APPLICATIONS OF COMPLEX VARIABLES
exp[±iφ(z) ] will be purely real. What is more, one such factor, known as the
dominantterm, will be exponentially large, whilst the other (thesubdominantterm)
will be exponentially small. AStokes lineis precisely where this happens.
We can now see how the change takes place without an observable discontinuity
occurring. Suppose thaty 1 (z) is very large andy 2 (z) is very small on a Stokes line.
Then a finite change inA 2 will have a negligible effect onY(z); in fact, Stokes
showed, for some particular cases, that the change is less than the uncertainty
iny 1 (z) arising from the approximations made in deriving it. Since the solution
with any particular asymptotic form is determined in a region bounded by two
Stokes lines to within an overall multiplicative constant and the original equation
is linear, the change inA 2 when one of the Stokes lines is crossed must be
proportional toA 1 ,i.e.A 2 changes toA 2 +SA 1 ,whereSis a constant (theStokes
constant) characteristic of the particular line but independent ofA 1 andA 2 .It
should be emphasised that, at a Stokes line, if the dominant term is not present
in a solution, then the multiplicative constant in the subdominant termcannot
change as the line is crossed.
As an example, consider the Bessel functionJ 0 (z)ofzeroorder.Itissingle-
valued, differentiable everywhere, and can be written as a series in powers ofz^2 .It
is therefore an integral even function ofz. However, its asymptotic approximations
for two regions of thez-plane, Rez>0andzreal and negative, are given by
J 0 (z)∼
1
√
2 π
1
√
z
(
eize−iπ/^4 +e−izeiπ/^4
)
, |arg(z)|<^12 π,|arg(z−^1 /^2 )|<^14 π,
J 0 (z)∼
1
√
2 π
1
√
z
(
eize^3 iπ/^4 +e−izeiπ/^4
)
, arg(z)=π,arg(z−^1 /^2 )=−^12 π.
We note in passing that neither of these expressions is naturally single-valued,
and a prescription for taking the square root has to be given. Equally, neither is
an even function ofz. For our present purpose the important point to note is
that, for both expressions, on the line argz=π/2 bothz-dependent exponents
become real. For large|z|the second term in each expression is large; this is the
dominant term, and its multiplying constanteiπ/^4 is the same in both expressions.
Contrarywise, the first term in each expression is small, and its multiplying
constant does change, frome−iπ/^4 toe^3 iπ/^4 ,asargzpasses throughπ/2 whilst
increasing from 0 toπ. It is straightforward to calculate the Stokes constant for
this Stokes line as follows:
S=
A 2 (new)−A 2 (old)
A 1
=
e^3 iπ/^4 −e−iπ/^4
eiπ/^4
=eiπ/^2 −e−iπ/^2 =2i.
If we had moved (in the negative sense) from argz= 0 to argz=−π, the relevant
Stokes line would have been argz=−π/2. There the first term in each expression
is dominant, and it would have been the constanteiπ/^4 in the second term that
would have changed. The final argument ofz−^1 /^2 would have been +π/2.