1 Piecewise Cubic Interpolation

Transcription

1 Piecewise Cubic Interpolation Typically the problem with piecewise linear interpolation is the interpolant is not differentiable as the interpolation points (it has a kinks at every interpolation point) A commonly used approach to obtain a smoother interpolation, is to instead replace the linear functions with cubic ones This means that we will have an interpolation function of the form s 0 (x) if x 0 x < x s (x) if x x < x 2 s(x) =, () s n (x) if x n x x n where s i (x) is a cubic polynomial of the form s i (x) = a i + b i (x x i ) + c i (x x i ) 2 + d i (x x i ) 3 (2) Since there are n polynomial functions with 4 coefficients each, we have 4n unknowns to solve for To satisfy the interpolation condition s(x i ) = y i requires that s i (x i ) = y i, i = 0,, n, and s n (x n ) = y n (3) Note that since s i (x i ) = a i, the first n conditions above immediately imply that a i = y i for i = 0, n Additionally we will impose that s(x) is twice continuously differentiable at the points x,, x n This means s i (x i ) = s i (x i ) = y i, i =,, n s i (x i ) = s i(x i ), i =,, n i (x i ) = i (x i ), i =,, n (4) The conditions (3) and (4) are the fundamental conditions that define a cubic spline Note that in addition to s i (x i ) = a i we also have s i(x i ) = b i and s i (x i ) = 2c i If we count up all the constraints, we see that the interpolation condition (3) imposes n + constraints, while twice continuous differentiability (4) imposes 3(n ) constraints This amounts to 4n 2 = (n + ) + 3(n ) total constraints, which is obviously 2 less than the 4n coefficients to be determined Typically this is corrected by imposing two conditions at the end points of the interval Some of the most common examples are the following Natural Spline: 0(x ) = n (x n ) = 0 This choice produces a spline with the natural property that it minimizes the total curvature of the the approximating spline (this will be discussed in more detail later)

2 2 Clamped Spline: s 0(x 0 ) = y 0 and s n (x n ) = y n Here y 0 and y n are either approximations of the derivatives of the function at that point or arbitrarily chosen to pin the slopes at a particular angle 3 Not-a-Knot Spline: 0 (x ) = (x ) and n 2(x n ) = n (x n ) Of course, in order to determine the value of s(x) for any one of these choices, one must solve a complicated linear system for the coefficients a i, b i, c i, d i, for i = 0,, n As it turns out, such a linear system can be written as a tridiagonal linear system and therefore solved in O(n) floating point operations Example: Lets consider an example Suppose we want to find the natural cubic spline that interpolates the points (0, ), (/2, ) and (, 2) The spline take the form { s 0 (x) if 0 x /2 s(x) = (5) s (x) if /2 x, where and s 0 (x) = a 0 + b 0 x + c 0 x 2 + d 0 x 3 s (x) = a + b 0 (x /2) + c 0 (x /2) 2 + d 0 (x /2) 3 From the interpolation conditions (3) and the continuity condition in (4) we have s 0 (0) = : a 0 =, s (/2) = : a =, s (/2) = 2 : a + 2 b + 4 c + 8 d = 2, s 0 (/2) = : a b c d 0 = Also the derivative conditions from (4) imply that s 0(/2) = s (/2) : b 0 + c d 0 = b, 0(/2) = (/2) : 2c 0 + 3d 0 = 2c Finally the condition of being a natural spline implies that 0(0) = 0 : c 0 = 0 (0) = 0 : 2c + 3d = 0 Now one simply need to solve the above linear system for the coefficients After doing this, one obtains s 0 (x) = 3 2 x + 0x3 and s (x) = + (x /2) + 5(x /2) 2 0(x /2) 3 2

