introduction to numerical analysisocw.snu.ac.kr/sites/default/files/note/lecture 10_0.pdf · 10.5...
TRANSCRIPT
Cho, Hyoung Kyu
Department of Nuclear EngineeringSeoul National University
Cho, Hyoung Kyu
Department of Nuclear EngineeringSeoul National University
INTRODUCTION TO NUMERICAL ANALYSIS
10. NUMERICAL INTEGRATION
10.1 Background10.2 Euler's Methods10.3 Modified Euler's Method10.4 Midpoint Method10.5 Runge‐Kutta Methods10.6 Multistep Methods10.7 Predictor‐Corrector Methods10.8 System of First‐Order Ordinary Differential Equations10.9 Solving a Higher‐Order Initial Value Problem10.10 Use of MATLAB Built‐In Functions for Solving Initial‐Value Problems
10.11 Local Truncation Error in Second‐Order Range‐Kutta Method
10.12 Step Size for Desired Accuracy10.13 Stability10.14 Stiff Ordinary Differential Equations
10.1 Background
Ordinary differential equation A differential equation that has one independent variable A first‐order ODE
The first derivative of the dependent variable with respect to the independent variable
Example Rates of water inflow and outflow
The time rate of change of the mass in the tank
Equation for the rate of height change
10.1 Background
Time dependent problem
Independent variable: time Dependent variable: water level
To obtain a specific solution, a first‐order ODE must have an initial condition or constraint that specifies the value of the dependent variable at a particular value of the independent variable.
In typical time‐dependent problems Initial condition Initial value problem (IVP)
10.1 Background
First order ODE statement General form
Ex)
Flow lines
Analytical solution In many situations an analytical solution is not possible!
Numerical solution of a first‐order ODE A set of discrete points that approximate the function y(x) Domain of the solution: , N subintervals
10.1 Background
Overview of numerical methods used/or solving a first‐order ODE Start from the initial value Then, estimate the value at a second nearby point third point …
Single‐step and multistep approach Single step approach: Multistep approach: ⋯ , , ,
Explicit and implicit approach Right hand side in explicit method: known values
Right hand side in implicit method: unknown value
– In general, non‐linear equation non‐linear equation solution method !
Implicit methods provide improved accuracy over explicit methods, but require more effort at each step.
10.1 Background
Errors in numerical solution of ODEs Round‐off errors Truncation errors
Numerical solution of a differential equation calculated in increments (steps)
Local truncation error: in a single step Propagated, or accumulated, truncation error
– Accumulation of local truncation errors from previous steps
Single‐step explicit methods
Euler’s explicit method: slope at , Modified Euler’s explicit method: average slope at , and , Midpoint method: slope at /2 Runge‐Kutta methods
A weighted average of estimates of the slope of at several points
10.2 Euler’s Method
Euler’s method Simplest technique for solving a first‐order ODE
Explicit or implicit
Euler's Explicit Method
The error in this method depends on the value of and is smaller for smaller h.
Derivation Numerical integration or finite difference approximation of the derivative
step size is exaggerated !
(rectangle method)
(forward Euler method)
10.2 Euler’s Method
Example 10‐1: Solving a first‐order ODE using Euler's explicit method.
10.2 Euler’s Method
Example 10‐1: Solving a first‐order ODE using Euler's explicit method.
% Solving Example 8-1clear alla=0; b=2.5; h=0.1; yINI = 3;[x, y] = odeEULER(@Chap8Exmp1ODE,a,b,h,yINI);xp=a:0.1:b;yp=70/9*exp(-0.3*xp)-43/9*exp(-1.2*xp);plot(x,y,'--b',xp,yp)xlabel('x'); ylabel('y')
function [x, y] = odeEULER(ODE,a,b,h,yINI)
x(1) = a; y(1) = yINI;N = (b-a)/h;for i = 1:N
x(i+1) = x(i) + h;y(i+1) = y(i) + ODE(x(i),y(i))*h;
end
function dydx = Chap8Exmp1ODE(x,y)dydx = -1.2*y + 7*exp(-0.3*x);
10.2 Euler’s Method
Analysis of truncation error in Euler's explicit method Local truncation error Global truncation error
Taylor series expansion at position 1
Numerical solution
Local truncation error
Total error The difference between the numerical solution
and the true solution.
10.2 Euler’s Method
Analysis of truncation error in Euler's explicit method Global truncation error
Truncation error is propagated or accumulated !
