Volume 555, July 2013
|Number of page(s)||9|
|Published online||24 June 2013|
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 single-valued branch of the integrand, we make the cuts of the complex s-plane 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 s-plane with cuts [−s−, −s+] and [s+, s−] on the strip −π < ℑ(w) < π in the complex w-plane, 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 left-hand side of Eq. (A.8) is of the order of ϵ2, 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 left-hand 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 right-hand 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 right-hand side is zero, while the imaginary part of the left-hand side is non-zero.
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 s-plane, 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 s-plane 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
Integration contour used to obtain Eq. (A.11) is shown by the thick solid line. The dashed lines show the cuts.
|Open with DEXTER|
of five intervals of the real axis, , , , , and , and four half-circles 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, Aj and Cj are constants of the order of unity, and Cj > 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 non-principal sheets of the Riemann surface attached to the principal sheet at the upper edges of the cuts. To define w(s) on the non-principal 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 = sr + isi, where s+ < sr < s− and |si| ≪ sr. 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 non-principal Riemann sheet we have to define w(s) on this non-principal sheet as (A.17)In a similar way we find that the analytic continuation of w(s) on the non-principal 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 non-principal Riemann sheet, and the horizontal line connecting the points s+ − ib and s− − ib, once again on the non-principal 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
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 dashed-dotted line indicates the parts of the contour on the non-principal Riemann sheet. The dashed line shows the cut.
|Open with DEXTER|
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 non-principal 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.
© ESO, 2013
Current usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform.
Data correspond to usage on the plateform after 2015. The current usage metrics is available 48-96 hours after online publication and is updated daily on week days.
Initial download of the metrics may take a while.