Numerical solving second order differential equation

I would like to solve a second order differential equation,please see the attachment. I used the following code:
b = 0.19;
a = 0.072;
beta = 0.72;
Pr = 0.72;
c = beta/Pr*a^(4/3);
xspan = [0.01 100];
g40 = [1 0];
opts = odeset('RelTol',1e-4,'AbsTol',1e-6);
[x4,g4] = ode45(@(x4,g4) model4ode(x4,g4,b,c), xspan, g40, opts);
plot(log10(x4),g4)
function dg4dx4 = model4ode(x4,g4,b,c)
dg4dx4 = zeros(2,1);
dg4dx4(1) = g4(2);
dg4dx4(2) = (2./(3*x4)).*(4+x4.^(2*b).*(x4.^(2*b)+1).^(-1)).*g4(2)+...
((22/3)*c*x4.^(-2/3).*(x4.^(2*b)+1).^(1/(3*b))-(22/9)*x4.^(2*b-2).*(x4.^(2*b)+1).^(-1)).*g4(1);
end
But the result is not true. Could someone help me?

Antworten (1)

Torsten
Torsten am 8 Jan. 2018
b = 0.19;
a = 0.072;
beta = 0.72;
Pr = 0.72;
c = beta/Pr*a^(4/3);
xspan = [0.01 100];
g40 = [(xspan(1)^(2*b)+1)^(-1/(3*b))*(-11/3) 22/3*c*xspan(1)^(1/3)];
opts = odeset('RelTol',1e-4,'AbsTol',1e-6);
[x4,g4] = ode45(@(x4,g4) model4ode(x4,g4,b,c), xspan, g40, opts);
f4(:,1)=3/22*1/c*x4(:).^(-1/3).*g4(:,2);
f4(:,2)=(g4(:,1).*(x4(:).^(2*b)+1).^(1/(3*b))+0.5*1/c*x4(:).^(-1/3).*g4(:,2))./x4(:);
plot(log10(x4),f4)
function dg4dx4 = model4ode(x4,g4,b,c)
dg4dx4 = zeros(2,1);
dg4dx4(1) = g4(2);
dg4dx4(2) = (22/3*c*x4^(1/3)*(x4^(2*b)+1)^(1/(3*b))*g4(1)+g4(2)*(11+x4)/3)/x;
end
Best wishes
Torsten.

4 Kommentare

Xiang Yi
Xiang Yi am 8 Jan. 2018
Bearbeitet: Jan am 9 Jan. 2018
Thank you Torsten. The code works well, but the results are still not identical to the literature.
there is a bump, but we cannot obtain. Could you help me to find where is not correct?
What differential equation solver does ‘the literature’ use?
I would trust the MATLAB solver results.
Could you include "the literature" as a pdf or a link ?
Best wishes
Torsten.
I'd like to solve the Eq.(20) in this paper. The parameters settings is the same as the Section 4, 1th paragraph. But we cannot obtain the results as Fig 1. and Fig. 2. You may note that there is a bump in the figure. x>0, and when x is very small and close to zero, f(x)=1; when x is very large and approaching positive infinity, f(x)=0.

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