Re: Re: Tricky differential equation
- To: mathgroup at smc.vnet.net
- Subject: [mg41531] Re: Re: Tricky differential equation
- From: Bobby Treat <drmajorbob+MathGroup3528 at mailblocks.com>
- Date: Sat, 24 May 2003 01:08:01 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
I think there's a better way to do these substitutions, and maybe
somebody in the group will enlighten us. Meanwhile, there's also this:
Differentiation:
(1/t) Dt[ t y'[t], t]
% // Simplify
And integration:
f[t]/t /.
First@DSolve[f'[t]/t == y'[t]/t + y''[t], f, t] /.
C[_] -> 0
Bobby
-----Original Message-----
From: Dr. Wolfgang Hintze <weh at snafu.de>
To: mathgroup at smc.vnet.net
Subject: [mg41531] Re: Tricky differential equation
Bobby,
thanks for the interesting chain of substitutions. I admit that to me
it
looks a bit heavy; but if that's Mathematica's way - so what? I
appreciate to see an example of substitutions in M which I usually do
on
paper before going to M.
If I may add //Simplify to your eq3 line then I find exactly my (**).
I wonder if it is possible to get M to reduce
y'' + y'/t
to the form (written conventionally)
(1/t) d/dt ( t dy/dt )
i.e. a kind of factoring differential operations.
Wolfgang
Bobby Treat wrote:
> I got your equation (*) this way, but I'm wondering if there's an
> easier way:
>
> eq0 = x''[r] + (1/r)*x'[r] + (p - 1/r^2)*Sin[x[r]]*Cos[x[r]] == 0
> eq1 = TrigReduce /@ eq0
> eq2 = FullSimplify[eq1 //. {x[r] :> y[t]/2,
> Derivative[n_][x][r] :> (1/2)*p^(n/2)*
> Derivative[n][y][t], r -> t/Sqrt[p]}]
> eq3 = (#1/p & ) /@ eq2
>
> There must be!
>
> Bobby
>
> -----Original Message-----
> From: Dr. Wolfgang Hintze <weh at snafu.de>
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
> Sent: Thu, 22 May 2003 06:57:59 -0400 (EDT)
> Subject: [mg41531] Re: Tricky differential equation
>
> Luiz,
>
> not a solution but just a hint:
>
> Make the differential equation dimensionless by setting
>
> r = t/Sqrt[p]
> y = 2x
>
> leading to
>
> (*) y'' + y'/t + (1-1/t^2) Sin[y] ==0
>
> Now we can see that
>
> (1) for y->0 we have Sin[y] -> y, the dgl is the Bessel-Dgl with the
> solution J_1[t]
>
> (2) for t>>1 (*) is the pendulum equation
>
> (**) y'' + Sin[y] ==0
>
> with a solution in terms of elliptic integrals.
>
> Hope this helps.
>
> Regards,
> Wolfgang
>
> Luiz Melo wrote:
>
>> Hello everyone,
>>
>> I'm trying to find the numerical solution of the following
>> differential equation (r is the independent variable):
>>
>> x''[r] + 1/r x'[r] + (p - 1/r^2)*Sin[x[r]]*Cos[x[r]] == 0 ,
>>
>> with boundary conditions: x'[1] == 0 , and x[0] -> "has to be
finite",
>>
>> but I'm having at least two problems:
>>
>> 1) I don't know how to submit the BC "finite" to Mathematica;
>> 2) The coefficient p is about 10^4. For this reason, it seems
>> that the Runge-Kutta method usually used for numerical
>> integration of ordinary differential equations turns out
>> to be unsuccessfull in our case. Do we need a special method
>> to solve this?
>>
>> The solution of this equation gives the internal magnetic structure
>> of a cylinder. The function x[r] is the angle between the
>> magnetization and the axial direction, and it depends on the radial
>> direction, r.
>>
>> I would like to plot the Cossine of the result as a function of r
>> (which varies from 0 to 1), for several values of p.
>>
>> Any help will be very appreciated!
>> Thank you
>>
>> Luiz Melo
>>
>> Ecole Polytechnique de Montreal,
>> Montreal, Quebec
>> luiz.melo at polymtl.ca
>>
>>
>>
>>
>