.

The Cauchy problem of the Ward equation

Derchyi Wu Institute of Mathematics Academia Sinica Taipei, Taiwan

§1. Introduction

⋄ 1.1 Background

The self-dual Yang-Mills equations are a fundamental tool in four-manifold topology. Taking a one dimension reduction, one obtains the three-dimensional \underline{monopole equation} [Hitchin,1988], whilst a two dimension reduction yields \underline{harmonic maps} into Lie groups [Uhlenbeck, 1989]. These equations have generated much interest and been extensively studied.

If we replace the background space by R^{2,2} , and take a two dimension reduction of the self-dual Yang-Mills equations on R^{1,1}, then the harmonic maps can be reduced to integrable systems - the \underline{sine}-\underline{Gordon} equation [Uhlenbeck, 1992] and the \underline{wave map} equation [Terng-Uhlenbeck, 2004]. These hyperbolic equations can be solved by the inverse scattering method.

⋄ 1.2 Ward Equation

The Ward equation [Ward,’88]

(J^{−1}J_{t})_{t}−(J^{−1}J_{x})_{x}−(J^{−1}J_{y})_{y}−[J^{−1}J_{t}, J^{−1}J_{y}] = 0,

J : R^{3} ⟶ SU(n),

is obtained by a dimension reduction and a gauge fixing from the self-dual Yang-Mills equation on R^{2,2}. It differs only slightly from the wave map equation and is an integrable system with a Lax pair

[λ∂_{x} − ∂_{ξ} − J^{−1}∂_{ξ}J, λ∂_{η} − ∂_{x} − J^{−1}∂_{x}J] = 0,

∀λ ∈ C, with ξ = \frac{t+y}{2} , η = \frac{t−y}{2} . Setting J^{−1}∂_{ξ}J = −∂_{x}Q, J^{−1}∂_{x}J = −∂_{η}Q and (η, x, ξ) → (x, y, t), then the Ward equation turns into [Fokas-Ioannidou,’01]

∂_{x}∂_{t}Q = ∂_{y}^{2}Q + [∂_{y}Q, ∂_{x}Q]  (2)

with a Lax Pair

(∂_{y} − λ∂_{x}) Ψ = (∂_{x}Q)Ψ,  (3)
(∂_{t} − λ∂_{y}) Ψ = (∂_{y}Q)Ψ.

⋄ 1.3 References

(1) Construction of rational solutions: [Ward, ’88] ..., [Dai-Terng, ’07].

• Ψ is mero. in λ ∈ C with finite many poles;

• Only discrete scattering data appears.

(2) Cauchy problem with small ini. data: [Villarroel, ’90], [Fokas-Ioannidou, ’01], [Dai-Terng-Uhlenbeck].

• Ψ has jump across R;

• Only continuous scattering data appears.

⋄ 1.4 Goal

(1) Solve the Cauchy problem of the Ward eq. with • large initial data;

• pure continuous scattering data by the Inverse Scattering Method:

operator, t = 0 → scattering data, t = 0

↓

operator, t = T ← scattering data, t = T.

(2) Solve the Cauchy problem of the Ward eq. with • mixed (finite) scattering data

by Backlund transformation.

§2. Direct Problem

⋄ 2.1 Theorem (Existence of the eigenfunction)

Let Q ∈ P_{∞,k,0}, k ≥ 2. Then there is a \underline{bounded} set Z ⊂ C such that Z ∩ (C\R) is \underline{discrete} in C\R, and

• For λ ∈ C\ (R ∪ Z), ∃!Ψ s.t. Ψ − 1 ∈ DH^{k},

(∂_{y} − λ∂_{x}) Ψ = (∂_{x}Q)Ψ,  (3)

and Ψ(x, y, λ) → 1, as |x|, |y| → ∞.

• Ψ(x, y, ·) is \underline{meromorphic} in λ ∈ C\R with \underline{poles} at Z\R, and Ψ(x, y, λ) → 1 uniformly, as |λ| → ∞.

•
det Ψ = 1, Ψ(λ)Ψ^{∗}(λ̄) = 1.

⋄ 2.2 Proof

• When Q ∈ P_{∞,k,0} and |Q̂_{x}(ξ, y)|_{L_{1}(dξdy)} < 1. Then by [V], [F-I], [D-T-U], there exist Ψ and Z = ϕ which fullfill the theorem.

