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Solve analytically a third oder coupled Ode?

Posted 6 years ago
POSTED BY: Umair Rashid
2 Replies

Just for a start, you can visualize the solution to the first equation in terms of M,f''[0]:

Manipulate[
 sol = NDSolveValue[{
    f'''[x] - (2 f'[x]^2 - 5/2 f[x] f''[x] + M*f''[x]) == 0,
    f[0] == 0, f'[0] == 1, f''[0] == a},
   f, {x, 0, xmax}];
 Plot[sol[x], {x, 0, xmax}],
 {{M, 1}, 0, 2},
 {{a, -1.049, f''[0]}, -2, 2},
 {xmax, 10, 100}]
POSTED BY: Gianluca Gorni
Anonymous User
Anonymous User
Posted 6 years ago

What in the world is "+Mf", what is Pr (two multiplied variables or one or a constant)? You want plots for Rd does that mean it's allowed to be a constant? It appears this is this a system of equations? in what variable?

It looks like a straight forward DSolve, but without knowing even what the variables all are I wouldn't say.

Perhaps if you look at the tutorials in Help it will give you an idea how to write the equations more fully and then show them to (me).

f(infinity)=0

Mathematica can solve such equations, but you can't choose the method (ie, it picks one sure to work it does not give back guesses). That being so - consider providing the solver information about your variables (boundaries, if they are real numbers or greater than zero say so) - since the more you tell it the less answers that will be blocked by being "less generic".

My point is, about infinity, that Mathematica does numeric approximation and returns special functions (you cannot tell it how to search for Taylor series, for example), or for symbolic solutions, you are not going to be able to simply state to Mathematica that f[inf]==0 because Mathematica, if it chooses to solve by series, will make that determination based on the type of variable and other factors (which you have some control of - specifying the variable type and if the variable has a range).

POSTED BY: Anonymous User
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