Re: a non-linear second-order PDE
From: Robert Israel (israel_at_math.ubc.ca)
Date: 09/24/04
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Date: Fri, 24 Sep 2004 14:00:06 +0000 (UTC)
In article <civrim$kmh$1@news.ks.uiuc.edu>,
andreas wagener <108076@gmx.net> wrote:
>I consider the following non-linear second-order partial differential
>equation with two non-negative variables x and y:
>(a+ b x) [ df/dx d^2f/(dx dy) - df/dy d^2f/(dx^2) ] + b y [ df/dx d^2
>f/d(y^2)- df/dy d^2f/(dx dy)] =0
>where a >0 and b is a real number. Additional conditions are
>f(x,0) = (a + b x)^(1-1/b) /(1/b (1-1/b))
>and
>df(x,0)/dy =0.
I assume you mean df/dy = 0 when y = 0.
>for all x.
>(For b=1, f(x,0) converges to logarithmic function ln(a+x), but that
>should not really matter).
>One solution to the PDE is:
>f(x,y) = 1/(1/b (1-1/b)) [ (a + b x)^(1- 1/b) + c2 b y^(1-1/b) ]
This doesn't satisfy your second additional condition in general, however:
df/dy has a singularity at y=0 if b > 0. It does work if c2 = 0 or if
b < 0. Of course, in general there's also a singularity when a+bx=0.
>with constants c1, c2.
You didn't have a c1.
>My question: Are there other solutions? If so, which? If no, how can
>one prove uniqueness?
Solutions of your PDE include
f(x,y) = c_0 + sum_{j=1}^n c_j (a+bx)^(k_j) y^(p-k_j)
for any constants c_j, k_j and p. For example, you could add to
your solution a term in (a+bx)^(-1-1/b) y^2 and get another solution
with the same values of f(x,0) and df/dy (x,0). So, no uniqueness
there.
BTW, an interesting feature of the PDE is that if f(x,y) is a solution, so
is f(x,y)^k for any k, or ln(f(x,y)), or exp(f(x,y)).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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