1(=t1 First derivative continuity: 2(C-A (6.12) t+2-t 2(B-C 2B (6.13 t+2-t hence, t2+1-t;t+2-t+1 2-t1+1 (6.14 1 4+2-1+ (6.15) t2+3-t where A, B= functions(C) Need one more condition(normalization) Nik(a)d We obtain A(t1+2-t)+B(t+3-t+1)+C(t+2-t+1)=t+3-t (6.16 From Equation 6.14 to 6.16, we obtain 4B1 t+1-t +3-1+2 t2+3-t1+1 Finall N23(u)=v1(u)N21()+y()M+11(x)+y()N+2(u) t t2+3 (6.18)at s = 0, s = 1(u = ti+1, u = ti+2). First derivative continuity: 2(C − A) 1 ti+2 − ti+1 = 2A 1 ti+1 − ti (6.12) 2(B − C) 1 ti+2 − ti+1 = −2B 1 ti+3 − ti+2 (6.13) hence, A " 1 ti+1 − ti + 1 ti+2 − ti+1 # = C 1 ti+2 − ti+1 (6.14) B " 1 ti+2 − ti+1 + 1 ti+3 − ti+2 # = C 1 ti+2 − ti+1 (6.15) where A, B = functions(C) Need one more condition (normalization): Z ti+k or +∞ ti or −∞ Ni,k(u)du = 1 k (ti+k − ti) We obtain A(ti+2 − ti) + B(ti+3 − ti+1) + C(ti+2 − ti+1) = ti+3 − ti (6.16) From Equation 6.14 to 6.16, we obtain A = ti+1 − ti ti+2 − ti , B = ti+3 − ti+2 ti+3 − ti+1 , C = 1 (6.17) Finally, Ni,3(u) = y1(u)Ni,1(u) + y2(u)Ni+1,1(u) + y3(u)Ni+2,1(u) = u − ti ti+2 − ti Ni,2(u) + ti+3 − u ti+3 − ti+1 Ni+1,2(u) (6.18) 6