Finite difference method to solve a nonlinear eqn?

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Onur Metin Mertaslan
评论： Onur Metin Mertaslan 2022-3-25
Hello,
I have a second order nonlinear question and I need to solve it for different times by using finite difference method but I don't know how to start it. I am quite new in Matlab. Is there anyone who can help me or at least show me a way to do this?
thank you
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Torsten
Torsten 2022-3-25
g = 9.81;
L = 1.0;
T = 1.0;
dT = 0.01;
y_0 = pi/2;
v_0 = 0;
f = @(t,y)[y(2);-g/L*sin(y(1))];
tspan = 0:dT:T;
y0 = [y_0;v_0];
[t,y] = ode45(f,tspan,y0);
plot(t,y)
Onur Metin Mertaslan
Onur Metin Mertaslan 2022-3-25
编辑：Onur Metin Mertaslan 2022-3-25
Thank you so much for this. And I have a question. If it is possible to add linear solution to the graph to compare the results?
The main question is to show the linear and nonlinear solution on the plot for 3 different delta t's, and I found the linear solution by my hand[y=y_0*cos(sqrts(g/l)t)]
I know I wanted so many things but I am really confused now
I mean is to show all results in one plot possible?

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Torsten
Torsten 2022-3-25
编辑：Torsten 2022-3-25
g = 9.81;
L = 1.0;
T = 1.0;
dT = 0.01;
y_0 = pi/2;
v_0 = 0;
f = @(t,y)[y(2);-g/L*sin(y(1))];
tspan = 0:dT:T;
y0 = [y_0;v_0];
[t,y] = ode45(f,tspan,y0);
y_linear = v_0/sqrt(g/L)*sin(sqrt(g/L)*t) + y_0*cos(sqrt(g/L)*t);
plot(t,[y(:,1),y_linear])
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Torsten
Torsten 2022-3-25
编辑：Torsten 2022-3-25
The graph won't change because the dt is not the actual stepsize of the solver, but prescribes the output times for ode45. The stepsize is chosen by the solver and computed internally - you can't influence it because it's adaptively chosen and different for each time step to meet a prescribed error tolerance.
If you want a fixed-step solver to solve your equations for which you can prescribe the stepsize, you will have to program it on your own. E.g. explicit Euler is a simple one.

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