introduction to standard and non-standard numerical methods

Outline One-step methods Runge-Kutta Methods Nonstandard Finite Difference Scheme Application of Polio model

Introduction to standard and non-standardNumerical Methods

Dr. Mountaga LAM

AMS : African Mathematic School 2018

May 23, 2018

One-step methods

Runge-Kutta Methods

Nonstandard Finite Difference Scheme

Application of Polio model

Solvingdx

dt= f (x)

Are numerical methods whose to forward a step, only the previous step information isneeded, ie step n+1 only depends on the step n.

• Euler’s idea

• I can’t solve the equation because I don’t know what dx is. So dt pick a smallnumber h > 0 and say that

x(t + h)− x(t)

• The differential equation then becomes

x(t + h)− x(t)

h≈ f (x)

• If you know x(t) and h then you can solve this equation for x(t + h).

Solvingdx

dt= f (x)

• Euler’s idea

x(t + h)− x(t)

h≈ f (x)

Solvingdx

dt= f (x)

• Euler’s idea

x(t + h)− x(t)

h≈ f (x)

Solvingdx

dt= f (x)

• Euler’s idea

x(t + h)− x(t)

h≈ f (x)

Solvingdx

dt= f (x)

•x(t + h)− x(t)

h≈ f (x)

has solutionx(t + h) ≈ x(t) + h · f (x)

• Exemple (t = 0): If we know x(0), then this equation allows us to computex(0 + h) = x(h).

• Exemple (t = h): If we know x(h), then this equation allows us to computex(h + h) = x(2h).

• and than x(2h + h) = x(3h), x(3h + h) = x(4h)...

Solvingdx

dt= f (x)

•x(t + h)− x(t)

h≈ f (x)

has solutionx(t + h) ≈ x(t) + h · f (x)

• Exemple (t = 0): If we know x(0), then this equation allows us to computex(0 + h) = x(h).

• Exemple (t = h): If we know x(h), then this equation allows us to computex(h + h) = x(2h).

• and than x(2h + h) = x(3h), x(3h + h) = x(4h)...

Solvingdx

dt= f (x)

• Pick small number of h and compute

x(h) = x(0) + h · f (0)↘ |

x(2h) = x(h) + h · f (h)↘ |

x(3h) = x(2h) + h · f (2h)↘ |

x(4h) = x(3h) + h · f (3h)...

How did Euler do this?R: By hand (main...), if yes ! How many times ?How do we do this in the 21st century?With computer

• Idea is that

New value = Old value+ step size*slope

xn+1 = xn + hnf (tn, xn)

• Slope is generally a function of t, hence x(t)

• Different methods differ in how to estimate φ

f (x) = 0.2 ∗ x2

ti ti+1

x(ti+1)

Euler’s Method (RK method with order 1)

A one-step method expresses xn+1 in terms of the previous value xn.The simplest example of a one-step method for the numerical solution of the initialvalue problem (IVP) is Euler’s method.Given that x(t0) = x0

We definexn+1 = xn + hnf (tn, xn)

The midpoint rule (RK method with order two)

xn+1 = xn + hf (tn +h

2, xn +

2f (xn, tn))

Second-Order Runge-Kutta MethodsLet

x′(t) = f (t, x(t)).

x(t + h) = x(t) + hx′(t) +

2!x′′

(t) +O(h3)

wherex′′

(t) = ft(t, x) + fx (t, x)x′(t) = ft(t, x) + fx (t, x)f (t, x)

with Jacobian fx

x(t + h) = x(t) + hf (t, x) +h2

[ft(t, x) + fx (t, x)f (t, x)

]+O(h3)

Second-Order Runge-Kutta Methods

x(t + h) = x(t) +h

2f (t, x) +

[ft(t, x) + hft(t, x) + hfx (t, x)f (t, x)

]+O(h3)

