In particular, we know from our previous calculations that for V(x) = t x^{2m}, t > 0, m = 1, 2, dots,

dmu^{eq} is supported on a single interval

I = (-a, a)

where

a = (mt prod_{l=1}^m frac{2l - 1}{2l})^{-frac{1}{2m}} (see (173.1))

and

dmu^{eq}(x) = frac{mt}{pi} sqrt{a^2 - x^2} h_1(x) chi_{(-a, a)} (x) dx = psi(x) dx

where h_1(x) is given by (172.2).

In particular, relations (202.2) - (202.5) hold for g(z) = int_I log (z - x) psi(x) dx with

l = 2 int_{-a}^a log (a - x) psi(x) dx - V(a).

For an explicit evaluation of l in the case m = 1, V = x^2, see ref [2], and in the general case m = 1, 2, dots, see Deift-Kriecherbauer-Mclaughlin-Venakides-Zhou, CPAM 52 (1999) 1491-1552.

Recall the RHP for pi_n(z): Sigma = mathbb{R}, v = egin{pmatrix} 1 & e^{-nV} \ 0 & 1 end{pmatrix}

(205.1)

  • Y(z) 2 imes 2 analytic in mathbb{C} ackslash mathbb{R}, continuous up to the boundary;
  • Y_+(z) = Y_-(z) egin{pmatrix} 1 & e^{-nV} \ 0 & 1 end{pmatrix}, z in mathbb{R};
  • Y(z) egin{pmatrix} z^{-n} & 0 \ 0 & z^n end{pmatrix} ightarrow mathbf{I} as z ightarrow infty in mathbb{C} ackslash mathbb{R}.

Bearing in mind the classical steepest descent method for scalar integrals, the question is this: which point in mathbb{C} plays the role of the critical point where the leading order contribution to the integral is located? Looking at (205.1), this is not at all clear.

The first step in the non-commutative steepest-descent method for RHPs of this type is to precondition (205.1) in such a way so as to transform the RHP into a standard RHP that is normalized at infty. This is done in the following way.

Let ilde{g}(z) be any function that is

(206.1)

  • analytic in mathbb{C} ackslash mathbb{R} and continuous up to the boundary;
  • ilde{g}(z) sim log z + mathcal{o}(1) as z ightarrow infty in mathbb{C} ackslash mathbb{R};

Set

U(z) = Y(z) egin{pmatrix} e^{-n ilde{g}(z)} & 0 \ 0 & e^{+n ilde{g}(z)} end{pmatrix} = Y(z) e^{-n sigma_3 ilde{g}(z)}

where sigma_3 = egin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix} is the third Pauli matrix.

Then

(206.2)

  • U(z) is analytic in mathbb{C} ackslash mathbb{R} and continuous up to the boundary;
  • U_+(z) = Y_+(z) e^{-n sigma_3 ilde{g}_+(z)} = Y_-(z) egin{pmatrix} 1 & e^{-nV(z)} \ 0 & 1 end{pmatrix} e^{-n sigma_3 ilde{g}_+(z)}

= U_-(z) ilde{v}(z)

where ilde{v}(z) = e^{nsigma_3 ilde{g}_-} egin{pmatrix} 1 & e^{-nV} \ 0 & 1 end{pmatrix} e^{-nsigma_3 ilde{g}_+}

= egin{pmatrix} e^{n( ilde{g}_- - ilde{g}_+)} & e^{n( ilde{g}_+ + ilde{g}_- - V)} \ 0 & e^{n( ilde{g}_+ - ilde{g}_-)} end{pmatrix}, z in mathbb{R};

  • U(z) = Y(z) egin{pmatrix} z^{-n} & 0 \ 0 & z^n end{pmatrix} egin{pmatrix} e^{n (log z - ilde{g}(z))} & 0 \ 0 & e^{-n (log z - ilde{g}(z))} end{pmatrix} ightarrow mathbf{I} as z ightarrow infty.

