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Solving with ode45 with a solution dependant variable in the ode

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Hugo
Hugo 2024-8-7，8:41
评论： Hugo 2024-8-7，10:31
Hi,
Let's say I want to solve :
with , with all following the same ode as y(z).
I can theoretically solve each ode with RK4 at step z+dz and then calculate N(z+dz). But how do I do it using the already implemented RK4 function implemented in Matlab? Would something like this work or is there a way better solution?
while(z<zLim)%%This is just an idea of the algorithm I would use
zspan=[0 dz];%Solving the ODE for a very small zspan
if(z~=0)
y0=y(length(y));
else
y0=0;
end
[z,y] = ode45(@(z,y) N*y, zspan, y0);
%Calculate c(z),g(z),....
%Calculate N at step dz
z=z+dz;
end
Thanks

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采纳的回答

Torsten
Torsten 2024-8-7，9:54
编辑：Torsten 2024-8-7，9:54
Ran in:
fun = @(t,y) sum(y)*y;
zspan = [0 3e-2];
y0 = [1 2 3 4 5];
[T,Y] = ode45(fun,zspan,y0);
plot(T,Y)

更多回答（1 个）

Aquatris
Aquatris 2024-8-7，9:57
编辑：Aquatris 2024-8-7，9:58
No, you should not do it that way. Instead you should create your function properly in a way that represents the equations. It will be easier to manipulate things later on.
[z,y] = ode45(@myfun, zspan, y0);
function dx = myfun(z,y)
% x = [y(z) N(z) g(z) c(z)]'
persistent z_prev
if isempty(z_prev)
z_prev = 0; % assuming zspan starts at 0
end
dz = z-z_prev;
dx = [x(1)*x(2); % dy = y(z)*N(z)
(x(1)+x(3)+x(4)-x(2))/dt; % dN = (y(z)+c(z)+g(z)-N)/dz so that N(z+dz) = N(z)+(y(z)+c(z)+g(z)-N(z))/dz*dz = y(z)+c(z)+g(z)
...% dc -> formulate similar to dN
...% dg -> formulate similar to dN]
end
  1 个评论
Hugo
Hugo 2024-8-7，10:30
I do not get how you reach this formula for dx? I tried to reach the same expression of N(z+dz) but I did not manage to do so :
Differentiating I get
Leading to :
But I can not get rid of the z+dz terms, wich are not defined at this step? The only I can see myself reach something close to your definition is by writing :
Leading to
Sorry but I am mathematically lost?
Also why not define dc and dg as dy instead of dN?

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