Smooth Solutions of the tt* Equation: A Numerical Aided Case Study

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Introduction
The tt* equations were introduced by Cecotti and Vafa when they studied the fusion of topological N = 2 supersymmetric quantum field theory with its conjugate, the anti-topological one [1].They also appeared in the extraction of exact results for supersymmetric σ models [2] and in the classification of the N = 2 supersymmetric theories [3].Dubrovin gave the zero-curvature representation of the tt* equations and studied their geometrical aspects [4].An important special class of the tt* equations are the tt*-Toda equations, which are the reduction of the two-dimensional (n + 1)-periodic Toda lattice with opposite sign 2(w i ) z z = −e 2(w i+1 −w i ) + e 2(w i −w i−1 ) , w i+n+1 = w i , where z denotes the complex conjugate of z ∈ C and w i = w i (z, z) ∈ R, constrained by both the l-anti-symmetry constraint w 0 + w l−1 = 0, w 1 + w l−2 = 0, . . ., w l + w n = 0, w l+1 + w n−1 = 0, . . ., where the fixed l ∈ {0, 1, . . ., n}, and the radial constraint The l = 0 case of tt*-Toda equations is called the A n type.They were first derived by Cecotti and Vafa when they deformed the superpotentials with the A n -minimal model of the Landau-Ginzburg approach [1].The existence of global solutions for any n for these A n type tt*-Toda equations can be established by the Higgs bundle method [10,11].Recently, the existence and uniqueness of these solutions were proved using the Riemann-Hilbert method [8].Almost all concrete example studies of the tt* equations were reduced to the third Painlevé equation before the work of Guest and Lin [9], where they initiated the direct study of a generalized tt* equation with two unknowns (1. 3) The tt*-Toda equations with two dependent variables are the cases a, b ∈ {1, 2}, exhausted in [9].In [5], Guest, Its and Lin proved the following property for equation (1.2) with boundary condition (1.3).
Theorem 1.1 establishes a map from the point (γ, δ) in the triangular region in Figure 1 to the smooth solution of equation (1.2).Thus, it characterizes a two-parameter family of smooth real solutions of the tt* equation in C * .Note that a result similar to Theorem 1.1 had been obtained by Guest and Lin in [9], where they required γ, δ > 0. But the difference is crucial since Theorem 1.1 characterizes all smooth radial solution of equation (1.2) [6].By the Riemann-Hilbert approach, Guest et al. obtained all connection formulae for the tt* cases, i.e., a, b ∈ {1, 2} [6].The complete picture of the monodromy data, holomorphic data, and asymptotic data were finally obtained in [7].
The case a = b = 2 of (1.2), which is the case 4a in their classification of the tt*-Toda equations, was studied more thoroughly.In [7], the fine asymptotics(see below for the exact definition) are all given for the class of solutions defined in Theorem 1.1.In this case, w 0 = 1 2 u and w 1 = 1 2 v were used as the proper independent variables.Then (1.2) becomes 2(w 0 ) z z = e 4w 0 − e 2w 1 −2w 0 , 2(w 1 ) z z = e 2w 1 −2w 0 − e −4w 1 . (1.4) According to the radical constraint (1.1), system (1.4) is written into an ordinary differential equation (ODE) with variable r = |z| where the prime denotes d dr .Near r = 0, by (1.3), w 0 and w 1 have properties 2w 0 (r) Near r = ∞, the asymptotics of w 0 and w 1 are expressed by the Stokes data s R 1 and s R 2 [6]: (1.7) The map from (γ 0 , γ 1 ) to s R 1 , s R 2 is the connection formula [6] s The r = ∞ asymptotics (1.7) is able to uniquely fix the solution of (1.5).This is an initial value problem from r = ∞.However, the rough asymptotics (1.6) itself is not enough to fix the solution.To fix the solution, it must be accompanied by the rough asymptotics at r = ∞: w 0 (r) r→∞ −−−→ 0, w 1 (r) r→∞ −−−→ 0. But this becomes a boundary value problem.To get an initial value problem from r = 0, one should start with a more detailed asymptotics near r = 0.In fact, it would be very appropriate to start with the fine asymptotics at r = 0. Definition 1.2.An asymptotics is said to be a fine asymptotics of a system of differential equations if it satisfies the system's truncation equation with respect to the asymptotics.
Practically, one can obtain the fine asymptotics from a rough one by the following way: first truncate and simplify the differential equation system according to the rough asymptotics, then solve the truncated system, and then fix the parameters of the solution by comparing it with the rough asymptotics.
As an example, let us find out the fine asymptotics of (1.5) at r = ∞ that coincides with asymptotics (1.7).The truncation equation for the solutions of (1.5) with respect to the asymptotics w 0 (r) → 0 and w 1 (r) → 0 is The exact solution of (1.9) that coincides with asymptotics (1.7) is where K 0 denotes the Bessel K 0 function.So (1.10) is the fine asymptotics for the solutions with asymptotics w 0 (r) → 0 and w 1 (r) → 0 at r = ∞, whereas asymptotics (1.7) should not be taken as a fine asymptotics since it is not an exact solution of (1.9).
The region map of the connection formula (1.8).
In [7], all fine asymptotics of (1.5) at r = 0 for the solutions described by Theorem 1.1 have been obtained.These fine asymptotics contain seven cases.For convenience, we list them in Section 2. Therefore, the fine asymptotics at r = ∞ and r = 0 are all known for the solutions described by Theorem 1.1, i.e., the situations at r = ∞ and r = 0 become symmetric.However, these fine asymptotics at r = 0 are complicated, especially that of the vertex case.An intuitive explanation is still lacking.Moreover, nothing is known for the general case outside of the triangle in Figure 1.This is our motivation to start the numerical study.The first part of this paper verifies these fine asymptotics numerically up to 100 digits for all the seven cases at r = 0.
Fine asymptotics are subject to the class of the solutions.If the solution class is enlarged, new fine asymptotics will appear.We will enlarge the solution class from the Stoke data side in the following way.The connection formula (1.8) maps the (γ 0 , γ 1 ) region to the s R 1 , s R 2 region.Coming down to equation (1.5), the region map can be represented by Figure 2. Any solution represented by a point s R 1 , s R 2 in the curved triangle (including the edges and the vertexes) in Figure 2 must have asymptotic (2.1) near r = 0, where (γ 0 , γ 1 ) is determined by s R 1 , s R 2 by the connection formula (1.8).So the class of solutions described by Theorem 1.1 are parameterized by the points in the curved triangle (including the edges and the vertexes).We enlarge the class of solutions to the ones parameterized by the points on the whole real s R 1 , s R 2 plane.Based on our numerical results, we will generalize the range and the explanation of the connection formula and obtain all the fine asymptotics of the enlarged class of solutions at r = 0. Of course, the solution class can also be generalized from the side of r = 0.However, the problem is much harder to solve.
The paper is organized as follows.In Section 2, we list all the seven fine asymptotics of (1.5) at r = 0 obtained in [7].In Section 3, we numerically verify these seven fine asymptotics.In Section 4, we study the cases where s R 1 , s R 2 is outside the curved triangle and obtain our main result.In Section 5, we present a numerical study from the r = 0 side.In Section 6, we give the conclusion and discussions.This paper can be seen as a complement to [5,6,7].
2 Fine asymptotics of (1.5) at r = 0 of the class of solutions defined by Theorem 1.1 The fine asymptotics of (1.5) at r = 0 of the class of solutions defined by Theorem 1.1 have all been obtained in [7].For convenience, we list them all here.We will use the following notations.
• s: s = ln(r) is used as an easy independent variable near r = 0.
The seven fine asymptotics of (1.5) at r = 0 obtained in [7] are the following.

