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Consider the PDE $U_{y} + UU_{x} = 1$

Show the PDE has no solution such that $U = (\frac{1}{2})y$ when $4x - y^{2} = 0$.

I want to do this by differentiating $U(\frac{y^{2}}{4},y)$ and comparing it with the PDE evaluated on the curve $x = \frac{y^{2}}{4}$.

Any help would be greatly appreciated. Thank you

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    what did you try? where are you stuck? – 2017-01-29
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    Typically I would just use method of characteristics and see where it might fail but the extra U in front of the U_x is tripping me up. Just getting started I guess is. – 2017-01-29
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    One of the ODE's for the characteristics leads to $x=\frac{1}{2}U^2+c_1$ The other to $y=U+c_2$ – 2017-01-29

1 Answers 1

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I don't agree. One can directly prove that $$U(x,y)=\frac{1}{2}\left(y\pm \sqrt{4x-y^2} \right)$$ is solution of the ODE $\quad U_y+UU_x=1$

and if $\quad 4x-y^2=0\quad $ then $\quad U=\frac{y}{2}\quad$ as expected.

So, in contradiction to the statement in the wording of the question, the PDE has a solution (above) such that $\quad U=\frac{y}{2}\quad$ when $\quad 4x-y^2=0$ .

HINT :

Solve the PDE thanks to the method of characteristics leading to the general solution on the form of implicit equation : $$2x-U^2=F(y-U)$$ where $F$ is any differentiable function.

Then determine the function $F$ so that $y=2U$ and $x=\frac{y^2}{4}=U^2$

The function is $F(X)=X^2$

This leads to $2x-U^2=(y-U)^2$ . Solving for U leads to the above particular solution.