%Title: DIFFERENTIAL EQUATIONS for the 2-LOOP EQUAL MASS SUNRISE
%Author: E. Remiddi
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\title{
DIFFERENTIAL EQUATIONS for the 2-LOOP EQUAL MASS SUNRISE
\thanks{Presented at the Ustron Conference, 15-21 September 2003;
to be published in {\tt Acta Physica Polonica B.}}
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\author{ E. Remiddi
\address{
Dipartimento di Fisica, Universit\`a di Bologna and
INFN, Sezione di Bologna, \\
via Irnerio 46, I-40126 Bologna, Italy \\
Institut f\"ur Theoretische Teilchen Physik,
Universit\"at Karslruhe, \\
D-76128 Karslruhe, Germany
}
}
\maketitle
\begin{abstract}
The differential equations for the 2-loop sunrise graph, at equal masses
but arbitrary momentum transfer, are used for the analytic evaluation
of the coefficients of its Laurent-expansion in the continuous dimension $d$.
\end{abstract}
PACS: 12.15.Lk
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\vspace*{9cm}
\section{Introduction}
The differential equations in the squared external momentum $p^2$
for the Master Integrals (MIs) of the 2-loop sunrise graph with
arbitrary masses $m_1, m_2, m_3$ of Fig.~\ref{fig1} were written
in~\cite{CCLR1}.
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\begin{figure}[h]
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\includegraphics*[2cm,0cm][10cm,4cm]{sunrise_1.epstex}
\caption{\label{fig1} The 2-loop sunrise graph.}
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They were used for obtaining analytically particular values and
behaviours at zero and infinite momentum transfer~\cite{CCLR1}, at
pseudothresholds~\cite{CCR1} and threshold~\cite{CCR2}, as well
as for direct numerical integration\cite{CCR3}.
In this contribution I will report on some progress in the anlytic study
of the solutions of the equations for arbitrary momentum transfer
in the equal mass limit $m_i=1$. A more complete account
will be given elsewhere~\cite{SLER}; while the algebraic burden
in the arbitrary mass case will surely be much heavier, there
are indications~\cite{HCER} that the approach can be extended to the
arbitrary mass case as well. \par
In the equal mass limit the 2-loop sunrise has two MIs, which in the
usual $d$-continuous regularization scheme can be written as
\begin{eqnarray}
S(d,p^2) &=& \frac{1}{\Gamma^2\left(3-\frac{d}{2}\right)}
\int\frac{d^dk_1}{ 4\pi^{ \frac{d}{2} } }
\int\frac{d^dk_2}{4\pi^{\frac{d}{2}}}
\frac{1}{(k_1^2+1)(k_2^2+1)[(p-k_1-k_2)^2+1]} \ , \nonumber\\
\label{MIdef} \\
S_1(d,p^2) &=& \frac{1}{\Gamma^2\left(3-\frac{d}{2}\right)}
\int\frac{d^dk_1}{4\pi^{\frac{d}{2}}}
\int\frac{d^dk_2}{4\pi^{\frac{d}{2}}}
\frac{1}{(k_1^2+1)^2(k_2^2+1)[(p-k_1-k_2)^2+1]} \ . \nonumber
\end{eqnarray}
Let us put $p^2=z$ ($z$ is positive when $p$ is Euclidean); the two
MIs then satisfy the following linear system of first order
differential equations in $z$
\begin{eqnarray}
z \frac{d}{dz} S(d,z) &=& (d-3) S(d,z) + 3 S_1(d,z) \ , \nonumber\\
z(z+1)(z+9) \frac{d}{dz} S_1(d,z) &=& \frac{1}{2}(d-3)(8-3d)(z+3)S(d,z)
\label{1stosys} \\
&+& \frac{1}{2} \left[ (d-4)z^2 + 10(2-d)z +9(8-3d) \right] S_1(d,z)
\nonumber\\
&+& \frac{1}{2} \frac{z}{(d-4)^2} \ . \nonumber
\end{eqnarray}
The system can be rewritten as a second order differential equation for
$S(d,z)$ only
\begin{eqnarray}
z(z+1)(z+9) & \frac{d^2}{dz^2}S(d,z) & \nonumber\\
+ \frac{1}{2}\left[ (12-3d)z^2 + 10(6-d)z + 9d \right] & \frac{d}{dz}S(d,z)
& \label{2ndoeq} \\
+ \frac{1}{2}(d-3) \left[ (d-4)z - d - 4 \right] & S(d,z) & =
\frac{3}{2} \frac{1}{(d-4)^2} \ . \nonumber
\end{eqnarray}
As the second MI $S_1(d,z)$ can be written in terms of $S(d,z)$ and its
first derivative
\begin{equation}
S_1(d,z) = \frac{1}{3}\left[ - (d-3) + z\frac{d}{dz} \right] S(d,z) \ ,
\label{S1}
\end{equation}
we can take from now on $S(d,z)$ and its derivative $dS(d,z)/dz$ as the
effective MIs.
