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 April 20th, 2013, 03:45 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory General Solution of a PDE Consider the following PDE : $\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}= 0$ I have manged to find out that $F= f(x - y)$ satisfies the above when $f$ is continues and differentiable everywhere on the xy plane. Is it possible to show that there are no more class of solutions to this one? Even if the RHS is is different in the given PDE, it seems AFAIHO, that the solution is dependent on this particular form. Thanks in advance, Balarka .
May 4th, 2013, 12:46 PM   #2
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Re: General Solution of a PDE

That is general solution! But, don't forget constant function. For real, this is general solution [attachment=0:261e79zq]PrtScr capture.jpg[/attachment:261e79zq].
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May 4th, 2013, 09:59 PM   #3
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Re: General Solution of a PDE

Quote:
 Originally Posted by limes5 For real, this is general solution : . . .
Absolutely not. It is a special case since another solution $\ln(x - y)$ can't be deduced from that. The general solution is just what I said, $F(x - y)$ where F is differentiable anywhere.

Quote:
 Originally Posted by limes5 Don't forget constant function
That is a special case too. Define F as a constant function and you'll get $F(x - y)= \bar{C}$

 May 5th, 2013, 02:19 AM #4 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Re: General Solution of a PDE Sorry. Mistake. I forgot f before (x-y). So z=a*f(x-y)+b, that's general solution. For real, it's something that we called "total (complete) integral". From total integral we can get any other integral (solution). If any other solution exist it can be derived using this solution.
May 5th, 2013, 03:38 AM   #5
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Re: General Solution of a PDE

Quote:
 Originally Posted by limes5 z=a*f(x-y)+b, that's general solution.
Which is the same as $F(x - y)$ since $a f(x - y) + b= ((x - y)^0 + (x - y)^0 + . . . + (x - y)^0) \cdot f(x - y) + (x- y)^0 + (x - y)^0 + . . . + (x - y)^0)$ in which everything is in terms of x - y.

Quote:
 Originally Posted by limes5 For real, it's something that we called "total (complete) integral".
I know, but I am not asking for the name. I want justification that this is the only total integral of the PDE. How could you guarantee that there are no other class of function for which another disjoint set of solution can generated; how can it be mathematically attacked ? Understood the question?

Quote:
 Originally Posted by limes5 If any other solution exist it can be derived using this solution.
I don't think this is very obvious, if you have a proof, please post it!

 May 5th, 2013, 10:37 AM #6 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Re: General Solution of a PDE I have understood what you tryed to say. When I learned PDE there wasn't any proof of that statement in my book. But i have another good book and I will see if there is any proof. But I haven't that book now with me, and I can do it in few days... :/ Note that a, b are real numbers... I'm not sure you can write as you wrote in previous message.
May 5th, 2013, 10:40 AM   #7
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Re: General Solution of a PDE

Quote:
 Originally Posted by limes5 When I learned PDE there wasn't any proof of that statement in my book.
You have a book that mentions the PDE $\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}= 0$ ? Would you care to mention the name?

 May 5th, 2013, 10:44 AM #8 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Re: General Solution of a PDE Not that equation. I'm talking about total (complete) integral. There is a proof of "If any other solution exist it can be derived using this solution" in that book, i think.
May 5th, 2013, 10:50 AM   #9
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Re: General Solution of a PDE

Quote:
 Originally Posted by limes5 If any other solution exist it can be derived using this solution
That's the definition for the total integrals no?

 May 5th, 2013, 10:54 AM #10 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Re: General Solution of a PDE [/quote] I don't think this is very obvious, if you have a proof, please post it![/quote] You ask me to proove the deffinition!

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