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Elisa  
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 More options Nov 20 2009, 3:58 pm
Newsgroups: sci.math
From: Elisa <elisa.d...@gmail.com>
Date: Fri, 20 Nov 2009 07:58:38 -0800 (PST)
Local: Fri, Nov 20 2009 3:58 pm
Subject: nonlinear PDE
Hi,
I have to solve this pde:
dW/dz + a(dW/dt)^2+g(t)(dW/dt)=f(t)
with
W(0,t)=0
any idea about possible solutions?

thanks a lot
elisa


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rancid moth  
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 More options Nov 21 2009, 1:28 am
Newsgroups: sci.math
From: "rancid moth" <rancidm...@yahoo.com>
Date: Sat, 21 Nov 2009 12:28:59 +1100
Local: Sat, Nov 21 2009 1:28 am
Subject: Re: nonlinear PDE

"Elisa" <elisa.d...@gmail.com> wrote in message

news:ae45891a-0e3e-4987-8ba4-e37b0215c246@j4g2000yqe.googlegroups.com...

> Hi,
> I have to solve this pde:
> dW/dz + a(dW/dt)^2+g(t)(dW/dt)=f(t)
> with
> W(0,t)=0
> any idea about possible solutions?

Yes, but you may need to be more specific with g(t) and f(t).  For example,
write the equation as follows

w_z+a(w_t)^2 +g(t)w_t-f(t)=0  such that w(0,t)=0 and a=constant

you then have the five Lagrange-Charpit equations to solve, namley

dz/ds = 1
dt/ds=2aw_t + g(t)
dw_z/ds=0
dw_t/ds=-g'(t)w_t + f'(t)
dw/ds=w_z+2a(w_t)^2+g(t)w_t

(i am omiting * signs for multiplication here becuase everything should be
obvious in this context and for what follows)

parametrise the initial data by choosing for s=0, z(0,r)=0 and t(0,r)=r, so
the initial condition is w(0,r)=0

by doing so I get w_t(0,r)=0 and w_z(0,r)=f(r)

now the first equation --> z=s, and the second that w_z(s,r) = w_z(0,r) =
f(r).

the fourth equation --> w_t(s,r) = f'(t)/g'(t) - f'(r)/g'(r) exp( -sg'(t) )

the second equation then -->

dt/ds = 2a [ f'(t)/g'(t) - f'(r)/g'(r) exp( -sg'(t) ) ] +g(t)

and this is where i run into trouble, because we need to solve this for
t(s,r) in order to successfully solve the last remaining eqauation.  i can
find no symmetries of this general equation and can see no obvious way it
can be integrated without specifying some sort of functional relationship on
f and g.

So we can keep going and illustrate the idea, let's assume, as an example,
that g(t)=f(t)=t then

w_t = 1 - exp(-s)  and

dt/ds = 2a(1-exp(-s)) + t

solving this we find t(s,r) = r exp(s) + 2a(cosh(s)-1)

From the very last equation...

dw/ds = r + 2a(1-exp(-s))^2 + (1-exp(-s)) (r exp(s) + 2a(cosh(s)-1))

solving this we have

w(s,r) = r (1+2s) - (2+s)a +exp(s) (a-r) +a exp(-s) ( 1+sinh(s) )

we know s=z and can find r from the equation for t(s,r), i.e. r = -exp(-z)
(-2a-t+2acosh(z))

putting it all into w we get after a bit of manipulation

w(z,t) = t + (sinh(z)-cosh(z)) ( 2a+t+a(z-2)cosh(z)+a(z-1)sinh(z) )

and this can indeed be verified to solve

w_z+a(w_t)^2 +g(t)w_t - f(t)=0   where g(t)=f(t)=t and w(0,t)=0

If you want to understand where the Lagrange-Charpit equations come from,
and why all this works - look at Copson's "Partial Differential Equations" -
i just found my dusty copy last night after a three day search.  it's a good
read.

cheers
moth


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Elisa  
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 More options Nov 22 2009, 3:16 pm
Newsgroups: sci.math
From: Elisa <elisa.d...@gmail.com>
Date: Sun, 22 Nov 2009 07:16:12 -0800 (PST)
Local: Sun, Nov 22 2009 3:16 pm
Subject: Re: nonlinear PDE
On 21 Nov, 02:28, "rancid moth" <rancidm...@yahoo.com> wrote:

Right, im my case f(t) = A sin(Wt) and g(t) = B cos(Wt). Haven't tried
the solution yet, hope it fits.
Thanks
elisa


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