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Solutions to ODEs. A general form for a first order ODE is Or alternatively. A general form for a first order ODE is Or alternatively We desire a solution y(x) which satisfies (1) and one specified boundary condition. .

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slide3
A general form for a first order ODE is
  • Or alternatively
  • We desire a solution y(x) which satisfies (1) and one specified boundary condition.
slide4
To do this we divide the interval in the independent variable x for the interval [a,b] into subintervals or steps.
slide5
To do this we divide the interval in the independent variable x for the interval [a,b] into subintervals or steps.
  • Then the value of the true solution is approximated at n+1 evenly spaced values of x.
  • Such that,
  • where,
slide7
We denote the approximation at the base pts by so that
  • The true derivation dy/dx at the base points is approximated by or just
  • where
slide8
We denote the approximation at the base pts by so that
  • The true derivation dy/dx at the base points is approximated by or just
  • where
  • Assuming no roundoff error the difference in the calculated and true value is the truncation error,
slide10
Numerical algorithms for solving 1st order odes for an initial condition are based on one of two approaches:
  • Direct or indirect use of the Taylor series expansion for the solution function
  • Use of open or closed integration formulas.
slide11
The various procedures are classified into:
  • One-step: calculation of given the differential equation and
  • Multi-step: in addition the previous information they require values of x and y outside of the interval under consideration
slide12
The first method considered is the Taylor series method.
  • It forms the basis for some of the other methods.
slide14
For this method we express the solution about some starting point using a Taylor expansion.
slide16
Where
  • And
  • etc.
slide17
Consider the example, where
  • And initial condition
slide18
Consider the example, where
  • And initial condition
  • Differentiating and then applying initial conditions,
slide19
Consider the example, where
  • And initial condition
  • Differentiating and then applying initial conditions,
slide20
Consider the example, where
  • And initial condition
  • Differentiating and then applying initial conditions,
slide21
Consider the example, where
  • And initial condition
  • Differentiating and then applying initial conditions,
slide24
Continuing,

. .

. .

Here the third derivative and higher are zero.

slide26
Substituting these values into Taylor expansion gives,
  • To determine the error lets consider compare this to the analytic solution.
slide27
Substituting these values into Taylor expansion gives,
  • To determine the error lets consider compare this to the analytic solution.
  • To do this we will integrate the differential equation.
slide31
Integrating,
  • Simplifying,
homework due thursday
Homework (Due Thursday)
  • Consider the function
  • Initial condition
  • Write the Taylor expansion and show that it can be written in the form,
  • Show if terms up to are retained, the error is,
slide35
Stepping from to follows from the Taylor expansion of about .
  • Equivalently stepping from to can be accomplished from the Taylor expansion of about .
slide36
Stepping from to follows from the Taylor expansion of about .
  • Equivalently stepping from to can be accomplished from the Taylor expansion of about .
slide37
Algorithms for which the last term in expansion is dropped are of order hn.
  • The error is or order hn+1.
  • The local truncation error is bounded as follows
  • where
slide39
However differentiation of can be complicated.
  • Direct Taylor expansion is not used other than the simplest case,
  • Big Oh is the order of the algorithm.
slide40
Usually only the value of is the only value of that is known, therefore must be replaced by .
  • Thus,
  • In general,

which is known as Euler’s method.

slide43

y

y1

y(x)

y(x0)

y(x1

h

x0

x1

x

  • The geometric interpretation is shown the diagram.
slide44

y

y1

y(x)

y(x0)

y(x1

h

x0

x1

x

  • The solution across the interval[x0,x1] is assumed to follow the line tangent to y(x) at x0.
slide45
When the method is applied repeatedly across several intervals in sequence, the numerical solution traces out a polygon segment with sides of slope fi, i=0,1,2,…,(n-1).
slide46
The simple Euler method is a linear approximation, and only works if the function is linear (or at least linear in the interval).
slide47
The simple Euler method is a linear approximation, and only works if the function is linear (or at least linear in the interval).
  • This is inherently inaccurate.
  • Because of this inaccuracy small step sizes are required when using the algorithm.
slide48
From the graph, as h -> 0, the approximation becomes better because the curve in the region becomes approx. linear.

