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can't solve this differential equation

Posted 10 years ago

Hi all i have problem with solving this eq: enter image description here

can u please help me

POSTED BY: Ali H
9 Replies
POSTED BY: Udo Krause

This is not an exact equation

In[160]:= Clear[a0, a1, a2, a3, a4]
a0[r_] := b
a1[r_] := 2 b/r
a2[r_] := (a - b/r)
a3[r_] := (b + a r^2)/r^3
a4[r_] := 0

In[167]:= (-1)^4 D[a0[r], {r, 4}] + (-1)^3 D[a1[r], {r, 3}] + (-1)^2 D[a2[r], {r, 2}] + (-1)^1 D[a3[r], {r, 1}] + a4[r] // Simplify
Out[167]= (15 b - 2 b r + a r^2)/r^4

unless a and b are both zero. So one cannot reduce the order of the equation. Nevertheless - in experimental mode - set v[r] = r^3 w'[r] to arrive at

(* Ansatz: w'[r] = r^3 v[r] *)
(b + a r^2) v[r] + (a - b/r) D[r^3 v[r], r] + 
  2 b/r D[r^3 v[r], {r, 2}] + b D[r^3 v[r], {r, 3}] // Simplify
(b (19 - 3 r) + 4 a r^2) v[r] + r ((30 b - b r + a r^2) v'[r] + b r (11 (v''[r] + r v'''[r]))

for this new equation it is possible to get some inhomogenous solutions (for simple inhomogenities) but it seems impossible to find one which is finite at r = 0.

(* inhomogen, transformiert *)
Clear[s, a, b, v, w, r]
s = With[{d = 2, \[Mu] = 1, \[Nu] = 1},
  DSolve[{(b (19 - 3 r) + 4 a r^2) v[r] + 
      r (30 b - b r + a r^2) v'[r] + 11 b r^2  v''[r] + 
      b r^3 v'''[r] == r^\[Mu] (d - r)^\[Nu], v[d] == 0, v'[d] == 2, 
    v''[d] == 1}, v, r]]

Clear[f]
f = s[[1, 1, 2]] /. {a -> 1, b -> 2}

Plot[f[x], {x, -1, 5}]

reduced equation

Things which did not work

  • prescribing a finiteness condition at r = 0
  • get solutions to the inhomogenous equation in less than an hour (computation aborted) with NDSolve
  • get solutions to the original equation for simple inhomogenities with DSolve
POSTED BY: Udo Krause
Posted 10 years ago

thank u

a,b are constant.

and my boundary conditions are:

w[d]==0 ,

w[0]= = limit (cant be infinite i.e if there is any "r" in denominator then the Corresponding Factor must be zero for example if we have a term like this: " C[1] /r + a*r^2 " then C[1] must be zero for the sake of not being infinite ,

D[D[w[r], r], r] + 1/r D[w[r], r]== limit ,

D[b (D[D[w[r], r], r] + 1/r D[w[r], r]), r]== limit

how can i solve it with this boundary conditions?

POSTED BY: Ali H

The homogeneous equation can be solved to give a DifferentialRoot object, to plot something one has to prescribe the values mentioned - done by chance here -

In[13]:= Clear[a, b, r]
DSolve[{(b + a r^2) w'[r] + (a r - b)  r^2 w''[r] + 2 b r^2 w'''[r] + b r^3 w''''[r] == 0, 
         w[1] == 1, w'[1] == 0, w''[1] == 1, w'''[1] == 2}, w, r]

then make it into a function which can be plotted

In[50]:= Clear[f]
f = (Out[14][[1, 1, 2]]) /. {a -> 2, b -> 3}

and plot it

sample plot a = 2, b = 3

the behaviour depends strongly on a and b. Experiment on your own.

Another thing to do (a.k.a. jobs for the weekend): Specify a and b in an attempt to reach the form of some known (named) differential equation. Check the results. Construct solutions from known solutions.

POSTED BY: Udo Krause
Posted 10 years ago

one of the boundary condition is (D[D[w[r], r], r] + 1/r D[w[r], r]) when r=0 ,cant be infinite

how can i define it in mathematica language!?

POSTED BY: Ali H
POSTED BY: Udo Krause

DSolve doesn't give a solution. This mean there probably isn't a general symbolic to the problem that you've specified.

You may want to try:

  • Using Assumptions to give more information about the problem.
  • Numerically Solving the differential equation

What you do of course very much on the problem you're trying to solve.

POSTED BY: Sean Clarke
Posted 10 years ago
Attachments:
POSTED BY: Ali H

That's an ordinary differential equation of fourth degree for w[r] with inhomogenity p[r]. To make this work you need to specify

  • an existence interval for the independent variable r
  • four boundary conditions for w, w', w'', w''', see OrdinaryDifferentialEquation
  • the inhomogenity p[r] as a function
  • a and b

Then most probably you have to use NDSolve because of p[r].

If you're a lover of step-wise thinking try to solve the homogenous equation (p[r] === 0) first.

Be aware of the singularity if r = 0 belongs to the existence interval.

POSTED BY: Udo Krause
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