Mean value theorem
10.2 Euler’s Method
Analysis of truncation error in Euler's explicit method Global truncation error
Suppose
Propagation of error– At the first point to the second point to the third point
– At the fourth point
– At the point
Suppose
10.2 Euler’s Method
Analysis of truncation error in Euler's explicit method Global truncation error
Suppose
– Difficult to determine the order of magnitude directly– Possible to determine the bound
∵
10.2 Euler’s Method
Euler's implicit method
In general, this equation is non‐linear! Must be solved with a numerical solution method
In the derivation Backward difference formula for the derivative backward Euler method
The local and global truncation errors Same as those in the explicit method
10.2 Euler’s Method
Example 10‐2: Solving a first‐order ODE using Euler's implicit method
To solve the non‐linear equationusing Newton method
Iteration function
10.2 Euler’s Method
Example 10‐2: Solving a first‐order ODE using Euler's implicit method% Solving First Order ODE with Euler's implicit Method.clear alla = 0; b = 0.5; h = 0.002;N = (b - a)/h;n(1) = 2000; t(1) = a;for i=1:N
t(i+1) = t(i) + h;x = n(i);
% Newton's method starts. for j = 1:20
num = x +0.800*x^(3/2)*h - 10.0*n(1)...*(1-exp(-3*t(i+1)))*h - n(i);
denom = 1 + 0.800*1.5*x^(1/2)*h;xnew = x - num/denom;if abs((xnew - x)/x) < 0.0001
breakelse
x = xnew;end
endif j == 20
fprintf('Numerical solution could no be calculated at t = %g s', t(i))break
end% Newton's method ends.
n(i+1) = xnew;endplot(t,n)axis([0 0.5 0 2000]), xlabel('t (s)'), ylabel('n')
10.3 Modified Euler’s Method
Modified Euler’s Method Main assumption in Explicit method
Constant derivative (slope) between , and , Equal to the derivative at point ,
Modified Euler method To include the effect of slope changes within the subinterval Uses the average of the slope at points , and ,
Slope at the beginning: at point ,
Using Euler’s explicit method
Estimating slope at the end of interval: at point ,
main source of error
10.3 Modified Euler’s Method
Modified Euler’s Method Better estimation of with new slope
Also derived by integrating ODE using the trapezoidal method
Algorithm1.2.3.4.
10.3 Modified Euler’s Method
Example 10‐3: Solving a first‐order ODE using the modified Euler method.
function [x, y] = odeModEuler(ODE,a,b,h,yINI)
x(1) = a; y(1) = yINI;N = (b-a)/h;for i = 1:N
x(i+1) = x(i) + h;SlopeEu = ODE(x(i),y(i));yEu = y(i) + SlopeEu*h;SlopeEnd = ODE(x(i+1),yEu);y(i+1) = y(i) + (SlopeEu+SlopeEnd)*h/2;
end
Explicit Euler Modified Euler
10.4 Midpoint Method
Midpoint method Another modification of Euler’s explicit method Slope at the middle point of the interval Two steps
1. Calculate
2. Estimate slope at the midpoint ,
3. Calculate numerical solution :
10.5 Runge‐Kutta Methods
Runge‐Kutta Methods A family of single‐step, explicit, numerical techniques for a first‐order ODE
Slope: obtained by considering the slope at several points within the subinterval Different orders of Runge‐Kutta method the number of points within the subinterval
– Second‐order RK: two points– Third‐order RK: three points– Fourth order RK (classical RK): four points
Global truncation error– Second‐order RK: second‐order accurate globally; third order accurate locally
More accurate than simple Euler’s explicit method
However, require several evaluations of function for the derivative
10.5 Runge‐Kutta Methods
Second‐order Runge‐Kutta Methods General form
The values of these constants vary with the specific second‐order method. Modified Euler method and the midpoint method
– Two versions of a second‐order RK method
Modified Euler method: , , 1, 1
Midpoint method: 0, 1, ,
10.5 Runge‐Kutta Methods
Second‐order Runge‐Kutta Methods Heun’s method
, , ,
Truncation error in second‐order RK methods Local truncation error: Global truncation error: A larger step size can be used for the same accuracy !
However, the function , is calculated twice.
