Solving integral equations based on radial basis function interpolation

2017 9 31 Ð 3 Sept. 2017 Communiction on Applied Mthemtics nd Computtion Vol.31 No.3 DOI 10.3969/j.issn.1006-6330.2017.03.003 ÀÆ Þµ Å Đ Ò Î Í Ï ( ÄÇ ÏÁ ±Å Ø Ä 400044) Â ÕÉ ¹ (rdil bsis function, RBF) º Æ Þ ÊÆ ¹ ±Ý» RBF Á Á ß»Á Á Á Æ Ï ÊÆ¹ Ð ±Ý RBF ß º Þ Ø ÜÃËµ (multiqudric, MQ) ¹ Ñ Í «Ð ÒÒ ÚÀ RBF Ü» ¹ ½ «½ ºÖ Û Ú ² «½ Æ RBF ºÚ² Ã ³ «ÉÆ ºÆ Ø Ô½ Ü½ Fredholm ¼ Volterr Æ ±¹Å ßÞ ØÔ ¼ÒÒ «ßÔ Ð Æ Ï ß Ü Á Á ÕÉ ¹ (rdil bsis function, RBF) º Ø ÜÃËµ (multi-qudric, MQ); º ß 2010 Ö ß 45A05; 45G10 È¾Ö ß O241.83 ÌÇ A ÃËß 1006-6330(2017)03-0275-15 Solving integrl equtions bsed on rdil bsis function interpoltion ZHANG Huiqing, CHEN Yu, NIE Xin (The Stte Key Lbortory of Trnsmission Equipment nd System Sfety nd Electricl New Technology, Chongqing University, Chongqing 400044, Chin) Abstrct In order to cquire numericl method with high precision chrcteristic nd multi-dimensionl dptbility for solving integrl equtions, the rdil bsis function (RBF) is introduced. Specificlly, the unknown function is firstly expressed s liner combintion of RBF, then the integrl eqution is trnsformed to discrete liner or nonliner equtions through the colloction method, nd finlly, n pproximte representtion for the unknown function is obtined fter determining the weights of RBF. The multi-qudric (MQ) function is dopted due to its superior interpoltion performnce. In ddition, solving three or higher dimensionl integrl equtions re esily implemented becuse the RBF is function of distnce nd the only chnging is the distnce formul. The Guss qudrture formul, the geometric splitting nd numericl integrl methods re employed ³Ú 2014-04-26; 2014-11-11 Ð ²Æ Ç½ (51377174) ¼ ÌÄ Â ĐÊ À»Ä E-mil: zhnghuiqing@cqu.edu.cn