\section{From near 4 to near 2 dimensions. }
We want to expand $S(d,z)$ around $d=4$ as Laurent series in $(d-4)$
and then to obtain analytically the values of the coefficients of
the expansion by solving the
relevant differential equations. It was found {\it a posteriori} that
all the formuale are much simpler when expanding around $d=2$.
To give the relations between the two expansions, let us recall that
acting on any scalar Feynman integral in $d$ dimensions with a suitable
differential operator, one obtains the same integral in $(d-2)$
dimensions. times a numerical factor depending on $d$ \cite{Tarasov}.
Acting on the MIs in $d$ dimensions (or, in our case, on the two
functions $S(d,z)$ and $dS(d,z)/dz$), one obtains the same MIs in $d-2$
dimensions in terms of mass derivatives of the MIs in $d$ dimensions,
which can be expressed again in terms of MIs in $d$ dimensions; solving
the linear system for the $d$-dimensional MIs and replacing finally
$d$ by $d+2$ one obtains
\begin{eqnarray}
S(2+d,z) &=& \frac{1}{3(d-1)(3d-2)(3d-4)} \times \nonumber\\
&& \left\{ - \frac{9}{(d-2)^2} + \frac{3z-63}{4(d-2)} \right.
\label{d+2} \\
&& {\kern10pt}
+ (z+1)(z+9) \left[ 1 + (z-3)\frac{d}{dz} \right] S(d,z) \nonumber\\
&& \left. \phantom{\frac{1}{1}} % to force a big \right\}
+ (d-2) (87+22z-z^2) S(d,z) \right\} \nonumber
\end{eqnarray}
Quite in general, if
\[ A(2+d) = B(d) \ , \]
one can set $d=2+\eta$ and Laurent-expand in $\eta$; one obtains
\[ \sum_k \eta^k A^{(k)}(4) = \sum_k \eta^k B^{(k)}(2) \ . \]
\par
The Laurent expansion in $\eta$ of $S(d,z)$ for $d=4+\eta$ begins with
a double pole in $\eta$ and reads
\[ S(4+\eta,z) = \frac{1}{\eta^2} S^{(-2)}(4,z)
+ \frac{1}{\eta} S^{(-1)}(4,z)
+ S^{(0)}(4,z) + \eta S^{(1)}(4,z) + ....... \]
while $S(2+\eta,z)$ has no singularities in $\eta$, and its expansion is
\[ S(2+\eta,z) = S^{(0)}(2,z) + \eta S^{(1)}(2,z) + ....... \]
By inserting the two expansions in
Eq.(\ref{d+2}), one gets the required coefficients $S^{(k)}(4,z)$ of the
Laurent expansion in $\eta$ of $S(4+\eta,z)$ around $4$ in terms of
the coefficients $S^{(k)}(2,z)$ of the expansion of $S(2+\eta,z)$
around $2$.
