# Thread: Determine a relationship bewteen x,y and u for this QL PDE

1. ## Determine a relationship bewteen x,y and u for this QL PDE

$
\displaystyle (x+y+u)u_x+(x+y-u)u_y=2u$
.Where u(x,x)=0

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

$
\displaystyle \frac{d(x+y)}{2x+2y}=\frac{du}{2u} \implies \frac{d(x+y)}{x+y}=\frac{du}{u}
$
. Integrating for u we get

$\displaystyle ln(u)=ln(x+y)+A' \implies u=A(x+y) \implies A=\frac{u}{(x+y)}=\phi (x,y,u)$

Check $\displaystyle a\frac{\partial \phi}{\partial x}+b\frac{\partial \phi}{\partial y}+h\frac{\partial \phi}{\partial u}=0$

$
\displaystyle \frac{-u(x+y+u)-u(x+y-u)+2u(x+y)}{(x+y)^2}=0
$

Determining $\psi(x,y,u)$.

Attempt A

$
\displaystyle \frac{\frac{du}{dt}}{\frac{dy}{dt}}=\frac{2u}{x+y-u}
$

Now

$
\displaystyle y=\frac{u}{A}-x \implies \frac{2u}{\frac{u}{A} -x+x-u} =\frac{2Au}{u(1-A)}
$

Attempt B

$
$
. Since u=Ax+Ay

Not sure of either of these attempts are correct or how to proceed?

2. Since you added the first two to get

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

why not try substracting the first two and see what that does for you.

3. Originally Posted by Danny
Since you added the first two to get

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

why not try substracting the first two and see what that does for you.

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

$
\displaystyle \implies \frac{u}{(x-y)}=\psi(x,y,u)
$

Differentiating this wrt x and y and substituting satisfies the original equation, but what about $\phi (x,y,u)$?

Both $\phi(x,y,u)$ and $\psi(x,y,u)$ must be = constant ie

$
\displaystyle \frac{u}{x-y}=f(\frac{u}{x+y})
$
. Using the IC u(x,x)=0 we get

$
\displaystyle \frac{0}{2x}=f(0)
$

I dont understand what this means assuming its correct?

4. Originally Posted by bugatti79
$
\displaystyle \frac{d(x-y)}{(x+y+u)-(x+y-u)}=\frac{du}{2u} \implies \frac{d(x-y)}{2u}=\frac{du}{2u} \implies
u=(x-y)$

$
\displaystyle \implies \frac{u}{(x-y)}=\psi(x,y,u)
$

Differentiating this wrt x and y and substituting satisfies the original equation, but what about $\phi (x,y,u)$?

Both $\phi(x,y,u)$ and $\psi(x,y,u)$ must be = constant ie

$
\displaystyle \frac{u}{x-y}=f(\frac{u}{x+y})
$
. Using the IC u(x,x)=0 we get

$
\displaystyle \frac{0}{2x}=f(0)
$

I dont understand what this means assuming its correct?
Correction on above, I forgot the constant of integration

Ie in the first line of post 3.

$\displaystyle u=x-y+c \implies c=u-x+y = \psi (x,y,u)$

$\psi (x,y,u)= \phi(x,y,u) \implies$

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

IC is u(x,x)=0 implies

$0=f(\frac{0}{0})$

How would you develop the particular solution based on this?