276 31 Ð in clculting the integrl in RBF interpoltion mtrix. Numericl exmples s Fredholm or Volterr equtions in one or two-dimension hve been crried out. The results show tht the proposed method is esily to implement, nd it hs the dvntge of high precision nd convenience. Therefore, it is new nd suitble method for solving integrl equtions. Key words liner integrl eqution; nonliner integrl eqution; rdil bsis function (RBF) interpoltion; multi-qudric (MQ); numericl integrtion method 2010 Mthemtics Subject Clssifiction 45A05; 45G10 Chinese Librry Clssifiction O241.83 0» º Ð º Ð ÂÔ Ð Ð Ï ¹ ¹ ÓÐ ¹ Â Ð ÂÔ Ë ¹ ßÙ Ï ÅÑÑ ¹ Ð Ó Úº f(p) = µ k(p, Q)F[f(Q)]dQ + g(p), P, Q Ω, (1) Ω Ú g(p) º Æ k(p, Q) ºº F[f(Q)] ºÉÅ Ö Ω º È µ ºÖ ¼ Û F[f(Q)] Ú ºÀ» À º Fredholm» Volterr ¹ Þ É Þ ¹Å Þ Â Þ Ô º Þ ¼ ÎÐÞ ¹Ð Ð ¹Å È Þ Þ ¾ Hr Ëµ Å Ó À Fredholm Ð [1] ; Hr ËµÅ Ó¼ [2]»Û¼ [3] À Ð ÌÂ Æ«Dubechie Ëµ ¾È ÞÅ Û À Volterr Ð [4] ; Đ Sinc ½ÞÅ Ó¼ Ó À [5]» À Fredholm Ð [6] ; Å± ÞÅ Ó Volterr Ð [7], Å±» ¾ÞÅ À Fredholm-Volterr Ð [8] ; Chebyshev ½ÐÞÅ Û¼À Ð [9] Đ Å± ÞÅ Û¼ Fredholm-Volterr Ð [10] ; ÆÚ [11] «Å º Û Fredholm Å» Ú Chebyshev ÆÚ Ë µ Ì ÆÚ È Ø ¹ «Ëµ ¹ Â Ô Æ Û¼ ¼ ¾ ¹ ßÁ ¾ ÑÓ ¹ÑÑÅ ¾ ¼ ÔÈ (rdil bsis function, RBF) Â Þ 1968 Hrdy ¼ Û ¹ Ô È¼ ¹ÑÑ Å ßÙ ¹ Ï Đ [12]. 1990 Kns [13] ÂÏ RBF Đ Ð Å ¹ Ó µ Þ ¼ RBF ¹ Ç RBF Å Ð ÓÝ»³ Ù Ï RBF Ð Ï Ð Å Ù [14-15] Ï Đ Å Ó¼ Û À Ð À - Ð [16] À

3 Á Å ÕÉ ¹ º Æ 277 Volterr-Fredholm-Hmmerstein Ð [17] À Ð [18-20]. Ò» Å Å Û¼ È Û Fredholm Ð [21]. Ð Ö ¾Ë Î ßÁ ĐÏ RBF ÐÞÐ Ï Ð Å RBF ¹ RBF Å À À Ð ÐÞ Ó Â ÐÆ µó¼ Û¼ ¹ ÐÞ Â 1 RBF ÆÁ RBF φ: R + R ( Û º R d ) Þ ÕÆ ß r = x x j [22]. RBF Û Ð φ(r) = r 2 lnr Å φ(r) = e βr2 Hrdy ÛÂÊ (multi-qudric, MQ) Frnke 1982 µ MQ ÐÞÞ 29 Û ¹ÐÞ ¾ Â ßÁ MQ Åº ¹ ÅÚº φ(x) = x c 2 + α 2, (2) Ú c Þ α º ¼ Ó «ÈßÂ Í Õ Ì α = βh, h º Èß β Ù ¼ RBF ¹ØÞ RBF À À¾ Ü¾ Ô ÓÀ Û RBF Üº (x m, f(x m )), m = 1, 2,, N, Ï ¹È Ú f(x) = λ i φ i (x) = λ i (x ci ) 2 + α 2, (3) f(x m ) = λ i φ i (x m ) = λ i (xm c i ) 2 + α 2, (4) Ù± Üº [Φ d ][λ] = [f d ], (5) Ú [f d ] = [f(x 1 ), f(x 2 ),, f(x N )] T º Û ¹ [λ] = [λ 1, λ 2,, λ N ] T ºÉ [Φ d ] º ¹Ù± Ù± ÆÂ Úº Φ mn = (x m c n ) 2 + α 2. Å Ú (5) ßÁ Ú x ¹ f(x) º [λ] = [Φ d ] 1 [f d ]. (6) f(x) = [Φ(x)][λ] = [Φ(x)][Φ d ] 1 [f d ]. (7)

