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Interpolating Solutions to IVPs
Douglas Wilhelm Harder, M.Math. LELDepartment of Electrical and Computer Engineering
University of Waterloo
Waterloo, Ontario, Canada
ece.uwaterloo.ca
dwharder@alumni.uwaterloo.ca
© 2012 by Douglas Wilhelm Harder. Some rights reserved.
2
Outline
Given a solution to an IVP in the form of two vectors tout and yout, how do we approximate the solution at a point t where
tout,k < t < tout,k + 1
We will look at:– Interpolation, and– Dormand Prince
Interpolating Solutions to IVPs
3
Outcomes Based Learning Objectives
By the end of this laboratory, you will:– Understand how to use piecewise polynomials to approximate
solutions of an IVP from discrete approximations– Understand how the mkpp and ppval functions in Matlab work
Interpolating Solutions to IVPs
4
Introduction
Suppose that we found an approximation to an initial-value problem:
[t_out, y_out] = dp45( @f, [a, b], y_init, ... );
The output vector gives us the information that tout,k ≈ yout,k
What happens if we want to approximate the solution u(t) at an arbitrary point that may fall between two of these t-values?
tout,k < t < tout,k + 1
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Introduction
We will deal with one 1st-order and one 2nd-order ODE:function [dy] = f8a( t, y ) dy = (y - 1).^2 .* (t - 1).^2;end
function [y] = y8a_soln( t ) y = (t.^3 - 3*t.^2 + 3*t)./(t.^3 - 3*t.^2 + 3*t + 3);end
function [dw] = f8b( x, w ) dw = [w(2); sin(x) - 4*w(2)*w(1) + 2*x*w(1)^2];end
function y = y8b_soln( t ) c = 5^(1/3); d = c/80; y = 1/c*exp( t/4 ).*(
... airy( 2, d )*airy( d*(1 - 80*t) )
... - airy( d )*airy( 2, d*(1 - 80*t)))/(airy( 3, d )*airy( d )
... - airy( 2, d )*airy( 1, d )
... );end
Interpolating Solutions to IVPs
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Approximating Solutions
We now know that
tk < t < tk + 1
and we know that
y(tk) ≈ uk and y(tk + 1) ≈ uk + 1
How do we approximate y(t)?
Interpolating Solutions to IVPs
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Linear Interpolation
As an initial idea, we could interpolate the two points
(tk, yk) and (tk + 1, yk + 1)
with a straight line...
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Linear Interpolation
We can define a separate linear polynomial between each consecutive pair of points from (t1, y1) to (tn, yn)
Interpolating Solutions to IVPs
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Linear Interpolation
This is no different than connecting the dots with straight lines...
[t6c, y6c] = dp45( @f6c, [0, 3], [0, 1]', 1, 1e-1 );
plot( t6c, y6c(1,:), '-b.' );
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Linear Interpolation
Such a structure is said to be a piecewise-defined polynomial– A different polynomial is defined on each line segment [tk, tk + 1]
– In this case, it would be a piecewise linear polynomial
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Linear Interpolation
This, however, would be very unsatisfying—we know the solution is both continuous and differentiable
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Linear Interpolation
Straight lines are not good approximations to differentiable functions– We learned how to draw lines in kindergarten—let’s come up
with something better....
