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Week 3.5 - Deriving 4th order approximation of a 2nd order derivative using Taylor Table method

Derive the $4^{t h}$ $4^(th)$ order approximation of the second-order derivative using central difference scheme let f be the function whose second derivative is approximated at $i^{t h}$ $i^(th)$ point using function values at (i-2), (i-1), i, (i+1), (i+2) points. let h be the distance between the consecutive points. let $\frac{d^{2} f}{{d x}^{2}} = a \cdot f_{i - 2} + b \cdot f_{i - 1} + c \cdot f_{i} + d \cdot f_{i + 1} + e \cdot f_{i + 2} + O (h^{n})$ $(d^2f)/dx^2=a*f_(i-2)+b*f_(i-1)+c*f_i+d*f_(i+1)+e*f_(i+2)+O(h^n)$ …

MATLAB

Jaswanth Kalyan Kumar Alapati
updated on 13 Jul 2021

Derive the $4^{t h}$ $4^(th)$ order approximation of the second-order derivative using central difference scheme

let f be the function whose second derivative is approximated at $i^{t h}$ $i^(th)$ point using function values at (i-2), (i-1), i, (i+1), (i+2) points. let h be the distance between the consecutive points.

let $\frac{d^{2} f}{{d x}^{2}} = a \cdot f_{i - 2} + b \cdot f_{i - 1} + c \cdot f_{i} + d \cdot f_{i + 1} + e \cdot f_{i + 2} + O (h^{n})$ $(d^2f)/dx^2=a*f_(i-2)+b*f_(i-1)+c*f_i+d*f_(i+1)+e*f_(i+2)+O(h^n)$

n is the order of the scheme and is to be determined.

By Taylor's series, $f_{i + 1} = f_{i} + (\frac{h}{1!}) f'_{i} + (\frac{h^{2}}{2!}) f''_{i} + (\frac{h^{3}}{3!}) f'''_{i} + ... . .$ $f_(i+1)=f_i+(h/(1!))f'_i+(h^2/(2!))f''_i+(h^3/(3!))f'''_i+.....$

Applying the above series to $f_{i - 2}, f_{i - 1}, and f_{i + 2}$ $f_(i-2), f_(i-1), and f_(i+2)$ and summing them appropriately gives

$\frac{d^{2} f}{{d x}^{2}} = (a + b + c + d + e) \cdot f_{i} + (- 2 a h - b h + d h + 2 e h) \cdot f'_{i} + (2 a h^{2} + \frac{b h^{2}}{2} + \frac{d h^{2}}{2} + 2 e h^{2}) \cdot f''_{i} + ... ...$ $(d^2f)/dx^2=(a+b+c+d+e)*f_i+(-2ah-bh+dh+2eh)*f'_i+(2ah^2+(bh^2)/2+(dh^2)/2+2eh^2)*f''_i+......$

On equating the corresponding terms of derivatives appropriately it gives

$a + b + c + d + e = 0$ $a+b+c+d+e=0$

$- 2 a h - b h + d h + 2 e h = 0$ $-2ah-bh+dh+2eh=0$

$2 a h^{2} + \frac{b h^{2}}{2} + \frac{d h^{2}}{2} + 2 e h^{2} = 1$ $2ah^2+(bh^2)/2+(dh^2)/2+2eh^2=1$

$\frac{- 8 a h^{3}}{6} - \frac{b h^{3}}{6} + \frac{d h^{3}}{6} + \frac{8 e h^{3}}{6} = 0$ $(-8ah^3)/6-(bh^3)/6+(dh^3)/6+(8eh^3)/6=0$

$\frac{16 a h^{4}}{24} + \frac{b h^{4}}{24} + \frac{d h^{4}}{24} + \frac{16 e h^{4}}{24} = 0$ $(16ah^4)/24+(bh^4)/24+(dh^4)/24+(16eh^4)/24=0$

Solving for the coefficients gives $a = - \frac{1}{12 h^{2}}, b = \frac{4}{3 h^{2}}, c = - \frac{5}{2 h^{2}}, d = \frac{4}{3 h^{2}}, and e = - \frac{1}{12 h^{2}}$ $a=-1/(12h^2), b=4/(3h^2), c=-5/(2h^2), d=4/(3h^2), and e=-1/(12h^2)$

substituting the above values gives n=4.

