HANDBOOK OFINTEGRAL EQUATIONS phần 2 ppt

... dt, where g(t)= π 8λ 2  d 2 dt 2 – λ 2  4  t a (t – τ) 3 /2 I 3 /2 [λ(t – τ)] f(τ) dτ. 51.  x a (x – t) 3 /2 I 3 /2 [λ(x – t)]y(t) dt = f(x). Solution: y(x)= √ π 2 3 /2 λ 5 /2  d 2 dx 2 – λ 2  3  x a sinh[λ(x ... dt, where g(t)= π 8λ 2  d 2 dt 2 + λ 2  4  t a (t – τ) 3 /2 J 3 /2 [λ(t – τ)] f(τ) dτ. 11.  x a (x – t) 3 /2 J 3 /2 [λ(x – t)]y(t) dt =...

Ngày tải lên: 23/07/2014, 16:20

... dx 1 sin(ax) a p 2 + a 2 2 |sin(ax)|, a >0 a p 2 + a 2 coth  πp 2a  3 sin 2n (ax), n =1 ,2, a 2n (2n)! p  p 2 +(2a) 2  p 2 +(4a) 2   p 2 +(2na) 2  4 sin 2n+1 (ax), n =1 ,2, a 2n+1 (2n + 1)!  p 2 + ... >1. 15.  xdx cos 2n x = n–1  k=0 (2n – 2) (2n – 4) (2n – 2k +2) (2n – 1)(2n – 3) (2n – 2k +3) (2n – 2k)x sin x – cos x (2n – 2k + 1)(2n – 2k) cos 2n–...

Ngày tải lên: 23/07/2014, 16:20

... + i 2  2 0 ϕ(ξ) dξ = f (x), (27 ) ϕ(x)=– 1 2  2 0 cot  ξ – x 2  f(ξ) dξ – i 2  2 0 f(ξ) dξ. (28 ) Equation (21 ), under the additional assumption (22 ), coincides with Eq. (27 ), and hence its solution ... =2t 2 + A(t + t –1 )+α 1 (1 – t 2 )+α 2 (t – t –1 ). The corresponding Riemann boundary value problem has the form Φ + (t)=t 2 Φ – (t)+t + 1 2 A(1 + t 2 )+ 1...

Ngày tải lên: 23/07/2014, 16:20

... obtain y 2 (x)=1+  1 /2 0  1 – x 2 t 2 + 1 2 x 4 t 4  y 2 (t) dt. ( 12) Therefore, y 2 (x)=1+A 1 + A 2 x 2 + A 3 x 4 , (13) where A 1 =  1 /2 0 y 2 (x) dx, A 2 = –  1 /2 0 x 2 y 2 (x) dx, A 3 = 1 2  1 /2 0 x 4 y 2 (x) ... i) 2 F + (u) (u – α – iβ)(u + α – iβ) = 1 1+K(u) F + (u)=F + (u)+R(u)F + (u). Here R(u)=– 2u 2 (a – b)+2a +2b [u 2 – (α + iβ) 2...

Ngày tải lên: 23/07/2014, 16:20

... solution is given by formula (3), where M(ξ)= 2 √ π µ Γ  µ 2  Γ  1 – µ 2  ξ µ , w(t, ξ)= 1 π cos  πµ 2   ξ 2 – t 2  µ–1 2 . • References for Section 10 .2: N. Kh. Arutyunyan (1959), I. C. Gohberg ... (f, λ 2 ). (27 ) Taking into account (24 ) and (26 ), we can rewrite Eq. (23 ) in the form  1 – λ 1 L  1 – λ 2 L  [y]=  1 – λ 2 L  [w], or, in view of the ident...

Ngày tải lên: 23/07/2014, 16:20

... =0. Solutions: y 1 (x)=0, y 2 (x)= 2 A(e 2 b – e 2 a ) e –λx . 23 . y(x)+A  b a e –λx–µt y 2 (t) dt =0. Solutions: y 1 (x)=0, y 2 (x)= 2 + µ A[e – (2 +µ)b – e – (2 +µ)a ] e –λx . 24 . y(x)+A  b a x λ e –µt y 2 (t) ... y 2 (x)=– 2 +1 A(b 2 +1 – a 2 +1 ) x λ . 21 . y(x)+A  b a x λ t µ y 2 (t) dt =0. Solutions: y 1 (x)=0, y 2 (x)=– 2 + µ +1 A(b 2 +µ+1 – a 2 +µ+1...

Ngày tải lên: 23/07/2014, 16:20

... s 3 +  (s 1 + s 3 ) 2 – 4s 2 2 2( s 2 2 – s 1 s 3 ) , λ 2 = s 1 – s 3 –  (s 1 + s 3 ) 2 – 4s 2 2 2( s 2 2 – s 1 s 3 ) , where s 1 =  b a g 2 (x) dx, s 2 =  b a g(x)h(x) dx, s 3 =  b a h 2 (x) dx. 1 ◦ . ... equation: λ 1 ,2 = A(b – a) ±  [A(b – a)+2Bh 1 ] 2 – 2Bh 0 [A(b 2 – a 2 )+2Bh 2 ] B{h 0 [A(b 2 – a 2 )+2Bh 2 ] – 2h 1 [A(b – a)+Bh 1 ]} , where h...

Ngày tải lên: 23/07/2014, 16:20

... A 2 ), where A 1 = 12f 1 +6λ ( f 1 ∆ 2 –2f 2 ∆ 1 ) λ 2 ∆ 4 1 + 12 , A 2 = –12f 2 +2 ( 3f 2 ∆ 2 –2f 1 ∆ 3 ) λ 2 ∆ 4 1 + 12 , f 1 =  b a f(x) dx, f 2 =  b a xf(x) dx, ∆ n = b n – a n . Page 24 7 © 1998 by CRC Press ... (z)=λf(z 1/λ ). 2 ◦ . Solution with a =0: y(x)=–Ax λ(k–1) 2 d dx   b x t λ(3–2k) 2 2 dt (t λ – x λ ) 1–k 2  t 0 s λ(k+1) 2 2 f(s ) ds (t λ –...

Ngày tải lên: 23/07/2014, 16:20

... µ i =  |z i |, where B 1 and B 2 are determined from the system B 1 µ 1 λ 2 1 – µ 2 1 + B 2 µ 2 λ 2 1 – µ 2 2 – 1=0, B 1 µ 1 λ 2 2 – µ 2 1 + B 2 µ 2 λ 2 2 – µ 2 2 – 1=0. Case 3.Ifz 1 > 0 and z 2 < 0, then ... B 2 sin  µ 2 (x – t)  f(t) dt, µ i =  |z i |, where B 1 and B 2 are determined from the system B 1 µ 1 λ 2 1 + µ 2 1 + B 2 µ 2 λ 2...

Ngày tải lên: 23/07/2014, 16:20

... Kernel 12. 2. The Cauchy Type Integral 12. 2-1. Definition of the Cauchy Type Integral 12. 2 -2. The H ¨ older Condition 12. 2-3. The Principal Value of a Singular Integral 12. 2-4. Multivalued Functions 12. 2-5. ... g 1 (x)=Ax 2 , h 1 (t)=t, g 2 (x)=B, and h 2 (t)=t 3 . 21 .  x a (Axt 2 + Bt 3 )y(t) dt = f(x). This is a special case of equation 1.9.15 with g 1 (x)=Ax, h 1 (t)...

Ngày tải lên: 23/07/2014, 16:20