78674566012860119002187 3503708621552410951680 x 4 + 7041702063821037891364399 74453808207988732723200 x 5 + 549895593575959030863563 9306726025998591590400 x 6 33087191796820586109271 141624091699978567680 x 7

392978766117540050705 5149966970908311552 x 8 + 2456170645710279487 7376767051530240 x 9 + 41780926473659473181 811444375668326400 x 10 306846492148853223323 998290120065638400 x 11 17826687959696195557 1331053493420851200 x 12 + 50400473203539207979 254466109036339200 x 13 549653451307296101 80966489238835200 x 14 104497168922928132917 1113289227033984000 x 15 + 4425289676013881327 508932218072678400 x 16 + 17404262491512559 511943651315712 x 17

23600088137477575 4991450600328192 x 18 20901216917575625 2155399122868992 x 19 + 682160726359375 391890749612544 x 20 + 5546125344453125 2514632310013824 x 21 96575453125000 216101214141813 x 22 644336083984375 1420093692931914 x 23

+ 3135924072265625 22721499086910624 x 24 + 212506103515625 5680374771727656 x 25 17364501953125 18461218008114882 x 26 209045410156250 11867925862359567 x 27 + 1125335693359375 193842789085206261 x 28 + 1907348632812500 1873813627823660523 x 29

This series has the closed form as m

u E x a c t ( x ) = { p 3 ( x ) , x < 1 2 p 4 ( x ) , x 1 2 (3.13)

where

p 3 ( x ) = 307 399 57000 e 1 2 x sin ( 399 2 x ) 51 1000 e 1 2 x cos ( 399 2 x ) + 51 1000 1 10 x ,

p 4 ( x ) = 307 399 57000 e 1 2 x sin ( 399 2 x ) 51 1000 e 1 2 x cos ( 399 2 x ) 51 1000 + 1 10 x + 1 500 e 1 2 x + 1 4 cos ( 399 4 ( 2 x 1 ) ) 199 399 199500 e 1 2 x + 1 4 sin ( 399 4 ( 2 x 1 ) ) ,

which is exactly the exact solution for the case 3.1.1.

In Table 1 show a comparison of the numerical results applying the HAM ( m = 15 ), Iteration of the Integral Equation (IIE) (3.9), and the numerical solution of (3.9) with Simpson rule (SIMP) with the exact solution (3.13). Twenty points have been used in the Simpson rule. In Table 2 we list the Maximum Absolute Error (MAE), 2 , the Maximum Relative Error (MRE), the Maximum Residual Error (MRR), obtained by the HAM with the exact solution (3.13) on the interval [ 0 , 1 ] . The Estimated Order of Convergence (EOC) for different values of the constant k are given in Table 3.

Figure 1 represents both the exact solution u E x a c t ( x ) and our approximation by HAM ( m = 14 ) within the interval 0 t 1 .

For k 13 , the application of the HAM requires approximants of order m > 15 if we want to arrive beyond the discontinuity (at x = 1 2 ).

Case 3.1.2 Taking β = 1 , k = 1 , λ = 1 and

f ( x , u , u ) = H ( x 1 ) = { 0 , if x < 1 1 , if x 1 , (3.14)

The Heaviside step function at x = 1 . We now successively obtain

Figure 1. Continuous line: u E x a c t ( x ) , + : HAM , λ = 10 , k = 10 .

Table 1. Numerical results for the case 3.1.1.

Table 2. MAE, 2 , MRE and MRR for the case 3.1.1.

Table 3. EOC for the case 3.1.1.

u 0 ( x ) = x ,

u 1 ( x ) = { 1 6 h x 3 + 1 2 h x 2 , x < 1 1 6 h x 3 + h x 1 2 h , x 1 ,

u 2 ( x ) = { 1 120 h 2 ) 5 h H ( x 1 ) ( x 2 2 x + 1 ) ,

u 2 ( x ) = 1 2520 h H ( x 1 ) ( 1260 h x 5 3150 h x 4 + 18900 h x 2 31500 h x + 12600 x 2 + 14490 h 25200 x + 12600 ) + 1 2520 h ( 10 h x 7 + 35 h x 6 + 21 h x 5 + 210 h x 4 + 420 h x 3 + 210 x 4 + 420 x 3 ) ,

and so on, in this manner the rest of the iterations can be obtained. Thus, the approximate solution in a series form when h = 1 is

u ( x ) = u 0 ( x ) + m = 1 8 u m ( x ) = H ( x 1 ) ( 41629462050061 11263435223040 2498253046769 563171761152 x 18764031409 4704860160 x 2 + 1761739129 274450176 x 3 502232959 181621440 x 4 + 1677121153 242161920 x 5 10243801115 581188608 x 6 + 341190523 13453440 x 7 150931763 6209280 x 8 + 11334138401 670602240 x 9 2673838147 304819200 x 10 + 2650717003 79833600 x 11 1572737 1905120 x 12 + 88586453 1089728640 x 13 + 213052381 8717829120 x 14

6247867 605404800 x 15 + 71250341 58118860800 x 16 + 48109 823350528 x 17 1322521 59281238016 x 18 + 74645 140792940288 x 19 + 157891 1126343522304 x 20 ) + ( x 1 6 x 3 1 12 x 4 + 1 120 x 5 + 1 72 x 6 + 19 5040 x 7 1 960 x 8 299 362880 x 9 43 362880 x 10 + 3239 39916800 x 11 + 869 21772800 x 12

+ 8201 6227020800 x 13 53 10644480 x 14 2150341 1307674368000 x 15 + 1660739 10461394944000 x 16 + 20675 79041650688 x 17 + 11819 209227898880 x 18 12799 750895681536 x 19 743 56458321920 x 20 803 252975550464 x 21 73 252975550464 x 22 ) (3.24)

In Table 10 we list the MAE, the MRE, and the MRR, obtained by the HAM with the numeric solution (rkf45) u N ( x ) on the interval [ 0 , 2 ] .

Figure 5 represents both the numeric solution (rkf45) u N ( x ) with a very small error and our approximation by HAM ( m = 8 ) within the interval 0 t 2 .

Figure 5. Continuous line: u N ( x ) , + : HAM , λ = 10 , k = 1 .

Table 10. MAE, MRE and MRR for the case 3.2.2.

4. Conclusions

In this work, the HAM has been successfully applied to solve IVPs of second order with discontinuities. The size of the jump (given by λ ) does not affect the convergence of the method, which behaves equally well on both sides of the discontinuity. In this IVPs, the application by the HAM with k, does not converge even for small values of the parameter like λ .

The proposed scheme of the HAM has been applied directly without any need for transformation formulae or restrictive assumptions. The solution process by the HAM is compatible with the method in the literature providing analytical approximation such as ADM. The approach of the HAM has been tested by employing the method to obtain approximate-exact solutions of the linear case. The results obtained in all cases demonstrate the reliability and the efficiency of this method.

Cite this paper
Al-Hayani, W. and Fahad, R. (2019) Homotopy Analysis Method for Solving Initial Value Problems of Second Order with Discontinuities. Applied Mathematics, 10, 419-434. doi: 10.4236/am.2019.106030.
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