All Courses
All Courses
Courses by Software
Courses by Semester
Courses by Domain
Tool-focused Courses
Machine learning
POPULAR COURSES
Success Stories
Objective Derive the following 4th order approximations of the second order derivative 1. Central difference 2. Skewed right sided difference 3. Skewed left sided difference 4. plot the graph for diffrence of error for above scheme Derivation of central diffrence For derive 4th order approximation of second order derivative.folllowing…
Dipakv Virkarwe
updated on 15 Dec 2019
Objective
Derive the following 4th order approximations of the second order derivative
1. Central difference
2. Skewed right sided difference
3. Skewed left sided difference
4. plot the graph for diffrence of error for above scheme
Derivation of central diffrence
For derive 4th order approximation of second order derivative.folllowing node are consider for central diffrence method. nodes are i,(i-2),(i-1),(i+2),(i+1). the value of af(i-2),b(i-1)..etc find out by taylor series expansion. like this there is use of taylor series expansion for skewed left & right 4th order approx.
af(i-2)=a[fi-2Δ
Taylor table for find out 4th order approximation
stensil |
f(i) |
dxf\'(i) |
dx^2f\'\'(i) |
dx^3f\'\'\'(i) |
dx^4f^iv(i) |
dx^5f^v(i) |
dx^6f^vi(i) |
af(i-2) |
a |
-2a |
4a/2 |
-8a/2 |
16a/24 |
-32a/120 |
64a/720 |
bf(i-1) |
b |
-b |
b/2 |
-b/6 |
b/24 |
-b/120 |
b/240 |
cf(i) |
c |
0 |
0 |
0 |
0 |
0 |
0 |
df(i+1) |
d |
d |
d/2 |
d/6 |
d/6 |
-d/120 |
d/720 |
ef(i+2) |
e |
2e |
4e/2 |
8e/6 |
8e/6 |
-32e/120 |
64e/720 |
sum |
0 |
0 |
1 |
0 |
0 |
? |
? |
after that made 5 equation for finding the value a,b,c,d,e
equation1: a+b+c+d+e=0
equation2: (-2a)-b+0+d+2e=0
equation3: 2a+b/2+0+d/2+4e/2=1
equaion4: (-4a)+(-b/6)+0+d/6+8e/6=0
equation5: (16a/24)+b/24+0+d/6+8e/6=0
the value of a,b,c,d,e is find out by use of matrix in matlab
A=[1 1 1 1 1;
-2 -1 0 1 2;
2 1/2 0 1/2 2;
-8/6 -1/6 0 1/6 8/6;
16/24 1/24 0 1/24 16/24];
B=[0;0;1;0;0];
x=A\\B;
disp(x)
command window shows the value for coefficent
a=-0.0833, b=1.3333, c=-2.5000, d=1.3333, e=-0.0833
after putting the value in in main equation we got final equation of 4th order approx.
(d^2f)/dx^2= (-0.8333f_(i-2)+1.333f_(i-1)-2.5f_(i)+1.33f_(i+1)-0.0833f_(i+2))/dx^2
Derivation for Skewed right side approximation
For derive 4th order approximation of second order derivative.folllowing node are consider for skewed right side approx. nodes are i,(i+1),(i+2),(i+3),(i+4).
|
f(i) |
dxf\'(i) |
dx^2f\'\'(i) |
dx^3f\'\'\'(i) |
dx^4f^iv(i) |
dx^5f^v(i) |
dx^6f^vi(i) |
af(i) |
a |
0 |
0 |
0 |
0 |
0 |
0 |
bf(i+1) |
b |
b |
b/2 |
b/6 |
b/24 |
b/120 |
b/720 |
cf(i+2) |
c |
2c |
4c/2 |
8c/2 |
16c/24 |
32c/120 |
64c/720 |
df(i+3) |
d |
3d |
9d/2 |
27d/6 |
81d/24 |
243d/120 |
729d/720 |
ef(i+4) |
e |
4e |
16e/2 |
64e/6 |
256e/24 |
1024e/120 |
4096e/720 |
sum |
0 |
0 |
1 |
0 |
0 |
? |
? |
after that made 5 equation for finding the value a,b,c,d,e
equation1: a+b+c+d+e=0
equation2: 0+b+2c+3d+4e=0
equation3: 0+b/2+4c/2+9d/2+16e/2=1
equaion4: 0+(b/6)+(8c/2)+27d/6+64e/6=0
equation5: 0+(b/24)+16c/24+81d/24+256e/24=0
a,b,c,d,e value find out by use of matrix in matlab
A=[1 1 1 1 1;
0 1 2 3 4;
0 1/2 2 9/2 8;
0 1/6 8/6 27/6 64/6;
0 1/24 16/24 81/24 256/24];
B=[0;0;1;0;0];
x=A\\B;
disp(x)
command window shows the value for coefficent
a=2.9167 b=-8.6667 , c=9.5000 , d=-4.6667 , e=0.9167
after putting the value in in main equation we got final equation of 4th order approx.
