0
$\begingroup$

I have a sequence of differential equations, which can be solved recursively, i.e, the solution of the first equation is necessary to solve the second equation.

Consider for example

$\frac{\partial}{\partial s}(\frac{1}{\epsilon^2}I_{-2}+\frac{1}{\epsilon} I_{-1} +I_0)=\frac{1}{\epsilon^2}f_0I_{-2}+\frac{1}{\epsilon}(f_0I_{-1}+f_1I_{-2})+(I_{-1}f_0+I_0f_1)$

Where $f_0\; \text{and}\; f_1$ are provided. The differential equations we will get from here are

$\frac{\partial}{\partial s}I_{-2}=f_0I_{-2}$

$\frac{\partial}{\partial s}I_{-1}=f_0I_{-1}+f_1I_{-2}$

and so on.

Is there any hope for solving these type of problem in mathematica? I could not try anything so far.

Improve this question
$\endgroup$
1
  • $\begingroup$ I{-1]'==f0 I[-1]+ f1 I[-2] and I[0]'== f1 I[0]+f0 I[-1] seems to be inconsiistent $\endgroup$ Commented Jul 9, 2023 at 18:59

2 Answers 2

Reset to default
1
$\begingroup$

This is not a recursive system of equations. It is a triangular one. Supposing we have to consider initial conditions as I2[0], I1[0], I0[0] and using the Laplace transform

sol = Solve[{LaplaceTransform[D[1/e^2 I2[s] + 1/e I1[s] + I0[s], s] - 1/e (f0 I1[s] + f1 I2[s]) - (f0 I1[s] + f1 I0[s]), s, z] == 0, LaplaceTransform[D[I2[s], s] - f0 I2[s], s, z] == 0, LaplaceTransform[D[I1[s], s] - f0 I1[s] - f1 I2[s], s, z] == 0}, {LaplaceTransform[I2[s], s, z], LaplaceTransform[I1[s], s, z], LaplaceTransform[I0[s], s, z]}][[1]] InverseLaplaceTransform[{ LaplaceTransform[I2[s], s, z], LaplaceTransform[I1[s], s, z], LaplaceTransform[I0[s], s, z]} /. sol, z, s]
Improve this answer
$\endgroup$
0
$\begingroup$

I changed the variables from {I[-2],I[-1],I[0]} to {y[0],y[1],y[2]}, mainly to avoid having to deal with minus signs in the variable names. If you only have the three terms to deal with:

y0 = First[y[0][t] /. DSolve[Derivative[1][y[0]][t] == f[0] y[0][t], y[0][t], t]] Out[10]= E^(t f[0]) C[1] y1 = First[y[1][t] /. DSolve[Derivative[1][y[1]][t] == f[0] y[1][t] + f[1] y0, y[1][t], t]] Out[11]= E^(t f[0]) C[2] + E^(t f[0]) t C[1] f[1]

And for your missing "and so on" equation, assuming I didn't screw this up (f0 and f1 seem reversed here, as noted in the comment, compared to the previous two equations):

y2 = First[y[2][t] /. DSolve[Derivative[1][y[2]][t] == f[0] y1 + f[1] y[2][t], y[2][t], t]] Out[13]= E^(t f[1]) C[3] + ( E^(t (f[0] - f[1]) + t f[1]) f[0] (C[2] (f[0] - f[1]) + C[1] (-1 + t (f[0] - f[1])) f[1]))/(f[ 0] - f[1])^2
Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.