General Solution of a PDE

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
Consider the following PDE :

\(\displaystyle \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} = 0\)

I have manged to find out that \(\displaystyle F = f(x - y)\) satisfies the above when \(\displaystyle f\) 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
.
 
Feb 2013
153
0
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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mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
limes5 said:
For real, this is general solution : . . .
Absolutely not. It is a special case since another solution \(\displaystyle \ln(x - y)\) can't be deduced from that. The general solution is just what I said, \(\displaystyle F(x - y)\) where F is differentiable anywhere.

limes5 said:
Don't forget constant function
That is a special case too. Define F as a constant function and you'll get \(\displaystyle F(x - y) = \bar{C}\)
 
Feb 2013
153
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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.
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
limes5 said:
z=a*f(x-y)+b, that's general solution.
Which is the same as \(\displaystyle F(x - y)\) since \(\displaystyle a f(x - y) + b = ((x - y)^0 + (x - y)^0 + . . . + (x - y)^0) \cdot f(x - y) + (x- y)^0 + (x - y)^0 + . . . + (x - y)^0)\) in which everything is in terms of x - y.

limes5 said:
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?

limes5 said:
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!
 
Feb 2013
153
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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.
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
limes5 said:
When I learned PDE there wasn't any proof of that statement in my book.
You have a book that mentions the PDE \(\displaystyle \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} = 0\) ? Would you care to mention the name?
 
Feb 2013
153
0
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.
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
limes5 said:
If any other solution exist it can be derived using this solution
That's the definition for the total integrals no? :lol:
 
Feb 2013
153
0
[/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! :D