W(0)=0; W(f)=2.8; d(P)/d(W)=((-0.3743)/(2*P))*(1-(0.15*X)); P(0)=1; d(X)/d(W)=0.5*((0.08*(0.75*(1-X)))/(1-(0.15*X))); X(0)=0; Trying to solve these two ODEs

 Akzeptierte Antwort

Star Strider
Star Strider am 24 Okt. 2018

1 Stimme

Try this:
PX_ODE = @(W,PX) [((-0.3743)./(2*PX(1))).*(1-(0.15*PX(2))); 0.5*((0.08*(0.75*(1-PX(2))))./(1-(0.15*PX(2))))];
[W,PX] = ode15s(PX_ODE, [0, 2.8], [1; 0]);
figure
plot(W, PX)
grid
Here ‘P’ is ‘PX(1)’, ‘X’ is ‘PX(2)’. The system encounters a singularity at 2.67, and the integration stops.

5 Kommentare

Kurt
Kurt am 24 Okt. 2018
Is there another way to better show both equations
Star Strider
Star Strider am 24 Okt. 2018
Show them in what sense?
Plotting them is the only way I can think of.
Kurt
Kurt am 24 Okt. 2018
I mean to keep them in their original forms
Kurt
Kurt am 24 Okt. 2018
I was trying to do it this way, syms P(W) X(W) eqn1 = diff(P, W) == ((-0.3743)/(2*P))*(1-(0.15*X)); eqn2 = diff(X, W) == (0.5*((0.08*(0.75*(1-X)))/(1-(0.15*X))));
[odes, vars] = odeToVectorField(eqn1, eqn2); fun = matlabFunction(odes,'Vars',{'t','Y'}); X0(0) = [0, 0]; tspan = [0, 2.8]; [t, sol] = ode45(fun,tspan,x0);
P = sol(:,2); X = sol(:,1);
Star Strider
Star Strider am 24 Okt. 2018
That certainly works, although ‘P(0)’ cannot be zero. If ‘X(0)’ is greater than 1, the system will integrate out to 2.8.
Also, ‘fun’ is essentially the same as my code, and the result is the same. (The system is ‘stiff’, so ode15s or another stiff solver is more appropriate.)

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