From: sridhar on
Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote in message <1084951456.35015.1272434321031.JavaMail.root(a)gallium.mathforum.org>...
> > Given the second order non linear BVP
> >
> > (&#402;')^n = 1 + &#947; &#952;
> > &#8230;..&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;
> > 230;&#8230;&#8230; 1
> >
> > &#952;" + (&#955; +n+1/ 2n +1) &#402; &#952;' - n (2
> > &#955; +1/2n+ 1) * &#402;' &#952; = 0
> > &#8230;&#8230;&#8230;&#8230; 2
> >
> > Prime in the above eqn&#8217;s describe partial
> > differentiation with respect to &#951;
> >
>
> In order to have a PDE, your equation must contain
> derivatives with respect to at least two independent
> variables.
> You say &#951; is one of them. What is the second one,
> and where does differentiation with respect to it occur ?
>
> > Boundary conditions are
> > &#402; (0) = 0, &#952;'(0) = -1
> > &#402;' (&#8734;) = 0, &#952;(&#8734;) = 0
> >
> > can any let me know how the above PDE equations are
> > converted to ODE's
>
> Best wishes
> Torsten.



Given the second order non linear BVP

(&#402;')^n = 1 + &#947; &#952; &#8230;..&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230; 1

&#952;" + (&#955; +n+1/ 2n +1) &#402; &#952;' - n (2 &#955; +1/2n+ 1) * &#402;' &#952; = 0 &#8230;&#8230;&#8230;&#8230; 2

Prime in the above eqn&#8217;s describe partial differentiation with respect to &#951;

Boundary conditions are
&#402; (0) = 0, &#952;'(0) = -1
&#402;' (&#8734;) = 0, &#952;(&#8734;) = 0

where &#951; is a function of x & y given by
Similarity variable, &#951; = x ^ (&#955;-n/2n+1) * y
&#920; is a dimensionless temperature
&#402; is a dimensionless stream function given by
Stream function, &#968; = x ^ (&#955;+n+1/2n+1) * &#402;(&#951;)

how to solve the above system of PD eqn&#8217;s