278 31 Ð 2 RBF ÓÐ Å ØÓ Á RBF Å Ð ÅÅ RBF ÞÅ Ð ÉÅ RBF À À¾ Ü È Ð Đ ½ÐÞ Î Ù ¾ Ð ½Ð Ù± ÆÞ RBF Â Ú Ù Þ Ð RBF ØÝ À» À Ð RBF Å µ 2.1 ÙÔ RBF Ó¼À Ð Ó Úº f(x) = µ b x k(x, t)f(t)dt + g(x), x [, b], (8) Ú b x Ü º b º x, ²ÔĐ Fredholm Volterr Ð b º Í» RBF Õ ÏÉÅ f(x) Üº RBF À¾ À ¾ [,b]»ó x m À Ð Íº b λ i φ i (x m ) µ λ i k(x m, t)φ i (t)dt = g(x m ), x m [, b]. (9) Ì M Æ Ð Ù± Úº [Φ µk][λ] = [G], (10) Ú Φ mn = (x m c n ) 2 + α 2, K mn = b k(x m, t) (t c n ) 2 + α 2 dt, [G] = [g(x 1 ), g(x 2 ),, g(x M )] T, x m, c n ²ÔĐº ½» RBF ½ x m» c n Ì Ó ßÁ Æ M» Æ N Â Φ Ù±ºÐ± Ù± K mn º» Å Å Ú 1 1 h(ξ)dξ Q A q h(ξ q ). (11) q=1 µ t=(b + )/2+(b )ξ/2=p(ξ) ÈÈ Â Úº K mn = b b 2 k(x m, t)φ n (t)dt = b 2 1 1 k[x m, p(ξ)]φ n [p(ξ)]dξ Q A q k[x m, p(ξ q )]φ n [p(ξ q )]. (12) q=1

3 Á Å ÕÉ ¹ º Æ 279 2.2 ÙÔ RBF Û¼À Ð Ó Úº f(x, y)=µ k(x, y, t, s)f(s, t)dsdt + g(x, y), (x, y) Ω, (13) RBF ¹ Úº λ i φ i (x m, y m ) = µ λ i Ω Ω k(x m, y m, s, t)φ i (s, t)dsdt + g(x m, y m ), (14) Ì M Æ ½ ¹Ù± [Φ µk][λ] = [G]. È² K mn = Ω k(x m, y m, s, t)φ n (s, t)dsdt ºÛ¼ Ú ² Ù» È Â ÐÆ (i) Ù È Â ÍÙ È Ω º [,b] [c,d] Õ Đ Û¼Å Å Ú µ t=(b+)/2+(b )ξ/2=p(ξ), s=(d + c)/2+(d c)η/2=q(η) ÈÈ K mn = d b c (b )(d c) 4 k(x m, y m, s, t)φ n (s, t)dsdt n 1 n 2 A i A j k[x m, y m, p(ξ i ), q(η j )]φ n [p(ξ i ), q(η j )]. (15) j=1 (ii) È Â Í È Ω º È Õ» Å Ú» ¹Å ÐÞ ³ Þ Óµ» ¹ ÐÞ Ï È À½ Â ²Æ È Å» ¹ µ K mn = k(x m, y m, s, t)φ n (t, s)dsdt = k(x m, y m, s, t)φ n (t, s)dsdt I i, (16) Ω i Ω i i Ú Ω i Ü i Æ È Û¼ È Ë Đ ßÙ» Í È Ô i Æ Ë Đ ÛÂ ¹ Ú I i = S 3 i k(x m, y m, x icj, y icj )φ n (x icj, y icj ), (17) 3 j=1 Ú Æ (x ij,y ij )(j = 1, 2, 3) Â Ë S i Æ (x icj, y icj ). 2.3 Õ ÙÔ RBF Ó¼ À Ð Ó Úº f(x) = µ b x k(x, t)f[f(t)]dt + g(x), x [, b]. (18)