Thus U(z) solves the normalized RHP (Sigma = mathbb{R}, ilde{v}(z)).

Now conjugate U(z) further; for any constant ilde{l}, set

(207.1)

W(z) = e^{-frac{n}{2} ilde{l} sigma_3} U(z) e^{frac{n}{2} ilde{l} sigma_3}.

Then W(z) solves the normalized RHP (Sigma, v_W) where

(207.2)

v_W = egin{pmatrix} e^{n( ilde{g}_- - ilde{g}_+)} & e^{n( ilde{g}_+ + ilde{g}_- - V - ilde{l})} \ 0 & e^{n( ilde{g}_+ - ilde{g}_-)} end{pmatrix}.

Step 2

Now choose

ilde{g}(z) = g(z) = int log (z - s) psi(s) ds

= log z int dmu^{eq} + mathcal{o}(frac{1}{z}) = log z + mathcal{o}(1) as z ightarrow infty

and

ilde{l} = l = 2 int_{-a}^a log (a - s) psi(s) ds - V(a).

Observe that with these choices, the relations (202.2) (202.4) and (202.5) imply that v_W takes the form

(208.1)

where

(208.2)

T = g_+ + g_- - V - l < 0, |z| > a

and

S = g_+ - g_-, -a < z < a.

At this point, as T(z) < 0 for z > a and z < -a, it is clear that as n ightarrow infty, the RHP for W localizes to the interval (-a, a) = I.

On the interval I,

S(z) = g_+ - g_- = 2 pi i int_z^a psi(s) ds in i mathbb{R}

and so e^{-nS} is oscillatory as n ightarrow infty. What we now try to do is to deform e^{-nS} into the complex plane so that it becomes exponentially decreasing, and at the same time deform e^{nS} into a different part of the plane, so that it is also decreasing. To do this we must clearly separate e^{-nS} and e^{nS} algebraically.

This is done in the following way: observe that for z in I,

S = 2 pi i int_z^a frac{mt}{pi} sqrt{a^2 - s^2} h_1(s) ds

= 2mt int_z^a (s^2 - a^2)_+^{frac{1}{2}} h_1(s) ds

so S(z) has an analytic continuation to mathbb{C}_+

S(z) = 2mt int_z^a (s^2 - a^2)^{frac{1}{2}} h_1(s) ds

where z ightarrow a is any path in mathbb{C}_+. Also S has an analytic continuation

S(z) = -2mt int_z^a (s^2 - a^2)^{frac{1}{2}} h_1(s) ds

where z ightarrow a is any path in mathbb{C}_-.

Write

S = alpha + i eta

where alpha, eta are the real and imaginary parts of S respectively. For z in I, we have alpha(z) = 0, and from (202.3) we see that i frac{partial}{partial x} i eta = i frac{partial}{partial x} S = i frac{partial}{partial x} (g_+ - g_-) > 0. Thus frac{partial eta}{partial x} < 0, and hence by the Cauchy-Riemann condition, frac{partial alpha}{partial y} = - frac{partial eta}{partial x} > 0. It follows that

(210.1)

Re S(x + i y) > 0 for -a < x < a, y > 0 small;

and

(210.2)

Re S (x + i y) < 0 for -a < x < a, y < 0 small.

Thus we want to deform e^{-nS} into mathbb{C}_+ and e^{nS} into mathbb{C}_-.

Now note that we can separate e^{-nS} and e^{nS} by using the following factorization

(210.3)

egin{pmatrix} e^{-nS} & 1 \ 0 & e^{nS} end{pmatrix} = egin{pmatrix} 1 & 0 \ e^{nS} & 1 end{pmatrix} egin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix} egin{pmatrix} 1 & 0 \ e^{-nS} & 1 end{pmatrix}.

