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## Main Question or Discussion Point

Let us consider the following partial differential equation:

[tex]{(}\frac{\partial z}{\partial x}{)}^2{+}{(}\frac{\partial z}{\partial y}{)}^{2}{=}{1}[/tex] ---------- (1)

The general solution[you will find in the texts: http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3201.pdf is given by:

[tex]{z}{=}{ax}{+}\sqrt{1-a^2}{y}{+}{B}[/tex]; a and B are constants.--------(2)

Now let me search for a solution in the form:

[tex]{z}{=}{ax}{f}{(}{z}{)}{+}\sqrt{1-a^2}{y}{+}{B}[/tex]

[tex]\frac{\partial z}{\partial x}{=}{ax}{f'}{(}{z}{)}\frac{\partial z}{\partial x}{+}{a}{f}{(}{z}{)}[/tex]

[tex]\frac{\partial z}{\partial y}{=}\sqrt{1-a^2}{+}{ax}{f'}{(}{z}{)}\frac{\partial z}{\partial y}[/tex]

Substituting the above values into (1) we get

[tex]\frac{{(}{a}{f}{(}{z}{)}{)}^{2}}{{(}{1}{-}{ax}{f'}{(}{z}{)}{)}^{2}}{+}\frac{1-a^2}{{(}{1-ax}{f'}{(}{z}{)}{)}^{2}}{=}{1}[/tex]

[tex]{a}^{2}{(}{f}{(}{z}{)}{)}^{2}{+}{1}{-}{a}^{2}{=}{(}{1}{-}{ax}{f'}{(}{z}{)}{)}^{2}[/tex]

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 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 many 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] by keeping x fixed at some arbitrary value. This will go in favor of my suggestion.

Suppose we have only one local maximum for 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.[One may consider the square:x=0,y=0,x=k,y=k]

This can change the whole picture of the problem.

[Equations of the form f(p,q)=0 do not have singular solutions."p" and "q" are the partial derivatives wrt to x and y respectively]

[tex]{(}\frac{\partial z}{\partial x}{)}^2{+}{(}\frac{\partial z}{\partial y}{)}^{2}{=}{1}[/tex] ---------- (1)

The general solution[you will find in the texts: http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3201.pdf is given by:

[tex]{z}{=}{ax}{+}\sqrt{1-a^2}{y}{+}{B}[/tex]; a and B are constants.--------(2)

Now let me search for a solution in the form:

[tex]{z}{=}{ax}{f}{(}{z}{)}{+}\sqrt{1-a^2}{y}{+}{B}[/tex]

[tex]\frac{\partial z}{\partial x}{=}{ax}{f'}{(}{z}{)}\frac{\partial z}{\partial x}{+}{a}{f}{(}{z}{)}[/tex]

[tex]\frac{\partial z}{\partial y}{=}\sqrt{1-a^2}{+}{ax}{f'}{(}{z}{)}\frac{\partial z}{\partial y}[/tex]

Substituting the above values into (1) we get

[tex]\frac{{(}{a}{f}{(}{z}{)}{)}^{2}}{{(}{1}{-}{ax}{f'}{(}{z}{)}{)}^{2}}{+}\frac{1-a^2}{{(}{1-ax}{f'}{(}{z}{)}{)}^{2}}{=}{1}[/tex]

[tex]{a}^{2}{(}{f}{(}{z}{)}{)}^{2}{+}{1}{-}{a}^{2}{=}{(}{1}{-}{ax}{f'}{(}{z}{)}{)}^{2}[/tex]

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 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 many 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] by keeping x fixed at some arbitrary value. This will go in favor of my suggestion.

Suppose we have only one local maximum for 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.[One may consider the square:x=0,y=0,x=k,y=k]

This can change the whole picture of the problem.

[Equations of the form f(p,q)=0 do not have singular solutions."p" and "q" are the partial derivatives wrt to x and y respectively]

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