280 31 Ð b º» RBF ÏÉÅ f(x) Üº RBF À¾ λ i φ i (x) = µ b N ] k(x, t)f λ i φ i (t) dt + g(x), x [, b]. (19) [, b]»ó x m º Ú Üº λ i φ i (x m ) µ b N ] k(x m, t)f λ i φ i (t) dt g(x m ) = 0, x m [, b]. (20) Ì M Æ À Ð À Ù±Úº [Φ][λ] µ[k] [G] = 0, (21) Ë K m = [ b ] k(x m, t)f λ i φ i (t) dt = b h(t)dt Þ Î [λ] À» ÈÐÞÅ Ô È ²Ó¹ ÛÎ [λ] = [λ 1, λ 2,, λ N ] T Í ¹ Ú (21) ÆÐ Â Å K m Å Ú K m = b 2 b 2 1 1 q=1 2.4 Õ ÙÔ RBF N ] k[x m, p(ξ)]f λ i φ i (p(ξ)) dξ Q N ] A q k[x m, p(ξ)]f λ i φ i (p(ξ)). (22) Û¼ À Ð Ó Úº f(x, y) = µ k(x, y, t, s)f[f(t, s)]dtds + g(x, y), (x, y) Ω, (23) RBF ¹ Úº Ω λ i φ i (x m, y m ) = µ k(x m, y m, t, s)f Ω [ N ] λ i φ i (t, s) dtds + g(x m, y m ). (24) Ì M Æ ½ À Ð À Ù± Ú [Φ][λ] µ[k] [G] = 0, K m Â Ô Ù È Đ Û¼Å Å Úº K m (b )(d c) 4 n 1 n 2 N A i A j k[x m, y m, p(ξ i ), q(η j )]F j=1 ] λ i φ i (p (ξ i ), q(η j )). (25)

3 Á Å ÕÉ ¹ º Æ 281 Ô È» ¹Å ÐÆ ÄÙ» K m = Ω = i N ] k(x m, y m, t, s)f λ i φ i (t, s) dtds N ] k(x m, y m, t, s)f λ i φ i (t, s) dtds, (26) Ω i N ] k(x m, y m, t, s)f λ i φ i (t, s) dtds Ω i = S i 3 3 N ] k(x m, y m, x icj, y icj )F λ i φ i (x icj, y icj ). (27) j=1 2.5 RBF «ÙÔ ± ÏÉ Ï RBF Å À Ð Ó¹ ¾ (i) RBF ½Ò½ ÉÅ Û Ò½ RBF c, ¼ α ½ x m ; (ii) Î Ð À Ï RBF ¹ ÚÈ Ð Đ Í Ð À À Ð Õ Ù± Úº [Φ µk][λ] = [G], À ÕÂĐº [Φ][λ] µ[k] [G] = 0; (iii) Â Ù± Æ Ô Ó¼ Û¼Ù È Đ Å ÞÂ «ÙÔ Û¼È È Â È Å»«(iv) Â Î ÔÀ Ð Đ SVD Å ¹Ù±Ð [Φ µk][λ] = [G], Ù Ô À Ð Đ Ö ÈÞÅ [Φ][λ] µ[k] [G] = 0; (v) Â ÉÅ f(x) RBF Ú f(x) f(x) = [Φ(x)][λ]. 3 Æ ¹ Ø ¹ Ó¼Ã Fredholm Ð» Volterr Ð Û¼Ù» È Ð º ÐÞ ÑÑ» Ð ¼ Ò f e (x k )» f cl (x k ) ² Ü x k ¹»Â ¹ Ð Ûº φ RMS = 1 [f e (x k ) f cl (x k )] N 2. (28) 3.1 ÙÔ «º 1 Ó¼ Volterr À Ð f(x) + x 0 k=1 (x t)f(t)dt = 1, 0 x 1,