So $\frac{d^{2} f}{{d x}^{2}} = \frac{- \frac{f_{i - 2}}{12} + \frac{4 f_{i - 1}}{3} - \frac{5 f_{i}}{2} + \frac{4 f_{i + 1}}{3} - \frac{f_{i + 2}}{12}}{h^{2}} + O (h^{4})$ $(d^2f)/(dx^2)=(-f_(i-2)/12+(4f_(i-1))/3-(5f_i)/2+(4f_(i+1))/3-f_(i+2)/12)/h^2+O(h^4)$

Thus the second derivative of a function using a central difference scheme (fourth-order) is derived using Taylor's table.

Following a similar approach to determine the second derivative using forward difference (right-sided) scheme.

let f be the function whose second derivative is approximated at $i^{t h}$ $i^(th)$ using function values at i, (i+1), (i+2), (i+3), (i+4), and (i+5) points. let h be the distance between the consecutive points.

let $\frac{d^{2} f}{{d x}^{2}} = a \cdot f_{i} + b \cdot f_{i + 1} + c \cdot f_{i + 2} + d \cdot f_{i + 3} + e \cdot f_{i + 4} + g \cdot f_{i + 5} + O (h^{n})$ $(d^2f)/dx^2=a*f_i+b*f_(i+1)+c*f_(i+2)+d*f_(i+3)+e*f_(i+4)+g*f_(i+5)+O(h^n)$

a, b, c, d, e, and g are constants to be determined.

Expanding the function values at (i+1), (i+2), (i+3), (i+4), and (i+5) using Taylor's series and summing them.

$\frac{d^{2} f}{{d x}^{2}} = (a + b + c + d + e + g) \cdot f_{i} + (b h + 2 c h + 3 d h + 4 e h + 5 g h) \cdot f'_{i} + (\frac{b h^{2}}{2} + \frac{4 c h^{2}}{2} + \frac{9 d h^{2}}{2} + \frac{16 e h^{2}}{2} + \frac{25 g h^{2}}{2}) \cdot f''_{i} + ... ...$ $(d^2f)/dx^2=(a+b+c+d+e+g)*f_i+(bh+2ch+3dh+4eh+5gh)*f'_i+((bh^2)/2+(4ch^2)/2+(9dh^2)/2+(16eh^2)/2+(25gh^2)/2)*f''_i+......$

On equating the corresponding coefficients of derivatives on both sides, it gives

$a + b + c + d + e + g = 0$ $a+b+c+d+e+g=0$

$b h + 2 c h + 3 d h + 4 e h + 5 g h = 0$ $bh+2ch+3dh+4eh+5gh=0$

$\frac{b h^{2}}{2} + \frac{4 c h^{2}}{2} + \frac{9 d h^{2}}{2} + \frac{16 e h^{2}}{2} + \frac{25 g h^{2}}{2} = 1$ $(bh^2)/2+(4ch^2)/2+(9dh^2)/2+(16eh^2)/2+(25gh^2)/2=1$

$\frac{b h^{5}}{120} + \frac{32 c h^{5}}{120} + \frac{243 d h^{5}}{2} + \frac{1024 e h^{5}}{120} + \frac{3125 g h^{5}}{120} = 0$ $(bh^5)/120+(32ch^5)/120+(243dh^5)/2+(1024eh^5)/120+(3125gh^5)/120=0$

The above 6 equations are solved to get the 6 constants.

They are $a = \frac{15}{4 h^{2}}, b = - \frac{77}{6 h^{2}}, c = \frac{107}{6 h^{2}}, d = - \frac{13}{h^{2}}, e = \frac{61}{12 h^{2}}, and g = - \frac{5}{6 h^{2}}$ $a=15/(4h^2), b=-77/(6h^2), c=107/(6h^2), d=-13/h^2, e=61/(12h^2), and g=-5/(6h^2)$

Substituting the above values gives n=4.

