Masujima M. Applied Mathematical Methods in Theoretical Physics

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196 6 Singular Integral Equations of the Cauchy Type

hence we obtain the homogeneous solution to be

H (z) = G(z)exp





b − z

a − z



= G(z)



b − z

a − z

. (6.1.21)

Suppose we choose

G(z) = 1(b − z),

so that

H (z) = 1



(b − z)(a − z). (6.1.22)

Therefore,

1H(z) =



(b − z)(a − z)

does not blow up at z = a or b. So, it is a good choice, since it does not make

f (x)H

(x)blowupanyfasterthanf (x)itself.HencewetakeEq.(6.1.22)forH(z).

Take that branch of H(z) f or which on the upper bank of the cut, we have

(x) = 1i



(b − x)(x − a),

where we have now the usual real square root. To do this, we set

z = a + r

iθ

and z = b + r

iθ

so that



(b − z)(z − a) =

√

exp(i

+ θ

On x ∈ [a, b], we have

√



(b − x)(x − a) which requires on the upper bank,

exp(i

+ θ

) = i.

We take on the upper bank θ

= 0andθ

= π.Thuswehave

(x) = 1i



(b − x)(x − a)andH

−

(x) =−1i



(b − x)(x − a ).

(6.1.23)

Step 2. Now look at the inhomogeneous problem:



(x) =−

−

(x) + f (x). (6.1.24)

6.2 Cauchy Integral Equation of the First Kind 197

Seek for the solution of the form



(

)

= H

(

)

(

)

. (6.1.25)

Then we have

(x)K

(x) =−H

−

(x)K

−

(x) + f (x) = H

(x)K

−

(x) + f (x),

where Eq. (6.1.19) has been used. Dividing both sides of the above equation by

(x), we have

(x) − K

−

(x) = f (x)H

(x) = i



(b − x)(x − a)f (x). (6.1.26)

By applying the discontinuity formula, we obtain

K(z) =

2πi





(b − x)(x − a)f (x)

x − z

dx + g(z), (6.1.27)

where g(z) is an arbitrary function with no cut on [a, b]. Thus the ﬁnal solution is

given by

(z) = H(z)K(z)



(b − z)(a − z)



2π





(b − x)(x − a)f (x)

x − z

dx + g(z)



or equivalently



(

)

= H (z)



2πi



f (x)

(x)(x − z)

dx + g(z)



. (6.1.28)

6.2

Cauchy Integral Equation of the First Kind

Consider now the inhomogeneous Cauchy integral equation of the ﬁrst kind:

πi



φ(y)

y − x

dy = f (x), a < x < b . (6.2.1)

Deﬁne



(

)

≡

2πi



φ(y)

y − z

dy for z not on [a, b]. (6.2.2)

198 6 Singular Integral Equations of the Cauchy Type

As z approaches the branch cut from above and below, we ﬁnd



(x) =

2πi



φ(y)

y − x

dy +

φ(x), (6.2.3a)



−

(x) =

2πi



φ(y)

y − x

dy −

φ(x). (6.2.3b)

Adding (6.2.3a) and (6.2.3b) and subtracting (6.2.3b) from (6.2.3a) results in



(x) + 

−

(x) =

πi



φ(y)

y − x

dy = f (x), (6.2.4)



(x) − 

−

(x) = φ(x). (6.2.5)

Our strategy is the following. Solve Eq. (6.2.4) (which is the inhomogeneous Hilbert

problem) to ﬁnd (z). Then use Eq. (6.2.5) to obtain φ(x). We remark that (z)

deﬁned above is analytic in C − [a, b], so it can have no other singularities. Also

note that it behaves as Az as |z|→∞.

Solution.



(x) =−

−

(x) + f (x), a < x < b.

This is the inhomogeneous Hilbert problem that we solved in Section 6.2. We know

(z) = H(z)



2πi



f (x)

(x)(x − z)

dx + g(z)



. (6.2.6)

However, we can say something about the form of g(z) in this case by examining

the behavior of (z)as|z|→∞. By our original deﬁnition of (z), we have

(z) ∼ Az as

→∞, (6.2.7a)

with

A =−

2πi



φ(y)dy. (6.2.8)

If we examine the above solution, upon noting that

H (z) ∼ 1z as

→∞,

weﬁndthatithastheasymptoticform

(z) ∼ O(1z

) + g(z)z as

→∞. (6.2.7b)

6.2 Cauchy Integral Equation of the First Kind 199

Now, since (z) is analytic everywhere away from the branch cut [a, b], we conclude

that g(z)H(z) must also be analytic away from the cut [a, b]. Other singularities

of g(z)on[a, b] can also be excluded. Hence g(z) must be entire. Comparing

the asymptotic forms (6.2.7a) and (6.2.7b), we conclude g(z) → A as |z|→∞.