282 31 Ð Ð º f(x)=cos(x).» Ì RBF Èß h=0.1, ¼ β=10, º 11; ¹ Â» 60 Å Ú«ÌÈß h t =0.001 1 001 Æ¾ ÐÞ Â ³º ¾ ÂÆ Ô º 3.302 2 10 6, Ð º 2.605 0 10 6, ÂÔ ÂÆ¹º 4.713 1 10 4. Ù [15] ÌÂ ¼» 1 000 Æ¾ Ð º 3.514 16 10 5. Ë ÐÞ Ð Ì Å ÑÑ [23] Ó¼À Ð x f(x) + e xs f(s)ds = g(x), 1 g(x) = e 4x + 1 x + 4 (ex(x+4) e (x+4) ), 1 x 1, º f(x) = e 4x. Å È [ 1,1] Èß h = 1 3 Ì ¼ α=8.8, ¹ Â» 7 Å Ú ¾Èßº h t =0.2. ÐÞ Â ³º ¾ ÂÆ Ô º 2.465 8 10 1, Ð º 9.311 43 10 1, ÂÔ ÂÆ¹º 1.563, ¹Ù± Í 6.050 9 10 14. Èß h=0.25 Ì» ¼ α=4.2,» 9 Å Ú ¾Èßº h t =0.2. ÐÞ Â ³º ¾ ÂÆ Ô º 2.258 8 10 2, Ð º 9.976 9 10 3, ÂÔ ÂÆ¹º 6.963 8 10 2, ¹ Ù± Í º 4.342 4 10 14. 1 Ë ÐÞ ( ¹» 7 9 Å )» [23] ÐÞ (Legendre- Guss-Lobtto ) ÂÆ Ô Ó» 6 8 Å Æ Ú Ù RBF ¹ ÓÉ Ø ± ÏÅ¼ Í RBF N=10 Õ Ð Å 1.329 5 10 3. Í 1 [23] Ñ É Ñ Ý Ê N=6 N=8 ß 2.465 8 10 1 2.258 8 10 2 [23] ß 3.660 0 10 1 1.880 0 10 2 2 Ó¼ Fredholm À Ð 1 e (x+1)t f(t)dt = g(x), 0 g(x) = 1 0 x 1, ex+1 (x + 1) 2 + ex+1 x + 1, Ð º f(x) = x. Ì RBF Èß h=0.1, ¼ β=20, Æ º 11;» 20 Å Â ³ 2 Ü [1]» Hr ËµÐÞ Í Ñ j=4

b3b 9I ^? qq?- 4.\B ~ N 283 q }:6. 3.809 9 10, JN~p:6. 0.035 0, JNJp:6. 8.248 9 10 ; U j=6 q J. 3.627 0 10, 0.012 8 / 3.009 0 10. u 3}z[`'p. 7.327 7 10, 2.133 1 10 / 6.300 8 10. U 2 48 [1] mk Q4mk yir K do# Z T*{ (j=4) T*{ (j=6) 4~{ 4 2 5 6 2 5 5 0.1 0.1 0.094 212 996 807 86 0.102 953 271 231 79 0.099 993 699 159 086 0.2 0.2 0.216 497 767 731 13 0.193 982 016 790 95 0.199 992 714 173 504 0.3 0.3 0.278 668 297 286 35 0.302 220 067 436 13 0.299 999 166 429 188 0.4 0.4 0.404 535 603 105 57 0.396 346 237 628 25 0.400 003 345 436 062 0.5 0.5 0.467 982 147 763 49 0.491 283 482 643 49 0.500 002 711 688 808 0.6 0.6 0.595 167 200 545 31 0.602 413 447 203 73 0.599 999 256 619 412 0.7 0.7 0.721 572 548 697 16 0.697 325 532 381 44 0.699 996 680 817 208 0.8 0.8 0.783 975 300 318 31 0.806 706 792 323 44 0.799 997 531 348 708 0.9 0.9 0.905 983 819 893 75 0.898 402 237 956 90 0.900 000 440 606 391 1.0 1.0 0.965 040 447 791 85 0.987 233 083 204 06 0.999 997 601 459 995 3 w0" P H A }? Fredholm Z 1Z f (x, y) = u(x, y) + 1 (t sin(s) + 1)f (s, t)dsdt, 0 0 u(x, y) = x cos(y) sin(1)(3 + sin(1))/6, 0 6 x, y 6 1, }?[?Yb. f (x,y) = xcos(y). #3 T RBF 3X P{ h=0.1, [>0 β=35, ; /3X Æ. 121 w0 A 3 x, y }P T 20 Æ A 2 P{. h =0.05,. 441. #`'$, 2 [JN~p:6. 5.530 9 10, } :6. 2.658 3 10, Jp :6JN-. 1.607 7 10. -b[:6r(/jp:6r(`$ 1 x 3E [3] Hr S)}z (α=4, m=32) q[jn~p:6. 8.32 10. t 6 6 4 5 2 6\ ;i6\ 2 1