Note that is just the upper/lower factorization of

egin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix} egin{pmatrix} e^{-nS} & 1 \ 0 & e^{nS} end{pmatrix} = egin{pmatrix} 1 & e^{nS} \ 0 & 1 end{pmatrix} egin{pmatrix} -1 & 0 \ 0 & 1 end{pmatrix} egin{pmatrix} 1 & 0 \ e^{-nS} & 1 end{pmatrix}.

Such factorization play a basic role in the steepest descent method.

Step 3

Now open up the "lenses", Sigma ightarrow hat{Sigma}

to obtain a new oriented contour hat{Sigma}. Note that mathbb{C} ackslash hat{Sigma} has 4 components mathrm{I}, mathrm{II}, mathrm{III}, mathrm{IV}.

Define hat{W}(z) as follows:

(211.1)

For z in mathrm{I} cup mathrm{IV}, hat{W}(z) equiv W(z);

(211.2)

For z in mathrm{II}, hat{W}(z) equiv W(z) egin{pmatrix} 1 & 0 \ e^{-nS(z)} & 1 end{pmatrix}^{-1};

(211.3)

For z in mathrm{III}, hat{W}(z) equiv W(z) egin{pmatrix} 1 & 0 \ e^{nS(z)} & 1 end{pmatrix}.

Clearly hat{W}(z) is analytic in mathbb{C} ackslash hat{Sigma} and continuous up to the boundary. Denote the arcs in hat{Sigma} as follows:

(212.1)

For z in overline{14}, we clearly have hat{W}_+(z) = hat{W}_-(z) v_W(z);

(212.2)

For z in overline{12}, hat{W}_+(z) = W_+(z) = W_-(z) = hat{W}_-(z) egin{pmatrix} 1 & 0 \ e^{-nS} & 1 end{pmatrix};

(212.3)

For z in overline{23}, hat{W}_+(z) = W_+(z) egin{pmatrix} 1 & 0 \ e^{-nS(z)} & 1 end{pmatrix}^{-1}

= W_-(z) egin{pmatrix} 1 & 0 \ e^{nS} & 1 end{pmatrix} egin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix} = hat{W}_- egin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix};

(212.4)

For z in overline{34}, hat{W}_+(z) = W_+(z) egin{pmatrix} 1 & 0 \ e^{nS(z)} & 1 end{pmatrix}

= W_-(z) egin{pmatrix} 1 & 0 \ e^{nS} & 1 end{pmatrix} = W_- egin{pmatrix} 1 & 0 \ e^{nS} & 1 end{pmatrix}.

Thus hat{W}(z) solves the normalized RHP (hat{Sigma}, hat{v}) where hat{v} has form:

As Re S underset{<}{>} 0 in mathbb{C}_+ / mathbb{C}_- respectively, and T < 0 for z in mathbb{R} ackslash I, we see that the RHP (hat{Sigma}, hat{v}) localizes to the constant coefficient normalized RHP (Sigma_infty = (-a, a), v_infty = egin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}).

It is crucial to note that all the RHPs in Steps 1, 2 and 3 are equivalent, in the sense that if we can solve any one of them, then we can obtain the solution of any of the others just by an algebraic transformation.

The solution of RHP (Sigma_infty, egin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}) can be obtained explicitly. Indeed observe that

egin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix} = egin{pmatrix} 1 & 1 \ i & -i end{pmatrix} egin{pmatrix} i & 0 \ 0 & -i end{pmatrix} egin{pmatrix} 1 & 1 \ i & -i end{pmatrix}^{-1}.

Hence if we seek W_infty(z) analytic in mathbb{C} ackslash (-a, a) s.t. (W_infty)_+ = (W_infty)_- egin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}, then (W_infty)_+ egin{pmatrix} 1 & 1 \ i & -i end{pmatrix} = (W_infty)_- egin{pmatrix} 1 & 1 \ i & -i end{pmatrix} egin{pmatrix} i & 0 \ 0 & -i end{pmatrix} .

