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Valentin Fadeev

Gunter's example

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Edited by Valentin Fadeev, Tuesday, 6 July 2010, 23:04

This one gave me some hard time:

z times open x plus z close times normal partial differential times z divided by normal partial differential times x minus y times open y plus z close times normal partial differential times z divided by normal partial differential times y equals zero

where z equals Square root of y when x equals one

The difference with "ordinary" linear equations is that coefficients here depend on z . At the same time right hand side is identically zero, so this is not strictly an inhomogeneous equations.

I made false starts trying to find some nice substitute to absorb z or alternatively pull out the missing term to construct an inhomogeneous equations. Finally, I found a hint in the book N.M. Günter, Integration of First-Order Partial Differential Equations, ONTI/GTTI, Leningrad/Moscow (1934). In the solution for an equation of a similar structure it was suggested to use the standard method and treat z as a constant when integrating the associated system.

So here we go. Searching for a solution in an implicit form:

cap v of z comma x comma y equals zero

z times open x plus z close times normal partial differential times cap v divided by normal partial differential times x minus y times open y plus z close times normal partial differential times cap v divided by normal partial differential times y equals zero

The associated system:

equation sequence d times x divided by z times open x plus z close equals negative d times y divided by y plus z equals d times z divided by zero equals d times t

In the same book it is hinted that one particular integral of this system is z equals cap c sub one which follows from the last identity. Another integral can be found using the first identity:

d times x divided by z times open x plus z close plus d times y divided by y plus z equals zero

Treating z as constant we can simplify the expressions:

d of x divided by z divided by one plus x divided by z plus d of y divided by z divided by y divided by z times open one plus y divided by z close equals zero

equation sequence x divided by z equals u times y divided by z equals v

d times u divided by one plus u plus d times v divided by v minus d times v divided by one plus v equals zero

equation left hand side u times open v plus one close divided by v plus one equals right hand side cap c sub two

equation left hand side open x plus z close times y divided by y plus z equals right hand side cap c sub two

Therefore the general solution can be written in the form:

equation sequence cap v of z comma x comma y equals normal cap phi of z comma open x plus z close times y divided by y plus z equals zero

It is already within reach of sheer guess to let normal cap phi of u comma v equals u minus v to establish the result, however we proceed with a more lengthy, yet rigorous way.

The system of the first integrals is written as follows:

psi sub one equals z

equation left hand side psi sub two equals right hand side open x plus z close times y divided by y plus z

Using the initial condition x equals one :

psi one macron equals z

equation left hand side psi two macron equals right hand side open one plus z close times y divided by y plus z

Solving for z and y :

z equals psi one macron

equation sequence y equals z times psi two macron divided by one plus z minus psi two macron equals psi one macron times psi two macron divided by one plus psi one macron minus psi two macron

Now following the standard method already mentioned below:

MathJax failure: TeX parse error: Extra open brace or missing close brace

equation left hand side psi sub one equals right hand side psi sub two divided by one plus psi sub one minus psi sub two

z equals open x plus z close times y divided by y plus z divided by one plus z minus open x plus z close times y divided by y plus z

equation left hand side sum with, 3 , summands z times y plus z squared plus open z squared minus x times y close times z equals right hand side x times y plus z times y

open z squared minus x times y close times open z plus one close equals zero

Suppressing the solution z equals negative one which does not satisfy intial conditions, finally we obtain:

z equals Square root of x times y

 

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