284 31 Ð 4 Û¼ È Volterr Ð f(x, y) (t sin(s) + 1)f(s, t)dsdt = xcos(y) 0.280 405 715 7, Ω Ω = {(x, y) R 2 : 0 y 1, 0.25 (y 0.5) 2 x } 1 4(y 0.5) 2, Ð º f(x,y) = xcos(y). Ì RBF» 2 Ü º 35, ¼ α=βh=2.5. Ω º È Đ ÍÏ º 2 272 Æ Ë» 1 225 Æ Ì Ë Åº¾ Ù± Æ Â È ¹» ÛÂ Ú ³ ¾ ÂÆ Ô º 4.011 7 10 5, Ð º 2.955 8 10 5, ÂÔ ÂÆ¹º 3.072 0 10 2. Ù [20] Ì 33 Â Õ Ð º 1.96 10 4,» Å Å Ð º 1.65 10 4. ¾ 2 RBF È ¾² 3.2 Õ Ô «º 5 Ó¼ Fredholm À Ð f(x) = x 1 Ð º f(x) = 2 x 2. Ì RBF Èß h= 1 9 0 t f(t)dt + 2 1 3 (2 2 1)x x 2, 0 x 1,, ¼ β=8, º 10; Â» 10 Å Ú«Ì¾Èß h t =0.05. Ð ÞÂ ¾ ÂÆ Ô º 5.883 1 10 6, Ð º 2.846 5 10 6, ÂÔ ÂÆ¹º 4.943 8 10 6 ; Ù [6] ÐÞ» ÐÞ ¾ Ô ³Ë 3. Ë RBF ÐÞ Ð Ì Å ÑÑ ÐÞ Å Æ

3 Á Å ÕÉ ¹ º Æ 285 Í 3 [6]» Ñ É Ñ Ý x t [6], N=10 ß 0.1 4.94 10 5 3.45 10 6 0.3 8.55 10 5 4.74 10 7 0.5 1.04 10 4 3.75 10 6 0.7 6.78 10 5 2.72 10 6 0.9 1.41 10 5 5.88 10 6 ÃÇĐÕ 2.18 10 4 5.88 10 6 6 Ó¼ Volterr À Ð f(x) = 3 2 1 2 e 2x x 0 [f 2 (t)+f(t)]dt, 0 x 1, Ð º f(x) = e x. Ì ¼ β=7,» 10 Å º» [2] ³Ä ÕÂÌ RBF Èß h= 1 3, h=1 7, h= 1 15» h= 1 31, ¾Èß h t=0.05. 4 Ì«À Æ Â ³ 3 Þ» ÐÞ ÂÆ Ô ³ Ô º» [8] ÄÔ Ì RBF Èß h = 1 5, Â ³ 5 Ü 3 4» 5 Ë Ç Æ ÅÂ ÑÑ RBF ÐÞ ÑÑ³ Í 4 ÊÎ½ Û» Ý N=4 N=8 N=16 N=32 RMS 5.02 10 4 1.06 10 5 2.40 10 6 9.89 10 7 ÃÇĐÕ 8.11 10 4 3.91 10 5 6.80 10 6 2.37 10 6 ÃÇÃÕ 1.54 10 3 4.11 10 5 1.76 10 5 4.72 10 6 ¾ 3 [2]» Ñ É Ñ Ë ¾