• Suppose |Q̂_{x}(ξ, y)|_{L_{1}(dξdy)} < (\frac{3}{2})^{N+1} and the theorem holds for N. Note

if Ψ(x, y, λ) is an e.f. of Q(x, y),
then Ψ(x, y + y_{0}, λ) is an e.f. of Q(x, y + y_{0}).

Hence wlog, we can assume

Q = Q^{−} + Q^{+},
Q^{−} = 0, for y ≥ 1,
Q^{+} = 0, for y ≤ −1,
|Q̂_{x}^{±}(ξ, y)|_{L_{1}(dξdy)} < (\frac{3}{2})^{N} .

So by induction, there exist Ψ^{±} of Q^{±} satisfying the theorem.

• Using the derivation property of ∂_{y} − λ∂_{x} and a change of variables x + λy = x̃ + iỹ = z, x̃, ỹ ∈ R. The existence of Ψ is eqivalent to solving the Riemann-Hilbert problem

∂z̄f(x, y, λ) = 0, for z ∈ C^{±},
f_{+}(x, 0, λ) = f_{−}(x, 0, λ)F (x, λ), for z ∈ R
F (x, λ) = Ψ^{−}(x, 0, λ)^{−1}Ψ^{+}(x, 0, λ).

⋄ 2.3 Remarks

• The RH problem is a problem with non-small data in general. To solve it, we will first approximate the solution by a piecewise rational function. Then the correction is made by a solution of a Riemann-Hilbert problem with small data and a solution of a finite linear system.

• Since in the induction scheme,

(1) the eigenfunction derived in each step consists the data of RH problems in the next step;

(2) the nice asymptote is required to assure the solvability of the system of linear equations.

Main estimates

∂_{x}^{i} (Ψ − 1) ∈ L_{2}(dx) unif., λ ∈ C\ (R ∪ D_{ϵ}(λ_{j})) ,

(4)

∂_{x}^{i} (Ψ − 1) → 0 unif. in L_{2}(dx) as |λ| → ∞. (5)

• The coefficents of the linear system are meromorephic by induction. Hence the solvability of the linear system outside a bounded discrete set in C\R follows from the asymptotic estimate (5).

⋄ 2.4 Definition

P_{∞,k_{1},k_{2}}

={q_{x}(x, y) : R × R → su(n)|

|ξ^{i}y^{s}q̂|_{L_{1}(dξdy)}, ||ξ^{h}q̂(ξ, y)|_{L_{1}(y)}|_{L_{2}(dξ)},

|∂_{x}^{j}∂_{y}^{l}q|_{L_{∞}}, sup_{y} |∂_{x}^{j}∂_{y}^{l}q|_{L_{1}(dx)}, |∂_{x}^{j}∂_{y}^{l}q|_{L_{1}(dxdy)} < ∞

1 ≤ i ≤ max{5, k_{1}}, 0 ≤ j, l ≤ max{5, k_{1}}, 1 ≤ h ≤ k_{1}, 0 ≤ s ≤ k_{2} }.

DH^{k}
={f |∂_{x}^{i}f(x, y) are uniformly bounded in L_{2}(R, dx),

0 ≤ i ≤ k.}

⋄ 2.5 Remark

Our goal is to prove the inverse problem when ♯(Z) < ∞ and Z ⊂ C\R. In this case, by (4), for λ ∈ R, Ψ_{±} exists, and

Ψ_{+}(x, y, λ) = Ψ_{−}(x, y, λ)v(x + λy, λ).

⋄ 2.6 Definition If Q ∈ P_{∞,k,0}, k ≥ 2 and Ψ_{±} exists, then v(x + λy, λ) is called the \underline{continuous scattering data} of Q.

⋄ 2.7 Theorem (Constraints on continuous SD)

For Q ∈ P_{∞,k,1}, k ≥ 7, if Ψ_{±} exist, then ∃ ! v(x, y, λ) s.t.

Ψ_{+}(x, y, λ) = Ψ_{−}(x, y, λ)v(x + λy, λ), λ ∈ R. Moreover,

det (v) ≡ 1, v = v^{∗} > 0,

and for i + j ≤ k − 4, ∂_{x}^{i}∂_{y}^{j} (v − 1) unif. bded in L_{∞} ∩ L_{2}(R, dλ) ∩ L_{1}(R, dλ) ∂_{x}^{i}∂_{y}^{j} (v − 1) → 0 unif. in L_{∞} ∩ L_{2}(R, dλ) ∩ L_{1}(R, dλ) ∂_{λ}v ∈ L_{2}(R, dλ) and the norms depend conti. on x, y.

