Cubic Spline. s(x) = s j (x) = a j + b j (x x j ) + c j (x x j ) 2 + d j (x x j ) 3, j = 0,... n 1 (1)

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1 Cubic Spline Suppose we are given a set of interpolating points (x i, y i ) for i = 0, 1, 2, n We seek to construct a piecewise cubic function s(x) that has the for x [[x j, x j+1 ] we have: s(x) = s j (x) = a j + b j (x x j ) + c j (x x j ) 2 + d j (x x j ), j = 0, n 1 (1) We also require that s(x) satisfies several properties, (i) s(x i ) = y i, or s(x) interpolates the at the x i s (ii) s(x) is continuous at x 1, x 2 x n 1 (iii) s (x) is continuous at x 1, x 2 x n 1 (iv) s (x) is continuous at x 1, x 2 x n 1 The conditions above allow one to determine the values of a j, b j, c j and d j for j = 0, n 1 (almost) The reason I say it almost allows us to compute them is that we are two conditions short Since there are n different s j (x) s and each s j (x) has four coefficients we have 4n unknowns to determine Property (i), is a condition at each of the n+1 points and Properties (ii), (iii) and (iv) are conditions at n-1 points This gives us n+1+(n 1) = 4n 2 equations Generally, you need 4n equations to solve for 4n unknowns which means we are two conditions short These two remaining conditions will be determined by some additional conditions that we will impose There are several different ways of doing this and they have different names: Boundary Conditions Natural Cubic Spline s (x 0 ) = s (x n ) = 0 Not-a-Knot s (x) is continuous at both x 1 and x n 1 Clamped If the spline is interpolating a function f(x) then one can require that s (x 0 ) = f (x 0 ) and s (x n ) = f (x n ) 1

2 Constructing the Spline To construct the spline let us first define h j = x j+1 x j or the spacings between each of the nodes and are known (or easily computable) given the x j sthe first condition (i) gives us that: s j (x j ) = y j or alternatively using (1) we have a j = y j for j = 0, 1, 2 n Note: There is some abuse of notation here as there isn t a function s n (x) but for notational purposes we let still define a n = y n We will also for notational ease allow, for notational ease to define b n, c n and d n to be s n 1(x n ), s n 1(x n ) and s n 1(x n ) However, we do not want these variables in our system when we solve at the end so they must be eliminated before solving Next we use conditions (ii) to recognize that we must have s j (x j+1 ) = y j+1 Using the definition of h his gives: y j+1 = a j + b j h j + x j h 2 j + d j h, j = 0, n 2 (2) Next, condition (iii) means that s j(x j+1 ) = s j+1(x j+1 ) If we take the derivatives of these based on (1) and evaluate at x j+1 we obtain: b j+1 = b j + 2c j h j + d j h 2 j j = 0, n 2 () Similarly, condition (iv) means that s j (x j+1 ) = s j+1(x j+1 ) If we take the derivatives of these based on (1) and evaluate at x j+1 we obtain: Using (4) we can solve for d j giving: 2c j+1 = 2c j + 6d j h j j = 0, n 2 (4) d j = c j+1 c j h j for j = 0, 1, 2, n 2 (5) Then one takes this expression for d j and substitutes it into both () and (2) This gives: and ( a j+1 = a j + b j h j + c j h 2 cj+1 c j j + h j = a j + b j h j + c j+1 + 2c j h 2 j ( b j+1 = b j + 2 cj+1 c j jh j + h j ) h j (6) ) h 2 j (7) = b j + (c j+1 + c j ) h j 2

3 If we solve this last equation (6) for b j we obtain: b j = a j+1 a j 2c j + c j+1 h j (8) h j Now take this expression for b j and substitute ist into (7) This will then leave an equation involving the c s and the a s After some algebra, one arrives at: h j 1 c j 1 + 2(h j 1 + h j )c j + h j c j+1 = h j (a j+1 a j ) h j 1 (a j a j 1 ) (9) which holds for j = 1, 2,, n 1 Assuming we know the a s then this is a system of n 1 equations for the n + 1 unknown c s This we either need to augment this system with two more relations or eliminate two of the c s With this, the general idea is as follows You are given the values {x j } n j=0 and {f j } n j=0 You can immediately find the values for the a js by setting a j = f j You construct a linear system (see below) using (9) to solve for the c s Knowing the c s and the a s you then back substitute for the b s using the boxed equation () and we can back substitute into () to solve for the d s The only issue left is that the linear system isn t a square system and it has c n in it Consequently we need to remove c n and we either need add one more equation (or remove one other variable) Not-A-Knot The not-a-knot boundary conditions require s 0 (x 1 ) = s 1 (x 1 ) and s n 1(x n 1 ) = s n 2(x n 1 ) Since these are cubic functions the third derivatives are constant Consequently, these imply that: d 0 = d 1, d n 2 = d n 1 Using the equations for the d j s given in (), we get: solving for c 0 we find: c 0 = c 1 c 0 h 0 = c 1 ( 1 + h ) 0 c 1 h 0

