Solve $$x(u_x^2 + u_y^2) = uu_x$$
with the initial condition $u(0,r^2) = 2r$. Later, solve the same PDE for a different initial condition, namely $u(2r,r) = \lambda r$ for $\lambda \in \Bbb N$.
Let $p := u_x, q := u_y,$ and $z = u(x,y)$. The PDE is $x(p^2 + q^2) = pz$. As usual, Charpit's equations give us
$$\frac{dx}{2xp-z} = \frac{dy}{2xq} = \frac{dz}{2x(p^2+q^2) – pz} = \frac{dp}{-q^2} = \frac{dq}{pq} \tag{1}$$
Integrating $\frac{dp}{-q^2} = \frac{dq}{pq}$, we get $p^2 + q^2 = c_1$. Substituting this back into the PDE, we have $xc_1 = pz$. As a result,
$$\frac{dz}{2x(p^2+q^2) – pz} = \frac{dz}{xc_1} = \frac{dz}{pz} \tag{2}$$
Using the third equality,
$$\frac{dz}{pz} = \frac{dp}{-q^2} = \frac{dp}{p^2 – c_1} \implies \frac{dz}{z} = \frac{2pdp}{2(p^2-c_1)} \tag{3}$$
Integrating,
$$\log z^2 = \log (p^2 – c_1) + c_2' \tag{4}$$
i.e.,
$$z^2 = c_2(p^2-c_1) = -c_2q^2 \tag{5}$$
What could be done from here? Is there a substitution that might simplify this PDE to a quasilinear one, for instance? Also, I haven't found a way to use the initial data, yet. Thanks a lot!
Best Answer
From $u_x^2+u_y^2=c$ we have that \begin{align} u_x=\frac{cx}{u},\quad u_y=\pm\sqrt{c-\left(\frac{cx}{u}\right)^2}. \end{align} Via the chain rule we have that \begin{align} \mathrm du&=u_x\mathrm dx+u_y\mathrm dy,\\\\ \mathrm du&=\frac{cx}{u}\mathrm dx\pm \sqrt{c-\left(\frac{cx}{u}\right)^2}\mathrm dy, \end{align} which is an exact equation. Rearranging we have \begin{align} \frac{cu}{\sqrt{cu^2-c^2x^2}}\mathrm du-\frac{c^2x}{\sqrt{cu^2-c^2x^2}}\mathrm dx&=\pm c\mathrm dy,\quad\longrightarrow\quad \sqrt{cu^2-c^2x^2}=k\pm cy,\\ &u^2=(cx)^2+(k+ cy)^2;\quad [\pm c^{1/2}\rightarrow c]. \end{align} Which we see satisfies the PDE. So we see our complete integral is \begin{align} \varphi(x,y,u,c,k)=u^2-(cx)^2-(k+cy)^2=0. \end{align} Do you know where to take it from here?