How to integrate linear system of vectorial equations?

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Anshuman Pal
Anshuman Pal am 20 Aug. 2021
Kommentiert: Star Strider am 22 Aug. 2021
Hello,
I have the following system of vectorial equations that describe a curve in space r, its tangent vector t and its normal $n$, parametrised by the arc length s:
Suppose I have the function given as a vector of values, as well as initial values of r, t and n. Then is there some simple way (or package) for numerically integrating this system of vectorial equations?
Thank you very much!

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Star Strider
Star Strider am 20 Aug. 2021
syms kappa n(s) r(s) s t(s) r0 t0 n0 Y
Eqn = [diff(r) == t; diff(t) == -r + kappa*n; diff(n) == -kappa*t];
rtn = dsolve(Eqn, r(0)==r0, t(0)==t0, n(0)==n0)
rtn = struct with fields:
t: [1×1 sym] r: [1×1 sym] n: [1×1 sym]
t = simplify(rtn.t, 500)
t = 
r = simplify(rtn.r, 500)
r = 
n = simplify(rtn.n, 500)
n = 
Alternatively:
[VF,Subs] = odeToVectorField(Eqn)
VF = 
Subs = 
rtnfcn = matlabFunction(VF, 'Vars',{s,Y,kappa})
rtnfcn = function_handle with value:
@(s,Y,kappa)[kappa.*Y(3)-Y(2);Y(1);-kappa.*Y(1)]
Use ‘rtnfcn’ with the approopriate numeric ODE solver (for example 0de45, ode15s) depending on the magnitude of κ.
For example:
sspan = linspace(0,10); % Vector Of 's' Values
initconds = rand(3,1); % Initial Conditions
[s,rtn] = ode15s(@(s,rtn,kappa), sspan, initconds); % Integrate
figure
plot(s, rtn)
grid
.
  4 Kommentare
Anshuman Pal
Anshuman Pal am 22 Aug. 2021
Thank you! A couple more comments:
1) Is there a typo in `ode15s(@(s,rtn,kappa), sspan, initconds)`? Shouldn't it be `ode15s(@rtnfcn, sspan, initconds)`?
2) Can I use `rtnfcn` with a boundary-value solver like `bvp4c`?
Star Strider
Star Strider am 22 Aug. 2021
As always, my pleasure!
Is there a typo in `ode15s(@(s,rtn,kappa), sspan, initconds)`?
There is. It should be:
[s,rtn] = ode15s(@(s,rtn) rtnfcn(s,rtn,kappa), sspan, initconds); % Integrate
Can I use `rtnfcn` with a boundary-value solver like `bvp4c`?
Probably. One addition would be to create ‘kappa’ as an anonymous function, for example with ‘s’ as the independent variable and ‘kapa’ as the dependent variable:
kapamtx = [s(:) kapa(:)];
kappa = @(s) interp1(kapamtx(:,1), kapamtx(:,2), s);
It would be necessary either to keep ‘s’ within the limits of ‘kapamtx(:,1)’ in order to avoid either extrapolating or returning NaN values for ‘s’ outside that range.
.

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