4 Similarly, we must have: c n = h ( n 1 c n h ) n 1 If we use these two relationships for c 0 and c n to eliminate both c 0 and c n from the equations given in (9) and we will then have a system of n 1 equations for the n 1 unknowns c 1,, We then add in the extra step of getting c 0 using the relations above prior to computing the b s and the d s The system becomes: This can be written in matrix form A c = b where, (h h2 0 h ( + h 2 ) h h 2 2(h 2 + h ) h 0 0 A = ( + ) h2 n 1 c = c 1 c c 4 c n 2 (a 2 a 1 ) h 0 (a 1 a 0 ) h 2 (a a 2 ) (a 2 a 1 ) h, b = (a 4 a ) h 2 (a a 2 ) (a n 1 a n 2 ) h n (a n 2 a n 1 ) (a n a n 1 ) (a n 1 a n 2 ) h2 n 1 (10) 4

5 Example Suppose we are given that our function must pass through the points: (1, 1), (2, 1/2), (, 1/), (4, 1/4), (5, 1/5) Each of our h j s is 1 which makes things a bit nicer First we know that Our linear system becomes c = c If we solve this matrix we obtain: a 0 = 1, a 1 = 1/2, a 2 = 1/, a = 1/4, a 4 = 1/5 (1/ 1/2) (1/2 1) (1/4 1/) (1/ 1/2) = (1/5 1/4) (1/4 1/) c 1 = 1/6, = 1/60, c = 1/60 1 1/4 1/10 We can then get c 0 and c 4 by: Now we can compute the b s c 0 = 2c 1 = 19/60 b 0 = (a 1 a 0 ) (2c 0 + c 1 )/ = (1/2 1) (2(19/60) + 1/6)/ = 2/0 b 1 = (a 2 a 1 ) (2c 1 + )/ = (1/ 1/2) (2(1/6) + 1/60)/ = 17/60 b 2 = (a a 2 ) (2 + c )/ = (1/4 1/) (2(1/60) + 1/60)/ = 1/10 b = b 2 + (c + ) = 1/10 + (1/60 + 1/60) = 1/15 Now we can compute the d s d 0 = (c 1 c 0 )/ = (1/6 19/60)/ = 1/20 d 1 = ( c 1 )/ = (1/60 1/6)/ = 1/20 d 2 = (c )/ = (1/60 1/60)/ = 0 d = d 2 = 0 Thus our spline is of the form: s 0 (x) = (x 1) + 0 (x 60 1)2 1 (x 20 1) 1 < x < 2 s 1 (x) = 1 s(x) = 17(x 2) + 1(x )2 1 (x 20 2) 2 < x < s 2 (x) = (x ) + (x )2 < x < 4 s (x) = (x 4) + (x )2 4 < x < 5 (11) 5

6 MatLab In practice this is not really very fun to do by hand MatLab has a built in command to compute the cubic spline for you The command is called spline(x,y) The default is for it to use the not-a-knot boundary conditions It calls two vectors, x which corresponds to the x j s and y which corresponds to the f j s The vectors should be exactly the same length It outputs the values in sort of an odd way that you have to unpack Below is an example of how to use the command The coef column is the d j, c j, b j and a j values You can alter the way you use spline(x,y) to give the Clamped boundary conditions too To do so you you change the input y vector by adding the value of f (x 0 ) and f (x n ) to the y vector as the first and last entry respectively Now the y will be two entries longer than x A second example of this is below >> x=[-1, -5, 0, 5, 1]; >> y=[ ]; >> pp=spline(x,y); >> [breaks, coef, l, k, d]=unmkpp(pp); >> coef coef = >> subplot(2,1,1) >> y2=[ ]; >> pp2=spline(x,y2); >> [breaks2, coesf2, l2, k2, d2]=unmkpp(pp2); >> coesf coesf = >>y_eval=ppval(spline(x,y),x_eval); >>Y_eval2=ppval(spline(x,y2),x_eval); >> subplot(2,1,1) >> plot(x_eval, y_eval1, x_eval, y_eval2, x_eval, (x_eval+1)*exp(-x_eval),x, y, bo ) >> subplot(2,1,2) >> plot(x_eval, y_eval1-(x_eval+1)*exp(-x_eval), x_eval, y_eval2-(x_eval+1) *exp(-x_eval),x, y-y, bo ) >>maxnak=max(abs(y_eval1-(x_eval+1)*exp(-x_eval)) >>maxclamp=max(abs(y_eval2-(x_eval+1)*exp(-x_eval)) 6

