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Willie Wong
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There's no way that your "solution for the PDE with another method" is correct.

Your equation can be rewritten as $$ u_y = \frac12 (x - u_x^2) $$ Evaluating on the set $\{x = 0\}$, you have $$ u_y(0,y) = - \frac12 u_x^2(0,y) \leq 0 $$

On the other hand, your initial data has $u_y(0,y) = -2y$, which is strictly positive when $y < 0$.

Your mistake is in the part you didn't show in the question.

Willie Wong
  • 39.1k
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  • 176