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From: sridhar on 26 Apr 2010 20:38
bvp4c is used for solving ODE
but i would like to solve the above PDE's
From: Torsten Hennig on 26 Apr 2010 22:26
> bvp4c is used for solving ODE
> but i would like to solve the above PDE's

I only see two ODEs for which boundary conditions
are prescribed at two different boundary points -
so bvp4c is the suitable integrator.

Best wishes
Torsten.
From: sridhar on 13 May 2010 16:40
Given the second order non linear BVP

(ƒ')^n = 1 + γ θ …..……………………… 1

θ" + (λ +n+1/ 2n +1) ƒ θ' - n (2 λ +1/2n+ 1) * ƒ' θ = 0 ………… 2

Prime in the above eqn’s describe partial differentiation with respect to η
the above equatins are PDE's since η is a function of x & y given by
Similarity variable, η = x ^ (λ-n/2n+1) * y

Boundary conditions are
ƒ (0) = 0, θ'(0) = -1
ƒ' (∞) = 0, θ(∞) = 0

Θ is a dimensionless temperature
ƒ is a dimensionless stream function given by
Stream function, ψ = x ^ (λ+n+1/2n+1) * ƒ( η)

how to solve the above system of eqn’s
From: Torsten Hennig on 14 May 2010 04:49
> Given the second order non linear BVP
>
> (ƒ')^n = 1 + γ θ
>
> …..………………&#
> 8230;…… 1
>
> θ" + (λ +n+1/ 2n +1) ƒ θ' - n (2
> λ +1/2n+ 1) * ƒ' θ = 0
> ………… 2
>
> Prime in the above eqn's describe partial
> differentiation with respect to η
> the above equatins are PDE's since η is a
> function of x & y given by
> Similarity variable, η = x ^ (λ-n/2n+1) * y
>
> Boundary conditions are
> ƒ (0) = 0, θ'(0) = -1
> ƒ' (∞) = 0, θ(∞) = 0
>
> Θ is a dimensionless temperature
> ƒ is a dimensionless stream function given by
> Stream function, ψ = x ^ (λ+n+1/2n+1) *
> ƒ( η)
>
> how to solve the above system of eqn's

The reason why a similarity variable eta is introduced
is to transform the underlying PDEs to a usual
boundary value problem with ODEs.
So again:
Write the second-order ODE in theta as two first-order
ODEs in theta and theta' as I suggested earlier.
Then use BVP4C to solve the boundary-value problem
of three ODEs.

Best wishes
Torsten.
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