(d^2f)/dx^2= (2.9167f_(i)-8.6667f_(i+1)+9.5f_(i+2)-4.6667f_(i+3)+0.9167f_(i+4))/dx^2
Derivation for Skewed left side approximation
For derive 4th order approximation of second order derivative.folllowing node are consider for skewed left side approx. nodes are i,(i-1),(i-2),(i-3),(i-4).
stensil |
f(i) |
dxf\'(i) |
dx^2f\'\'(i) |
dx^3f\'\'\'(i) |
dx^4f^iv(i) |
dx^5f^v(i) |
dx^6f^vi(i) |
af(i-4) |
a |
4a |
16a/2 |
-64a/6 |
256a/24 |
-1024a/120 |
4096a/720 |
bf(i-3) |
b |
-3b |
9b/2 |
-27b/6 |
81b/24 |
-243b/120 |
729b/720 |
cf(i-2) |
c |
-2c |
4c/2 |
-8c/6 |
16c/24 |
-32c/120 |
64c/720 |
df(i-1) |
d |
-d |
d/2 |
-d/6 |
d/24 |
-d/120 |
d/720 |
ef(i) |
e |
0 |
0 |
0 |
0 |
0 |
0 |
sum |
0 |
0 |
1 |
0 |
0 |
? |
? |
after that made 5 equation for finding the value a,b,c,d,e
equation1: a+b+c+d+e=0
equation2: 4a-3b-2c-d+0=0
equation3: 16a/2+9b/2+4c/2+d/2+0=1
equaion4: (-64a/6)+(-27b/6)+(-8c/6)+(-d/6)+0=0
equation5: (256a/24)+81b/24+16c/24+d/24+0=0
A=[1 1 1 1 1;
-4 -3 -2 -1 0;
8 9/2 4/2 1/2 0;
-64/6 -27/6 -8/6 -1/6 0;
256/24 81/24 16/24 1/24 0];
B=[0;0;1;0;0];
x=A\\B;
disp(x)
command window shows the value for coefficent.
a=2.9167 , b=-8.6667 , c=9.5000 , d=-4.6667 ,e=0.9167
after put the value in main equation we got final equation of 4th order approx.
(d^2f)/dx^2= (2.9167f_(i)-8.6667f_(i-1)+9.5f_(i-2)-4.6667f_(i-3)+0.9167f_(i-4))/dx^2_
Function out code for central diffrence
function out= central_fourth_order_approximation(x,dx)
% of analyatical function=exp(x)*cos(x)
% of analyatical derivative
% f\'\'(x)=-2*e(x)*sin(x)
second_order_analytical_derivative= -2*exp(x)*sin(x);
%numerical derivative
% use of cenrtal diffrencing
% fourth order approximation
central_diffrence=(((-0.08333)*((exp(x-(2*dx)))*(cos(x-(2*dx)))))+((1.3333)*((exp(x-dx))*(cos(x-dx))))-((2.5)*((exp(x))*(cos(x))))+((1.3333)*((exp(x+dx))*(cos(x+dx))))+((-0.0833)*((exp(x+(2*dx)))*(cos(x+(2*dx))))))/(dx^2);
out=abs(central_diffrence - second_order_analytical_derivative);
end
Function out code for skewed right side
function out= skewed_right_side_approximation(x,dx)
% of analyatical function=exp(x)*cos(x)
% of analyatical derivative
% f\'\'(x)=-2*e(x)*sin(x)
second_order_analytical_derivative= -2*exp(x)*sin(x);
%numerical derivative
% use of skewed right side
% fourth order approximation
skewed_right_side =(((2.9167)*((exp(x))*(cos(x))))-((8.6667)*((exp(x+dx))*(cos(x+dx))))+((9.5)*((exp(x+(2*dx))*(cos(x+(2*dx))))))-((4.6667)*((exp(x+(3*dx)))*(cos(x+(3*dx)))))+((0.9167)*((exp(x+(4*dx)))*(cos(x+(4*dx))))))/(dx^2);
out=abs(skewed_right_side- second_order_analytical_derivative);
end
Function out code for skewed left side
function out= skewed_left_side_approximation(x,dx)
% of analyatical function=exp(x)*cos(x)
% of analyatical derivative
% f\'\'(x)=-2*e(x)*sin(x)
second_order_analytical_derivative= -2*exp(x)*sin(x);
%numerical derivative
% use of skewed left side
% fourth order approximation
skewed_left_side =(((0.9167)*((exp(x-(4*dx))*(cos(x-(4*dx))))))-((4.6667)*((exp(x-(3*dx))*(cos(x-(3*dx))))))+((9.5)*((exp(x-(2*dx))*(cos(x-(2*dx))))))-((8.6667)*((exp(x-dx))*(cos(x-dx))))+((2.9167)*((exp(x))*(cos(x)))))/(dx^2);
