Week 3.5 - Deriving 4th order approximation of a 2nd order derivative using Taylor Table method
Objective :To derive the 4th order approximations scheme of the second order derivative for
1. Central difference
2. Skewed right sided difference
3. Skewed left sided difference
and also compare these errors with forward and backward difference of first order derivative and second order approximation .
Deriving the 4th order approximations scheme of the second order derivative for,
1. Central difference :
To compute fourth order central difference approximation of second order derivative at a point x we need 4 extra points with the current point i.e; 2 on the left and 2 on the right.
Now, we are considering the hypothesis equation as
Then calculate the values of a,b,c,d,e ,then you get the required approximation.These values can be found out by using taylors table.
Since we are trying to compute second order derivative the only terms you should have on the right hand side is f''(x) that means sum of columns in the table for the first 2 columns should be equal to 0 and third column =1 and 4th and 5th column =0 and in the orginal hypothesis you have 5 unknowns which means you need only 5 equations to solve the system and remaining what they sum upto we dont know we just ignore higher order terms.
then we get , a+b+c+d+e =0 -----(1)
-2a-b+0+d+2e=0 -----(2)
4a+b+0+d+4e=2 -------(3)
-8a-b+0+d+8e=0 ------(4)
16a+b+0+d+16e=0 -----(5)
now writing these equations in matrix form A X = B and solving these using matlab by X =inv(A)*B or X =A\B we get the values for a,b,c,d,e
we get, a=-0.0833;
b=1.333;
c=-2.50;
d=1.333;
e=-0.0833;
As we have to equate the coefficients of dx^2 .f''(u) to 1 ,then our hypothesis equation should be divided by dx^2 to get the required
To show that the derived scheme is of fourth order approximation,we use the formula,
The order of accuracy P = Number of nodes in the stencil N - order of derivative Q + 1 for symmetric/central schemes
p = 5-2+1 =4
Hence the order of derived scheme is fourth order approximation. we can also verify this by expanding the orginal hypothesis function,
Hence here a =b and b=d the coefficient terms of the fifth derivative get cancelled and =0, then the remaining terms are
Then divide the equations by dx2 on both sides we get, then we can see that in first term of higher order terms, the power of dx will become 4. Then we will ignore the higher order terms i.e., order of 4. Then,
Hence, we can conclude that our derived scheme is of 4th order accurate.
2. Skewed right sided difference :
To compute fourth order Skewed right sided difference approximation of second order derivative at a point x we need 5 extra points with the current point i.e; 6 on the right.
Now, we are considering the hypothesis equation as
Then calculate the values of a,b,c,d,e,g ,then you get the required approximation.These values can be found out by using taylors table.
Since we are trying to compute second order derivative the only terms you should have on the right hand side is f''(x) that means sum of columns in the table for the first 2 columns should be equal to 0 and third column =1 and 4th and 5th and 6th column =0 and in the orginal hypothesis you have 6 unknowns which means you need only 6 equations to solve the system and remaining what they sum upto we dont know we just ignore higher order terms.
then we get , a+b+c+d+e+g =0 -----(1)
0-b+2c+3d+4e+5g=0 -----(2)
0+b+4c+9d+16e+25g=2 -------(3)
0+b+8c+27d+64e+125g=0 ------(4)
0+b+16c+81d+256e+625g=0 -----(5)
0+b+32c+243d+1024e+3125g =0 -------(6)
now writing these equations in matrix form A X = B and solving these using matlab by X =inv(A)*B or X =A\B we get the values for a,b,c,d,e,g
we get , a= 3.7500;
b= -12.8333;
c = 17.8333;
d = -13.0000;
e = 5.0833;
g = -0.8333;
As we have to equate the coefficients of dx^2 .f''(u) to 1 ,then our hypothesis equation should be divided by dx^2 to get the required
To show that the derived scheme is of fourth order approximation,we use the formula,
The order of accuracy P = Number of nodes in the stencil N - order of derivative Q for one-sided/skewed schemes
p = 6-2 =4
Hence the order of derived scheme is fourth order approximation.we can also verify this by expanding the orginal hypothesis function,
Then divide the equations by dx2 on both sides we get, then we can see that in first term of higher order terms, the power of dx will become 4. Then we will ignore the higher order terms i.e., order of 4. Then,
Hence, we can conclude that our derived scheme is of 4th order accurate.
3. Skewed left sided difference :
To compute fourth order Skewed left sided difference approximation of second order derivative at a point x we need 5 extra points with the current point i.e; 6 on the left .
Now, we are considering the hypothesis equation as
Then calculate the values of a,b,c,d,e,g ,then you get the required approximation.These values can be found out by using taylors table.
