Page 574 - Handbook Of Integral Equations

Example. Consider Eq. (16) for which
√ √
–(1 + α) 2πe –x for x >0, –(1 + β) 2πe –x for x >0, 0 for x >0,
K 1 (x)= K 2 (x)= f(x)= √
0 for x <0, 0 for x <0, – 2πe x for x <0,
where α and β are real constants. In this case, K 1 (x – t) = 0 for x < t and K 2 (x – t) = 0 for x < t. Hence, the equation
under consideration has the form
x 0

y(x) – (1 + α) e –(x–t) y(t) dt – (1 + β) e –(x–t) y(t) dt =0, x >0,
0 –∞
x √

x
y(x) – (1 + β) e –(x–t) y(t) dt = – 2πe , x <0.
–∞
Let us calculate the Fourier integrals
∞ i(1 + α) i(1 + β) i u – iβ

–x iux
K 1 (u)= –(1 + α) e e dx = – , K 2 (u)= – , F(u)= , D(u)= .
0 u + i u + i u – i u – iα
The boundary condition can be rewritten in the form
u – iβ i(u + i)
–
+
Y (u)= Y (u)+ . (30)
u – iα (u – i)(u – iα)
The solution of the Riemann problem depends on the signs of α and β.
◦
1 . Let α > 0 and β > 0. In this case we have ν = Ind D(u) = 0. The left-hand side and the right-hand side of the boundary
condition contain functions that have analytic continuations to the upper and the lower half-plane, respectively. On applying
the theorem on the analytic continuation directly and the generalized Liouville theorem (Subsection 10.4-3), we see that
z – iβ i(z + i)
+
–
Y (z)=0, Y (z)+ =0.
z – iα (z – i)(z – iα)
Hence,

1 ∞ i(u + i)
y +(x)=0, y(x)= –y –(x)= √ e –iux du.
2π –∞ (u – i)(u – iβ)
On calculating the last integral, under the assumption that β ≠ 1, by the Cauchy residue theorem (see Subsections 7.1-4
and 7.1-5) we obtain

 0 √ for x >0,
y(x)= 2π x βx
 – [2e – (1 + β)e ] for x <0.
1 – β
In the case β = 1,wehave
" 0 for x >0,
y(x)= √ x
– 2πe (1+2x) for x <0.
–
–1
+
2 . Let α < 0 and β < 0. Here we again have ν =0, X (z)=(z – iβ)(z – iα) , and X (z) = 1. On grouping the
◦
terms containing the boundary values of functions that are analytic in each of the half-planes and then applying the analytic
continuation theorem and the generalized Liouville theorem (Subsection 10.4-3), we see that
+
–
Y (z) β +1 1 Y (z) 2 1
+ = + =0.
+
–
X (z) i(β – 1) z – iβ X (z) i(β – 1) z – i
Hence,
β +1 i 2i 1
+
–
Y (z)= , Y (z)= ,
β – 1 z – iα β – 1 z – i
√ β +1

 2π e αx for x >0,


1 ∞ + – –iux  β – 1
y(x)= √ Y (u) – Y (u) e du = √
2π –∞  2 2π x

 e for x <0.
β – 1
3 . Let α < 0 and β > 0. In this case we have ν = 1. Let us rewrite the boundary condition (30) in the form
◦
i(1 + α) 1 u – iβ 2i 1
–
+
Y (u)+ = Y (u) – .
1 – α u – iα u – iα 1 – α u – i
On applying the analytic continuation theorem and the generalized Liouville theorem (Subsection 10.4-3), we see that
i(1 + α) 1 z – iβ 2i 1 C
+
–
Y (z)+ = Y (z) – = .
1 – α z – iα z – iα 1 – α z – i z – iα
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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