Therefore, by Liouville’s theorem, we must have g(z) = A identically. Therefore,

we have

(z) = H(z)



2πi



f (x)

(x)(x − z)

dx + A



. (6.2.9)

Thus we have



(x) = H

(x)



2πi



f (y)

(y)(y − x)

dy +

f (x)

(x)

+ A



, (6.2.10a)



−

(x) =−H

(x)



2πi



f (y)

(y)(y − x)

dy −

f (x)

(x)

+ A



. (6.2.10b)

Hence φ(x)isgivenby

(

)

= 

(x) − 

−

(x) = H

(x)

πi



f (y)

(y)(y − x)

dy + 2AH

(x).

(6.2.11)

Note that

(x) = 1



(b − x)(a − x) = 1i



(b − x)(x − a). (6.2.12)

The second term on the right-hand side of φ(x) turns out to be the solution of the

homogeneous problem. So A is arbitrary. In order to understand this point more

completely, it makes sense to examine the homogeneous problem separately.

Consider the homogeneous Cauchy integral equation of the ﬁrst kind:

πi



φ(y)

y − x

dy = 0. (6.2.13)

Deﬁne



(

)

≡

2πi



φ(y)

y − z

dy.

(z) is analytic everywhere away from [a, b], and it has the asymptotic behavior



(

)

∼ Az as

→∞,

with

A =−

2πi



φ(y)dy. (6.2.14)

200 6 Singular Integral Equations of the Cauchy Type

We have



(x) =

2πi



φ(y)

y − x

dy +

(

)



−

(

)

2πi



φ(y)

y − x

dy −

(

)

From these, we obtain



(

)

+ 

−

(

)

= 0, (6.2.15 a)



(x) − 

−

(x) = φ(x). (6.2.15b)

Solve



(x) =−

−

(x),

which is the homogeneous Hilbert problem whose solution (z)wealreadyknow

from Section 6.2. Namely, we have



(

)

= G(z)



b − z

a − z

, (6.2.16)

with G(z)havingnobranchcutson[a, b]. But in this case, we can determine the

form of G(z).

(i) We know that (z)isanalyticawayfrom[a, b], so G(z)canonlyhave

singularities on [a, b], but has no branch cuts there.

(ii) We know



(

)

∼ Az as

→∞,

so G(z) must also behave as

G(z) ∼ Az as

→∞,

taking that branch of



(b − z)(a − z)whichgoesto+1as|z|→∞.

(iii) (z) can only be singular at the end points and then not as bad as a pole,

since it is of the form

2πi



φ(y)

y − z

dy.

So, G(z) may only be singular at the end points a or b.

6.3 Cauchy Integral Equation of the Second Kind 201

The last condition (iii), together with the asymptotic form of G(z)as|z|→∞,

suggests functions of the form

(a − z)

(b − z)

n+1

(b − z)

(a − z)

n+1

The only choice for which the resulting (z) does not blow up as bad as a pole at

the two end points is

G(z) = A(b − z)withA = arbitrary constant. (6.2.17)

Therefore, the form of (z) is d etermined to be



(

)

= A



(b − z)(a − z). (6.2.18)

Then the homogeneous solution is given by

(

)

= 

(x) − 

−

(x) = 2

(x) = 2A



(b − x)(a − x ) = 2AH

(x),

(6.2.19)

where H

(x) is given by Eq. (6.2.12). To verify that any A works, recall Eq. (6.2.14),

A =−

2πi



φ(y)dy.

Substituting φ(y) just obtained, Eq. (6.2.19), into the above expression for A,we

have

A =−

2πi





(b − y)(a − y)





(b − y)(y − a)



−1

dξ



1 − ξ

= A,

which is true for any A.

6.3

Cauchy Integral Equation of the Second Kind

We now consider the inhomogeneous Cauchy integral equation of the second kind,

φ(x) = f (x) +



φ(y)

y − x

dy,0< x < 1. (6.3.1)

Deﬁne (z)

(z) ≡

2πi



φ(y)

y − z

dy. (6.3.2)

202 6 Singular Integral Equations of the Cauchy Type

The boundary values of (z)asz approaches the cut [0, 1] from above and below

are given, respectively, by



(x) =

2πi



φ(y)

y − x

dy +

φ(x), 0 < x < 1,



−

(x) =

2πi



φ(y)

y − x

dy −

φ(x), 0 < x < 1,

so that



(x) − 

−

(x) = φ(x), 0 < x < 1, (6.3.3)



(x) + 

−

(x) =

πi



φ(y)

y − x

dy,0< x < 1. (6.3.4)

Equation (6.3.1) now reads



(x) − 

−

(x) = f (x) + iλ(

(x) + 

−

(x)),



(x) −

1 + iλ

1 − iλ



−

(x) =

1 − iλ

f (x). (6.3.5)

Werecognizethisastheinhomogeneous Hilbert problem.