10.5 Runge‐Kutta Methods
Second‐order Runge‐Kutta Methods and Talyor series expansion
First and second derivatives: given by the differential equation
Then,
General form of RK
10.5 Runge‐Kutta Methods
Second‐order Runge‐Kutta Methods and Talyor series expansion Two different equations for
Three equations with four unknowns
Modified Euler: , , 1, 1
Midpoint method: 0, 1, ,
Heun’s: , , ,
10.5 Runge‐Kutta Methods
Example 10‐4: Solving by hand a first‐order ODE using the second‐order RungeKutta method
10.5 Runge‐Kutta Methods
Third‐order RK methods General form
, , , , , , , Four term Talyor series expansion
Classical third‐order RK method
, , , , 1, , 1, 2
10.5 Runge‐Kutta Methods
Third‐order RK methods Truncation error
Local truncation error: Global truncation error:
Other third‐order RK method
10.5 Runge‐Kutta Methods
Fourth‐order RK methods General form
13 constants: , , , , , , , , , , , ,
Classical fourth‐order RK method
10.5 Runge‐Kutta Methods
Fourth‐order RK methods
10.5 Runge‐Kutta Methods
Fourth‐order RK methods Truncation error
Local truncation error: Global truncation error:
10.5 Runge‐Kutta Methods
Example 10‐5: Solving by hand a first‐order ODE using the fourth‐order Runge‐Kutta method
Example 10‐6: A user‐defined function for solving a first‐order ODE using the fourth‐order Runge‐Kutta method.
10.5 Runge‐Kutta Methods
Example 10‐6: A user‐defined function for solving a first‐order ODE using the fourth‐order Runge‐Kutta method.
clear alla=0; b=2.5;h=0.5; yIni=3;[x,y] = odeRK4(@Chap8Exmp6ODE,a,b,h,yIni)xp=a:0.1:b;yp=70/9*exp(-0.3*xp)-43/9*exp(-1.2*xp);plot(x,y,'*r',xp,yp)yExact=70/9*exp(-0.3*x)-43/9*exp(-1.2*x )error=yExact-y
function [x, y]… = odeRK4(ODE,a,b,h,yIni)
x(1) = a; y(1) = yIni;n = (b-a)/h;for i = 1:n
x(i+1) = x(i) + h;K1 = ODE(x(i),y(i));xhalf = x(i) + h/2;yK1 = y(i) + K1*h/2;K2 = ODE(xhalf,yK1);yK2 = y(i) + K2*h/2;K3 = ODE(xhalf,yK2);yK3 = y(i) + K3*h;K4 = ODE(x(i+1),yK3);y(i+1) = y(i) + ...(K1 + 2*K2 + 2*K3 + K4)*h/6;
end
function dydx = Chap8Exmp6ODE(x,y)dydx = -1.2*y + 7*exp(-0.3*x);
10.6 Multistep Methods
Single‐step method Use only the value of and at the previous point
Multi‐step method Two or more previous points Explicit and implicit methods
Explicit:– The first few points can be determined by single‐step methods or by multistep methods that use fewer
prior points.
Implicit method: appears on both sides. non‐linear equation solution method!
Adams‐Bashforth Method Adams‐Moulton Method
10.6 Multistep Methods
Adams‐Bashforth Method Explicit multistep method for solving a first‐order ODE Second‐order formula uses
, and , Third‐order formula uses
, , , , ,
Integration of ODE
Integration is carried out with a polynomial that interpolates the value of , at , and at previous points.
10.6 Multistep Methods
Adams‐Bashforth Method Second‐order AB method
First order polynomial
Integration
Third‐order AB method
Fourth‐order AB method
Euler’s explicit method
10.6 Multistep Methods
Adams‐Moulton Method Implicit multistep method for solving first‐order ODEs
Second‐order formula
Implicit form of the modified Euler method
Third‐order and fourth formulas
Two ways to use AM method If they are used by themselves, they have to be solved numerically. Usually, however, they are used in conjunction with other equations in methods that are called
predictor‐corrector methods.
10.7 Predictor‐Corrector Methods
Predictor‐corrector methods Solving ODEs using two formulas; predictor and corrector formulas
Predictor step Explicit formula Used first to determine an estimate of the solution is calculated from the known solution at the previous point , . Single‐step method or multistep methods
Corrector step Uses the estimated value of on the right‐hand The corrector equation, which is usually an implicit equation, is being used in an explicit manner. This scheme utilizes the benefits of the implicit formula while avoiding the difficulties associated
with solving an implicit equation directly. Furthermore, the application of the corrector can be repeated several times such that the new
value of is substituted back on the right‐hand side of the corrector formula to obtain a more refined value for .