Rightarrow Thus W_infty(z) egin{pmatrix} 1 & 1 \ i & -i end{pmatrix} and hence egin{pmatrix} 1 & 1 \ i & -i end{pmatrix}^{-1} W_infty(z) egin{pmatrix} 1 & 1 \ i & -i end{pmatrix}, solves a pair of scalar RHPs and hence can be solved explicitly by Plemeljs formula (cf 193.x: note that a scalar RHP phi_+ = phi_- v becomes an additive RHP, if we take logarithms, which we can do only in the scalar case, (log phi)_+ = (log phi)_- + log v). Hence

egin{pmatrix} 1 & 1 \ i & -i end{pmatrix}^{-1} W_infty(z) egin{pmatrix} 1 & 1 \ i & -i end{pmatrix} = egin{pmatrix} e^{frac{1}{2 pi i} int_{-a}^a frac{log i}{s - z} ds} & 0 \ 0 & e^{frac{1}{2 pi i} int_{-a}^a frac{log (-i)}{s - z} ds} end{pmatrix}

= egin{pmatrix} e^{frac{1}{4} log frac{z - a}{z + a}} & 0 \ 0 & e^{-frac{1}{4} log frac{z - a}{z + a}} end{pmatrix}

= egin{pmatrix} eta & 0 \ 0 & eta^{-1} end{pmatrix}

where

(214.1)

eta(z) = (frac{z - a}{z + a})^{frac{1}{4}}, z in mathbb{C} ackslash [-a, a]

eta(z) is analytic in mathbb{C} ackslash [-a, a] and eta(z) > 0 for z > a.

Thus

W_infty(z) = egin{pmatrix} 1 & 1 \ i & -i end{pmatrix} egin{pmatrix} eta & 0 \ 0 & eta^{-1} end{pmatrix} egin{pmatrix} -i & -1 \ -i & 1 end{pmatrix} frac{1}{-2i}

= egin{pmatrix} eta & eta^{-1} \ i eta & -ieta^{-1} end{pmatrix} egin{pmatrix} +i & +1 \ +i & -1 end{pmatrix} frac{1}{+ 2i}

i.e.

(215.1)

W_infty(z) = egin{pmatrix} frac{eta + eta^{-1}}{2} & frac{eta - eta^{-1}}{2i} \ frac{eta^{-1} - eta}{2i} & frac{eta + eta^{-1}}{2} end{pmatrix}.

Note that W_infty(z) does not solve the RHP (Sigma_infty, v_infty) in the classical sense: it has a fourth root singularity at z = pm a. However, W_infty(z) is the unique solution of the RHP (Sigma_infty, v_infty) in an appropriate L^2 sense. Indeed, if hat{W}_infty is a second solution with an at-worst L^2 singularity at z = pm a, then as before, hat{W}_infty W_infty^{-1}(z) has no jumps across (-a, a) cup (a, infty) cup (-infty, -a) (note from (215.1) that det W_infty = 1: this of course also follows from the fact that det v_infty = 1, as before, but of course can be seen directly from (215.1)).

Hence hat{W}_infty W_infty^{-1} has at worst isolated singularities at pm a. But as hat{W}_infty(z) and W_infty^{-1}(z) have L^2 singularities at pm a, oint (s pm a)^k hat{W}_infty W_infty^{-1}(s) ds = 0, k ge 0 for any small circle about pm a.

Remark 216+ ightarrow

Hence hat{W}_infty W_infty^{-1}(z) has a remarkable singularity at pm a, which implies that hat{W}_infty W_infty^{-1}(z) is in fact entire, and so as hat{W}_infty(z), W_infty(z) ightarrow mathrm{I}, by Liouville, we must have hat{W}_infty = W_infty. Uniqueness!