Preliminary for the numerical experiments:
an approximation proper for calculations near r = ∞ Note that (3.1) is written in a form that better preserves the significant digits in the numerical integration near r = ∞.The errors in the approximation of (w p , w m ) by the fine asymptotics (1.10) are caused by the nonlinear terms in the expansion of (3.1).In general, the most significant correction to w p is proportional to w These results are sufficient for the rough numerical investigations for smooth solutions of the tt* equation.They are called rough simply because they can be refined.For high precision numerical integration of (3.1) from the r = ∞ side, the relative error will not enlarge too much when r is still large.For w m (r), the relative error is about O r − 1 2 e −4( √ 2−1)r .If we give the initial values by the fine asymptotics (1.10) with r = 45, the relative error of the initial values are of order 10 −33 , which is not so satisfactory.If we want to reach a relative error of order 10 −100 by this way, r = 138 is needed to give the initial values.We will see, after considering the most significant contribution of the nonlinear terms, the starting r can be greatly reduced.Suppose where Then w p and w where I 0 is the Bessel I 0 function.Then The relative errors are both of order r −1 e −4 √ 2r .To acquire a relative error of order 10 −100 , it is enough to start the numerical integration from r = 45.Higher-order nonlinear terms should not be considered, otherwise we will run into high-dimensional integrations that are time-consuming to compute to high accuracy, for example, an accuracy of 10 −100 .
The truncation of (3.3) will be used to give initial values for the numerical integration of (3.1) near r = ∞ for all of the following cases.

