# $u_x(x , y) + u_y(x , y) = 0 .for (x , y) \in \mathbb{R^2}$

$u_x(x , y) + u_y(x , y) = 0 .for (x , y) \in \mathbb{R^2}$

u(x,x) =

has

a) unique solution

b) a family of straight lines as characteristics.

c)solution which vanishes at (2 , 1)

4)infinitely many solutions

(It can have multiple solutions)

My try - $\frac{dx}{1} = \frac{dy}{1} = \frac{dz}{0}$

$x - y =c_1$ and $z = c_2$

Let $f(x):=u(x,x)$ for $x \in \mathbb R$. Then we have

$f'(x)=u_x(x,x)+u_y(x,x)=0$ for all $x$. Hence $f$ is constant.

• Sir actually $u_{xx}$ has not been given in the question. I have written exactly as it was in the question...Can u guess what will be $u_{xx}$?@Fred – sani Jun 9 '17 at 6:19

Sani : Your results are correct concerning the characteristic equations $x-y=c_1$ and $z=c_2$.

The general solution of the PDE can be expressed on various equivalent manners :

Implicit equation where $\Phi$ is any differentiable function of two variables : $\quad \Phi(x-y\:,\:z)=0$

Or explicit equation where $F$ is any differentiable function : $z=F(x-y)$

Or other equivalent forms, for example $\quad z=f(y-x)\quad$ or $\quad y-x=g(z)\quad$ , etc.

So, you can conclude in the case of $y=x$.

• I did not get u..Can u please be little more elaborate?@JJACQUELIN – sani Jun 9 '17 at 7:09
• With your notations $\quad u(x,y)=z(x,y)$ – JJacquelin Jun 9 '17 at 7:21
• that is not a problem....I did not get this line "you can conclude in the case of y=x " specially.. – sani Jun 9 '17 at 7:23
• If $x=y\quad\to\quad u(x,x)=F(x-x)=F(0)=$any constant since $F$ is any function. – JJacquelin Jun 9 '17 at 7:26
• so answer will be infinitely many solutions?? – sani Jun 9 '17 at 7:33