Recalling the multivariate Taylor expansion

f (t + h, x + k) = f (t, x) + hft(t, x) + fx (t, x)k + · · ·

x(t + h) = x(t) +h

2f (t, x) +

[f (t + h, x + hf (t, x))

]+O(h3)

we get the numerical method :

xn+1 = xn + h(1

k1 = f (tn, xn) and k2 = f (tn + h, xn + hk1)

Fourth-Order Runge-Kutta MethodsThe classical method is given by

xn+1 = xn +h

(k1 + 2k2 + 2k3 + k4

)withk1 is the increment based on the slope at the beginning of the interval, using x(Euler’s method);k2 is the increment based on the slope at the midpoint of the interval, using x and k1

k3 is again the increment based on the slope at the midpoint, but now using x and k2

k4 is the increment based on the slope at the end of the interval, using x and k3

Butcher tableauThe family of explicit Runge-Kutta methods is a generalization of the RK4 methodmentioned above. It is given by

xn+1 = xn + hs∑

k1 = f (tn + 0h, xn),k2 = f (tn + c2h, xn + h(a21k1)),k3 = f (tn + c3h, xn + h(a31k1 + a32k2)),k4 = f (tn + c4h, xn + h(a41k1 + a42k2 + a43k3)),

...ks = f (tn + csh, xn + h(as1k1 + as1k2 + · · ·+ as,s−1ks−1)),

And the final step to combine these intermediate steps like this:

xn+1 = xn + b1k1 + b2k2 + b3k3 + b4k4 + · · ·+ bsks

The Runge-Kutta method is consistent if

i−1∑j=1

aij = ci , for i = 2, · · · , s.

Exemple : The explitcit RK4 method falls in this framework

1 0 0 116

It is not difficult to construct s-stage implicit methods which are A-stable.For example, this can be done by choosing the coefficients ci and bi to be thequadrature points and weights respectively in the Gauss quadrature formula for theevaluation of ∫ 1

0f (t)dt ≈

s∑i=1

bi f (ci )

The numbers aij can then be chosen so that the method has order 2s, and is A-stable.

Let us observe that on expanding x(tn+1) = x(tn + h) into a Taylor series

x′(tn) = f (tn, x(tn))

we have that :

x(tn+1) = x(tn) + hf (tn, x(tn)) +O(h2)

More generally, a one-step method may be written in the form

x(tn+1) = x(tn) + hφ(tn, xn, h), n = 0, 1, · · · ,N − 1, x(t0) = x0,

where we assume that φ : [t0, t0 + T ]× R× R→ R is a continuous function.In pratical case, the function φ(t, x , h) can be defineFor example : Euler’s Method is given by φ(tn, xn, h) = f (tn, xn)

Global error

In order to assess the accuracy of the numerical method, we define the global error,en, by

en = x(tn)− xn

The truncation error, Tn, is define by

Tn =x(tn+1)− x(tn)

h− φ(tn, x(tn); h).

Convergence

Consider the general one-step method where, in addition to being a continuousfunction of its arguments, φ is assumed to satisfy a Lipschitz condition with respect toits second argument, that is, there exists a positive constant Lφ such that, for0 ≤ h ≤ h0 and for (t, u) and (t, v) in the rectangle

D = {(t, x) : t0 ≤ x ≤ tM , |x − x0| ≤ C}

we have that|φ(t, u; h)− φ(t, v ; h)| ≤ Lφ|u − v |.

Then, assuming that |x − x0| ≤ C , n = 1, 2, · · · ,N it follows that

|en| ≤T

(eLφ(tn−t0) − 1

)where T = max0≤n≤N−1 |Tn|

Consistance

Let us apply this general result in order to obtain a bound on the global error inEuler’s method. The truncation error for Euler’s method is given by

Tn =x(tn+1)− x(tn)

h− f (tn, x(tn)) =

x(tn+1)− x(tn)

h− x′(tn)

Taylor series :

x(tn+1) = x(tn) + hx′(tn) +

2!x′′

(ζn), tn < ζn < tn+1.

