Computational Biology, Part 17 Biochemical Kinetics IIGeneral form of ordinary differential equationsEuler’s methodSlide 4Slide 5Slide 6Midpoint methodMidpoint method (2nd order Runge-Kutte)Slide 9Slide 10Fourth-order Runge-KuttaSlide 12Interactive demonstrationPowerPoint PresentationSlide 15Solving differential equations using MapleSolving a single differential equation using MapleSlide 18Solving a set of differential equations numerically using MapleSlide 20Other systems for explorationBiochemical System 2Slide 23Biochemical System 3Slide 25Computational Biology, Part 17Biochemical Kinetics IIComputational Biology, Part 17Biochemical Kinetics IIRobert F. MurphyRobert F. MurphyCopyright Copyright  1996, 1999-2007. 1996, 1999-2007.All rights reserved.All rights reserved.General form of ordinary differential equationsGeneral form of ordinary differential equationsFor a set of For a set of nn unknown functions unknown functions yyii (for (for ii=1 =1 to to nn) we are given a set of ) we are given a set of nn functions functions ffii that specify the derivatives of each that specify the derivatives of each yyii with with respect to some independent variable respect to some independent variable xxdyi( x )dx= ′ f i(x, y1,...,yn) for i =1,...,nEuler’s methodEuler’s methodThe simplest numerical integration method The simplest numerical integration method is is Euler’s methodEuler’s method. It simply converts each . It simply converts each differential to a differencedifferential to a differenceand then calculates the value of and then calculates the value of yyii by by multiplying the right hand side of each multiplying the right hand side of each differential equation by the step size differential equation by the step size xxΔyiΔx= ′f i( x , y1, ..., yn)Δyi= Δx ⋅ ′f i( x , y1, ..., yn)Euler’s methodEuler’s methodWe can rewrite this as a recursion formula We can rewrite this as a recursion formula that allows us to calculate values of the that allows us to calculate values of the functions functions yyii at a series of at a series of xx values. We values. We introduce a second subscript introduce a second subscript jj to indicate to indicate which which xx value we refer to (note that value we refer to (note that yyi,ji,j now now refers to a refers to a valuevalue not a not a functionfunction).).yi , j +1= yi , j+Δx ⋅ ′ f i(xj, y1, j,...,yn, j)xj +1= xj+ΔxEuler’s methodEuler’s methodNote the asymmetry of this method: the Note the asymmetry of this method: the derivative (derivative (ffii ) that is used to span the ) that is used to span the xx is is calculated calculated onlyonly for the for the xx value at the value at the beginningbeginning of the interval. In regions where of the interval. In regions where ffii is increasing with is increasing with x,x, this leads to this leads to underunderestimation of estimation of y,y, and, in regions and, in regions where where ffii is decreasing with is decreasing with x,x, to to overoverestimation of estimation of y.y.Euler’s methodEuler’s methodConsider Consider dy/dxdy/dx=6=6xx+2+2. The analytical . The analytical solution is solution is yy=3=3xx22+2+2xx..The graph shows the Euler’s method The graph shows the Euler’s method approximation for approximation for yy and the analytical and the analytical solution for solution for yy (along with (along with dy/dx).dy/dx).Euler's Method vs. Analytical Solution051015202530350 1 2 3xy02468101214161820y(Euler)y(Analyt)dy/dxNote how the approximation always underestimates y since dy/dx is increasingMidpoint methodMidpoint methodA better estimate would come from evaluating A better estimate would come from evaluating the the ffii at the at the midpointmidpoint of the of the x x interval. The interval. The problem: we know problem: we know x x at the midpoint but we at the midpoint but we don’t know the don’t know the yyii at the midpoint (yet). The at the midpoint (yet). The solution is to use Euler’s method to estimate solution is to use Euler’s method to estimate yy and then and then rere-estimate -estimate yy using the derivatives using the derivatives evaluated halfway along the line segment evaluated halfway along the line segment encompassing the original encompassing the original y.y.Midpoint method (2nd order Runge-Kutte)Midpoint method (2nd order Runge-Kutte)This is called the This is called the midpoint method midpoint method or the or the second-order Runge-Kutte methodsecond-order Runge-Kutte method..Δyj= Δx ⋅ ′f ( xj, yj)Δyj + 0.5= Δx ⋅ ′f ( xj+Δx2, yj+Δyj2)yj +1= yj+ Δyj + 0.5Midpoint method (2nd order Runge-Kutte)Midpoint method (2nd order Runge-Kutte)Again consider Again consider dy/dxdy/dx=6=6xx+2+2..The graph shows the midpoint The graph shows the midpoint approximation for approximation for yy and the analytical and the analytical solution for solution for yy ( (along with along with dy/dx dy/dx at at x x andand x+ x+0.50.5).).Midpoint Method vs. Analytical Solution for dy/dx =6x +2051015202530350 0.5 1 1.5 2 2.5 3xyy(2°RK)y(Analyt)dy/dx(x)dy/dx(x+Dx/2)Note that the approximation and the analytical solution are identical in this case.Midpoint method (2nd order Runge-Kutte)Midpoint method (2nd order Runge-Kutte)Now consider Now consider dy/dxdy/dx=3=3yy. The analytical . The analytical solution is solution is yy=e=e33xx..The graph shows the Euler, Midpoint and The graph shows the Euler, Midpoint and analytical solutions (along with derivatives).analytical solutions (along with derivatives).Three methods for dy/dx =3y01020304050607080901000 0.5 1 1.5xy050100150200250300y(2°RK)y(Analyt)y(Euler)dy/dx(x)dy/dx(x+Dx/2)Note that the midpoint method is better than Euler’s method but it does not give results identical to the analytical solution since dy/dx now depends on y.Fourth-order Runge-KuttaFourth-order Runge-KuttaThe midpoint method can be extended by The midpoint method can be extended by considering other intermediate estimates. considering other intermediate estimates. The most frequently used variation is the The most frequently used variation is the fourth-order Runge-Kutta fourth-order Runge-Kutta method which method which considers one estimate at the initial point, considers one estimate at the initial point, two estimates at