Since we are trying to compute second order derivative the only terms you should have on the right hand side is f''(x) that means sum of columns in the table for the first 2 columns should be equal to 0 and third column =1 and 4th,5th and 6th column =0 and in the orginal hypothesis you have 6 unknowns which means you need only 6 equations to solve the system and remaining what they sum upto we dont know we just ignore higher order terms.
then we get , a+b+c+d+e+g =0 -----(1)
-5a-4b-3c-2d-e+0=0 -----(2)
25a+16b+9c+4d+e+0=2 -------(3)
-125a-64b-27c-8d-e+0=0 ------(4)
625a+256b+81c+16d+e+0=0 -----(5)
-3125a-1024a-243c-32d-e+0=0 -------(6)
now writing these equations in matrix form A X = B and solving these using matlab by X =inv(A)*B or X =A\B we get the values for a,b,c,d,e
we get , a= -0.8333;
b= 5.0833;
c = -13.0000;
d = 17.8333;
e = -12.8333;
g = 3.7500;
As we have to equate the coefficients of dx^2 .f''(u) to 1 ,then our hypothesis equation should be divided by dx^2 to get the required,
To show that the derived scheme is of fourth order approximation,we use the formula,
The order of accuracy P = Number of nodes in the stencil N - order of derivative Q for one-sided/skewed schemes
p = 6-2 =4
Hence the order of derived scheme is fourth order approximation.we can also verify this by expanding the orginal hypothesis function,
Then divide the equations by dx2 on both sides we get, then we can see that in first term of higher order terms, the power of dx will become 4. Then we will ignore the higher order terms i.e., order of 4. Then,
Hence, we can conclude that our derived scheme is of 4th order accurate.
Now let's compare how the error varies for these schemes when dx is decreased using matlab code.
Matlab code :
clear all
close all
clc
% defining the function
f=(@(x)exp(x)*cos(x));
% assuming value of x
x=pi;
%range of dx(grid sizing)
dx = linspace(pi/10,pi/10000,50);
%Analytical Derivative of function exp(x)*cos(x)
Analytical_derivative = -2*exp(x)*sin(x);
% computing the value of 4th order approximations of different schemes of
% the second order derivative for given function.
for i = 1 :length(dx)
central_difference(i) = (-0.0833*f(x-2*dx(i))+1.3333*f(x-dx(i))-2.5*f(x)+1.3333*f(x+dx(i))-0.0833*f(x+2*dx(i)))/dx(i)^2;
Skewed_right_sided_difference(i) =(3.75*f(x)-12.8333*f(x+dx(i))+17.8333*f(x+2*dx(i))-13*f(x+3*dx(i))+5.0833*f(x+4*dx(i))-0.8333*f(x+5*dx(i)))/dx(i)^2;
Skewed_left_sided_difference(i) =( -0.8333*f(x-5*dx(i))+5.0833*f(x-4*dx(i))-13*f(x-3*dx(i))+17.8333*f(x-2*dx(i))-12.8333*f(x-dx(i))+3.75*f(x))/dx(i)^2;
end
% calculating the absolute error
central_difference_error =abs(Analytical_derivative-central_difference);
Skewed_right_sided_difference_error =abs(Analytical_derivative-Skewed_right_sided_difference);
Skewed_left_sided_difference_error =abs(Analytical_derivative-Skewed_left_sided_difference);
% plotting graph dx vs value of error for derived fourth order schemes
loglog(dx,central_difference_error)
hold on
loglog(dx,Skewed_right_sided_difference_error)
hold on
loglog(dx,Skewed_left_sided_difference_error)
title('variation of error as a factor of dx for various fourth order schemes for given function at x=pi')
xlabel('dx(grid size)')
ylabel('value of error')
legend('central difference error','Skewed right sided difference error','Skewed left sided difference error')Conclusion :
Finally we can conclude that by seeing the above plots that how error is varying as a factor of given dx for various fourth order schemes,The error is decresasing as grid size is decreasing and also central difference scheme is more accurate than skewed right side and skewed left side scheme.The higher the order of the method the lower is the error is going to be for the same grid size.But in reality when the geometry become very complex and when the flow becomes very complex you cannot use higher order method always, and inorder for the simulation to run the commercial codes always go to first order schemes inorder to get a solution.
Importance of skewed scheme :
When we want to calculate the value of a function at the boundary we cannot use central difference schemes because it requires the value at the nodes on both sides whereas in skewed schemes only one sided values are used .This is where skewed schemes come in to play.This is what can a skewed scheme can do that a central difference scheme cannot, but when it comes to accuracy point of view as we have seen above central difference is more accurate than skewed schemes for the same order of error.