ for r in a small neighborhood of $r_{n+1}$ , except for the $(n+1)$ -th and $(n+2)$ -th coefficients, which are $O_{\beta _{n+1}} \left ( \frac {1}{n+1-k} \right )$ . We can thus write

(4.8) $$ \begin{align} D_Z (b^{\mathrm{nr}}_{n+1} (t)) = -D_{Z, 1}t + \frac{\partial_Z D_Z (P_s) Z_{n+1} t^{n+1}}{(n+1)!} + O(t^2) + O \left( \frac{t^{n+2} \beta_{n+1}}{n+1-k} \right). \end{align} $$

We have that $-D_{Z, 1} < 0$ as a consequence of Lemma A.11 and Lemma A.7 and

$$\begin{align*}\partial_Z D_Z (P_s) Z_{n+1} = \frac{1+\alpha}{2} Z_{n+1}> 0\,\end{align*}$$

by Corollary 5.8 and Lemma A.31. Therefore, the polynomial $D_Z (b^{\mathrm {nr}}_{n+1} (t))$ is initially negative.

Taking $t=C (n+1-k)^{1/n}$ and using Corollary 2.4, we can choose C sufficiently large so that $|D_{Z, 1} t| \leq \frac {1}{3} \left | \frac {\partial _Z D_Z (P_s) Z_{n+1} t^{n+1} }{(n+1)!} \right |$ for $r \in (r_{n+1}-\varepsilon , r_{n+1})$ . This constant C is allowed to depend on n. We can also choose $\varepsilon $ sufficiently small so that the error $O(t^2) + O(t^{n+2}\beta _{n+1} /(n+1-k))$ in (4.8) is smaller than $\frac {1}{3} \left | \frac {\partial _Z D_Z (P_s) Z_{n+1} t^{n+1} }{(n+1)!} \right |$ . With those choices, using (4.8) yields

$$ \begin{align*} D_Z (b^{\mathrm{nr}}_{n+1} (t)) \geq \frac{\partial_Z D_Z (P_s) Z_{n+1} t^{n+1}}{(n+1)!} - \frac{2}{3} \left| \frac{\partial_Z D_Z (P_s) Z_{n+1} t^{n+1}}{(n+1)!} \right|> 0 \end{align*} $$

because $Z_{n+1}> 0$ . The proof of $Z_{n+1}> 0$ is in Lemma A.31 for $n = 3$ and will be done in Corollary 5.8 for the case $\gamma = 7/5$ .

As $D_Z(b^{\mathrm {nr}}_{n+1}(t))$ is initially negative and is positive at $t=C (n+1-k)^{1/n}$ , by continuity there exists a first time $t_D \in (0, C(n+1-k)^{1/n} )$ at which $D_Z (b^{\mathrm {nr}}_{n+1} (t_D)) = 0$ and $D_Z(b^{\mathrm {nr}}_{n+1} (t))$ is negative up to $t_D$ .

Proof. First note that by definition, the Taylor expansion of $b^{\mathrm {nr}}_{n+1} (-t)$ agrees with the smooth solution up to order $n+1$ at $P_s$ . In particular, we have

$$\begin{align*}b^{\mathrm{nr}}_{n+1} (t)-(W^{(r)}(-t),Z^{(r)}(-t))=\tfrac{ t^{n+2}}{(n+2)!} \left( (0, -\beta_{n+1} Z_{n+1} )- (-W_{n+2}, -Z_{n+2})\right)+O_{\beta_{n+1}}(t^{n+3}).\end{align*}$$

Both $b^{\mathrm {nr}}_{n+1}(t)$ and $(W^{(r)}(-t), Z^{(r)}(-t))$ start with slope $(-W_1, -Z_1)$ . Then, in order for $b^{\mathrm {nr}}_{n+1} (t)$ to begin below the smooth solution, it suffices to check the geometric condition

$$\begin{align*}(-W_1, -Z_1) \wedge \left( (0, -\beta_{n+1} Z_{n+1} )- (-W_{n+2}, -Z_{n+2})\right) < 0, \end{align*}$$

or equivalently,

$$\begin{align*}\beta_{n+1} Z_{n+1} W_1 < W_1 Z_{n+2} - W_{n+2} Z_1. \end{align*}$$

Note that both sides are of order $\frac {1}{|k-n-1|}$ by Corollary 2.4 and $n \geq 3$ . The case $n = 3$ follows from the fact that $Z_{4}> 0$ (Lemma A.31) and $W_1 < 0$ (Lemma A.26), taking $\beta _4$ sufficiently large. The case $\gamma = 7/5$ with n odd and sufficiently large would follow as long as we have $Z_{n+1}> 0$ , since $W_1<0$ is guaranteed by Lemma A.7. The proof that $Z_{n+1}> 0$ will be delayed to Section 5 – specifically, this will be a consequence of Corollary 5.8).

Proof. By continuity of $P^{\mathrm {fr}}$ with respect to r, we just need to prove this for $r = r_3$ in the case of $n = 3$ and for r close enough to $r^\ast (7/5)$ for $\gamma = 7/5$ . We prove the statement via a computer-assisted proof. The code can be found in the supplementary material, and we refer to Appendix B for details about the implementation.

Proof. We consider the $(2n+2)$ -th degree polynomial $B^{\mathrm {fr}} (b^{\mathrm {nr}}_n (t))$ . We have that ${B^{\mathrm {fr}} (b^{\mathrm {nr}}_n (0)) = B^{\mathrm {fr}} (P_s) = 0}$ and $\nabla B^{\mathrm {fr}} (P_s) (W_1, Z_1) = 0$ due to our choice of $F_1$ . Therefore, the terms in the polynomial have $t^2$ as a common factor. Taking into account Corollary 2.4, all the coefficients in $b^{\mathrm {nr}}_n (t)$ are $O(1)$ around $r\approx r_n$ except the n-th and $(n+1)$ -th coefficients of $b_Z^{\mathrm {nr}}(t)$ . We get