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

u(x,x) =


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 $z = c_2$

Now what to do? Can anyone please help me out?


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.

  • $\begingroup$ 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 $\endgroup$ – 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 $y=x$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.