out=abs(skewed_left_side- second_order_analytical_derivative);
end
Main code
clear all
close all
clc
x=pi/2;
dx=linspace(pi/4,pi/40,10);
second_order_analytical_derivative=-2*sin(x)*exp(x);
for i=1:length(dx)
central_diffrence_error(i)= central_fourth_order_approximation(x,dx(i));
skewd_right_side_error(i)=skewed_right_side_approximation(x,dx(i));
skewd_left_side_error(i)=skewed_left_side_approximation(x,dx(i));
end
figure(1)
loglog(dx,central_diffrence_error,\'b\')
hold on
loglog(dx,skewd_right_side_error,\'r\')
hold on
loglog(dx,skewd_left_side_error,\'g\')
legend(\'central_diffrence_error\',\'skewd_right_side_error\',\'skewd_left_side_error\')
xlabel(\'dx value\')
ylabel(\'error value\')
Plot for comparison of absolute error between CDS, SRS,SLS
result is found at x=pi/2 & dx=linspace(pi/4,pi/40,10). so from graph the it is found that if we use central diffrence method for 4th order approximation there is error is less one compare with the other two method of skewed left & right side method. the central diffrence error gives less error because it take the data from symeterical node point from left & side right side.
why skewed system is useful?
in case of skewed for right side & left side system. it take inforamtion from current point & neighbouring point on left & right side so we can find out any order approximation. but in case of Cental diffrence system if we want to find out higher order approximation at boundry node, in that case there is no node is available on any side of node. , so it is not useful for Cental diffrence system.
Leave a comment
Thanks for choosing to leave a comment. Please keep in mind that all the comments are moderated as per our comment policy, and your email will not be published for privacy reasons. Please leave a personal & meaningful conversation.
Other comments...
Week 10: Project 1 - FULL HYDRO case set up (PFI)
Title: Simulation of Full Hydro 4-Stroke PFI(Port fuel injection) Objective: 1. To understand the entire case setup 2. Simulation of PFI Engine in paraview 3. To calculate the engine…
09 Sep 2020 12:40 PM IST
No-hydro simulation of PFI (Port fuel injection) Engine in CONVERGE CFD
Title: No-hydro simulation by Surface preparation & boundry flagging in CONVERGE CFD Objective : 1. To Understand the boundry flagging & surface preparation on PFI 2. To perform the No-Hydro simulation of PFI Engine …
01 Sep 2020 01:40 PM IST
Literature review on RANS derivation & its Analysis
Title: Literature review on RANS derivation & its Analysis Objective: 1. Derive the RANS derivation 2. Importance of RANS Theory: The Reynolds-averaged Navier–Stokes equations (or RANS equations) are time-averaged equations…
21 Aug 2020 09:19 AM IST
Related Courses
0 Hours of Content
Skill-Lync offers industry relevant advanced engineering courses for engineering students by partnering with industry experts.
© 2025 Skill-Lync Inc. All Rights Reserved.