Team:St Andrews/project/modelling/models/RK4

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	The y-value iteration contains a much more elegant set of relationships. It should be clear to you that the y-iteration formula is in fact a weighted average of the four value k values, k<sub>1</sub>, k<sub>2</sub>, k<sub>3</sub> and k<sub>4</sub>. Even at a first glance it should also be apparent that a 'weight' of 2/6 is given to k<sub>2</sub> and k<sub>3</sub> while a smaller weighting of 1/6 is attributed to k<sub>1</sub> and k<sub>2</sub>. What then do these k-value correspond to geometrically?		The y-value iteration contains a much more elegant set of relationships. It should be clear to you that the y-iteration formula is in fact a weighted average of the four value k values, k<sub>1</sub>, k<sub>2</sub>, k<sub>3</sub> and k<sub>4</sub>. Even at a first glance it should also be apparent that a 'weight' of 2/6 is given to k<sub>2</sub> and k<sub>3</sub> while a smaller weighting of 1/6 is attributed to k<sub>1</sub> and k<sub>2</sub>. What then do these k-value correspond to geometrically?
-	Look first at k<sub>1</sub>. It is equal to the function f(x<sub>n</sub>,y<sub>n</sub>) multiplied by the step-size h, which is just the Euler prediction for Δy. It is the vertical distance from the previous point to the next Euler predicted point.	+	Look first at k<sub>1</sub>. It is equal to the function f(x<sub>n</sub>,y<sub>n</sub>) multiplied by the step-size h, which is just the Euler prediction for Δy. It is the vertical distance from the previous point to the next Euler-predicted point.
-	Then consider ksub>2</sub>. It should be noted that the x-value at which the function is being evaluated is half-way across the prediction interval and the y-value consists of the current y-value plus half of the Euler predicted Δy.	+	Then consider ksub>2</sub>. It should be noted that the x-value at which the function is being evaluated is halfway across the prediction interval and the y-value consists of the current y-value plus half of the Euler predicted Δy. So then the function is evaluated at a point lying halfway between the current point and the Euler-predicted next point. When we evaluate the function at this point what is produced is an estimate of the slope of the solution curve at this halfway point on curve. Multiplying this result by h gives a prediction of the y-jump made by the solution across the entire interval. However, this prediction is made not on the slope of the solution at the left end of the interval but on the estimated slope halfway to the next Euler-predicted point.

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Revision as of 10:55, 13 September 2010

University of St Andrews iGEM 2010

Welcome!

University of St Andrews iGEM 2010

Our first year at iGEM!

Foruth Order Runge-Kutta Method

Introduction

Our method of solving differential equations is based on Fourth Order Runge-Kutta Method. This technique is the most widely used way of numerically solving differential equations and various methods of implementation were looked at. Most of our coding has been based on the work of Aberdeen 2009 iGEM team who used the same method in their modelling. We would like to thank the team for their work and making it available to others like us for future use.

Mathematical basis

While the strict mathematical derivation of the Runge-Kutta method is available( see [http://www.ss.ncu.edu.tw/~lyu/lecture_files_en/lyu_NSSP_Notes/Lyu_NSSP_AppendixC.pdf here for details] ), we omit it here and instead give a brief explanation of the principles behind the technique.

There are several key formulae in the 4th Order Runge-Kutta algorithm, which are:

The iteration of the x-values is done very simply adding a fixed step-size (h) at each iteration, thus resulting in a constant increase in the x-value according to the value of h chosen for a particular purpose.

The y-value iteration contains a much more elegant set of relationships. It should be clear to you that the y-iteration formula is in fact a weighted average of the four value k values, k₁, k₂, k₃ and k₄. Even at a first glance it should also be apparent that a 'weight' of 2/6 is given to k₂ and k₃ while a smaller weighting of 1/6 is attributed to k₁ and k₂. What then do these k-value correspond to geometrically?

Look first at k₁. It is equal to the function f(x_n,y_n) multiplied by the step-size h, which is just the Euler prediction for Δy. It is the vertical distance from the previous point to the next Euler-predicted point.

Then consider ksub>2</sub>. It should be noted that the x-value at which the function is being evaluated is halfway across the prediction interval and the y-value consists of the current y-value plus half of the Euler predicted Δy. So then the function is evaluated at a point lying halfway between the current point and the Euler-predicted next point. When we evaluate the function at this point what is produced is an estimate of the slope of the solution curve at this halfway point on curve. Multiplying this result by h gives a prediction of the y-jump made by the solution across the entire interval. However, this prediction is made not on the slope of the solution at the left end of the interval but on the estimated slope halfway to the next Euler-predicted point.