3 Lets see how this approach works in general We will assume in what follows that the interpolation points x 0 < x < < x n (knots) are equally spaced with h = x i+ x i being the spacing between points As remarked already, the first n conditions in (3) imply that a i = y i We still need to determine the remaining 3n coefficients b i, c i and d i for i = 0,, n If we include the last condition of (3) with the continuity condition of (4), then we have s i (x i ) = y i, i =, n Using the fact that a i = y i, this can be written as y i + b i h + c i h 2 + d i h 3 = y i+, i = 0, n (6) When considering the derivative conditions in (4) we may use the fact that s i(x i ) = b i and i (x i ) = 2c i to conclude that the first derivative conditions s i (x i ) = s i(x i ), i =,, n can be written as b i + 2c i h + 3d i h 2 = b i+ i = 0, n 2 (7) while the second derivative conditions i (x i ) = i (x i ), i =,, n can be written as c i + 3d i h = c i+ i = 0, n 2 (8) If we consider the natural cubic spline condition, then we also have c 0 = 2 s 0(x 0 ) = 0 and c n = 2 s n (x n ) = 0 In this case it will make the most sense to attempt to reduce solving the linear systems (6), (7), (8) to solving for the coefficients c i, i =, n 2, first (This is because the natural cubic spline condition are on c 0 and c n If we were to consider clamped splines, then it would make more sense to solve for b i first) To begin, we solve for d i in equation (8) d i = c i+ c i (9) 3h We may also solve for b i in (6), which after using the above expression for d i gives b i = f[x i, x i+ ] c i h d i h 2 = f[x i, x i+ ] 2 3 c ih 3 c i+h (0) where we have used the divided difference notation f[x i, x i+ ] = (y i+ y i )/h Finally we can substitute the expression for b i and the expression for d i into equation (7) to obtain f[x i, x i+ ] 2 3 c ih 3 c i+h + 2c i h + (c i+ c i )h = f[x i+, x i+2 ] 2 3 c i+h 3 c i+2h Rearranging this and collecting terms gives an equation for c i, i =, n 2 3 c i c i+ + 3 c i+2 = f[x i+, x i+2 ] f[x i, x i+ ] h 3 = f[x i, x i+, x i+2 ]

4 Finally, using the fact that for the natural spline c 0 = 0 and c n = 0 gives the following linear system c f[x 0, x, x 2 ] c 2 f[x, x 2, x 3 ] = () c n 3 f[x n 3, x n 2, x n ] c n 2 f[x n 2, x n, x n ] Note that this is a tridiagonal linear system and is strictly diagonally dominant since 4 > + Therefore the matrix is positive definite and non-singular and can be solved quickly using the Thomas algorithm with O(n) flops Once you solve this system, for c i, i =, n the other coefficients b i and d i can be obtained from equations (0) and (9), each costs O(n) flops Cubic B-Splines As another approach to the computation of cubic splines, we will find useful follow an approach similar to piecewise linear interpolation or Lagrange interpolation approach, introducin a family of basis functions Our goal is to write the spline as s(x) = i a i B i (x) where B i (x) is a certain piecewise cubic spline The functions B i (x) are called cubic B- Splines Although similar to the formulas for piecewise linear interpolation and Lagrange interpolation formulas, there is an important difference The coefficients a i are not necessarily equal to the function output values y i The main benefit of this approach is that the interpolation procedure can be broken into two pieces First construct the B-splines B i (x) and then find the a i s The has the benefit that, since the construction of the B-splines will not in any way depend on the y i data, meaning, once you have computed the B-spines you do not need to compute them again for different sets of y i values Computing the B-splines We will assume that each B i (x) is piecewise cubic function of the following form 0 if x x i 2, q i 2 (x) if x i 2 x i, q i (x) if x i x i, B i (x) = q i+ (x) if x i x i+, q i+2 (x) if x i+ x i+2, 0 if x i+2 x, 4