7 Clamped For the clamped boundary conditions one must have information about the values of the derivatives at the endpoints of the interval Namely, we must have that s 0(x 0 ) = b 0 = f 0 is a given value and s n 1(x n ) = b n = f n is also a given value The result of these assignments is that we have: 2h 0 c 0 + h 0 c 1 = h 0 (a 1 a 0 ) f 0 which will be added to our linear system and 2 c n = f n (a n a n 1 ) which can be added as an extra equation for c n from the linear system The matrix equation we have then for the clamped boundary conditions is A c = b where, 2h 0 h h 0 2(h 0 + ) ( + h 2 ) h h 2 2(h 2 + h ) h A = ( + ) h n 2(h n + ) c = c 0 c 1 c c 4 c n 2 c n h 0 (a 1 a 0 ) f 0 (a 2 a 1 ) h 0 (a 1 a 0 ) h 2 (a a 2 ) (a 2 a 1 ) h, b = (a 4 a ) h 2 (a a 2 ) (a n 1 a n 2 ) h n (a n 2 a n 1 ) (a n a n 1 ) (a n 1 a n 2 ) f n (a n a n 1 ) (12) 7

8 ERROR THEOREM Let f(x) be continuous, with four continuous derivatives, on the interval [a, b], and let s be the clamped cubic spline interpolant of f relative to the partition: Then where h = max (x i+1 x i ) 0 i n 1 a = x 0 < x 1 < x 2 x n = b 5 max f(x) s(x) x [a,b] 84 h4 max f (4) (x) x [a,b] 8

9 Notes for Exams Given the x j s and corresponding y j s Not-A-Knot a j = y j for j = 0, 1, 2 n (1) First find the a j s and the h j s Then for the Not-a-Knot boundary conditions find the c 1, by solving the linear system: A c = b where, (h h2 0 h ( + h 2 ) h h 2 2(h 2 + h ) h 0 0 A = ( + ) h2 n 1 c = Then find c 0 using: and set c 1 c c 4 c n h2 n 1 (a 2 a 1 ) h 0 (a 1 a 0 ) h 2 (a a 2 ) (a 2 a 1 ) h, b = (a 4 a ) h 2 (a a 2 ) (a n 1 a n 2 ) h n (a n 2 a n 1 ) (a n a n 1 ) (a n 1 a n 2 ) c 0 = Then find the b j s and the d j s using: ( 1 + h ) 0 c 1 h 0 c n = h ( n 1 c n h ) n 1 d j = c j+1 c j h j for j = 0, 1, 2, n 1 b j = a j+1 a j 2c j + c j+1 h j h j 9

10 Given the x j s and corresponding y j s and f 0 and f n Clamped a j = y j for j = 0, 1, 2 n (14) First find the a j s and the h j s Then, for the Clamped boundary conditions find the c 1, by solving the linear system A c = b where, 2h 0 h h 0 2(h 0 + ) ( + h 2 ) h h 2 2(h 2 + h ) h A = ( + ) h n 2(h n + ) c = c 0 c 1 c c 4 c n 2 c n Then find the b j s and the d j s using: h 0 (a 1 a 0 ) f 0 (a 2 a 1 ) h 0 (a 1 a 0 ) h 2 (a a 2 ) (a 2 a 1 ) h, b = (a 4 a ) h 2 (a a 2 ) (a n 1 a n 2 ) h n (a n 2 a n 1 ) (a n a n 1 ) (a n 1 a n 2 ) f n (a n a n 1 ) d j = c j+1 c j h j for j = 0, 1, 2, n 1 b j = a j+1 a j 2c j + c j+1 h j For both equations you then write down: s 0 (x) = a 0 + b 0 (x x 0 ) + c 0 (x x 0 ) 2 + d 0 (x x 0 ) x [x 0, x 1 ] s 1 (x) = a 1 + b 0 (x x 1 ) + c 1 (x x 1 ) 2 + d 1 (x x 1 ) x [x 1, x 2 ] s(x) = s n 1 (x) = a n 1 + b n 1 (x x n 1 ) + (x x n 1 ) 2 + d n 1 (x x n 1 ) x [x n 1, x n ] h j 10