So $\frac{d^{2} f}{{d x}^{2}} = \frac{\frac{15 f_{i}}{4} - \frac{77 f_{i + 1}}{6} + \frac{107 f_{i + 2}}{6} - 13 f_{i + 3} + \frac{61 f_{i + 4}}{12} - \frac{5 f_{i + 5}}{6}}{h^{2}} + O (h^{4})$ $(d^2f)/dx^2=((15f_i)/4-(77f_(i+1))/6+(107f_(i+2))/6-13f_(i+3)+(61f_(i+4))/12-(5f_(i+5))/6)/h^2+O(h^4)$

Thus fourth-order right-sided difference scheme is derived.

Following a similar approach, a fourth-order left-sided difference scheme is derived.

So $\frac{d^{2} f}{{d x}^{2}} = \frac{\frac{15 f_{i}}{4} - \frac{77 f_{i - 1}}{6} + \frac{107 f_{i - 2}}{6} - 13 f_{i - 3} + \frac{61 f_{i - 4}}{12} - \frac{5 f_{i - 5}}{6}}{h^{2}} + O (h^{4})$ $(d^2f)/dx^2=((15f_i)/4-(77f_(i-1))/6+(107f_(i-2))/6-13f_(i-3)+(61f_(i-4))/12-(5f_(i-5))/6)/h^2+O(h^4)$

A function f for $e^{x} \cdot \cos (x)$ $e^x*cos(x)$ is created to calculate its values at different input values. It is as follows.

function y=f(x)

    y=exp(x)*cos(x);

end

It is then used to calculate the second derivative for different dx values using various numerical schemes.

The MATLAB code is as follows.

close all
clear all
clc

% f=exp(x)*cos(x)
% Analytical derivative, f'=exp(x)*exp(x)*(-2*sin(x))

% let x=pi/4 and dx varies from pi/1000 to pi/10

x=pi/3;
dx=linspace(pi/4000,pi/10,20);

exact_value=exp(x)*(-2*sin(x));

% 1 row and 20 columns, each column for different value of dx
numerical_forward=zeros(1,20);   
forward_error=zeros(1,20);

numerical_central=zeros(1,20);
central_error=zeros(1,20);

numerical_backward=zeros(1,20);
backward_error=zeros(1,20);

for i=1:length(dx)
    
    numerical_forward(i)=((15/4)*f(x)-(77/6)*f(x+dx(i))+(107/6)*f(x+2*dx(i))-13*f(x+3*dx(i))+(61/12)*f(x+4*dx(i))-(5/6)*f(x+5*dx(i)))/(dx(i)^2);
    forward_error(i)=abs(exact_value-numerical_forward(i));
    
    numerical_central(i)=(-(1/12)*f(x-2*dx(i))+(4/3)*f(x-dx(i))-2.5*f(x)+(4/3)*f(x+dx(i))-(1/12)*f(x+2*dx(i)))/(dx(i)^2);
    central_error(i)=abs(exact_value-numerical_central(i));
    
    numerical_backward(i)=((15/4)*f(x)-(77/6)*f(x-dx(i))+(107/6)*f(x-2*dx(i))-13*f(x-3*dx(i))+(61/12)*f(x-4*dx(i))-(5/6)*f(x-5*dx(i)))/(dx(i)^2);
    backward_error(i)=abs(exact_value-numerical_backward(i));
    
end
    
plot(dx,forward_error,dx,central_error,dx,backward_error);
legend('Forward','Central','Backward');
xlabel('Grid-length(dx)');
ylabel('Error');
title('Plot of error for different numerical schemes vs dx');

The generated plot is as follows.

In general, central difference schemes have better accuracy (or higher-order) than single-sided schemes, visible from the above plot where the central difference has less error than other schemes at all values of dx. But at boundaries, a central difference scheme can't be employed.

Single-sided schemes perform better in cases where the flow (or information) flows in a single particular direction.