2hx′′

M = maxζ∈[t0,tM ]

|x′′|

|Tn| ≤ T =1

2M[ eL(tn−t0) − 1

], n = 0, 1, · · · ,N.

Remark : In practice, for such h we shall have |x(tn)− xn| = |en| ≤ Tol for anyn = 0, 1, · · · ,N. So h ≤ (expression ∗ Tol)

Bref definition

• A Scheme is said explicit if x(ti+1) can be write as a linear combination ofx(ti ), f (xk , ti1 ), ... ∀k. The scheme is implicite if it others values are necessar.

• The scheme is said one-step scheme if we use only two values of times (i.e. tiand ti+1), otherwise it will be a multi step scheme.

• The convergence means that the numerical solution of the numerical schemetend to the solution of the ordinary differentiale equation.

• The discretisation scheme Mh,∆t of operator L is consistant if the function φ issmooth enought and

limh,∆t→0

(Mφ−Mh,∆tφ) = 0

• A linear scheme consistant is convergent if only if it is stable.

Bref definition

limh,∆t→0

Bref definition

limh,∆t→0

Bref definition

limh,∆t→0

Bref definition

limh,∆t→0

Nonstandard Finite Difference Scheme (NFDS)

Exact schemes : ExamplesLet

dt= f (t, x) = −λx and x(t0) = x0

The general solution is given by :

x(t) = x0e−λ(t−t0).

then the exact scheme is

uk+1 = e−λhuk .

NFDS : ExamplesLet

dt= f (t, x) = −λx and x(t0) = x0

xk+1 = e−λhxk .

Combining the previous exact schemes, the NFDS scheme is given by :

xk+1 − xk = (e−λh − 1)xk = −λ(

1− e−λh

)xk , (1)

thanxk+1 − xk

φ(h)= −λxk avec φ(h) =

1− e−λh

The Lotka-Volterra system: Euler’s Method{ xk+1−xkφ1(h)

= axk − bxk+1yk ,yk+1−ykφ2(h)

= −cyk+1 + dxk+1yk ,

φ1(h) = −1− e−λh

λand φ2(h) = −

1− e−λh

Figure: Phase plane and time history : Euler explicit method and NFDS method ofLotka-Volterra system (a=1;b=0.01;c=1;d=0.02)

The Lotka-Volterra system: Euler’s Method{ xk+1−xkφ1(h)

= axk − bxk+1yk ,yk+1−ykφ2(h)

= −cyk+1 + dxk+1yk ,

φ1(h) = −1− e−λh

λand φ2(h) = −

1− e−λh

Figure: Phase plane and time history : Euler explicit method and NFDS method ofLotka-Volterra system (a=1;b=0.01;c=1;d=0.02)

Mikens Rules

Pr. Ronald Mickens (Atlanta, USA) used the construction of exact schemes topropose ”simple” rules to develop nonstandard finite difference schemes for differentialequations and even partial differential equations.

• Rule 1 The orders of the discrete derivatives should be equal to the orders of thecorresponding derivatives of the differential equation.

• Rule 2 Denominator functions for the discrete derivatives must, in general, beexpressed in terms of more complicated functions of the step-size than thoseconventionally used

• Rule 3 Nonlinear terms should be, in general, be replaced by nonlocal discreterepresentations.

• Rule 4 Special conditions that hold for the solutions of the differential equationsshould also hold for the solutions of the finite difference scheme.

• Rule 5 The scheme should not introduce extraneous or spurious solutions.

• Rule 6 For N differential system, it could be useful to construct nonstandardschemes for subsystems of M < N differential equations and to combine themto obtain a consistant scheme.

