Question

0.19 0.. 1. (Natural Splines) Find the natural cubic spline S(x) satisfying S(0) = 0, S(1/2) = 1, S(1) = 0. Your answer will be 2 cubic polynomials, S.(x), S1(x). Verify that your answer satisfies all the necessary conditions (interpolation, continuity of 1st and 2nd derivatives, boundary conditions).

Answer 1

Let the two cubic polynomials be So(x) = 20x3 + box? + Cox + do, x € [0,1/2] and S1(x) = a1x3 + b1.c? +212 +d1, x € [1/2, 1] Let us try to find the former first. The naturality of spline condition at x = 0 gives $0) = 0 (690x + 2b0].x=0 =0) = bo = 0 The boundary conditions give: S. (0) = 0 = do= 0 bo So (1/2) =1 = 20 8 + + CO 2 1300 + 40 = 8 4 Hence the first cubic now is characterised by one unknown: So(x) = (8 – 4c0)x3 + cox At this juncture, we could proceed to go on finding similar relations for the unknown coefficients of the second cubic. But we note that we need not do so, as x=1/2 being the midpoint of the [0,1] suggests that the two polynomials are symmetric about the line y=1/2: S1(x) = So (1 – x) Let us assume this to be consistent for now (this is completely OK, and we shall see why soon: it is because we can now solve for the three coefficients of S_{0} uniquely). Taking derivatives of the above equation on both sides at x=1/2, we get S: (1/2) = -S(1/2) But we also have the condition

get S (1/2) = -S (1/2) But we also have the condition S: (1/2) = S. (1/2) which forces 3 Só (1/2) = 0 [3(8 – 400 )x2 + co]!=1/2 = 0 = (8 – 4c0) + C0 = 0 which gives Co = 3 Hence, plugging back, the first cubic is S. (x) = -4x3 + 3.0 S1(x) = -4(1 – x)3 + 3(1 – x) = 4x3 – 12x2 + 9x - 1 To verify that this is indeed the constructed spline, the boundary conditions: S(0) = S(0) = (-4x3 + 3x)=0 = 0 S(1) = Si(1) = [4x3 – 12x2 + 9x – 1]x=1 = 4 – 12+9-1=0 The interpolation at x=1/2: 4 3 So (1/2) + = 1 8 2 4 12 Si (1/2) + 9 2 -1=1 = So (1/2) = S(1/2) = 1 8 4 The continuity of first derivatives at x=1/2: 12 Só (1/2) = (–12x² + 3] x=1/2 +3=0 4 12 Si(1/2) = [1222 – 24x + 9]<=1/2 24 2 +9 = 0) = S(1/2) 4 The continuity of second derivatives at x=1/2: S% (1/2) = (–24x)=1/2 = -12 S: (1/2) = [24x – 24)=1/2 = -12 = S(1/2) Naturality of spline at x=0,x=1: S"(0) = S0) = (–24x]c=0 = 0

The interpolation at x=1/2: 4 So (1/2) 3 +-=1 2 8 4 12 9 Si(1/2) + -1=1= S. (1/2) = S(1/2) = 1 8 2 The continuity of first derivatives at x=1/2: 12 Só(1/2) = (-1222 + 3)=1/2 +3=0 12 24 S[(1/2) = [1222 – 24x +9]=1/2 +9= 0) = S(1/2) 2 The continuity of second derivatives at x=1/2: S (1/2) = [-24x]c=1/2 S'(1/2) = [24x – 24],=1/2 = -12 = S (1/2) = -12 Naturality of spline at x=0,x=1: S"(0) = S(0) = [-24x)=0 = 0 S"(1) = S(1) = [24x – 24]=1 = 0
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