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Cubic Interpolation
Recall that we have more information—we always have exact values or approximations of
y(1)(tk) and y(1)(tk + 1)
If it is a 1st-order ODE, we can calculatey(1)(tk) = f(tk, yk) and y(1)(tk + 1) = f(tk + 1, yk + 1)
If it is a 2nd- or higher-order ODE, the approximation of the derivatives is in the second row of the output
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Cubic Interpolation
We can find an interpolating cubic polynomial p(t) that:– Matches the values p(tk) = yk and p(tk + 1) = yk + 1
– Matches the derivatives p(1)(tk) = f(tk, yk) and p(1)(tk + 1) = f(tk + 1, yk + 1)
The general form of a cubic polynomial is
at3 + bt2 + ct + d
and its derivative is
3at2 + 2bt + c + d
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Solutions to 1st-order IVPs
This creates a system of equations: p(tk) = yk atk
3 + btk2 + ctk + d = yk
p(1)(tk) = f(tk, yk) 3atk2 + btk + c = f(tk,
yk)
p(tk + 1) = yk + 1 atk + 13 + btk + 1
2 + ctk + 1 + d = yk + 1
p(1)(tk + 1) = f(tk + 1, yk + 1) 3atk + 12 + btk + 1 + c = f(tk + 1, yk + 1)
3 2
2
3 211 1 1
21 11 1
1
,3 2 1 0
1
,3 2 1 0
kk k k
k kk k
kk k k
k kk k
yat t t
f t ybt t
yct t t
f t ydt t
Interpolating Solutions to IVPs
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Solutions to 1st-order IVPs
Thus, if the two points that bracket t are at k and k + 1, the code would be
M = [ t(k)^3 t(k)^2 t(k) 1; 3*t(k)^2 2*t(k) 1 0; t(k + 1)^3 t(k + 1)^2 t(k + 1) 1; 3*t(k + 1)^2 2*t(k + 1) 1 0];
p = M \ [y(k); f(t(k), y(k)); y(k+1); f(t(k+1), y(k+1))];
Interpolating Solutions to IVPs
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The Matlab mkpp Function
The mkpp function returns a piecewise polynomial data structure in Matlab
The arguments are:– A vector t of n break points—the points defining the sub-intervals– An (n – 1) × 4 matrix where each row is the interpolating cubic
polynomial defined on [0, tk + 1 – tk] interpolating the points
(0, yk) and (tk + 1 – tk, yk)
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The dpinterp Function
Thus, we could define the function dpinterp:function [pp] = dpinterp( t, y, f ) n = length(t) - 1; P = zeros( n, 4 );
for k=1:n dt = t(k + 1) - t(k);
M = [ 0 0 0 1; 0 0 1 0; dt^3 dt^2 dt 1; 3*dt^2 2*dt 1 0];
p = M \ [y(k); f(t(k), y(k)); y(k+1); f(t(k+1), y(k+1))]; P(k, :) = p'; end
pp = mkpp( t, P );end
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Matching Derivatives or Splines?
An alternate means of interpolating points are cubic splines:– Rather than matching the derivatives at the end points, we
simply state that at each of the interior points:• The adjacent piecewise cubic polynomials must equal yk,
• The adjacent polynomials must have the same derivative at tk, and
• Adjoining polynomials must have the same second derivative at tk.
The result is an interpolating polynomial that has a twice-differentiable value but may not match the derivatives at the points
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Matching Derivatives or Splines?
For our example, we have:[t, y] = dp45( @f8a, [0, 2], 0', 0.1, 1e-5 )pp = dpinterp( t, y, @f8a ); pps = spline( t, [f8a(t(1), y(1)) y
f8a(t(end),y(end))] );hold onts = 0:0.001:2;plot( t, y, 'bo' ); plot( ts, ppval( pp, ts ), 'b' )plot( ts, ppval( pps, ts ), 'r' ); plot( ts, y8a_soln( ts ), 'k' )
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Matching Derivatives or Splines?
However, if you plot the errors, we see that our interpolating polynomial has less overall error and matches the derivativeshold onplot( t, y - y8a_soln(t), 'bo' ); plot( ts, ppval( pp, ts ) - y8a_soln(ts), 'b' )plot( t, t*0, 'ko' ); plot( ts, ppval( pp2, ts ) - y8a_soln(ts),
'r' )
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Solutions to 2nd-order IVPs
What if you have both derivative and second-derivativeinformation that is easily accessible:– Find an interpolating a quintic polynomial
1,1
2,2
35 4 3 2
1, 144 3 2
2, 153 2
1 16
0 0 0 0 0 1
0 0 0 0 1 0
,0 0 0 1 0 0
1
5 4 3 2 1 0
,20 12 6 2 0 0
k
k
k k
kk k k k k
kk k k k
k kk k k
ya
ya
f ta
yt t t t t a
yt t t t a
f tt t t a
y
y
Interpolating Solutions to IVPs