The general case: in the triangular
This subsection is devoted to the verification of (2.1).
To be specific, we fix (γ 0 , γ 3) means that we can start our numerical integration from r = 45 for moderate s R 1 , s R 2 to get a relative error of order less than 10 −100 .Recall that in Section 2, we have mentioned is a proper independent variable near r = 0. Therefore, the numerical integration is naturally divided into two parts: on r ∈ [r m , 45] and on s ∈ [s f , s m = ln r m ].For convenience, we always choose r m = 1.s f varies with s R 1 , s R 2 and will be determined after we solve the associated truncation of (1.5) for the fine asymptotics.

Numerical integration from r = 45 to r = 1
By the truncation of (3.3), the initial values for the numerical integration of (3.1) are calculated up to more than 100 digits To save space, we list only the first 50 digits in (3.5).It is not surprising that w m (45) in (3.5) coincides with w √ 2 for the first 33 digits and that w p (45) in (3.5) coincides with w (0) p (45) = − 2 π K 0 (180) for all the listed 50 digits.Formula (3.3) gives only the order of the error, not the actual value.We obtain the errors of (3.5) by comparing the initial values (3.5) with a more accurate numerical solution starting from r = 55.Table 1 shows both the absolute error and the relative error of the initial values at r = 45.In this paper, we use the Gauss-Legendre method, which is an implicit Runge-Kutta method suitable for high-precision numerical integration, to numerically integrate ODEs.Integrating (3.1) numerically from r = 45 to r = 1 by a 100-stage Gauss-Legendre method with step size 1 100 , we obtain the numerical values of w p , w ′ p , w m and w ′ m at r = 1: Note that (3.6)only lists the first 50 digits of the numerical solution.Numerical experiments show that the errors caused by the numerical integration are all negligible.This is easy to understand because the precision order of the numerical integration, which is twice the stage number, is large and the step size is small.Comparing (3.6) with the more accurate solution starting from r = 55, we obtain the errors of (3.6) as Table 2. Inspired by the form of (2.1), we use independent variable s and dependent variables Please recall that s = ln(r) is defined by (3.4).From the numeric point of view, the advantage of using s rather than r is that it can avoid the frequent adjustment of the step size when we solve (1.5) numerically near r = 0.The equations for w0 and w1 are 1 4 So, all terms in the right of (3.8) can be ignored at first.Thus, is the associated truncation of (3.8) for the fine asymptotics of the general case.
The initial values of w0 , d w0 ds , w1 and d w1 ds at s = 0 can be derived from w p , w ′ p , w m and w ′ m at r = 1: In the truncation of equation (3.8) to (3.9), the ignored terms are of order O e 2(γ 0 +1)s , order O e (γ 1 −γ 0 +2)s and order O e 2(1−γ 1 )s .Now, we have fixed (γ 0 , γ 1 ) = 1, 1  3 .Thus, w0 , w1 will approach (ρ 0 , ρ 1 )| γ 0 =1,γ 1 = 1 3 with a distance of order O e 4 3 s , where Table 3 shows that the numerical solution is as accurate as we expected.The relative error of d w0 ds or d w1 ds in Table 3 seems to be large.But this is really nothing since it is only another demonstration of the fact that d w0 ds and d w1 ds are small.Table 4 shows how good the asymptotic solution (2.1) is.Table 4 not only numerically verifies the asymptotics of the general case for (γ 0 , γ 1 ) = 1, 1  3 , but also confirms our estimate that w0 , w1 is close to its asymptotics (ρ 0 , ρ 1 )| γ 0 =1,γ 1 = 1 3 with a distance of order O e