As $S(d,z)$ is regular at $d=2$, the poles in $\eta$
are not hidden in $S(d,z)$ but are explicitly exhibited by the
$1/(d-2)$ factors in the r.h.s. of Eq.(\ref{d+2}).
Working out the algebra, one finds at once
\begin{eqnarray}
S^{(-2)}(4,z) &=& - \frac{3}{8} \ , \nonumber\\
\label{polesat4} \\
S^{(-1)}(4,z) &=& \frac{9}{16} + \frac{z}{32} \ . \nonumber
\end{eqnarray}
\section{The expansion at $d=2$ of the differential equation.}
By expanding systematically in $(d-2)$ all the terms appearing in
Eq.(\ref{2ndoeq}), one obtains a set of chained equations of the
form
\begin{eqnarray}
\biggl\{ \ \ \frac{d^2}{dz^2} +
\left[\frac{1}{z} + \frac{1}{z+1} + \frac{1}{z+9}
\right] \frac{d}{dz} & & \nonumber\\
\label{chain} \\
+ \left[ \frac{1}{3z} - \frac{1}{4(z+1)} - \frac{1}{12(z+9)}
\right] & \biggr\} S^{(k)}(2,z) = N^{(k)}(2,z) \nonumber
\end{eqnarray}
where the homogeneous part is the same for any order $k$, and the first
few inhomogeneous terms are
\begin{eqnarray}
N^{(0)}(2,z) &=& \frac{1}{24z} - \frac{3}{64(z+1)} + \frac{1}{192(z+9)}
= \frac{3}{8z(z+1)(z+9)} \ , \nonumber\\
N^{(1)}(2,z) &=& \left( - \frac{1}{2z} + \frac{1}{z+1} + \frac{1}{z+9}
\right)\frac{dS^{(0)}(2,z)}{dz} \label{Nchained} \\
&+& \left( \frac{5}{18z} - \frac{1}{8(z+1)}
- \frac{11}{72(z+9)} \right) S^{(0)}(2,z) \nonumber\\
&+& \frac{1}{24z} - \frac{3}{64(z+1)} + \frac{1}{192(z+9)}
\ , \nonumber\\
N^{(2)}(2,z) &=& ..... \ . \nonumber
\end{eqnarray}
The equations Eq.(\ref{Nchained}) are chained, in the sense that the
inhomogeneous term of order $k$ involves lower terms, of order $(k-1)$
(for $k>0$) and $(k-2)$ (for $k>1$) in the expansion of $S(2,z)$,
as can be seen from Eq.(\ref{2ndoeq}) and is shown explicitly in
Eq.s(\ref{Nchained}).
\par
The system Eq.(\ref{chain}) is
to be solved bottom up in $k$, starting from $k=0$ (in which
case the inhomogeneous term is completely known) and then proceeding
to higher values increasing $k$ by one, so that at each step the
inhomogeneous term is known from the solution of the previous equations.
The chained equations can then be solved by using Euler's method of the
variation of the constants. The homogeneous equation is the same for
all the values of $k$,
\begin{eqnarray}
\biggl\{ \ \ \frac{d^2}{dz^2} +
\left[\frac{1}{z} + \frac{1}{z+1} + \frac{1}{z+9}
\right] \frac{d}{dz} & & \nonumber\\
\label{homo} \\
+ \left[ \frac{1}{3z} - \frac{1}{4(z+1)} - \frac{1}{12(z+9)}
\right] & \biggr\} \Psi(z) = 0 \ ; \nonumber
\end{eqnarray}
if $\Psi_1(z), \Psi_2(z)$ are two independent solutions of the homogeneous
equation, $W(z)$ the corresponding Wronskian
\begin{equation}
W(z) = \Psi_1(z) \frac{d\Psi_2(z)}{dz}
- \Psi_2(z) \frac{d\Psi_1(z)}{dz}
\label{Wronskian}
\end{equation}
according to Euler's method the solutions of Eq.s(\ref{chain}) are
given by the integral representations
\begin{eqnarray}
S^{(k)}(2,z) &=& \Psi_1(z) \left( \Psi^{(k)}_1
- \int_0^z \frac{dw}{W(w)} \Psi_2(w) N^{(k)}(2,w) \right) \nonumber\\
\label{Euler} \\
&+& \Psi_2(z) \left( \Psi^{(k)}_2
+ \int_0^z \frac{dw}{W(w)} \Psi_1(w) N^{(k)}(2,w) \right) \ , \nonumber
\end{eqnarray}
where \( \Psi^{(k)}_1, \Psi^{(k)}_2 \) are two integration constants.