286 31 Ð Í 5 [8]» Ñ É Ñ Ý Ê x t [8], M=3, N=6 ß N=6 0 0 7.361 2 10 12 0.1 1.626 8 10 4 8.865 8 10 5 0.2 2.438 2 10 4 3.278 2 10 5 0.3 1.275 4 10 4 5.689 3 10 5 0.4 2.858 7 10 4 1.034 1 10 5 0.5 3.999 3 10 4 1.150 7 10 5 0.6 2.250 4 10 4 1.191 4 10 5 0.7 3.594 6 10 4 2.969 1 10 5 0.8 1.043 8 10 4 3.180 2 10 6 0.9 2.968 3 10 4 3.209 5 10 5 [23] À Ð x f(x) + e x 3s [f 2 (s)]ds = g(x), 1 1 g(x)= 2(1+36π 2 ) [(1+36π2 )e x e x cos(6πx)+6πe x sin(6πx) 36eπ 2 ]e x +e x sin(3πx), Ð º f(x) = e x sin(3πx). Å È [ 1, 1] Èß h = 1 3 Ì» «¼ α=0.5; ¹ Â ²» 7 Å Ú«¾Èßº h t =0.1; ÐÞ¾ ÂÆ Ì» ¼ α=1.3,» 9 Ô º 2.662 9 10 1. Èß h = 1 4 Å Ú ¾Èßº h t =0.1, ÐÞ¾ ÂÆ Ô º 2.024 8 10 3. Í 6 É Ñ [23] Ñ Ë Ê N=6 N=8 ß 2.662 9 10 1 2.024 8 10 3 [23] ß 2.330 0 10 2 7.220 0 10 4 6 ÐÞ ÂÆ Ô À Ð Å Õ [23] Ñ ß Å RBF À Ð ÈÐÞÐ Æ ÄÙ Â ÑÑ Å Ù» MATLAB ÑÑ Û ÜÅ Ð 7 Û¼ È Volterr À Ð y x f(x, y) = g(x, y) + (x + y t s)f 2 (t, s)dtds, 0 0 g(x, y) = x + y 1 12 xy(x3 + 4x 2 y + 4xy 2 + y 3 ), Ð º f(x, y) = x + y. 0 x, y 1, Ì RBF Èß h=0.2, ¼ β=5, Æ º 36 x, y ÐÈ Ì 10 ÆÅ ¾Èßº h t =0.1, º 121. ³ ¾

3 Á Å ÕÉ ¹ º Æ 287 ÂÆ Ô º 2.703 4 10 4, Ð º 6.684 7 10 5, ÂÔ ÂÆ¹º 1.351 7 10 3. ¹ Ô Ê»ÂÔ Ê 4 Ü [3] Hr ËµÐÞ (α = 4, m = 32) Õ ÂÆ Ô º 4.58 10 4, ÐÞ ÑÑ ¾ 4 ¾ 8 Û¼ È Volterr À Ð x f(x, y) Ω 1 + y (1 + t + s)f2 (t, s)dtds = g(x, y), 1 0.027 138 047 1x g(x, y) =, (1 + x + y) 2 1 + y Ð º f(x, y) = 1/(1 + x + y) 2. È Ω 5 Ü Ì RBF» Æ º 38, ¼ α = βh = 0.45. Đ ÍÏ È º 288 Æ Ë» 185 Æ Ó¹Ñ Ï 657 Æ Åº¾ Â Ô Ë È ¹» ÛÂ Ú ³ ¾ ÂÆ Ô º 3.755 8 10 3, Ð º 2.834 9 10 4, ÂÔ ÂÆ¹º 2.347 4 10 2. ¾ 5 RBF È ¾² 4 Ï RBF Ð Ï Ð ½ RBF ÐÞÅ Ð Ô ½ Ð À Ù± Æ Â ÐÆ Ð ÀÅ ÐÞ Ì

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