Case E1
This subsection is devoted to the verification of the fine asymptotics of the E1 case.Note that the E1 case is parameterized by −1 < γ 0 < 3 and γ 1 = 1.To fix the problem, we take γ 0 = 1 as an example to verify the E1 case.Substituting (γ 0 , γ 1 ) = (1, 1) to the connection formula (1.8), we immediately get s R 1 , s R 2 = (2, −2).Similar to the general case of Section 3.2, the numerical integration is divided into two parts: for r ∈ [1, 45] and for s ∈ [s f , 0].  5.  Comparing with the more accurate solution starting from r = 55, the errors of (3.12) are obtained as shown by Table 6.
where s = ln(r) as defined by (3.4).Then the differential equations for w0 and w1 are 1 4 Note that (3.14) can also be obtained from (3.8) by substituting γ 1 = 1 to it.We expect w0 is of order O(1) and that w1 is of order O(ln(−s)).Also considering −1 < γ 0 < 3, we obtain the associated truncation of (3.14) near s = −∞ for the fine asymptotic of the E1 case 1 4 The general solution of (3.15) is w(0) By (2.3) and (3.13), we know the fine asymptotics of (1.5) in the E1 case corresponds to and the "±" sign chosen to be minus.
Remark 3.1.It is obvious that k 1E1 = 0 and k 2E1 → 0, or else w0 and w1 will have order O(s) at s = −∞, which is in contradiction with our assumption that w0 and w1 are of order O(1) and O(ln(−s)) respectively.Therefore, the consistent solution of (3.15) is , which is distinguished by that w 0 (r) and w 1 (r) are smooth on r ∈ (0, ∞) and that they have asymptotics (1.7).
In the truncation from (3.14) to (3.15), the ignored term for the differential equation of w1 is e w1 − w0 +(3−γ 0 )s , which is of order O se (3−γ 0 )s .Similarly, the ignored terms for the differential equation of w0 are of order O se (3−γ 0 )s and order O e 2(γ 0 +1)s .In the current numerical experiment, γ 0 = 1.Therefore, the difference between the asymptotic solution and the exact solution is of order O se 2s .So, we should do high-precision numerical integration from s = 0 to about s = s f = −120 since 120 × e 2×(−120) ≈ 7.055 × 10 −103 .Similar to the general case of Section 3.2, the values of w0 , d w0 ds , w1 and d w1 ds at s = 0 are obtained by formula (3.10).Then, numerically integrating (3.14) by the Gauss-Legendre method, the high-precision numerical solution is obtained.Comparing it with the more accurate numerical solution starting from r = 55, the errors of the numerical solution are obtained.Table 7 shows that the numerical solution is as accurate as we expected.The large relative error of d w0 ds is nothing but the fact that d w0 ds | s=−120 ≈ −1.29 × 10 −102 is small.Table 8 shows how good the asymptotic solution (2.3) is.Table 8 not only numerically verifies the asymptotics of the E1 case for γ 0 = 1, but also confirms our estimate that w0 , w1 differs with its asymptotic solution by an order of O se 2s .

Case E2
In this case, γ 0 = −1 and −3 < γ 1 < 1.As explained in the beginning of Section 3, the fine asymptotics of the E2 case can be obtained from the E1 case.So we omit the numerical verification for this case.
Let us take γ 0 = 1 3 as an example to verify (2.4) numerically.Then s R 1 , s R 2 = (−2, −3).Similar to the general case of Section 3.2, the numerical integration is divided into two parts: for r ∈ [1, 45] and for s ∈ [s f , 0].It is not surprising that w p (45) and w ′ p (45) of (3.17) coincide with that of (3.11) with many digits since s R 1 = −2 in the example for this case and s R 1 = 2 in the example for the E1 case.Comparing with the more accurate solution starting from r = 55, the errors of the initial values (3.17) are obtained as shown by Table 9. Comparing with the more accurate solution starting from r = 55, the errors of (3.18) are obtained as shown by Table 10.Near r = 0, we still use the transformation (3.7).So the differential equations for w0 and w1 are also (3.8).
We expect w0 and w1 are of order o(s).Also considering −1 < γ 0 < 3 and γ 1 = γ 0 − 2, we get the associated truncation of (3.8) near s = −∞ for the E3 case: The solution of (3.19) is Because we expect w0 and w1 are of order o(s), we should take k 1E3 = 0 and k 3E3 → 0 or else w0 and w1 will be of order O(s).So the consistent solution of (3.19) is By (2.4) and (3.7), we know that the fine asymptotics of the E3 case is fixed by k 0E3 = a E3 and k 3E2 = b E3 .