\par
Eq.(\ref{Euler}) at this moment is just a formal representation of the
solutions for the coeffcients $S^{(k)}(2,z)$; it becomes a
substancial (not just formal!) formula only when all the ingredients
-- the two solutions of the homogeneous equation $\Psi_i(z)$, their
Wronskian $ W(z) $ and the two integration constants $ \Psi^{(k)}_i $
are known explicitly.
\par
Although the Wronskian is defined in terms of the $\Psi_i(z)$, it can be
immediately obtained (up to a multiplicative constant) from
Eq.(\ref{homo}). An elementary calculation using the definition
Eq.(\ref{Wronskian}) and the value of the second derivatives of the
$\Psi_i(z)$, as given by Eq.(\ref{homo}) of which they are solutions,
leads to the equation
\[ \frac{d}{dz} W(z) = - \left( \frac{1}{z} + \frac{1}{z+1}
+ \frac{1}{z+9} \right) W(z) \ , \]
which gives at once
\begin{equation}
W(z) = \frac{9}{z(z+1)(z+9)} \ , \label{Wvalue}
\end{equation}
where the multiplicative constant has been fixed anticipating later
results.
\par
Finding the two $\Psi_i(z)$ requires much more work.
\section{Solving the homogeneous equation at the singular points.}
By inspection, the singular points of Eq.(\ref{homo}) are found to be
\[ z=0, -1, -9, \infty \ ; \]
at each of those points one has two independent solutions, the first
regular and the second with a logarithmic singularity.
The expansions of the solutions around each of the singular points is
immediately provided by the differential equation itself.
\par
Around $z=0$ the two solutions of Eq.(\ref{homo}) can be written as
\begin{eqnarray}
\Psi_1^{(0)}(z) &=& \psi_1^{(0)}(z) \nonumber\\
\Psi_2^{(0)}(z) &=& \ln{z} \ \psi_1^{(0)}(z) + \psi_2^{(0)}(z) \ ,
\label{Psiat0}
\end{eqnarray}
where the $\psi_i^{(0)}(z)$ are power series in $z$.
Imposing $\psi_1^{(0)}(0)=1$, one finds
\begin{eqnarray}
\psi_1^{(0)}(z) &=& 1 - \frac{1}{3}z + \frac{5}{27}z^2 + ... \ ,\nonumber\\
\psi_2^{(0)}(z) &=& - \frac{4}{9}z + \frac{26}{81}z^2 + ... \ ;
\label{Psiat0a}
\end{eqnarray}
the coefficients are given recursively (hence up to any reuired order)
by the equation. The radius of convergence is 1 (the next singularity is
at $z=-1$) and the two solutions are real for positive $z$ (spacelike
momentum transfer). The continuation to the timelike region
is done by giving to $z$ the value $z=-(u+i\epsilon)$;
for $ 0< u < 1\, $ one has $\ln{z} = \ln{u} - i\pi $ and
$ \Psi_2^{(0)}(z) $ develops an imaginary part $-i\pi \psi_1^{(0)}(z)$.