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Solutions to 2nd-order IVPs
You can also find interpolating heptic polynomial andwe begin to see a pattern:
1
2
3
47 6 5 4 3 2
56 5 4 3 2
65 4 3 2
74 3 2
8
0 0 0 0 0 0 0 1
0 0 0 0 0 0 1 0
0 0 0 0 0 1 0 0
0 0 0 0 1 0 0 0
1
7 6 5 4 3 2 1 0
42 30 20 12 6 2 0 0
210 120 60 24 6 0 0 0
k k k k k k k
k k k k k k
k k k k k
k k k k
a
a
a
a
t t t t t t t a
t t t t t t a
t t t t t a
t t t t a
1,
2,
3,
1, 1
2, 1
3, 1
1 1
,
,
k
k
k
k k
k
k
k
k k
y
y
y
f t
y
y
y
f t
y
y
Interpolating Solutions to IVPs
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The dpinterp Function
This allows us to define a general dpinterp function:function [pp] = dpinterp( t, y, f ) [m, n] = size( y ); P = zeros( n - 1, 2*m + 2 ); M = zeros( 2*m + 2, 2*m + 2 ); for i = 1:(m + 1) M(i, 2*m + 3 - i) = factorial( i - 1 ); end for k=1:(n - 1) dt = (t(k + 1) - t(k)).^((2*m + 1):-1:0); for i = 1:(m + 1) M(i, 2*m + 3 - i) = factorial( i - 1 ); M(m + 1 + i, 1:(end + 1 - i)) = dt; dt = dt(2:end).*((2*m + 1):-1:i); end dyk = f( t(k), y(:, k)); dyk1 = f(t(k + 1), y(:, k + 1)); P(k, :) = (M \ [y(:, k); dyk(end); y(:, k + 1); dyk1(end)])'; end pp = mkpp( t, P );end
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The dpinterp Function
This piecewise quintic is more accurate than a cubic spline:[t, y] = dp45( @f8b, [0, 2], [0 1]', 0.1, 1e-2 );pp = dpinterp( t, y, @f8b );dy = f8b(2, y(:, end)); pps = spline( t, [1 y(1,:) dy(1)] );hold on; plot( ts, ppval( pp, ts ) - y8b_soln( ts ), 'r' );plot( ts, ppval( pps, ts ) - y8b_soln( ts ), 'b' );plot( t, y(1,:) - y8b_soln( t ), 'ro' );
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The dpinterp Function
What happens if we do not want a 7th degree polynomial?– Can we use less information to make a interpolating polynomial?– We can provide a 4th argument—if the user provides this
argument m, it will uses a polynomial of 2m – 1
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The dpinterp Function
We now have a dpinterp function that gives maximum choice to the user:
Interpolating Solutions to IVPs
function [pp] = dpinterp( t, y, f, m ) [mp, n] = size( y ); if nargin == 3 m = mp + 1; else m = max( min( mp + 1, m ), 1 ); end P = zeros( n - 1, 2*m ); M = zeros( 2*m, 2*m ); for i = 1:m M(i, 2*m + 1 - i) = factorial( i - 1 ); end
for k=1:(n - 1) dt = t(k + 1) - t(k); dt = dt.^((2*m - 1):-1:0); for i = 1:m M(i, 2*m + 1 - i) = factorial( i - 1 ); M(m + i, 1:(end + 1 - i)) = dt; dt = dt(2:end).*((2*m - 1):-1:i); end if m == mp + 1 dyk = f( t(k), y(:, k)); dyk1 = f(t(k + 1), y(:, k + 1)); P(k, :) = (M \ [y(:, k); dyk(end); y(:, k + 1); dyk1(end)])'; else P(k, :) = (M \ [y(1:m, k); y(1:m, k + 1)])'; end end pp = mkpp( t, P );end
28
The dpinterp Function
Even now, a cubic polynomial is better than a spline:[t, y] = dp45( @f8b, [0, 2], [0 1]', 0.1, 1e-2 );pp = dpinterp( t, y, @f8b, 2 );dy = f8b(2, y(:, end)); pps = spline( t, [1 y(1,:) dy(1)] );hold on; plot( ts, ppval( pp, ts ) - y8b_soln( ts ), 'r' );plot( ts, ppval( pps, ts ) - y8b_soln( ts ), 'b' );plot( t, y(1,:) - y8b_soln( t ), 'ro' );
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The dpinterp Function
As may be noted, quintics are very good approximations but in this case, clamped cubics are also better than cubic splines
pp = dpinterp( t, y, @f8b, 3 ); pp = dpinterp( t, y, @f8b, 2 );
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Summary
We have looked approximating solutions to IVPs between the points returned by functions such as dp45:– We discussed piecewise-defined polynomials– We found piecewise cubic polynomials for 1st-order IVPs– We determined that we could get even more accurate
approximations for an Nth-order IVP– The clamped polynomials produce a better result than splines
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References
[1] Glyn James, Modern Engineering Mathematics, 4th Ed., Prentice Hall, 2007, p.778.
[2] Glyn James, Advanced Modern Engineering Mathematics, 4th Ed., Prentice Hall, 2011, p.164.
[3] J.R. Dormand and P. J. Prince, "A family of embedded Runge-Kutta formulae," J. Comp. Appl. Math., Vol. 6, 1980, pp. 19-26.
Interpolating Solutions to IVPs
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