Y. Li
In the truncation from (3.8) to (3.19), the ignored terms for the differential equation of w0 + w1 are e 2 w0 +2(γ 0 +1)s and e −2 w1 +2(1−γ 1 )s , which are of order O s 2 e 2(γ 0 +1)s and order O s −2 e 2(γ 0 +1)s .Similarly, the ignored terms for the differential equation of w1 are also of order O s −2 e 2(3−γ 0 )s and order O s −2 e 2(γ 0 +1)s .In the current numerical experiment, γ 0 = 1 3 .Therefore, the difference between the asymptotic solution and the exact solution is of order O s 2 e 8 3 s .So, we should do high-precision numerical integration from s = 0 to about s = s f = −90 since 90 2 × e Table 11 shows that the numerical solution is as accurate as we expected.Table 12 not only numerically verifies the asymptotics of the E3 case for γ 0 = 1 3 , but also confirms our estimate that w0 + w1 and w1 − w0 deviate from their asymptotics by an order of O s 2 e 8 3 s .More detailed analysis shows that w0 and w1 deviate from their asymptotics by an order of O s 2 e 8 3 s and an order of O e 8 3 s , respectively.

Case V1
This subsection is devoted to the verification of the fine asymptotics of the V1 case.Note that γ 0 = 3 and γ 1 = 1 in this case.
s R 1 , s R 2 = (4, −6) by (1.8).Similar to the general case of Section 3.2, the numerical integration is done on two intervals: r ∈ [1, 45] and s ∈ [s f , 0].Comparing with the more accurate solution starting from r = 55, the errors of the initial values (3.20) are obtained as shown by Table 13.
Numerically integrating (3.1) from r = 45 to r = 1 by the Gauss-Legendre method with the same parameters as the ones in Section 3. Comparing with the more accurate solution starting from r = 55, the errors of (3.21) are obtained as shown by Table 14.Near r = 0, the transformation is still (3.7).Hence, the differential equations for w0 and w1 are also (3.8).Now, (γ 0 , γ 1 ) = (3, 1) and the expected w0 and w1 are of order o(s).So the associated truncation of (3.8) near s = −∞ for the V1 case is 1 4 Then, we have Y.Li There are only two sets of solutions that has form (3.24).Set A: The fine asymptotic solution of the V1 case is in Set B with The error of the truncation from (3.8) to (3.22) is caused by the term e 2 w0 +8s , which is of order O s 6 e 8s .So we set s f = −32 since (−32) 6 e 8×(−32) ≈ 7.1 × 10 −103 has been smaller than 10 −100 .The high-precision numerical solution is obtained by numerically integrating (3.8) by the Gauss-Legendre method.Comparing it with the more accurate numerical solution starting from r = 55, the errors of the numerical solution are obtained.Table 15 shows that our numerical solution is as accurate as we expected.Table 16 shows how good the asymptotic solution (2.5) is.Table 16 not only numerically verifies the asymptotics of the V1 case, but also confirms our estimate that w0 and w1 + w0 differ from their asymptotics by an order of O s 6 e 8s .
Let w = w 0 = −w 1 .Then, the differential equation for w is 1 2 and the numerical experiments show that 2w(r) differs from its asymptotics by an order of O s2 e 4s near r = 0.

Case V3
In this case, (γ 0 , γ 1 ) = (−1, −3).Thus, s R 1 , s R Y. Li use variables v 0 = e 2w 0 and v 1 = e 2w 1 , then v 0 and v 1 will have no cuts for r > 0. v 0 or v 1 may still have singularities, i.e., in general, v 0 and v 1 are not the final smooth variables.Fortunately, we were able to find two smooth variables for each part of Figure 3, see Conjecture 4.1.From this point of view, Theorem 1.1 studies only those solutions that have "positivity" property so that they are still real after taking logarithm.

The conjecture
The fine asymptotics for the cases of Ω 0 , E1, E2, E3, V1, V2 and V3 have been rigorously proved in [7] and numerically verified in Section 3.So the following conjecture only deals with the other remaining 12 cases: 3 and E L 3 .Similar to the explanation at the beginning of Section 3, the formulas of 1 and E R 3 , respectively.But for convenience, we will list all formulas for the 12 cases.
By the truncation of (3.Comparing with the more accurate solution starting from r = 55, the errors of the initial values (4.3) are obtained as shown by Table 17.Comparing with the more accurate solution starting from r = 55, the errors of (3.21) are obtained as shown by Table 18.When r < 1, w 0 and w 1 may be complex.As Conjecture 4.1 suggests, we use v 0 and v 1 as dependent variables for the Ω 1 case.Then, v 0 and v 1 will be real for r > 0.

