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I need to solve the Burgers' equation

$uu_{x} - u_{y} = 0$ with initial data $u(x,0) = f(x)$ where $f$ is given to be analytic

I have solved this one with method of characteristics and the solution is :

$u(x,y) = f(x + yu)$

Now, I have been asked to prove the following statements

(A) No analytic solution exists for all positive values of $y$ provided $f$ is monotonically increasing.

(B) No continuous solution exists for all positive values of $y$ provided $f$ is monotonically decreasing.

I don't get how to even begin proving these claims? I know that a real analytic function on a domain is constant but how do I use this information here?

Please suggest some ideas

Thank you.

night_crawler
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1 Answers1

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I would suggest to differentiate the implicit solution $u = f(x+yu)$ w.r.t. $x$ to get $$ u_x = \frac{f'(x+yu)}{1-yf'(x+yu)} \, . $$ See what happens for increasing positive values of $y$ if $f'\geq 0$?

EditPiAf
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  • Ok, so I solved for $y$ and I get $y = \dfrac{1}{f'(x+yu)} - \dfrac{1}{u_{x}}$ , $y > 0$ gives $f'(x+yu) < u_{x}$ , using this in original equation $1 - yf'(x+yu) < 1$ , I don't see how this leads to any contradiction. – night_crawler Sep 10 '21 at 15:35
  • @night_crawler look at the denominator of $u_x$. When does $u_x$ become undefined? – EditPiAf Sep 11 '21 at 10:41