Issue 
A&A
Volume 555, July 2013



Article Number  A27  
Number of page(s)  9  
Section  The Sun  
DOI  https://doi.org/10.1051/00046361/201220195  
Published online  24 June 2013 
Online material
Appendix A: Asymptotic solution for large time
In this appendix we study the asymptotic behaviour of the solution to Eq. (29) for ωt ~ ϵ^{1}. We introduce the Laplace transform (A.1)Applying the Laplace transform to Eq. (29) and using the theorem about the Laplace transform of convolution, we obtain (A.2)Using the standard table of Laplace transforms, we find (A.3)We do not give the expression for ℒ [F(t)] (s) because it is not used in what follows. The solution to Eq. (29) with the initial condition (17) is given by (A.4)where ς is any positive constant. When calculating the inverse Laplace transform, we have made the substitution s = ωs′ and then dropped the prime.
Now we use Eq. (A.4) to obtain the asymptotic expression for η(t) valid for ωt ~ ϵ^{1}. The ratio of the second term in the numerator of the integrand in Eq. (A.4) to the first one is of the order of ϵ. Hence, we can neglect the second term in the numerator and use the approximate expression (A.5)to calculate the asymptotic behaviour of η(t) at large time. The integrand in this expression has four logarithmic branch points at s = ±s_{+} and s = ±s_{−}, where (A.6)To obtain a singlevalued branch of the integrand, we make the cuts of the complex splane along the intervals [−s_{−}, −s_{+}] and [s_{+}, s_{−}] . The multivaluedness of the integrand is related to the presence of logarithm in Eq. (A.3). We define the principal sheet of the Riemann surface of the integrand by the condition that the imaginary part of this logarithm is between −πi and πi. In what follows we use the notation “ln” for this value of logarithm. All other values differ from this value by 2πni, where n is an integer number.
The function (A.7)maps the complex splane with cuts [−s_{−}, −s_{+}] and [s_{+}, s_{−}] on the strip −π < ℑ(w) < π in the complex wplane, where ℑ indicates the imaginary part of a quantity. We find the zeros of the denominator of the integrand in Eq. (A.5), which are determined by the equation (A.8)Let s_{∗} be a root of this equation. If it is not close to ±s_{±} then it follows that . This implies that either or . In the first case we immediately see that the lefthand side of Eq. (A.8) is of the order of ϵ^{2}, so that the only possible solution of this kind is s_{∗} = 0. Since ℒ [G(t)] (0) = κ, it is straightforward to see that the denominator of the integrand in Eq. (A.5) does not equal zero when s = 0. Hence, s_{∗} = 0 is a spurious root.
Considering the second possibility, namely, , we look for the root to Eq. (A.8) in the form s_{∗} = ±1 + ϵδ. By substituting this expression in Eq. (A.8) we obtain (A.9)The imaginary part of the lefthand side of this equation is close to −πi when ℑ(±δ) < 0 and to πi when ℑ(±δ) > 0. This implies that the imaginary parts of the left and righthand side of Eq. (A.9) have different signs, so that this equation cannot have a solution with ℑ(δ) ≠ 0. It also cannot have a real solution because in that case the imaginary part of the righthand side is zero, while the imaginary part of the lefthand side is nonzero.
Hence, Eq. (A.8) can only have solutions close to ±s_{±}. We look for the solution in the form s_{∗} = ±s_{±} + δ, where δ ≪ 1. Substituting this expression in Eq. (A.8) we obtain four roots, and , where (A.10)Hence, there are four poles on the real axis, at and .
The integration contour in Eq. (A.5) is a straight line parallel to the real axis. Its equation in ℑ(s) = ς > 0. Since the integrand in Eq. (A.5) is an analytical function in the upper part of the complex splane, we can deform the integration contour arbitrarily without changing the value of the integral. The only conditions that have to be satisfied are that the new contour is completely in the union of the upper part of the complex splane and the real axis and that the part of the new contour that is on the real axis does not contain singularities of the integrand. It is convenient to use the new integration contour shown in Fig. A.1. This contour consists
Fig. A.1
Integration contour used to obtain Eq. (A.11) is shown by the thick solid line. The dashed lines show the cuts. 

