Can someone help me to correct the code ​​for this problem using ode45 solver?

10 次查看(过去 30 天)
Question is:
Governing equations:
Boundary conditions:
f'(10)= 0, (assume )
As tried here, I want to plot graphs for velocity profile using parameters like Pr, beta, and N.
I tried giving 'for' loop, but it failed.
How to enter the code command to plot a graph for velocity profile with different beta parameter?
Trial code
function dy=vj_Casson()
clc; clear all;
betava=[0.5 1 2 5];
for i=1:4
beta=betava(i);
tspan = [0 10];
S = 0.5;
y0 = [S 0 0 1 1];
% y0 = [S 1 0 1 0];
[eta,y] = ode45(@fun,tspan,y0);
plot(eta, y(:,1));
xlabel('\bf\eta','FontSize',20,'FontWeight','bold');
ylabel('f(\eta)','FontSize',10,'FontWeight','bold');
legend('\beta=0.5','\beta=1','\beta=2','\beta=5')
hold on
end
end
function dy = fun(eta,y)
dy = zeros(5,1);
Pr=0.3; N=0.1; beta=0.5;
dy(1) = y(2);
dy(2)=y(3);
dy(3)=(((2*y(2))^(2))-y(1)*y(3))./((1+(1./beta)));
dy(4) =y(5);
dy(5)=-(((3*Pr)*(y(1)*y(5)-y(2)*y(4)))/(4*N+3));
end
  1 个评论
Alex Sha
Alex Sha 2023-3-12
The results below are what you want?
1: f' = df/dt = f'
2: f'' = df'/dt = f''
3: theta' = dtheta/dt = theta'
4: f''' = df''/dt = (2*(f')^2-f*f'')/(1+1/0.5)
5: theta'' = dtheta'/dt = 0.3*(f'*theta-f*theta')/(1+4/3*0.1)
Objective Function: 1.41854343905794E-26
Boundary Values Estimated:
f''(t=0): -0.887446621710334
theta'(t=0): -0.36495059823582

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回答(1 个)

Torsten
Torsten 2023-3-9
This is a boundary value problem, not an initial value problem since conditions on f and theta are given on both ends of the integration interval (eta = 0 and eta = 10). You have to use bvp4c or bvp5c instead of ode45 to solve.
  3 个评论
Torsten
Torsten 2023-3-10
编辑:Torsten 2023-3-10
You cannot use conditions at eta = 10 if you use ode45. You must assume two further conditions at eta = 0 to make ode45 work and try to adjust these conditions in several runs such that you arrive at your two conditions at eta = 10. Look up "shooting method" for more details.
I can assure you: the simpler way for you to go is to use bvp4c or bvp5c.

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