Y. Li
To improve computation efficiency, we use s = ln(r) as independent variable.Then the equations for v 0 and v 1 are The associated truncation of (4.5) for the fine asymptotics of the Ω 1 case should be In fact, after substituting (4.2) to the Ω 1 case of Conjecture 4.1, it becomes obvious which terms of (4.5) should be ignored.The solution of (4.6) is known Numerical results show that v 0 (s) has no zero for s ∈ (−∞, 0] but v 1 (s) has, just as Conjecture 4.1 predicts.For the sake of numerical integration, it is better to integrate around the zeros of v 1 (s).In order to keep away from the zeros of v 1 (s), we first compute v 1 (s + iϵ) with ϵ = 10 −2 to determine the approximate zeros of v 1 (s) by solving Re(v 1 (s + iϵ)) = 0. Then we get the approximate zeros s i of v 1 (s) within the range −87 ≤ s ≤ 0.  Obviously, the distance between two adjacent zeros in Table 19 is about 2.5.To avoid the numerical instabilities caused by those zeros, we use a contour in the complex plane of s, as shown in Figure 4.The radii of the circles around the zeros are set to 1  5 .The values of v i (s) for s on the contour can be obtained directly from the numerical integration.Then we should supplement the values of v i (s) in the circles in order to complete the numerical solution of v i (s).In principle, the values of v i (s) can be evaluated using the Cauchy integral formula v i (s) = 1 2πi v i (ξ) ξ−s dξ.But here v i (ξ) is only a numerical solution, which has high-precision value only at some points on the circle.This restricts our choice of high-precision numerical integration method to calculate the Cauchy integral efficiently.Since v i are periodic functions on the circle, we use the trapezoidal rule to calculate them Re(s)  where R = 1 5 denotes the radius of the circle, and ṽi (θ j ) the value of v i at θ j on the circle.The distance between the adjacent θ j is π n .Obviously, formula (4.8) is not appropriate for a point near the circle.Therefore, the contour has 2 line segments in each circle.We use line segments of length 1  10 .Altogether, for s ∈ s j − 1 10 , s j + 1 10 we obtain the numerical solution of v 0 (s) and v 1 (s) by (4.8) rather than solving (4.5) numerically.In our numerical experiments, n is equal to 1000, which is far more than enough to guarantee an accuracy better than 10 −100 .
The plots of v 0 and v 1 are shown in Figure 5.
Table 20 shows that the numerical solution is as accurate as we expected.This section is concerned with what the solution looks like when (2.2) is not satisfied, i.e., we drop the assumption of solutions of (1.4) being smooth on C * .First, we derive a better asymptotics near r = 0, which is suitable to give initial values for the numerical integration.Then, (1.5) is integrated numerically from r = 0 to r = ∞.The integration contour on the complex plane of r is used to surround the singularities.We will find that the singularities are regularly distributed.
where δc 0 and δc 1 can not be 0 simultaneously.In the following numerical experiment, we use To solve (4.5) numerically, the initial values of v 0 , dv 0 ds , v 1 , dv 1 ds must be given.We start from s 1 = −100 and give the initial values by (5.4).Since it is easy to compute the initial values by (5.4), the details of the initial values are omitted.We only list the errors of the initial value by Table 22.
Table 22.Errors of the numerical solution at s = −100 with (γ 0 , γ 1 , c 0 , c 1 )) = 1, 1  3 , e ρ0 + Table 23 gives the errors of (5.5).Again, the errors are evaluated by comparing the two numerical solutions starting from s = −140 and from s = −100, respectively.For s > 0, i.e., r > 1, it is convenient to use the variable r itself instead of s: the pattern of the singularities is more transparent with respect to r than with respect to s.Then (4.Then, we compute the numerical solution of (5.6), for which the initial values are given by (5.5).Near r ≈ 1.539167317, the numerical solution blows up.Figures 6 and 7 show the plots of v 0 and v 1 on the circle with a radius of about 0.239167317 around the singular point.
Obviously, v 0 and v 1 are smooth functions on the circle.Numerical results show that the singularity at r ≈ 1.539167317 is a simple pole of v 1 .By (5.6)To show the pattern of the singularities of v 0 and v 1 , we plot v i r+10 −2 i , i = 0, 1 as Figures 8  and 9.Although we cannot give a precise description of Figures 8 and 9, we can still make several heuristic observations from the two figures.First, we can observe that both v 0 (r) and v 1 (r) have infinitely many singularities since some adjacent singularities are almost equidistant.Second, v 0 (r) and v 1 (r) should be real since the imaginary parts of v 0 r + 10 −2 i and v 1 r + 10 −2 i are small except near the singularities.Third, v 1 (r singular + 0 − ) > 0 and v 1 (r singular + 0 + ) < 0 and the imaginary part of v 1 (r singular + 0 + i) is always positive.Fourth, the singularities of v 0 (r) have two frequencies: the class of singularities with v 0 (r singular + 0 + i) < 0 have one frequency and the class of singularities with v 0 (r singular + 0 + i) > 0 have another frequency.The first two observations should be general for cases deviating from (2.2).It seems that there is no simple combination of v 0 and v 1 such that the composite variable is smooth for r ∈ (0, ∞).