Consider the case λ>0andsetλ equal to

λ = tan πγ,0<γ ≤

. (6.3.6)

Then we have

1 + iλ

1 − iλ

= e

2πiγ

First, we solve a homogeneous problem:

(x) = H

−

(x)e

2πiγ

,0< x < 1. (6.3.7)

In terms of h(z)deﬁnedby

H (z) = e

h(z)

we have

(x) − h

−

(x) = 2π iγ.

6.3 Cauchy Integral Equation of the Second Kind 203

Hence we obtain

h(z) =

2πi



2πiγ

y − z

dz = γ ln



1 − z

−z



and

H (z) =



1 − z

−z



,0<γ ≤ 1 2. (6.3.8)

Since we can add any function with no cut on [0, 1] to h(z), we can multiply any

function with no cut on [0, 1] onto H(z). By multiplying the above equation by

iπγ

(1 − z), we choose

H (z) = 1[(1 − z)

1−γ

], 0 <γ ≤ 12. (6.3.9)

Returning to the inhomogeneous Hilbert problem, Eq. (6.3.5), we write

(z) = H(z)G(z). (6.3.10)

Then Eq. (6.3.5) reads

(x)G

(x) − H

(x)G

−

(x) =

1 − iλ

f (x).

Dividing through the above equation by H

(x), we have

(x) − G

−

(x) =

1 − iλ

f (x)H

(x) =

1 − iλ

f (x)(1 − x)

1−γ

. (6.3.11)

Since, for our choice of H(z), Eq. (6.3.9), we have

1H

(0) = 1H

(1) = 0,

we did not bring in extra singular behavior at x = 0 and 1; rather we made

f (x)H

(x) better behaved than f (x)atx = 0 and 1. By the discontinuity formula,

we have

G(z) =

2πi

1 − iλ



[f (y)(1 − y)

1−γ

(y − z)]dy + g(z). (6.3.12)

The integral on the right-hand side of Eq. (6.3.12) takes care of the discontinuity of

G(z) across the cut on [0, 1] so that g(z) does not have any cut on [0, 1]. We know

furthermore that

204 6 Singular Integral Equations of the Cauchy Type

(1) G(z) is analytic in the cut z plane.

(2) As z → 0, G(z)isboundedby



G(z)



· z



and similarly for z → 1.

(3) As |z|→∞, G(z) is bounded by constant. Thus we know from Eq. (6.3.12)

that g(z) is analytic everywhere and

g(z) ∼ constant as

→∞.

By Liouville’s theorem, we obtain

g(z) = constant = k. (6.3.13)

Henceweobtain

(z) =

(1 − z)

1−γ

2πi

1 − iλ



(1 − y )

1−γ

y − z

f (y)dy +

(1 − z)

1−γ

(6.3.14)

Then by choosing the branch of H(z) as indicated in Figure 6.1, we have from the

Plemelj formula



(x) =

2πi

1 − iλ

(1 − x)

1−γ



(1 − y )

1−γ

y − x

f (y)dy +

1 − iλ

f (x)

(1 − x)

1−γ

, (6.3.15a)



−

(x) =

2πi

1 − iλ

−2πiγ

(1 − x)

1−γ



(1 − y )

1−γ

y − x

f (y)dy −

−2πiγ

1 − iλ

f (x)

−2πiγ

(1 − x)

1−γ

. (6.3.15b)

By Eq. (6.3.3),

φ(x) = 

(x) − 

−

(x)

2πi

1 − e

−2πiγ

1 − iλ

(1 − x)

1−γ



(1 − y )

1−γ

y − x

f (y)dy

1 + e

−2πiγ

1 − iλ

f (x) +

(1 − x)

1−γ

. (6.3.16)

6.4 Carleman Integral Equation 205

Fig. 6.1 The branch of H(z) chosen for Eqs. (6.3.15a) and (6.3.15b).

Sincewehave

1 − e

−2πiγ

2iλ

1 + iλ

,1+ e

−2πiγ

1 + iλ

we ﬁnally obtain

φ(x) =

1 + λ

(1 − x)

1−γ



(1 − y )

1−γ

y − x

f (y)dy

1 + λ

f (x) +

(1 − x)

1−γ

, (6.3.17)

where c is an arbitrary constant. We note that solution (6.3.17) involves one

arbitrary constant and the spectrum of the eigenvalue λ of Eq. (6.3.1) with f (x) ≡ 0

is continuous. It is noted that the homogeneous solution of Eq. (6.3.1) with f (x) ≡ 0

comes from the entire function g(z).

6.4

Carleman Integral Equation

Consider the inhomogeneous Carleman integral equation

a(x )φ(x) = f (x) + λP



−1

φ(y)

y − x

dy, − 1 < x < 1, (6.4.1)

which has a Cauchy kernel and is a generalization of the Cauchy integral equation

of the second kind. Assume that λ is real and that f (x)anda(x) are prescribed real

functions. Without loss of generality, we can take λ>0(otherwise,wejustchange

the sign of a(x)andf (x)).