§3. Inverse Problem for ♯(Z) = 0 ⋄ 3.1 Theorem (Solvability of the RH problem) If v(x + λy, λ) satis. constraints in Theorem 2.7, then

• ∃ ! Ψ(x, y, ·) for the RH problem

∂λ̄Ψ = 0, λ ∈ C\R,  (6)
Ψ_{+} = Ψ_{−}v, λ ∈ R.  (7)

• Ψ − 1, ∂_{x}Ψ, ∂_{y}Ψ unif. bded in L_{2}(R, dλ), and for λ ∉ R, and i + j ≤ k − 4,

∂_{x}^{i}∂_{y}^{j}Ψ ∈ L_{∞}(dxdy),

∂_{x}^{i}∂_{y}^{j} (Ψ − 1) → 0 in L_{∞}(dxdy)
•
det Ψ(x, y, λ) ≡ 1,
Ψ(x, y, t, λ)Ψ(x, y, t, λ̄)^{∗} ≡ 1.

⋄ 3.2 Proof of Theorem 3.1

For |x|, |y| large, the RH problem is a \underline{small data} problem. Hence it can be solved. For |x|, |y| not large, as before, we will first approximate the solution by a piecewise rational function. Then the correction is made by a solution of a Riemann-Hilbert problem with small data and a solution of a finite linear system. Note we obtain the global solvability by applying the \underline{Fredholm property} and the \underline{reality condition} v = v^{∗} > 0.

⋄ 3.3 Theorem (Solv. of the inverse problem)

If v(x + λy, λ) satis. the constraints in Theorem 2.7, then

(∂_{y} − λ∂_{x}) Ψ = (∂_{x}Q)Ψ

with

Q(x, y) = \frac{1}{2πi} ∫_{R} Ψ_{−}(v − 1)dξ

and ∂_{x}Q(x, y) ∈ su(n), for i + j ≤ k − 4, i > 0,

∂_{x}^{i}∂_{y}^{j}Q, ∂_{y}Q, Q ∈ L_{∞},
∂_{x}^{i}∂_{y}^{j}Q, ∂_{y}Q, Q → 0 as x or y → ∞.

⋄ 3.4 Proof of Theorem 3.3 By Theorem 3.1,

Ψ =1 + \frac{1}{2πi} ∫_{R} \frac{Ψ_{−}(x, y, ζ) (v(x + ζy, ζ) − 1)}{ζ − λ}dζ
=1 + CΨ(v − 1)
=1 + C_{v}Ψ.

Hence L_{λ}v = 0 yields

L_{λ}Ψ =L_{λ}CΨ(v − 1)
= [L_{λ}, C] Ψ(v − 1) + CL_{λ}(Ψ(v − 1))
=∂_{x} (\frac{1}{2πi} ∫_{R} Ψ_{−}(v − 1)dξ) + C (L_{λ}Ψ) (v − 1)
=∂_{x}Q + C_{v}(L_{λ}Ψ),

with

Q(x, y) = \frac{1}{2πi} ∫_{R} Ψ_{−}(v − 1)dξ.

Finally

L_{λ}Ψ =(1 − C_{v})^{−1}(∂_{x}Q)
=(∂_{x}Q)(1 − C_{v})^{−1}1
=(∂_{x}Q)Ψ

⋄ 3.7 The spectral flow

From previous observation, we find the algebraic constraint v(x, y, λ) = v(x + λy, λ) is key to derive (3). Hence if we let

v(x, y, t, λ) = v(x + λy + λ^{2}t, λ).

By analogy, we will obtain

(∂_{t} − λ∂_{y}) Ψ = (∂_{y}Q)Ψ.

Theorem (Solvability of the Cauchy problem)

If Q_{0} ∈ P_{∞,k,1}, k ≥ 7, and there are no poles of the eigenfunction Ψ_{0} of Q_{0}, then the Cauchy problem of the Ward equation with initial condition Q(x, y, 0) = Q_{0}(x, y) admits a global solution where for i + j + h ≤ k − 4, i^{2} + j^{2} > 0,

∂_{x}Q(x, y, t) ∈ su(n),
∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}Q, ∂_{t}Q, Q ∈ L_{∞},
∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}Q, ∂_{t}Q, Q → 0, as x, y, t → ∞.