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of five intervals of the real axis, , , , , and , and four halfcircles of radius ε centred at and . Taking ε → +0 we obtain (A.11)where , , indicates the principal Cauchy part of an integral, and in the integrals along the intervals [−s_{−}, −s_{+}] and [s_{+}, s_{−}] the integrand is calculated at the upper edges of the cuts.
First of all, we note that (A.12)where j = 1,2,3,4, A_{j} and C_{j} are constants of the order of unity, and C_{j} > 0. This implies that the contribution of the last term in Eq. (A.11) is exponentially small and can be neglected. Using the integration by parts we obtain (A.13)We see that the contribution of the integral along the interval (−∞, −s_{−}] is of the order of 1/ωt. Similarly we can show that the contribution of the integrals along the intervals [−s_{+}, s_{+}] and [s_{−}, ∞) are also of the order of 1/ωt. Hence, we obtain (A.14)To evaluate the asymptotic behaviour of the integrals in this expression, we need to make the analytic continuation of the function w(s) on its Riemann surface. This Riemann surface consists of an infinite set of complex planes with cuts [−s_{−}, −s_{+}] and [s_{+}, s_{−}] properly attached to each other at the cuts. In what follows we use only two nonprincipal sheets of the Riemann surface attached to the principal sheet at the upper edges of the cuts. To define w(s) on the nonprincipal sheet of the Riemann surface attached to the principal sheet at the upper edge of the right cut, we consider a point s near this cut on the principal sheet. We write s = s_{r} + is_{i}, where s_{+} < s_{r} < s_{−} and s_{i} ≪ s_{r}. Then (A.15)and we obtain (A.16)We assume that a point s on the principal Riemann sheet moves in such a way that it crosses the right cut from the upper to the lower part of the complex plane, i.e. ℑ(s) changes the sign from positive to negative. Then Eq. (A.16) implies that ℑ(w) jumps by −2πi during this crossing. This means that to have the continuous change of w(s) when s moves from the upper part of the principal Riemann sheet to the lower part of the nonprincipal Riemann sheet we have to define w(s) on this nonprincipal sheet as (A.17)In a similar way we find that the analytic continuation of w(s) on the nonprincipal Riemann sheet attached to the principal sheet at the upper edge at the left cut is defined by (A.18)To calculate the integral along the interval [s_{+}, s_{−}] in Eq. (A.14), we use the contour shown in Fig. A.2. It consists of the interval [s_{+}, s_{−}] at the upper edge of the right cut on the principal Riemann sheet, two vertical lines from s_{+} to s_{+} − ib and from s_{−} to s_{−} − ib on the nonprincipal Riemann sheet, and the horizontal line connecting the points s_{+} − ib and s_{−} − ib, once again on the nonprincipal Riemann sheet. Here b > 0. To calculate the integral along this contour we need to find the poles of the integrand inside this contour. These poles coincide with the zeros
Fig. A.2
Integration contour used to calculate the integral along the interval [s_{+}, s_{−}] in Eq. (A.14). The solid line indicates the part of the contour that coincides with the upper edge of the right cut on the principal Riemann sheet. The dasheddotted line indicates the parts of the contour on the nonprincipal Riemann sheet. The dashed line shows the cut. 

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of Eq. (A.8). Since the area enclosed by the contour is on the nonprincipal Riemann sheet, we look for zeros of Eq. (A.8) on this sheet.
Although w(s) is now not given by Eq. (A.7) but by Eq. (A.17), the roots of Eq. (A.8) are still close to either 0 or 1 or to one of the numbers ±s_{±}. Obviously, the root close to 0 cannot be inside the contour. Simple analysis shows that there are roots close to ±s_{−} and ±s_{+}, but their real parts are equal to and , respectively, so these roots are also not inside the contour. Finally, the root close to 1 is given by (A.19)This root is inside the contour. Hence, (A.20)We estimate the integrals along the parts of the closed contour that are on the nonprincipal Riemann sheet. Using the integration by parts we have (A.21)where the limit is taken as b → ∞. In the same way we prove that the integral along the vertical line connecting the points s_{+} − ib and s_{+} is of the order of 1/ωt. Finally, the integral along the horizontal line connecting the points s_{−} − ib and s_{+} − ib tends to zero as b → ∞. As a result we have (A.22)}where is a complex constant. In a similar way we can obtain a similar expression for the integral along the interval [−s_{−}, −s_{+}] .
It is worth explaining why we cannot prove in the same way that the integral along the interval [s_{+}, s_{−}] is of the order of 1/ωt.
The reason is that there is a pole at , which is close to the interval [s_{+}, s_{−}] . After the integration by parts we obtain 1/ωt times an integral. The integrand in this integral is very big, in the vicinity of s = 1, so the whole product is not of the order of 1/ωt.
Summarising the analysis of this section, we arrive at the asymptotic expression (A.23)where C is a complex constant and c.c. denotes complex conjugate.
© ESO, 2013
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