# QL PDE's - Checking IC's

• January 12th 2011, 04:00 AM
bugatti79
QL PDE's - Checking IC's
All,

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

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

to which the general solution is defined implicitly by

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

Using IC's , I reduce it to

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

reducing further, I get

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

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

Thanks
• January 12th 2011, 06:37 AM
Jester
How did you obtain that general solution? I obtained

$x^2 + y^2 + u^2 = f(x+y+u)$.
• January 12th 2011, 07:16 AM
bugatti79
Quote:

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

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

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

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

I can write the derivation later if you require.

Thanks
• January 12th 2011, 07:27 AM
Jester
Quote:

Originally Posted by bugatti79
All,

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

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

to which the general solution is defined implicitly by

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

Using IC's , I reduce it to

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

reducing further, I get

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

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.
• January 12th 2011, 07:37 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

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

typo on my part,

Thanks
• January 12th 2011, 07:54 AM
Jester
Quote:

Originally Posted by bugatti79
Sorry, the general solution should be

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

typo on my part,

Thanks

How did you end with this part

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

• January 12th 2011, 09:46 AM
bugatti79
Quote:

Originally Posted by Danny
How did you end with this part

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

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
• January 12th 2011, 10:21 AM
Jester
I don't get your last line - why did you do that?
• January 12th 2011, 10:48 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?
• January 17th 2011, 02:27 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...

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

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