HANDBOOK OFINTEGRAL EQUATIONS phần 3 potx

HANDBOOK OFINTEGRAL EQUATIONS phần 3 potx

HANDBOOK OFINTEGRAL EQUATIONS phần 3 potx

... then u 1 (x)=ξ 1/2 J 1 /3  2 3 √ Aλ ξ 3/ 2  , u 2 (x)=ξ 1/2 Y 1 /3  2 3 √ Aλ ξ 3/ 2  , W =3/ π, ξ = x +(λ/A); if Aλ < 0, then u 1 (x)=ξ 1/2 I 1 /3  2 3 √ –Aλ ξ 3/ 2  , u 2 (x)=ξ 1/2 K 1 /3  2 3 √ –Aλ ξ 3/ 2  , W ... f(x). This is a special case of equation 2.9.62 with K(x)=A sinh  λ √ –x  . 2 .3- 3. Kernels Containing Hyperbolic Tangent 30 . y(x) – A  x a tanh(...

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HANDBOOK OFINTEGRAL EQUATIONS phần 9 potx

HANDBOOK OFINTEGRAL EQUATIONS phần 9 potx

... x, y 2 (x)=  x 0 1 + arctan 2 t 1+t 2 dt = arctan x + 1 3 arctan 3 x, y 3 (x)=  x 0 1 + arctan t + 1 3 arctan 3 t 1+t 2 dt = arctan x + 1 3 arctan 3 x + 2 3 5 arctan 5 x + 1 7⋅9 arctan 7 x. On continuing ... Press LLC 13. 3. Complete Singular Integral Equations Solvable in a Closed Form In contrast with characteristic equations and their transposed equations, complete s...

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HANDBOOK OFINTEGRAL EQUATIONS phần 10 pptx

HANDBOOK OFINTEGRAL EQUATIONS phần 10 pptx

... sin(bx)+b 2 cos(bx)  33 1 p 3 + a 3 1 3a 2 e –ax – 1 3a 2 e ax/2  cos(kx) – √ 3 sin(kx)  , k = 1 2 a √ 3 34 p p 3 + a 3 – 1 3a e –ax + 1 3a e ax/2  cos(kx)+ √ 3 sin(kx)  , k = 1 2 a √ 3 Page 722 © ... – 1 2  x 2 2 – a b x  . 13.  x 2 ln(a + bx) dx = 1 3  x 3 – a 3 b 3  ln(a + bx) – 1 3  x 3 3 – ax 2 2b + a 2 x b 2  . 14.  ln xdx (a + bx) 2 = – ln x...

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HANDBOOK OFINTEGRAL EQUATIONS phần 8 docx

HANDBOOK OFINTEGRAL EQUATIONS phần 8 docx

... unknowns; to the fourth decimal place, the solution is A 1 = 0.9 930 , A 2 = –0.0 833 , A 3 = 0.0007. Hence, y(x) ≈ y 2 (x) = 1.9 930 – 0.0 833 x 2 + 0.0007 x 4 ,0≤ x ≤ 1 2 . (15) An error estimate for ... (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) dx. (14) From ( 13) and (14) w...

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HANDBOOK OFINTEGRAL EQUATIONS phần 7 pps

HANDBOOK OFINTEGRAL EQUATIONS phần 7 pps

... A 2 = A 3 = ···= A n–1 = h, h = b – a n – 1 , x i = a + h(i – 1) (i =1, , n). (3) Simpson’s rule (or prizmoidal formula): A 1 = A 2m+1 = 1 3 h, A 2 = ···= A 2m = 4 3 h, A 3 = ···= A 2m–1 = 2 3 h, h ... (33 ) The Laplace transform can be applied to construct a solution of systems of Volterra equations of the first kind and of integro-differential equations as well. • References...

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HANDBOOK OFINTEGRAL EQUATIONS phần 6 docx

HANDBOOK OFINTEGRAL EQUATIONS phần 6 docx

... case of equation 6.8 .32 with f(t, y)=A coth(βy). 35 . y(x)+A  b a cos(λx + µt) coth[βy(t)] dt = h(x). This is a special case of equation 6.8 .33 with f(t, y)=A coth(βy). 36 . y(x)+A  b a sin(λx ... f(x)t. 1 ◦ . Solutions: y 1 (t)= √ A, y 3 (t)= √ A (3t – 2), y 5 (t)= √ A (10t 2 – 12t + 3) , y 2 (t)=– √ A, y 4 (t)=– √ A (3t – 2), y 6 (t)=– √ A (10t 2 – 12t + 3) . 2 ◦ . The integral...

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HANDBOOK OFINTEGRAL EQUATIONS phần 5 ppsx

HANDBOOK OFINTEGRAL EQUATIONS phần 5 ppsx

... equation: λ 1,2 = s 1 + s 3 ±  (s 1 – s 3 ) 2 +4(b – a)s 2 2[s 1 s 3 – (b – a)s 2 ] , where s 1 =  b a g(x) dx, s 2 =  b a g(x)h(x) dx, s 3 =  b a h(x) dx. Page 30 4 © 1998 by CRC Press LLC © ... LLC 13. y(x) – λ  b a [g(x)g(t) – h(x)h(t)]y(t) dt = f (x). The characteristic values of the equation: λ 1 = s 1 – s 3 +  (s 1 + s 3 ) 2 – 4s 2 2 2(s 2 2 – s 1 s 3 ) , λ 2 = s...

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HANDBOOK OFINTEGRAL EQUATIONS phần 4 docx

HANDBOOK OFINTEGRAL EQUATIONS phần 4 docx

... f  z 1 /3  . 36 .  ∞ –∞ y(t) |x 3 – t 3 | 1–λ dt = f (x), 0 < λ <1. The transformation z = x 3 , τ = t 3 , w(τ)=τ –2 /3 y(t) leads to an equation of the form 3. 1 .34 :  ∞ –∞ w(τ) |z – τ| 1–λ dτ = F (z), F (z)=3f  z 1 /3  . 37 .  ∞ –∞ sign(x ... Marichev (19 93) . 35 .  ∞ –∞ y(t) |x 3 – t| 1–λ dt = f (x), 0 < λ <1. The substitution z = x 3 leads to an...

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HANDBOOK OFINTEGRAL EQUATIONS phần 2 ppt

HANDBOOK OFINTEGRAL EQUATIONS phần 2 ppt

... 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 – t)] f(t) dt. 52.  x a (x – t) 5/2 I 3/ 2 [λ(x ... f(x). Solution: y(x)=  2 πλ d 2 dx 2  x a cos  λ √ x – t  √ x – t f(t) dt. 64.  x a (x – t) 3/ 4 I 3/ 2  λ √ x – t  y(t) dt = f(x). Solution: y(x)= 2 3/ 2 √ πλ...

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HANDBOOK OFINTEGRAL EQUATIONS phần 1 pdf

HANDBOOK OFINTEGRAL EQUATIONS phần 1 pdf

... Functions 3. 3. Equations Whose Kernels Contain Hyperbolic Functions 3. 3-1. Kernels Containing Hyperbolic Cosine 3. 3-2. Kernels Containing Hyperbolic Sine 3. 3 -3. Kernels Containing Hyperbolic Tangent 3. 3-4. ... Degenerate Kernel 3. 8-2. Equations Containing Modulus 3. 8 -3. Equations With Difference Kernel: K(x, t)=K(x – t) 3. 8-4. Other Equations of the Form  b a K(x, t...

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