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$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 $z = c_2$

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

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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.

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  • $\begingroup$ 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
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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 $y=x$.

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  • $\begingroup$ 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

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