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On PDEs [First Order Non-linear]
Posted:
Sep 10, 2011 10:47 AM
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Resent-From: <bergv@illinois.edu>
From: Anamitra Palit <palit.anamitra@gmail.com>
Date: September 10, 2011 7:56:57 AM MDT
To: "sci-math-research@moderators.isc.org"
<sci-math-research@moderators.isc.org>
Subject: On PDEs [First Order Non-linear]
Let us consider the following partial differential equation:
[del_z/del_x]^2+[del_z/del_y]^2=1 ---------- (1)
The general solution[you will find in the texts:
http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3201.pdf] is given by:
z=ax+sqrt[1-a^2]y+B; a and B are constants.--------(2)
Now let me search for a solution in the form:
z=axf(z)+sqrt[1-a^2]y+ B
del_z/del_x=axf'(z)del_z/del_x+af(z)
del_z/del_y=sqrt[1-a^2]+axf'(z)del_z/del_y
Substituting the above values into (1) we get
[af(z)/[1-axf'(z)] ]^2+(1-a^2)/[1-axf'(z)]^2 = 1
Or,
a^2 [f(z)]^2+1-a^2=[1-ax f'(z)]^2 ------------- (3)
If the above differential equation is solvable we should get a broader
range of general values.Are we missing some solutions if the
conventional general solution is considered?
Now , the above equation[ie,(3) contains x.
We write,
z=F(x,y)
For any particular value of z say, z=k, we have,
k=F(x,y)
Keeping x fixed at some arbitrary value we can obtain k be changing y
only[I believe that this may be possible in most situations or in may
situations]. [We could have achieved the same effect,i.e, getting z=k,
by changing both x and y in some manner]
So we may try to solve our differential equation[the last one--(3)] by
keeping x fixed at some arbitrary value. This will go in favor of my
methodology.
Suppose we have only one local maximum for the entire range of our
function z=F(x,y) at
the point (x0,y0) . For this point,
k=F(x0,y0)
The value of k in this situation, will not be accessible for any
arbitrary x. One has to use x=x0.
To surmount this difficulty one may think of dividing the the domain
of definition of the function F,into sub-domains so that in any
particular sub-domain, the value of z may be accessible for an
arbitrary x in that sub-domain.
[Rather we would look for a function,z=F(x,y), of this type]
On Boundary Conditions]
The general solution[relation (2) seems to be too restrictive with
simple boundary conditions. We may consider a square domain:
x=0,x=k,y=0,y=k [k: some constant]
Let us take the line:x=k
It is perpendicular to the x-axis
If we use the value x=k in (2) ,then z changes linearly wrt y
The general solution[conventional one] talks of plane surfaces given
by (2). I can always take small pieces of such surfaces and sew them
into a large curved surface ,z=F(x,y).Along the boundary z may be a
non-linear function of x or y.
Alternatively:
Along the line x=constant[=k],we have from (1),
m(x=k,y)+[del_z/del_y]^2=1
Where m(x=k,y) represents the value of [del_z/del_x]^2 along the line
x=k
=>[del_z/del_y]=+Sqrt[1-m(k,y)] or -Sqrt[1-m(k,y)]
---------- (3)
Now we divide the line x=k into small strips[one may consider
infinitesimal strips] each of length h [from y=0 to y=k]
For one [or more ] strip we take the positive value[+Sqrt[1-m]] in
relation
(3) and for some other[or others] we take the negative value
[-Sqrt[1-m] in (3)
As a result at the boundary x=k we may have a nonlinear function
z=g(y) instead of a linear one as predicted by the conventional
general solution.The linearity is produced by the change of sign and
by changes in the value of m. This can change the whole picture of the
problem
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