Mikens Rules

Some examples of NFDS of ODE

Equations NFDS Denominatorsdxdt

= −λx xk+1−xkφ(h)

= −λxk φ(h) = 1−e−λh

= −λx xk+1−xkφ(h)

= −λxk+1 φ(h) = eλh−1λ

= x2 xk+1−xkφ(h)

= xkxk+1 φ(h) = h

= λ1x − λ2x2 xk+1−xkφ(h)

= λ1xk − λ2xkxk+1 φ(h) = eλ1h−1λ1

= λ1x − λ2x2 xk+1−xkφ(h)

= λ1xk+1 − λ2xkxk+1 φ(h) = 1−e−λ1h

d2xdt2 + ω2x = 0

xk+1−2xk+xk−1

φ2(h)+ ω2xk = 0 φ(h) = 2

ωsin(

)d2xdt2 = λ dx

xk+1−2xk+xk−1

φ1(h)= λ

xk+1−xkφ2(h)

φ1(h) = ( eλh−1λ

)h;φ2(h) = h

Fundamental Rules?

• it is fundamental to follow rules 2 and 3. The others rules then follow...

• to build the best numerical scheme we need a deep theoretical study of thecontinuous problem is helpful to capture the properties of the solution and theproblem

Re-formulation for applying GAS in dynamical systems: Kamgang-Sallet

dt= A(x)x + f , x(0) = x0,

= (1− π)ρ− (τ + ϑ+ µ)S ,dVdt

= πρ+ τS − (µ+ γ)V ,dIdt

= ϑS − (α+ δ + µ+ κ)I ,dJdt

= δI − µJ,dGdt

= ξI − ηG ,dRdt

= γV + αI − µR,

ϑ = βhI

N+ βG

G + K, y = (S ,V ,R), z = (I , J,G)

= A1(x)(y − y∗) + A12(x)z,dzdt

= A2(x)z,

Application of GAS with numerical dynamical systems : Kamgang and Sallet

φ(h) ≤ min

µ+ γ,

α+ δ + µ+ κ,

∃Q such that Q ≥ max1

2{−λi}, i = 1, · · · , 5,

chose of φ(h) is given by

φ(h) =ϕ(Qh)

Qavec ϕ(z) = 1− e−z , ∀ z ∈ R+.

Implicit scheme

Sk+1−Sk

φ(h)= (1− π)ρ− (τ + βh

Nk + βGGk

Gk+K+ µ)Sk+1,

V k+1−V k

φ(h)= πρ+ τSk+1 − (µ+ γ)V k+1,

I k+1−I k

φ(h)= (βh

Nk + βGGk

Gk+K)Sk+1 − (α+ δ + µ+ κ)I k+1,

Jk+1−Jk

φ(h)= δI k+1 − µJk+1,

Gk+1−Gk

φ(h)= ξI k+1 − ηG k+1,

Rk+1−Rk

φ(h)= γV k+1 + αI k+1 − µRk+1,

yk+1−yk

φ(h)= A1(xk )(yk+1 − y∗) + A12(xk )zk+1,

zk+1−zk

φ(h)= A2(xk )zk+1,

explicit scheme

Sk+1−Sk

φ(h)= (1− π)ρ− (τ + βh

Nk + βGGk

Gk+K+ µ)Sk+1,

V k+1−V k

φ(h)= πρ+ τSk − (µ+ γ)V k ,

I k+1−I k

φ(h)= (βh

Nk + βGGk

Gk+K)Sk+1 − (α+ δ + µ+ κ)I k ,

Jk+1−Jk

φ(h)= δI k − µJk ,

Gk+1−Gk

φ(h)= ξI k − ηG k ,

Rk+1−Rk

φ(h)= γV k + αI k − µRk ,

yk+1−yk

φ(h)= A1(xk )(yk − y∗) + A12(xk )zk ,

zk+1−zk

φ(h)= A2(xk )zk ,

Simulations

Figure: R0 > 1.

Simulations

Figure: R0 < 1.

Comparison

Figure: Comparison between NFDS and RK4

Comparison

Figure: Comparison between NFDS and RK4

introduction to standard and non-standard numerical methods

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