10.7 Predictor‐Corrector Methods
Predictor‐corrector methods Algorithm
1. Calculate using an explicit method2. Substitute from Step 1, as well as any required values from the already known solution at
previous points, in the right‐hand side of an implicit formula to obtain a refined value for .3. Repeat Step 2 by substituting the refined value of back in the implicit formula, to obtain an
even more refined value for .4. Step 2 can be repeated as many times as necessary to produce the desired level of accuracy, that
is, until further repetitions do not change the answer for to a specified number of decimal places.
The simplest example of a predictor‐corrector method: modified Euler method
10.7 Predictor‐Corrector Methods
Predictor‐corrector methods Modified Euler predictor‐corrector method
1. Calculate a first estimate for using Euler's explicit method as a predictor:
2. Calculate better estimates for repetitively as a corrector:
3. Stop the iterations when
10.7 Predictor‐Corrector Methods
Predictor‐corrector methods Adams‐Bashforth and Adams‐Moulton predictor‐corrector methods
AB method: explicit method AM method: implicit method
Can be used together in a predictor‐corrector method!
Predictor equation with the third‐order formula
Corrector equations
10.8 System of First‐order ODE System of coupled first‐order ODEs
In many of these instances there is a need to solve a system of coupled first‐order ODEs. Initial value problems involving ODEs of second and higher orders
Solved by converting the equation into a system of first‐order equations
Predator‐prey problem Suppose a community consists of lions (predators) and gazelles (prey).
In chemical reactions
10.8 System of First‐order ODE General form of a system of first‐order ODEs
With a single‐step explicit method
Euler's explicit method The modified Euler method The fourth‐order Runge‐Kutta method
10.8 System of First‐order ODE General form of a system of first‐order ODEs
Euler's explicit method
Initial conditions: and z
The modified Euler method
10.8 System of First‐order ODE Example 10‐7: Solving a system of two first‐order ODEs using Euler's explicit method
and second‐order Runge‐Kutta method.
10.8 System of First‐order ODE Example 10‐7: Solving a system of two first‐order ODEs using Euler's explicit method
and second‐order Runge‐Kutta method.
cleara = 0; b = 3; yINI = 3; zINI = 0.2; h = 0.1;[x, y, z] = Sys2ODEsRK2(@odeExample7dydx,@odeExample7dzdx,a,b,h,yINI,zINI);% Data from part (a)xa = [0 0.25 0.5 0.75];ya = [3 1.472 1.374 1.427];za = [0.2 0.94 1.087 1.135];% Data from part (b)xb = [0 0.25 0.5];yb = [3 2.187 1.903];zb = [0.2 0.6436 0.9230];plot(x,y,'-k',x,z,'-r',xa,ya,'*k',xa,za,'*r',xb,yb,'ok',xb,zb,'or')axis([0 3 0 4])
function dzdx = odeExample7dzdx(x,y,z)dzdx = y-z^2;
function dydx = odeExample7dydx(x,y,z)dydx = (-y+z)*exp(1-x)+0.5*y;
10.8 System of First‐order ODE Example 10‐7: Solving a system of two first‐order ODEs using Euler's explicit method
and second‐order Runge‐Kutta method.
function [x, y, z] = Sys2ODEsRK2(ODE1,ODE2,a,b,h,yINI,zINI)
x(1) = a; y(1) = yINI; z(1) = zINI;N = (b - a)/h;for i = 1:N
x(i+1) = x(i) + h;Ky1 = ODE1(x(i),y(i),z(i));Kz1 = ODE2(x(i),y(i),z(i));Ky2 = ODE1(x(i+1),y(i)+Ky1*h,z(i)+Kz1*h);Kz2 = ODE2(x(i+1),y(i)+Ky1*h,z(i)+Kz1*h);y(i+1) = y(i) + (Ky1 + Ky2)*h/2;z(i+1) = z(i) + (Kz1 + Kz2)*h/2;
end
10.8 System of First‐order ODE Solving a System of First‐Order ODEs Using the Classical Fourth‐Order RK Method
Same as the second‐order method except the number of constants
13 constants: , , , , , , , , , , , ,
Initial conditions: , z ,
10.8 System of First‐order ODE Solving a System of First‐Order ODEs Using the Classical Fourth‐Order RK Method