# QL PDE's - Checking IC's

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• Jan 12th 2011, 03:00 AM
bugatti79
QL PDE's - Checking IC's
All,

I wish to check that $\displaystyle u(x,2x)=0$ for the following problem

$\displaystyle (y-u)\frac{\partial u}{\partial x}+(u-x)\frac{\partial u}{\partial y}=(x-y)$,

to which the general solution is defined implicitly by

$\displaystyle x+y+u=3 \sqrt { 2( \frac {(x-y)^2}{2}-\frac{2xu +2yu-u^2}{2})$

Using IC's , I reduce it to

$\displaystyle 3x+u=3 \sqrt{u^2-6xu-x^2}$

reducing further, I get

$\displaystyle 3x+u=3 \sqrt {3x+- \sqrt {2} 2x}$

I am having difficulty showing u=0 from here. Mathematica shows 2 solutions u=0 and I think the other is 15x/6.

Thanks
• Jan 12th 2011, 05:37 AM
Jester
How did you obtain that general solution? I obtained

$\displaystyle x^2 + y^2 + u^2 = f(x+y+u)$.
• Jan 12th 2011, 06:16 AM
bugatti79
Quote:

Originally Posted by Danny
How did you obtain that general solution? I obtained

$\displaystyle x^2 + y^2 + u^2 = f(x+y+u)$.

I get x+y+u like you do on the RHS but the LHS is based on the ratio

$\displaystyle \frac{d(x-y)}{(x+y-2u)}=\frac{du}{(x-y)}$

I can write the derivation later if you require.

Thanks
• Jan 12th 2011, 06:27 AM
Jester
Quote:

Originally Posted by bugatti79
All,

I wish to check that $\displaystyle u(x,2x)=0$ for the following problem

$\displaystyle (y-u)\frac{\partial u}{\partial x}+(u-x)\frac{\partial u}{\partial y}=(x-y)$,

to which the general solution is defined implicitly by

$\displaystyle x+y+u=3 \sqrt { 2( \frac {(x-y)^2}{2}-\frac{2xu +2yu-u^2}{2})$

Using IC's , I reduce it to

$\displaystyle 3x+u=3 \sqrt{u^2-6xu-x^2}$

reducing further, I get

$\displaystyle 3x+u=3 \sqrt {3x+- \sqrt {2} 2x}$

I am having difficulty showing u=0 from here. Mathematica shows 2 solutions u=0 and I think the other is 15x/6.

Thanks

I ask because your "general" solution has no arbitrary function in it.
• Jan 12th 2011, 06:37 AM
bugatti79
Quote:

Originally Posted by Danny
I ask because your "general" solution has no arbitrary function in it.

Sorry, the general solution should be

$\displaystyle x+y+u=f ( \frac {(x-y)^2}{2}-\frac{2xu +2yu-u^2}{2})$

typo on my part,

Thanks
• Jan 12th 2011, 06:54 AM
Jester
Quote:

Originally Posted by bugatti79
Sorry, the general solution should be

$\displaystyle x+y+u=f ( \frac {(x-y)^2}{2}-\frac{2xu +2yu-u^2}{2})$

typo on my part,

Thanks

How did you end with this part

$\displaystyle \dfrac {(x-y)^2}{2}-\dfrac{2xu +2yu-u^2}{2}$

in your solution?
• Jan 12th 2011, 08:46 AM
bugatti79
Quote:

Originally Posted by Danny
How did you end with this part

$\displaystyle \dfrac {(x-y)^2}{2}-\dfrac{2xu +2yu-u^2}{2}$

in your solution?

Danny,

Find attached the furthest I can go with it. Difficulty seeing how to getu=0 on last line.
Hope I havent made any mistakes!

Thanks
• Jan 12th 2011, 09:21 AM
Jester
I don't get your last line - why did you do that?
• Jan 12th 2011, 09:48 AM
bugatti79
Quote:

Originally Posted by bugatti79
Danny,

Find attached the furthest I can go with it. Difficulty seeing how to getu=0 on last line.
Hope I havent made any mistakes!

Thanks

I was looking for the roots of u under the square root sign in order to simplify it further. Did I need to?
• Jan 17th 2011, 01:27 PM
bugatti79
Quote:

Originally Posted by bugatti79
I was looking for the roots of u under the square root sign in order to simplify it further. Did I need to?

Ok, I think I have it. My above comment and the last line of attachment is incorrect. Here goes...from the second last line...

$\displaystyle 3x+u-3\sqrt{u^2-6xu+x^2}=0$
$\displaystyle u^2-6xu+x^2=\frac{u^2}{9}+\frac{2xu}{3}+x^2}$
$\displaystyle \frac{8u^2}{9}-\frac{20xu}{3}=0$

Unifying u^2 and solving the quadratic yields 2 solutions
u=0 and u=15x/2 which is what mathematica outputs.