Conclusion and discussion
This paper numerically studies equation (1.5), the case 4a of the tt*-Toda equation.The fine asymptotics of the solutions described by Theorem 1.1 are verified with an accuracy of order 10 −100 .We enlarge the class of the solutions described by Theorem 1.1 from the Stokes data side by assuming that they have asymptotics (1.7) for s R 1 , s R 2 ∈ R 2 but may have singularities for r ∈ (0, ∞).For the enlarged class of solutions, we construct the proper dependent variables (smooth for r ∈ (0, ∞)) for every case, and find all the fine asymptotic formulas for these proper dependent variables.The associated truncation equations of (1.5) are crucial for the realization of the high-precision verifications and are indeed useful in the search for the new fine asymptotics.Some preliminary numerical studies are also made to investigate what happens when the fine asymptotics is broken at the r = 0 side.However, the studies in Section 5 are far from complete in investigating the deviation from (2.2).The first problem is whether we can find two proper dependent variables that are smooth near r = ∞.It can be shown that the singularity of 1 v 0 (r) coincides with v 1 (r), differing only in amplitude.But this does not help much in determining what are the proper variables.Without proper variables it will be almost impossible to talk about the asymptotics near r = ∞.The second problem is to find out the r = ∞ asymptotics of (5.6) beyond v 0 (r) Y. Li from the r = 0 side and we see no classification near r = 0 for the cases parameterized by points in the triangle.We expect the behavior of these solutions to separate near r = ∞ and provide a natural classification of the r = ∞ asymptotics of (5.6).

8 3 ×
(−90) ≈ 4.76 × 10 −101 .Just as the general case, the values of w0 , d w0 ds , w1 and d w1 ds at s = 0 are obtained by formula (3.10).Then, the high-precision numerical solution is obtained by numerically integrating (3.8) by the Gauss-Legendre method.Comparing it with the more accurate numerical solution starting from r = 55, the errors of the numerical solution are obtained.

Figure 4 .
Figure 4. Contour in the complex plane of s to compute v 0 (s) and v 1 (s).

Figure 6 .Figure 7 .
Figure 6.Plots of v 0 r 1 + R 1 e iθ .r 1 ≈ 1.539167317 is the location of the first singularity and R 1 ≈ 0.239167317 is the radius of the circle around the singularity.

Table 5 .
Errors of the initial values of case E1 with γ 0 = 1.

Table 8 .
Approximate derivation from the asymptotic solution for the E1 case with γ 0 = 1.

Table 12
shows how good the asymptotic solution (2.4) is.

Table 12 .
Approximate derivation from the asymptotic solution for the E3 case with γ 0 = 1 3 .

Table 14 .
Errors of the numerical solution at r = 1 of case V1.

Table 15 .
Errors of the numerical solution at s = −32 for the V1 case.

Table 16 .
Approximate derivation from the asymptotic solution for the V1 case.

Table 17 .
Errors of the initial values of case Ω1 with s R 1 , s R 2 = (2, 1).

Table 19 .
The first few approximate zeros s i of v 1 (s) for the Ω 1 case with s R