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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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Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes

  • We denote the approximation at the base pts by so that variable x for the interval [a,b] into subintervals or steps.

  • 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,



Solutions to odes

  • Numerical algorithms for solving 1 the numerical methods.st 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.


Solutions to odes

  • The various procedures are classified into: the numerical methods.

  • 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


Solutions to odes


Taylor expansion

Taylor Expansion the numerical methods.


Solutions to odes


Solutions to odes

  • where starting point using a Taylor expansion.


Solutions to odes

  • Where starting point using a Taylor expansion.

  • And

  • etc.


Solutions to odes


Solutions to odes

  • Consider the example, where starting point using a Taylor expansion.

  • And initial condition

  • Differentiating and then applying initial conditions,


Solutions to odes

  • Consider the example, where starting point using a Taylor expansion.

  • And initial condition

  • Differentiating and then applying initial conditions,


Solutions to odes

  • Consider the example, where starting point using a Taylor expansion.

  • And initial condition

  • Differentiating and then applying initial conditions,


Solutions to odes

  • Consider the example, where starting point using a Taylor expansion.

  • And initial condition

  • Differentiating and then applying initial conditions,


Solutions to odes


Solutions to odes


Solutions to odes

  • Continuing, starting point using a Taylor expansion.

    . .

    . .

    Here the third derivative and higher are zero.


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes

  • Integrating, starting point using a Taylor expansion.

  • Simplifying,



Homework due thursday
Homework (Due Thursday) there is no truncation error.

  • 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,



Solutions to odes


Solutions to odes


Solutions to odes



Solutions to odes


Solutions to odes


Euler s method

Euler’s Method that is known, therefore must be replaced by .



Solutions to odes

y inaccuracy in the formula.

y1

y(x)

y(x0)

y(x1

h

x0

x1

x

  • The geometric interpretation is shown the diagram.


Solutions to odes

y inaccuracy in the formula.

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.


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes

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




Solutions to odes


Solutions to odes


Solutions to odes

  • However it is not possible to implement this directly. region.

  • 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.


Solutions to odes

Corrector curve using region.

values of predictor

Predictor curve using

the Simple Euler

True value



Solutions to odes


Solutions to odes

  • 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.


Solutions to odes

  • However we have methods which use only 1 derivatives of the Taylor expansion is not practical.st order derivates while simulating higher order(producing equivalent results).

  • These one step methods are called Runge-Kutta methods.


Solutions to odes

  • However we have methods which use only 1 derivatives of the Taylor expansion is not practical.st 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).



Solutions to odes


Solutions to odes

  • The Runge-Kutta methods have algorithms of the form, at more than m nearby points.

  • 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.


Derivation of the 2 nd order runge kutta

Derivation of the 2 at more than m nearby points.nd order Runge-Kutta





Solutions to odes


Solutions to odes


Solutions to odes

  • Recall: variables and dropping terms higher than h,

  • Substituting for ,


Solutions to odes

  • Recall: variables and dropping terms higher than h,

  • Substituting for ,

  • Expanding the function about


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes

  • Assuming that we compare coefficients. So that, variables and dropping terms higher than h,

  • 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


Solutions to odes

  • For . and we get, variables and dropping terms higher than h,

  • which can be written as,

  • where

  • This improved Euler method or Heun’s Method.


Solutions to odes


Solutions to odes

  • For . and we get, variables and dropping terms higher than h,

  • where

  • This improved polygon method or the modified Euler’s Method.


Solutions to odes


Higher orders

Higher Orders variables and dropping terms higher than h,


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes


Solutions to odes



Solutions to odes