# partial differential equation

• Oct 30th 2012, 03:01 PM
MAX09
partial differential equation
The question I'm solving is as follows.

The differential equation given by,
$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 2u$

satisfying the initial conditions y = xg(x) and u = f(x) with

a) f(x) = 2x, g(x)= 1 has no solution.
b) f(x) = 2x2, g(x) = 1 has infinite number of solutions
c) f(x) = x3, g(x) = x has a unique solution
d) f(x) = x4, g(x) = x has a unique solution.

My initial line of approach was that the give equation was similar to the Euler's theorem for a homogenous function 'u' whose degree was 2. Since u = f(x), f(x) must be also homogenous function of degree 2, and the options had only one of them with degree 2.

Am i right in my conclusion? If so can someone guide me on concluding the no. of solutions as infinite?
If not, can someone help me get started?

P.S. I tried using LATEX but i kept getting an image saying that my tag was more than 25 characters. Hence I had to revert to attaching a .jpeg file.

Thanks
• Oct 31st 2012, 12:40 AM
JJacquelin
Re: partial differential equation
Hi !

This is the opportunity to show you how to solve this kind of PDE.
About the conditions : Two examples are shown. Then, I suppose that you will answer by yourself to the other cases.
• Nov 3rd 2012, 05:13 AM
MAX09
Re: partial differential equation
I have a doubt with one of the steps shown in the image...

When I add up $\frac {\partial u}{\partial X} and \frac {\partial u}{\partial Y}$ , i don't get 2u...

Am I missing something?

Thanks JJacquelin
• Nov 3rd 2012, 07:31 AM
JJacquelin
Re: partial differential equation
Are you sure to correctly apply the chain rule for the derivation of a function of function ?
You should publish on the forum your calculus. So, we could see where is the mistake.
See in attachment the final verification.
• Nov 3rd 2012, 06:45 PM
MAX09
Re: partial differential equation
Jjacquelin, i guess i made a minor error when it came down to the chain rule ... i didn't observe something which was quite obvious and before me... now i'm able to see the elegance of your solution..

Thanks
• Nov 5th 2012, 08:08 AM
MAX09
Re: partial differential equation
Jjacquelin.. i've followed your instructions and i just wanna verifythe other two cases as well... here they are..

option (i) is true because 2x can never be made proportional to x2 for any definition of f(x).
option (iii) is true because equality is obtained for a unique definition of f(1/x) = 1/x itself.

Option (ii) f(x) = 2x2, g(x) = 1. This implies y = x(1)= x u = f(x) = 2x2.
This means $2x^2 = x * x * f( \frac {x}{x})$
= x2 * f(1)

Equality can be arrived in this case for any definition for f(1) such that it is a constant.

Option (iv) f(x) = x4, g(x) = x. This implies y = x(x)= x2 ; u = f(x) = x4.
This means $x^4 = x * x^2 * f( \frac {x}{x^2})$
$= x^3 * f(\frac {1}{x})$ which can never be true as x3 can never give x4 for any definition of f(x).

Please let me know if there 's any mistake in my understanding of the solution provided by you.

Thanks
• Nov 5th 2012, 11:15 AM
JJacquelin
Re: partial differential equation
Option (ii) : OK.
Option (iv): << ... which can never be true as x^3 can never give x^4 for any definition of f(x). >> is incorrect.
There is one function f(x) so that x^4 = (x^3) f(1/x). What is this function ?
• Nov 5th 2012, 05:49 PM
MAX09
Re: partial differential equation
I guess i missed it... so x^4 may equal x^3 if f(1/x) is defined to be x, isn't it?
• Nov 5th 2012, 09:47 PM
JJacquelin
Re: partial differential equation
Quote:

Originally Posted by MAX09
I guess i missed it... so x^4 may equal x^3 if f(1/x) is defined to be x, isn't it?

Yes.
But then, what is f(x) ?
• Nov 10th 2012, 07:56 AM
MAX09
Re: partial differential equation
Will the definition of f(x) be as follows ?

f(x) = 1/x , x is not equal to 0.