\par
Similarly, around $z=-1$ the 2 independent solutions can be written as
\begin{eqnarray}
\Psi_1^{(1)}(z) &=& \psi_1^{(1)}(z) \nonumber\\
\Psi_2^{(1)}(z) &=& \ln{(z+1)} \psi_1^{(1)}(z) + \psi_2^{(1)}(z) \ ,
\label{Psiat1}
\end{eqnarray}
with Eq.(\ref{homo}) providing recursively the coefficients of the
expansions in powers of $(z+1)$ of the two $\psi_i^{(1)}(z)$ once the
initial condition is given. If $\psi_1^{(1)}(0)=1 $ one has
\begin{eqnarray}
\psi_1^{(1)}(z) &=& 1 + \frac{1}{4}(z+1) + \frac{5}{32}(z+1)^2 + ... \ ,
\nonumber\\
\psi_2^{(1)}(z) &=& + \frac{3}{8}(z+1) + \frac{33}{128}(z+1)^2 + ... \ ,
\label{Psiat1a}
\end{eqnarray}
with radius of convergence 1 (up to the singularity at $z=0$) etc.
\par
The other two singular points $z=-9$ and $z=\infty$ can be treated in the
same way, the corresponding formualae are not given due to lack of space.
\par
\section{The interpolating solutions.}
Having the solutions piecewise is not sufficient, one must build two
solutions in the whole $ -\infty < z < \infty $ range by suitably
joining the above expressions of the solutions at the singular points.
A hint is provided by the knowledge of the imaginary part of the
original Feynman integral $S(d,p^2)$ Eq.(\ref{MIdef}) at $d=2$
dimensions; as already observed in~\cite{GP},
the Cutkosky-Veltman rule gives for the imaginary part of $S(d,p^2)$
at $d=2$ and $ u = -z \ge 9 $ (and up to a multiplicative constant) the
integral representation
\begin{equation}
J(u) = \int_4^{(\sqrt{u}-1)^2} \frac{db}{\sqrt{R_4(u,b)} } \ ,
\label{ImS}
\end{equation}
where $ R_4(u,b) $ stands for the polynomial (of 4th
order in $b$ and 2nd order in $u$)
\[ R_4(u,b) = b(b-4)(b-(\sqrt{u}-1)^2) (b-(\sqrt{u}+1)^2) \ , \]
and the $b$ integration runs between two adjacent zeros of $ R_4(u,b) $.
As the inhomogeneous part of Eq.(\ref{2ndoeq}) cannot develop an imaginary
part, the imaginary part of the Feynman integral in $d=2$ dimensions,
$J(u)$ of Eq.(\ref{ImS}), is necessarily a solution of the associated
homogeneous equation at $d=2$, \ie of Eq.(\ref{homo}) -- a fact which
can also be checked explicitly.
\par
One is then naturally lead to consider all the $b$-integrals of
$1/\sqrt{R_4(u,b)}$ between any two adjacent roots for all possible values
of $u$.
The details of the analysis cannot be reported here again for lack of
space. As a result, one finds for instance that when $u$ is in the range
$0__0$,
\begin{eqnarray}
\Psi_1(z) &=& \Psi_1^{(0)}(z) \ , \nonumber\\
\Psi_2(z) &=& \Psi_2^{(0)}(z) \ . \nonumber
\end{eqnarray}
That fixes the multiplicative constant in the Wronskian as well,
giving the result already anticipated in Eq.(\ref{Wvalue}).
From Eq.s(\ref{Psiat0},\ref{Psiat0a}) we easily read the behaviours
of the $\Psi_i(z)$ for
$u=-z$ small and positive; but in the range $0____z>-1$ the argument
$w$ of the inhomogeneous term $N^{(k)}(w)$ varies in the interval
$0>w>z>-1$, and is therefore real ($N^{(k)}(w)$, Eq.(\ref{Nchained})
involves either real algebraic fractions or lower order terms of the
expansion in $(d-2)$ of $S(d,z)$, which are real in that region) --
therefore an imaginary part, if any, can come only from the $\Psi_i(z)$
and the $\Psi_i(w)$. By using the values of the $\Psi_i(z)$ as given
by Eq.s(\ref{PsiasJ01}), one finds for $u=-z$ in the range $0____ z > -9,\ $ \ie $\ 1__