Let to < t1 < t2 < t3 and wo, w₁, w2, and w³ be constants. (a) S[w] = [ -t3 the stationary path w(t) of the functional [{"ʻdt w" (t)², w(to) = wo, w(t) = wr, w(tz)=wz, w(ts)=wa, is given by w(t) = co+c₁t + c2t² + c3t³ where co, €1, c2 and c3 satisfy the matrix equation Vc = = W where Wo co 1 to W1 €1 and w = 1 t₁ c = 7 W2 €2 1 t₂ 03 C3 1 t3 You may use the fact that w(t) satisfies the equation OF d OF ď² OF + = 0, მი dt ǝw dt² ǝw" where F(t, w, w', w") = w" (t)². (b) let Pi = (xi, Yi, Zi), i = 0, 3 be four points in three-dimensional space. Let paths x(t), y(t) and z(t), parametrized by t, interpolate Po, … … …, P3 at t = to,……, t3 respectively, so that (x(ti), y(ti), z(ti)) = (xi, Yi, Zi) for i = 0, … … …, 3. A spline is a choice of interpolating functions x(t), y(t) and z(t) such x, y and z are each stationary paths of the functional S[w] given in (a) above. (i) x(t) = €2,0 + €2,1t+Cz,2t² + Cz,3t³, y(t) = Cy,0 + Cy,1t + Cy,2t² + Cy,3t³, z(t) = €2,0 + cz,1t+Cz,2t² + Cz,3t³, VC = x, where (CÃ,0 €₂,0 €2,0\ xo Yo Zo C= Cz,1 Cy,l Cz,1 X1 Y1 21 and X = = Cz,2 Cy,2 Cz,2 Cz3 Cy3 Cz,3/ X2 Y2 Z2 3 Y3 Z3, (ii) For the case to = 0, t₁ = 1, t2 = 2 and t3 3 and = - Po = (0,0,0), P₁ = (1,0,0), P₂ = (1,1,0), and P3 = (1,1,1), find the interpolating functions x(t), y(t) and z(t). You may find the following matrix result useful: 1 1 0 0 0 1 0 0 0 1 1 1 1 1 2 4 8 1 3927 = -11/6 3 -3/2 1/3 1 -5/2 2 -1/2 −1/6 1/2 −1/2 1/6 02@ knew to get stationary point which minimizers is given functional, we consider associated Euler- Lagrange's equation : = = Cy,ot 裴一起()+()= = 0 зст = 0261 x(t) = a + a₁ (t-to) + a₂ (t-to) (t-t₁) + a3 (t-to) (t-t₁) (t-t>) Similarly, y Cx,0 + Cx,, •t + Cx₁₂ •t² + Cx,³t³ Cy,1 .t Cz,o + G₁₁ • t + G₁₂²±² + G₁st³ +Cyz →d (20") =0 ⇒ =0 (integrating wir.to it) اطمه dt = (again integrating) Art + A1 → dn = Aof² + Alt + A₂ → Wits = Aof ³ + AL + ² + Art + A3 (again integracting) For t = to, Xx₁ = x(to)=Cx,0 + Cx,1 to + Cx12 to² + Cx,3 to³ Yo = y(to) = Cyro + Cyll to + Cyr 2 to² + Cy, 3 to ³ 30 = 3 (to) = C₂,0 + C₂,1 •to + C₂₁₂ to ² + C₂13 to ³ Lagain integrating) Let Co = A 3 Cl = Az Cz = > = A and C3 = to where Ao, A1, A2, As are For t = ti (i.e. 1, 2, 3) in general, So we get wit Co + c₁t + c₂t² + c₂t³ arbitrary constants. xi = Cx10 + Cx, ti + Cx, 2 t₁ + Cx,3 ti Yi = Cyro + Cyll ti + Cy,zti + Cysti Zi = G₁o + C₁₁te + G₁zt² + G₁3 ti Now w(to) = Wo ⇒ Co + Cito + C₂ to² + C3 to ³ wlti) = Wo = W₁ = Co + at₁ + cati² + as t₁³ = W₁ wct₂) = W₂ Co + C₁tz +C₂t₂² + c₂+33 = W₂ Wcb3) = W3 ⇒ Co + Cits + C₂tz² + Czt = W3 For X In • matrix form, * * * = V Cx, o Cx,1 655 می کی کی می Co = हु हु हु W₂ Cx,2 -Cx, 3 LW3. for y, = vx, For 31 30 31 on on A or CE 32 Cy1o yz Cy Cy1z Cy,3] = = VG,1 4,2 C310 = √x3 Gis ⇒ VC = W where v= b₁ fox for t +ty of of C = Co and w= Wo G W₁ Cz W₂ LW3. Thus we ' get the stationary functional Co, C., C2,C3 obtained from matrix equation VC=w. path is given is Web) = Co + C₁t + C₂t² + Cst³ where Схіо Сую X Yo 30 vc = v Cxpl Cy₁' G₁ = [VX √x₂ VX3] 81 = X Cx, 2 Cy, 2 C₁z Уг 82 Cx13 Cy.3 48.31 X3 Уз 83

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