⋄ 3.6 Remarks on examples

• Schwartz potentials with |Q̂_{x}|_{L_{1}(dξdy)} < 1;

• Let v(x, y, λ) = v(x + λ, λ) satisfy

det(v) = 1, v = v^{∗} > 0, v − 1 ∈ Schwartz, v satisfies (4′), (5′), (8).

Then we can solve Ψ, and justify that

∂_{x}^{i}∂_{y}^{j}∂_{λ}^{h}(Ψ − 1) is Schwartz.

Moreover, by using the fomula

Q(x, y) = \frac{1}{2πi} ∫_{R} Ψ_{−}(v − 1)dξ,

one obtains that Q is Schwartz.

§4. The Ward Solitons

⋄ 4.1 Definition (Loop Groups)

D = {f : C → M_{n}(C) satisfies :

• f(λ̄)^{∗}f(λ) = 1, lim_{λ→∞} f = I.

• f is mero. in C\R w. finitely many poles.

• f_{±} exist.}; D_{c} = {f ∈ D s.t. f is holo. in C\R and f_{±} exist.}; D_{r} = {f ∈ D s.t. f is rational. }.

⋄ 4.2 Theorem (Uhlenbeck ’89) For ψ ∈ D_{r} and Z(f) ⊂ C^{+}, we can factorize

ψ = f_{1} ⋯ f_{p}.

Where

f_{i} = g_{z_{i},π_{i_{1}}} ⋯ g_{z_{i},π_{im}}

and g_{z_{i},π_{i_{k}}} is a simplest element of the form

g_{z_{i},π_{i_{k}}} = 1 + \frac{z_{i} − z̄_{i}}{λ − z_{i}} π_{i_{k}}^{⊥},

π_{i_{k}}^{2} = π_{i_{k}}, π_{i_{k}}^{∗} = π_{i_{k}}, z_{i} ∈ C^{+} and z_{i} ≠ z_{j} if i ≠ j.

⋄ 4.3 Theorem (Dai-Terng ’07 )

Suppose ∃A_{0}(x, y), Ψ_{0}(x, y, λ) such that

(∂_{y} − λ∂_{x})Ψ_{0} = A_{0}Ψ_{0},

with

Ψ_{0}(x, y, ·) ∈ D_{r} with only pole z of multi. k,
∂_{x}^{i}∂_{y}^{j}Ψ_{0}(x, y, λ) ∈ L_{∞},
lim_{|x|, |y|→∞} ∂_{x}^{i}∂_{y}^{j}(Ψ_{0}(x, y, λ) − C_{1}) = 0,

for i+j ≤ N, λ ∈ C\{z}. Then there exist Ψ(x, y, t, λ), A(x, y, t), B(x, y, t) s.t.

(∂_{y} − λ∂_{x})Ψ = AΨ, (9)
(∂_{t} − λ∂_{y})Ψ = BΨ, (10)
Ψ(x, y, 0, λ) = Ψ_{0}, A(x, y, 0) = A_{0},

and

Ψ(x, y, t, ·) ∈ D_{r} with only pole z of multi. k,
∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}Ψ(x, y, t, λ) ∈ L_{∞},
lim_{|x|, |y|, |t|→∞} ∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}(Ψ(x, y, t, λ) − C_{1}) = 0,

for i+j +h ≤ N, λ ∈ C\{z}. Where C_{1} is independent of x, y, t.

⋄ 4.4 Ideas of the proof • If k = 1, one can set

Ψ_{0} = 1 + \frac{z − z̄}{λ − z}π_{0}^{⊥}(x, y), π_{0}^{∗} = π_{0}, π_{0}^{2} = π_{0}.

Using the initial condi. (∂_{y} −λ∂_{x})Ψ_{0} = A_{0}Ψ_{0}, one can justify that

A_{0}(x, y) = (z − z̄)∂_{x}π_{0},

Imπ_{0} is spanned by columns of V (x + zy),
V : C → M_{n×r} is holo. in C/{p_{1}, ⋯, p_{s}},

and M_{n×r} denotes the space of rank r complex n × r matrices. Therefore, define

Ψ = 1 + \frac{z − z̄}{λ − z}π^{⊥}, π^{∗} = π, π^{2} = π,

Imπ is spanned by columns of V (x + zy + z^{2}t). One can justify Ψ solve the Cauchy problem with A(x, y, t) = (z − z̄)∂_{x}π, B(x, y, t) = (z − z̄)∂_{y}π.

