How do I prepare the following ODE for ode45?

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Sergio Manzetti 2018-2-20

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https://ww2.mathworks.cn/matlabcentral/answers/383733-how-do-i-prepare-the-following-ode-for-ode45

评论： Torsten 2018-3-9

Hello, I would like to solve the following ODE in ode45, but the example's on the site are not describing using higher order derivatives with non-linear terms.

The ODE is:

y''' = y(2+x^2)

initial conditions are: y(0)=0 y'(0)=0 y''(0)=0

Thanks!

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Torsten 2018-2-20

Google is your friend:

matlab & higher order odes

Best wishes

Torsten.

Sergio Manzetti 2018-2-20

编辑：Sergio Manzetti 2018-2-20

在 MATLAB Online 中打开

I got so far:

dYdX = @(X,Y) [Y(3) +  (x^2+2)*Y(1)];  % Differential equation
res = @(ya,yb) [ya(1);  ya(2);  yb(2)-1];                   % Boundary conditions
SolYinit = bvpinit([0 1E+1], [1; 1; 1]);
Fsol = bvp4c(dYdX, res, SolYinit);
X = Fsol.x;
F = Fsol.y;
figure(1)
plot(X, F)
legend('F_1', 'F_2', 'F_3', 3) 
grid

But the first line is not correct. Can you see what is missing?

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采纳的回答

Torsten 2018-2-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/383733-how-do-i-prepare-the-following-ode-for-ode45#answer_306099

在 MATLAB Online 中打开

fun = @(x,y)[y(2);y(3);y(1)*(2+x^2)];
y0 = [0 0 0];
xspan = [0 5];
[X,Y] = ode45(fun,xspan,y0);
plot(X,Y(:,1),X,Y(:,2),X,Y(:,3));

Best wishes

Torsten.

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Steven Lord 2018-2-21

在 MATLAB Online 中打开

Your initial conditions state that y, its first derivative, and its second derivative are all 0 at x = 0. If you think about this from a physics standpoint where y is the position of your solution (let's think of this as a race car), y' is the velocity, and y'' is the acceleration this corresponds to you being at the starting line at a dead stop with your foot off the gas pedal.

Speaking very roughly, at the first time step (I'm treating x as time here) y'' is 0*something, so you never accelerate. If you don't accelerate you can't go faster than 0. If you don't go faster than 0, you can't move away from 0. Torsten's solution (correctly) demonstrates that the solution to your system of equations with that set of initial conditions is y(x) = 0.

Now if you were to tweak the initial conditions, say by stomping on the gas pedal as the race starts:

>> y0 = [0 0 1];
>> [X,Y] = ode45(fun,xspan,y0);
>> plot(X,Y(:,1),X,Y(:,2),X,Y(:,3));

You get something that looks a bit more interesting. For most of the time span your curves looks like they are at 0, but that's because of the scaling on the Y axis. If you zoom in you can see that y, its first derivative, and its second derivative take off very rapidly. [Given that the acceleration increases as the square of x, you get going REALLY quickly.]