# A Challenging PDE question

• Aug 29th 2011, 05:07 AM
MarcoPolo
A Challenging PDE question
Dear users of MHF, I've found two PDE question very challenging, would you guys mind to give some hints?

First question:

1) (-x) Ux + (2y) Uy = x^2 + y^2

a) find the characteristic curve (I solved)

b) solve the above equation by conditions U(x,1)= x^2 , u(x,-1) = x.

second question:

2)Ux + Uy= x - y + U

condition : U(0,y)= e^(-y^2).

for the second question I tried U = x^2 - (xy) + f(x-y) e^(x)
but I yielded a result like x - y + f(x-y)e^(x)....

the questions just sound so impossible for me...

and the condition in 1b was so tricky...
• Aug 29th 2011, 12:29 PM
Jester
Re: A Challenging PDE question
Did you obtain tbe general solutions

(1) $u = -\frac{1}{2}x^2 + \frac{1}{4} y^2 + f(x^2y)$?

In (2) you mentioned you tried $u = x^2 - x y + e^x f(x-y).$ Where did this come from?
• Aug 29th 2011, 02:07 PM
MarcoPolo
Re: A Challenging PDE question
Quote:

Originally Posted by Danny
Did you obtain tbe general solutions

(1) $u = -\frac{1}{2}x^2 + \frac{1}{4} y^2 + f(x^2y)$?

In (2) you mentioned you tried $u = x^2 - x y + e^x f(x-y).$ Where did this come from?

Thanks danny, definitely I obtained (1), but when I try to use the condition given, I can't figure out how can I get u(x,-1)=x .

for 2) I only know that $u= f(x-y)e^y$or $u=f(x-y)e^x$usually results in $ux+uy= u$, how ever the x-y parts I find it so hard...

if i let $Ux= x$, $x= x^2 /2$
$Uy= -y$, $y= -y^2/2$
but then applying the condtion given I can't get the answer $u(x,0)= e^(-(y^2))$
• Aug 29th 2011, 02:38 PM
Jester
Re: A Challenging PDE question
For the first part, subs $y =1$ into your solution, set this equation the the IC and solve for $f$. For example,

$u(x,1) = - \frac{x^2}{2} + 1 + f(x^2) = x^2$ so $f(x^2) = \frac{3}{2}x^2- 1$.

Let $r = x^2$ so $f(r) = \frac{3}{2}r- 1$. This then defines $f$. Similarly for the second IC.

For the second question, do you know the method of characteristics?
• Aug 29th 2011, 07:37 PM
MarcoPolo
Re: A Challenging PDE question
With method of characteristic, I know how to get for Ux + Uy=0 u= f(x-y)

(at work, sorry for short reply)
• Aug 30th 2011, 05:18 AM
Jester
Re: A Challenging PDE question
OK. Your answer above kinda makes sense. You can do one of two things. If you let

$u = v - x + y$ then your PDE

$u_x + u_y = x - y + u$ becomes $v_x + v_y = v$. Then do what you did.

OR you can solve the charactistic equations

$\frac{dx}{1} = \frac{dy}{1} = \frac{du}{x-y+u}$.