• k > 1, the loop group factorization theorem implies

Ψ_{0}(x, y, λ) = g_{z,π_{k+1,0}}Ψ_{k,0}(x, y, λ),

g_{z,k+1,0}(x, y, λ) = 1 + \frac{z − z̄}{λ − z}π_{k+1,0}^{⊥}(x, y),

(∂_{y} − λ∂_{x})Ψ_{k,0} = A_{k,0}(x, y)Ψ_{k,0},

Ψ_{k,0} ∈ D_{r}, Z(Ψ_{k,0}) = {z} with multiplicity k. By induction, ∃Ψ_{k}(x, y, t, λ) such that (∂_{y} − λ∂_{x})Ψ_{k}(x, y, t, λ) = A_{k}(x, y, t)Ψ(x, y, t, λ), (∂_{t} − λ∂_{y})Ψ_{k}(x, y, t, λ) = B_{k}(x, y, t)Ψ(x, y, t, λ), Ψ_{k}(x, y, 0, λ) = Ψ_{k,0}(x, y, λ), A_{k}(x, y, 0) = A_{k,0}(x, y), and Ψ(x, y, t, ·) ∈ D_{r} with only pole z of multiplicity k, ⋯ On the other hand, by the Backlund transformation, one has

τ_{1,0}(x, y), ⋯, τ_{n_{k+1},0}(x, y) span Im π_{k+1,0},
(∂_{y} − z∂_{x})τ_{j,0} = A_{k,0}(x, y)τ_{j,0}.

Hence τ_{j,0}(x, y) = η_{k,0}(x, y)H_{j}(x + zy). Where η_{k,0} is a fundmental solution of (∂_{y} − z∂_{x})f = A_{k,0}(x, y)f.

Dai and Terng show that ∃η_{k}(x, y, t) such that

(∂_{y} − z∂_{x})η_{k} = A_{k}(x, y, t)η_{k},

(∂_{t} − z∂_{y})η_{k} = B_{k}(x, y, t)η_{k},

η_{k}(x, y, 0) = η_{k,0}(x, y), A_{k}(x, y, 0) = A_{k,0}(x, y).

So τ_{j,0} can be extended by

τ_{j}(x, y, t) = η_{k}(x, y, t)H(x + zy + z^{2}t),

then g_{z,π_{k+1},0} = g_{z,π_{k+1,0}}Ψ_{k,0} can extended, too. Finally, let

Ψ(x, y, t, λ) = g_{z,π_{k+1}}Ψ_{k}(x, y, t, λ).

Applying the theory of Backlund transformation, one will show that Ψ satisfies the Cauchy problem of the Theorem.

⋄ 4.5 Theorem (Dai-Terng ’07) Suppose

• f(x, y, t, λ) and g(x, y, t, λ) satisfy (9), (10) with different A, B.

• Z (g) = {z} with multiplicity of k, and f is holomorphic and non-degenerate at z, z̄. Then there exist unique f̃ and g̃ s.t.

• Ψ = f̃g = g̃f satisfies (9), (10) w. new A, B.

• Z \binom{}{g̃} = \{z\} with mult. of k, f̃ is holo. and non-degenerate at z, z̄.

Moreover, if for i + j + h ≤ N, λ ∈ C\{z},

∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}f, ∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}g ∈ L_{∞}
lim_{|x|, |y|, |t|→∞} ∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}(f − C_{1}) = 0,
lim_{|x|, |y|, |t|→∞} ∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}(g − C_{2}) = 0,

then .......

⋄ 4.6 Ideas of the proof

• If k = 1, then by assumption, ∃π such that g = 1 + \frac{z−z̄}{λ−z}π^{⊥}(x, y). To find Ψ, we first need to find g̃,

g̃ = 1 + \frac{z − z̄}{λ − z}π̃^{⊥}(x, y)

and

f̃ = (g̃f) g^{−1} is idependent of λ.  (11)

Computing the residue of (g̃f) g^{−1}, we find that (11) is equiv. to that π̃ is the hermitian projection of C^{n} onto f(x, y, t, z) (Im (π(x, y))).