y

y1

y(x)

y(x0)

y(x1

h

x0

x1

x

slide51
One modification is to determine the average slope in the region.
  • The average slope may be approximated by the mean of the slopes at the beginning and the end of the interval.
slide52
One modification is to determine the average slope in the region.
  • The average slope may be approximated by the mean of the slopes at the beginning and the end of the interval.
  • This modified Euler may be written as the arithmetic average,

or

slide53
However it is not possible to implement this directly.
  • Instead we first “predict” a value for using the simple Euler method.
  • Then use this value to determine the gradient of the slope at . This gives a “corrected” value for .
  • However since the predicted value is not usually accurate, is also inaccurate.
slide54

Corrector curve using

values of predictor

Predictor curve using

the Simple Euler

True value

slide56
The solution of a differential equation using higher order derivatives of the Taylor expansion is not practical.
slide57
The solution of a differential equation using higher order derivatives of the Taylor expansion is not practical.
  • Since for only the simplest functions, these higher orders are complicated. Also there is no simple algorithm which can be developed.
  • This is because each series expansion is unique.
slide58
However we have methods which use only 1st order derivates while simulating higher order(producing equivalent results).
  • These one step methods are called Runge-Kutta methods.
slide59
However we have methods which use only 1st order derivates while simulating higher order(producing equivalent results).
  • These one step methods are called Runge-Kutta methods.
  • Approximation of the second, third and fourth order (retaining h2, h3, h4 respectively in the Taylor expansion) require estimation at 2, 3 , 4 pts respectively in the interval (xi,xi+1).
slide61
The Runge-Kutta methods have algorithms of the form,
  • where is the increment function.
  • The increment function is a suitably chosen approximation to on the interval .
slide62
The Runge-Kutta methods have algorithms of the form,
  • where is the increment function.
  • The increment function is a suitably chosen approximation to on the interval .
  • Because of the amount of algebra involved, on the simplest case(2nd order) will be derived in detail.
slide67
Let be the weighted average of two derivative evaluations and on the interval ,
  • Then,
  • Let,
  • The constants p and q will be established.
slide68
Expanding in a Taylor series for a function of two variables and dropping terms higher than h,
  • NB: 1st few terms in two-variable Taylor series,
slide69
Recall:
  • Substituting for ,
slide70
Recall:
  • Substituting for ,
  • Expanding the function about
slide72
Using the chain rule can be written as,
  • Equating like terms in equation 1 and 2,
slide75
Assuming that we compare coefficients. So that,
  • Therefore,
  • However we have four unknowns and only three equations.
  • We have the variable b which can be chosen arbitrarily.
  • Two common choices are and
slide76
For . and we get,
  • which can be written as,
  • where
  • This improved Euler method or Heun’s Method.
slide79
For . and we get,
  • where
  • This improved polygon method or the modified Euler’s Method.
slide83
The higher orders are developed in the way as the 2nd order Runge-Kutta.
  • Consider the 3rd order Runge-Kutta,
  • approximate the derivative at the points on the interval.
slide84
The higher orders are developed in the way as the 2nd order Runge-Kutta.
  • Consider the 3rd order Runge-Kutta,
  • approximate the derivative at the points on the interval.
  • For this case,
slide86
Third order Runge-Kutta algorithms are of the form,
  • To determine a,b,c,p,r and s,
  • First expand about
slide87
Third order Runge-Kutta algorithms are of the form,
  • To determine a,b,c,p,r and s,
  • First expand about
  • Expand as a Taylor series.
slide88
Third order Runge-Kutta algorithms are of the form,
  • To determine a,b,c,p,r and s,
  • First expand about
  • Expand as a Taylor series.
  • Compare coefficients of the Runge-Kutta and Taylor series.
slide89
The equations formed are,
  • Two of the constants a,b,c,p,r,s are arbitrary.
slide91
For one set of constant, selected by Kutta the third order algorithm is,
  • If is a function of x only then this reduces to the Simpson’s rule.
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