Now we anticipate that hat{W} ightarrow W_infty as n ightarrow infty. The situation is a familiar one: suppose we are trying to solve an elliptic problem in a region Omega in the plane:

(216.1)

abla ullet a(epsilon) abla f = 0 in Omega

f = F on partial Omega

Remark 216+

The argument is as follows:

Suppose Z(z) = hat{W}_infty(z) W_infty^{-1}(z) is in L^prime near z = a in the sense that int_Sigma |Z(s)| |ds| le c < infty for small contours Sigma near a. Then for any k ge 0, any for Sigma = { |z - a| = epsilon }, we have

(216+.1)

left| int_{|s - a| = epsilon} Z(s) (s - a)^k ds ight| le epsilon^k int_{|s - a| = epsilon} |Z(s)| ds le c epsilon^k.

But as Z(s) (s - a)^k is analytic in the punched disk |z - a| > 0, the LHS in (216+.1) is independent of epsilon. Letting epsilon downarrow 0, we conclude that int_{|s - a| = epsilon_0} Z(s) (s - a)^k ds = 0 for any fixed, small epsilon_0 > 0, and k ge 1. Hence Z(z) is of the form frac{c^prime}{z - a} + analytic for some constant c^prime. Now for Sigma = { a + i t+ x: c_0, t > 0, -c_0 < x < c_0 }

we have

int_Sigma |Z(s)| ds

= |c^prime| int_{-c_0}^{c_0} |frac{1}{x + it}| dx + bounded

= |c^prime| int_{-c_0/t}^{c_0/t} frac{dy}{sqrt{1 + y^2}} + bounded

Keeping c_0 fixed and letting t downarrow 0 we conclude that

int_Sigma |Z(s)| |ds| sim |c^prime| log t^{-1} ightarrow infty

which contradicts int_Sigma |Z(s)| ds le c < infty, unless c^prime = 0.

Thus the singularity of hat{W}_infty W_infty^{-1}(z) at z = pm a is remarkable.

Now suppose a(epsilon) ightarrow a(0) in some sense as epsilon downarrow 0. For example a(epsilon) may have oscillations, and the convergence is in the weak sense. Question: Does the solution f = f_epsilon of (216.1), converge in some sense to the solution f = f_0 of

abla ullet a(0) abla f_0 = 0 in Omega

f_0 = F on partial Omega

as epsilon downarrow 0 ?

The answer is sometimes "no": e.g. suppose we try to solve the simple equation

H_epsilon f = (1 + frac{1}{2} sin(frac{x}{epsilon})) f(x) = 1, x in mathbb{R}.

Clearly, H_epsilon converges weakly to 1 as epsilon downarrow 0.

But

f = f_epsilon(x) = frac{1}{1 + frac{1}{2} sin frac{x}{epsilon}}

= 1 - frac{1}{2} sin frac{x}{epsilon} + (frac{1}{2} sin^2 frac{x}{epsilon}) + dots

= 1 - frac{1}{2} sin frac{x}{epsilon} + frac{1}{4}(1 - cos frac{2x}{epsilon}) + dots

underset{epsilon downarrow 0}{ ightharpoonup} 1 + frac{1}{4} + dots e 1 = frac{1}{H_0} = f_0.

More abstractly, if we are solving an equation A_epsilon f = g, and A_epsilon ightarrow A_0 in norm, A_0 invertible, then f = f_epsilon = frac{1}{A_epsilon} g ightarrow frac{1}{A_0} g = f_0. But if A_epsilon ightarrow A_0 weakly, or even strongly, then f_epsilon may not converge to f_0. This is precisely the situation we are facing with our RHP (hat{Sigma}, hat{v}). Although hat{v}(z) ightarrow v_infty(z), z in hat{Sigma} ackslash {pm a}, the convergence is not uniform: it clearly becomes slower and slower as z approaches pm a. Thus ||hat{v} - v_infty||_{L^infty (hat{Sigma})} ot ightarrow 0 as n ightarrow infty, and as we will see shortly, it is precisely the L^infty norm of hat{v} - v_infty that controls the convergence of the RHPs. (See 220+1, ..., 220+7, below for more details.)

Nevertheless, it turns out that indeed hat{W}(z) ightarrow W_infty(z) as n ightarrow infty.


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