Secondly, we need to justify

((∂_{y} − λ∂_{x})Ψ) Ψ^{−1}, ((∂_{t} − λ∂_{y})Ψ) Ψ^{−1}

is independent of λ. Since Ψ = g̃f,

[(∂_{y} − λ∂_{x})Ψ]Ψ^{−1}
=[(∂_{y} − λ∂_{x})g̃]g̃^{−1} + g̃[((∂_{y} − λ∂_{x})f)f^{−1}]g̃^{−1}

So [(∂_{y} − λ∂_{x})Ψ]Ψ^{−1} is holo. on C\{z, z̄} and is bdd at λ = ∞. On the other hand, using Ψ = f̃g, one has

[(∂_{y} − λ∂_{x})Ψ]Ψ^{−1}
=[(∂_{y} − λ∂_{x})f̃]f̃^{−1} + f̃[((∂_{y} − λ∂_{x})g)g^{−1}]f̃^{−1}

and can show that [(∂_{y} − λ∂_{x})Ψ]Ψ^{−1} is holo. at {z, z̄}

• If k > 1, ......

§5. The Cauchy problem w. mixed scattering data

⋄ 5.1 Theorem

If Q_{0} ∈ P_{∞,k,1}, k ≥ 7, Z(Ψ_{0}) ⊂ C\R, ♯(Z(Ψ_{0})) < ∞, then the Cauchy problem of the Ward equation with initial condition Q(x, y, 0) = Q_{0}(x, y) admits a global solution where for i + j + h ≤ k − 4, i^{2} + j^{2} > 0,

∂_{x}Q(x, y) ∈ su(n),
∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}Q, ∂_{t}Q, Q ∈ L_{∞},
∂_{x}^{i}∂_{y}^{j}∂_{t}^{h}Q, ∂_{t}Q, Q → 0, as x, y, t → ∞.

⋄ 5.2 Remarks on the proof • Ψ_{0,±} exists and ∃v such that Ψ_{0,+} = Ψ_{0,−}v.

• Solve the RH problem with data v, ∃f s. t.

∂λ̄f = 0, λ ∈ C\R,
f_{+} = f_{−}v.

Hence

Ψ_{0} = g̃f, g̃ ∈ D_{r}.

• By analogy,

∃f̃ ∈ D_{c}, f̃_{+} = ṽf̃_{−},

and

Ψ_{0} = f̃g, g ∈ D_{r}.

• Since

L_{λ}Ψ_{0} = A_{0}Ψ_{0}, Ψ_{0} = g̃f = f̃g.

We conclude

L_{λ}f = A_{0,c}f, f ∈ D_{c}  (12)
L_{λ}g = A_{0,r}g, g ∈ D_{r}  (13)

• (Induction) If Z(g) = {z}, then:

Using the solvability of the Cauchy problem with pure. conti. data, ∃F,

L_{λ}F = A_{c}F, M_{λ}F = B_{c}F, F (x, y, 0, λ) = f(x, y, λ), A_{c}(x, y, 0) = A_{0,c}(x, y).

Using the solvability of the Cauchy problem with pure. discrete data, ∃G,

L_{λ}G = A_{r}G, M_{λ}G = B_{r}G, G(x, y, 0, λ) = g(x, y, λ), A_{r}(x, y, 0) = A_{0,r}(x, y).

• Using the Backlund transformation, ∃! F̃, G̃ s.t.

Ψ = F̃G = G̃F

and

L_{λ}Ψ = AΨ, M_{λ}Ψ = BΨ,
Ψ(x, y, 0, λ)
=F̃(x, y, 0, λ)g(x, y, λ) = G̃(x, y, 0, λ)f(x, y, λ)
=f̃(x, y, λ)g(x, y, λ) = g̃(x, y, λ)f(x, y, λ)
=Ψ_{0}(x, y, λ).

• (Induction) If Z(g) = {z_{1}, ⋯, z_{k}}, k > 1, then by the loop group factorization theorem, we have

Ψ′_{0} = f̃_{k−1,0}g′′_{0} = g̃′′_{0}f_{k−1,0},
Z(g′′_{0}) = {α}, α ∉ Z(f̃_{k−1,0}),
Z(g̃′′_{0}) = {β}, β ∉ Z(f_{k−1,0}).

Where

f̃_{k−1,0} = f̃_{0}g′_{0,1} ⋯ g′_{0,k−1},
f_{k−1,0} = g̃′_{0,2} ⋯ g̃′_{0,k}f_{0}.

Using the induction hypo.(solv. of the Cauchy problem with mixed scattering data with distinct k−1 poles), and the solv. of the Cauchy problem with pure. discrete data, we obatin G′′(x, y, t, λ), F_{k−1}(x, y, t, λ) ......

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