Thank you very much Prof. Bangerth.

So, by multiplying with an additional weight factor it simply becomes a 1D
coupled system with homogeneous Neumann.

Thanks & Regards

Pawan

On Sat, Apr 21, 2018 at 4:33 PM, Wolfgang Bangerth <[email protected]>
wrote:

> On 04/21/2018 03:06 AM, Pawan Kumar wrote:
>
>>
>> $$u_t = u_{rr}+ \frac{1}{r} u_r -  \frac{1}{r} \frac{\partial}{\partial
>> r} (r \lambda(u) \frac{\partial v}{\partial r})  +f(u,v)$$
>> $$v_t  = v_{rr} + \frac{1}{r} v_r + g(u,v)  $$
>>
>> But at $r=0$, it becomes :
>>
>> $$u_t = 2u_{rr}-2uv_{rr}-u_r h_r +f(u,v)$$
>> $$v_t = 2v_{rr}+g(u,v)$$
>>
>> So, my doubts are:
>>
>> 1. Since, its a time dependent, vector valued \& non-linear set of PDEs,
>> which tutorials should I follow to solve these? I found 21, 26, 31 \& 31
>> related  but some of them are too complicated, is there any relatively
>> simpler way !! \\
>>
>
> You just have a system of coupled heat equations. 21, 26, 22 seems like a
> reasonably choice.
>
>
> 2. How to handle the condition at $r=0$ with the homogeneous Neumann BC?
>>
>
> You don't have to do anything special if you don't just multiply by a test
> function (say, w(r)) and then integrate over r=0...R, but if you also
> multiply by a weight factor 2*pi*r. So, for example, out of the u_t term,
> you'll get
>
>   \int_{r=0}^R  w(r)  u_t(r) 2*pi r  dr
>
> This way, the singularity disappears in all relevant terms.
>
> Best
>  W.
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth          email:                 [email protected]
>                            www: http://www.math.colostate.edu/~bangerth/
>
>
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