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I have the following PDE: $$u_y + u^2 u_x = 0 \\ u(x, 0) = \sqrt{x}$$

I tried to solve it with lagrange's method: $$\tag 1 \frac{du}{dt} = 0 \quad \implies \quad \phi_1 = u$$ Second eq: $$\tag 2 \frac{dx}{u^2} = \frac{dy}{1}$$ since $u=\phi_1 = \text{Constant}$ I thought I can integrate it as a constant: $$\int \frac{dx}{u^2} = \frac{x}{u^2} $$ With this approach I get $$\tag 3 \phi_2 = \frac{x}{u^2} - y \quad \implies \quad u = \sqrt{\frac{x}{y+\phi_2}}$$ which doesn't get me much closer to the solution: $u(x, y) = \sqrt{\frac{x}{y+1}}$

Edit:

Perhaps I can say: $u(x,0) = \sqrt{x} \quad \implies \quad \phi_2 = 1$

Which then will give the correct solution? Any help\hints would be appreciated!

Edward Josef
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1 Answers1

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I would try separation of variables: $$u(x,y)=f(x)g(y)$$ So the equation becomes $$-f(x)\frac{dg(y)}{dy}=f^2(x)g^3(y)\frac{df(x)}{dx}$$ Dividing by $f(x)g^3(y)$ you get:$$-\frac{1}{g^3(y)}\frac{dg(y)}{dy}=f(x)\frac{df(x)}{dx}$$ Since the left hand side only depends on $y$ and the right hand side only depend on $x$, both sides must be constant. $$f(x)df(x)=Cdx\\f(x)=\sqrt{2Cx+D}$$ Similarly $$-\frac{dg(y)}{g^3(y)}=Cdy$$ This gives $$g(y)=\frac1{\sqrt{2Cy+E}}$$ So $$u=f(x)g(y)=\sqrt\frac{2Cx+D}{2Cy+E}=\sqrt\frac{x+\frac D{2C}}{y+\frac E{2C}}$$ From the initial condition, $D=0$, $E=2C$, so $$u=\sqrt\frac{x}{y+1}$$

Andrei
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