5 where q i (x), i = 0, n are the usual cubic polynomials of the form q i (x) = A i + B i (x x i ) + C i (x x i ) 2 + D i (x x i ) 3 The functions q i will be determined by the requirement that B i (x i 2 ) = B i(x i 2 ) = B i (x i 2 ) = 0 (2) and B i (x i+2 ) = B i(x i+2 ) = B i (x i+2 ) = 0 and that B i (x) is twice continuously differentiable at x i, x i, x i+ Since we are assuming the interpolation points are equally spaced, this will imply that B i (x) is symmetric about x = x i Note that the two continuous derivatives and the symmetry automatically imply that B i(x i ) = 0 Using symmmetry, it follows that we only need to ensure equation (2) holds,that the two continuous derivatives at x i match and that B i(x i ) = 0 This is accomplished by requiring the following q i 2 (x i 2 ) = q i 2(x i 2 ) = q i 2(x i 2 ) = 0 q i 2 (x i ) = q i (x i ), q i 2(x i ) = q i (x i ), q i 2(x i ) = q i (x i ) as well as q i (x i ) = 0 From the conditions at x i 2 we can easily conclude (by integration) that q i 2 = D i 2 (x x i 2 ) 3 Additionally, from the remaining smoothness conditions at x i and the zero derivative condition at x i one finds that q i (x) = D i 2 ( h 3 + 3h 2 (x x i ) + 3h(x x i ) 2 3(x x i ) 3) Of course, since we needed to determine 8 coefficients for q i 2 and q i with only 7 constraints, we have an undetermined constant D i Typically the convention is to take B i (x i ) = 2/3 which sets D i = /6h 3 Using the symmetry of B i (x) about x = x i we automatically obtain q i+ (x) and q i+2 (x) by the formulas q i+ (x) = q i (2x i x), q i+2 (x) = q i 2 (2x i x) After doing this, we find that B i (x) is completely determined by q i 2 = 6h 3 (x x i 2) 3 q i = 6 + 2h (x x i ) + 2h 2 (x x i ) 2 2h 3 (x x i ) 3 q i+ = 6 2h (x x i+) + 2h 2 (x x i+) 2 2h 3 (x x i+) 3 q i+2 = 6h 3 (x x i+2) 3 5

6 By factoring the polynomials in the above formula, it is possible to show that ( ) x xi B i (x) = B h where 2 ( 3 x2 x ) if x <, 2 B(x) = 6 (2 x )3 if x 2, 0 if 2 x Finding the coefficients Now we write our spline as s(x) = n+ i= a i B i (x) Note that we have taken the sum over i =,, n + even though we do not have points x 3, x 2, x, x n+, x n+2, x n+3 This is because we want to sum over all B i which are non-zero on the interval [x 0, x n ] When a piece of B i (x) lies outside of the interval [x 0, x n ], we simply ignore that piece Any piece of B i (x) that lies inside [x 0, x n ] will, by the piecewise polynomial nature of B i (x), necessarily only depend on the points x i that belong to x 0,, x n The task of determining the coefficients a i is now simple once you realize that at any point x i only B i (x i ), B i (x i ) and B i+ (x i ) are non-zero (ie they are the only functions turned on at x i ) Using the fact that for each i =, n we have the exact values B i (x i ) = 6, B i(x i ) = 2 3, B i(x i+ ) = 6 We see that s(x i ) = 6 (a i + 4a i + a i+ ) and therefore, the interpolation requirement s(x i ) = y i for i = 0, n implies that a i + 4a i + a i+ = 6y i for i =, 2,, n In order to solve for a i we need to know the values of a and a n+ Again, this is where the additional endpoint conditions are used It is not difficult to show that for the natural spline, we have (x i ) = a i B i (x i ) + a i B i (x i ) + a i B i+(x i ) = h 2 (a i 2a i + a i+ ) Solving the natural spline conditions (x 0 ) = 0 and (x n ) = 0 using the above formula give a = 2a 0 a, a n+ = 2a n a n 6

7 This means when i = 0, we get a 0 = y 0 and when i = n, we get a n = y n Now the remaining linear system for a i, i =,, n can be solved by a tridiagonal system a a 2 a n 2 a n = 6y y 0 6y 2 6y n 2 6y n y n Again we find that the remaining coefficients can be computed by inverting a tridiagonal, positive definite system The main difference between this system and the tridiagonal system obtained () lies in the right-hand side In () the right-hand side involves computing n 2 2nd order divided differences, while the right-hand side of the above linear system only involves the values y i, i = 0,, n In some sense we have already done this computation in computing the B-splines B i (x) 7