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Solutions to ODEs

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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 ... – PowerPoint PPT presentation

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Title: Solutions to ODEs

1
Solutions to ODEs
2

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.

To do this we divide the interval in the
independent variable x for the interval a,b
into subintervals or steps.

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 n1 evenly spaced values of x.
Such that,
where,

We denote the approximation at the base pts by
so that

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

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,

The only errors which will be examined are those
inherent to the numerical methods.

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.

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

The first method considered is the Taylor series
method.
It forms the basis for some of the other methods.

13
Taylor Expansion
14

For this method we express the solution about
some starting point using a Taylor expansion.

where

Where
And
etc.

Consider the example, where
And initial condition

Consider the example, where
And initial condition
Differentiating and then applying initial
conditions,

Consider the example, where
And initial condition
Differentiating and then applying initial
conditions,

Consider the example, where
And initial condition
Differentiating and then applying initial
conditions,

Consider the example, where
And initial condition
Differentiating and then applying initial
conditions,

Continuing,

Continuing,

Continuing,
. .
. .
Here the third derivative and higher are zero.

Substituting these values into Taylor expansion
gives,

Substituting these values into Taylor expansion
gives,
To determine the error lets consider compare this
to the analytic solution.

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.

Integrating,

Integrating,

Integrating,

Integrating,
Simplifying,

In this case the two expressions are identical,
therefore there is no truncation error.

33
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,

Stepping from to follows from the
Taylor expansion of about .

Stepping from to follows from the
Taylor expansion of about .
Equivalently stepping from to can be
accomplished from the Taylor expansion of
about .

Stepping from to follows from the
Taylor expansion of about .
Equivalently stepping from to can be
accomplished from the Taylor expansion of
about .

Algorithms for which the last term in expansion
is dropped are of order hn.
The error is or order hn1.
The local truncation error is bounded as follows
where

However differentiation of can be
complicated.

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.

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 Eulers method.

41
Eulers Method
42

The problem with the simple Euler method is the
inherent inaccuracy in the formula.

The geometric interpretation is shown the diagram.

The solution across the intervalx0,x1 is
assumed to follow the line tangent to y(x) at x0.

When the method is applied repeatedly across
several intervals in sequence, the numerical
solution traces out a polygon segment with sides
of slope fi, i0,1,2,,(n-1).

The simple Euler method is a linear
approximation, and only works if the function is
linear (or at least linear in the interval).

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.

From the graph, as h -gt 0, the approximation
becomes better because the curve in the region
becomes approx. linear.

The modified Euler method improves on the simple
Euler method.

50
Modified Euler
51

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.

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

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.

54
Corrector curve using values of predictor
Predictor curve using the Simple Euler
True value
55
Runge-Kutta Methods
56

The solution of a differential equation using
higher order derivatives of the Taylor expansion
is not practical.

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.

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.

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,xi1).

Method of order m (where m gt4) require derivate
evaluation at more than m nearby points.

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
.

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.

63
Derivation of the 2nd order Runge-Kutta
64