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How to solve differential equation with variable in differential term?

3 vues (au cours des 30 derniers jours)
Tien Ching Ma
Tien Ching Ma le 28 Fév 2023
I'm trying to solve differential equations with polar coordinate, the equation looks like the following:
r*dy/dr = A*r*exp(B*(C-x(r))) - y(r)
dx/dr = y(r)./D
I'm using ode45 solver with boundary conditions that y(r=0) = 0, and x(r=R) = E.
Since in ode45 tutorial guide of Matlab, I didn't see any example with "r" included in dy/dr term, I simply divide the first equation with r.
It becomes:
dy/dr = A*exp(B*(C-x(r))) - y(r)/r
And I would have a problem that when r = 0, y(0)/0 = Nan.
I understant that from the mathematical view 0/0 doesn't make sence, but from my physical view, I need y(0) = 0 at the this point. Therefore, I just ignore the last term when r = 0 by using if function:
if r == 0
dy/dr = A*exp(B*(C-x(r)));
else
dy/dr = A*exp(B*(C-x(r))) - y(r)/r;
end
Could someone tell me if there is any other solver/method can solve this r*dy/dr = A*r*exp(B*(C-x(r))) - y(r) equation? Or could the method I'm using so far would lead to some problems?
Thanks a lot!
  1 commentaire
Sam Chak
Sam Chak le 28 Fév 2023
With these info,
Can you transform the differential equations to and ?

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Réponses (1)

DUY Nguyen
DUY Nguyen le 2 Mar 2023
You could try to define a new variable u = r + eps ( where eps is a small positive number close to zero, should be carefully chosen) and solving the differential equation on the interval [eps, R] instead of [0, R]. This approach will effectively shift the singularity at r = 0 to u = eps, where it can be avoided.

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