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Below is Exercise $3.6$ (p.$91$) of "Applied Partial Differential Equations" by Ockendon et al., $2^\mathrm{nd}$ ed.:

Show that, if $$\big(x+\alpha y\big)\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0,\tag{1}\label{1}$$ where $\alpha\neq 0$, the characteristics are given by $$ \left( z\pm \alpha \right)^2 \mp 4\alpha z + 2\alpha^2\log\left(z\pm \alpha \right) + x = \operatorname{constant}, \tag{2}\label{2} $$ where $z^2 = x+\alpha y$. Show that, for small $z$, the characteristics through the origin are $$ \mp \big( x+\alpha y\big)^{\frac32} = \dfrac{3\alpha x}{2}. \tag{3}\label{3} $$ What happens if $\alpha =0$?


First, I do not see how did authors derive expression $\eqref{2}$ for characteristics. For example I tried to carry out proposed change of variables, but the results do not seem to agree with $\eqref{2}$:

Assuming $x$ and $y$ are independent, then the new variable $z = z\left(x,y\right)$ is given implicitly by $z^2 = x+\alpha y$. Differentiating the last expression w.r.t $x$ and $y$ yeilds $$\begin{aligned} \left\lbrace\begin{aligned} z_x = \frac{1}{2z} \\ z_y = \frac{\alpha}{2z} \end{aligned}\right. &\implies \left\lbrace\begin{aligned} \dfrac{\partial u}{\partial{x}} &= \frac{d u}{d z} \!\cdot\!\frac{\partial z}{\partial x} = \frac{1}{2}\frac{ u_z }{z} \\ \dfrac{\partial u}{\partial{y}} &= \frac{d u}{d z}\!\cdot\!\frac{\partial z}{\partial y} = \frac{\alpha}{2} \frac{ u_z }{z} \end{aligned}\right. \\ &\implies \left\lbrace\begin{alignedat}{4} \dfrac{\partial^2 u}{\partial{x^2}} &= \dfrac{\partial }{\partial{x}} \left(\dfrac{\partial u}{\partial{x}} \right) &&= \dfrac{\partial }{\partial{z}} \left(\frac{1}{2} \frac{ u_z }{z}\right) \!\cdot\!\dfrac{\partial z}{\partial x} = \dfrac12\dfrac{zu_{zz} - u_z}{z^2}\!\cdot\! \dfrac{1}{2z} = \dfrac{1}{4z}\dfrac{zu_{zz} - u_z}{z^2} \\ \dfrac{\partial^2 u}{\partial{y^2}} &= \dfrac{\partial }{\partial{y}} \left(\dfrac{\partial u}{\partial{y}} \right) &&= \dfrac{\partial }{\partial{z}} \left(\frac{\alpha}{2} \frac{ u_z } {z}\right) \!\cdot\!\dfrac{\partial z}{\partial y} = \dfrac\alpha2 \dfrac{zu_{zz} - u_z}{z^2}\!\cdot\! \dfrac{\alpha}{2z} = \dfrac{\alpha^2}{4z}\dfrac{zu_{zz} - u_z}{z^2} \end{alignedat}\right.\end{aligned}$$ In this way, assuming $u\left(x,y\right) = u\left(z\right)$ is the solution of $\eqref{1}$, $$\begin{alignedat}{7} z^2=x+\alpha y &\implies {\ \alpha^2 u_{xx} = u_{yy} = \alpha^2z^{-3} \left(zu_{zz} - u_z\right)/4}\\ \eqref{1} & \iff \big(x+\alpha y\big) u_{xx} + u_{yy} = \left(z^2+\alpha^2\right)u_{xx} = \dfrac{z^2+\alpha^2}{4z^3}\big(zu_{zz} - u_z\big) = 0 \\ & \implies u_z=zu_{zz}\implies u_z=Az\implies {u=Az^2+B=A\left(x+\alpha y\right)+B} \end{alignedat} $$

I appreciate any hint regarding where the expression $\eqref{2}$ comes from, as it bothers me most.

Second, can we assume $x\approx y\approx 0$ when talking about characteristics through origin, or does the phrase "for small $z$" imply $ x\approx -\alpha y$ only? Either way I do not see how to get $\eqref{3}$ from $\eqref{2}$.

Third, if $\alpha=0$ then $z^2 = x$. Wouldn't then $\eqref{2} \iff z^2 +x = c$ imply characteristics $x=\operatorname{const}$?

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  • $\begingroup$ Maybe I think wrong, but something confuses me: On the one hand we have $z^2=-\frac{u_{yy}}{u_{xx}}|_{u_{xx}\neq 0}$ because of $(1)$ with $z^2=x+\alpha y$, on the other hand your calculations show $\frac{u_{yy}}{u_{xx}}|_{u_{xx}\neq 0}=\alpha^2$ and it follows $x+\alpha y=z^2=-\frac{u_{yy}}{u_{xx}}|_{u_{xx}\neq 0}=-\alpha^2$ and therefore one gets a dependency between $x$ and $y$, which contradicts the assumption, that $x$ and $y$ are independend. I think this must first be resolved before someone can answer your questions. $\endgroup$ – user90369 Apr 11 '17 at 13:45
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This answer is not complete (the answer to the first question has an extra $i$), but I am sure it is useful. I hope it can also bring some discussions which I can benefit from. So I decide to post it.

The equation of the characteristics is (refer to Equation 3.20 in Ockendon's book)

$$ \frac{dy}{dx} = \pm \sqrt{-\frac{1}{x + a y}} $$

I only worked on the minus sign. Let $x + a y = z^2$, differentiate w.r.t $x$ gives $1 + a y' = 2 z z'$. The prime means $d/dx$ here. Substitute: $$ 2 z z' = 1 - \frac{a i}{z} $$ Or $$ z' = \frac{z - a i}{2 z^2} $$ This form screams for a solution in terms of $x(z)$, the equation for which is just (the prime means $d/dz$ here, sorry for the abuse) $$ x' = \frac{2 z^2}{z - a i} $$ With help from Mathematica, the result is $$ x(z) = C-a^2 \log \left(a^2+z^2\right)+2 i a^2 \tan ^{-1}\left(\frac{a}{z}\right)+2 i a z+z^2 $$ With $\tan^{-1}(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)$ and some manual simplification we arrive at

$$ x + (a - i z)^2 + 2 a^2 \ln(a+i z) = C $$ This is close to the answer, actually by the replacement $i z \rightarrow z$ we have the answer, unfortunately I don't know how to deal with the $i$.

The second question is much easier, it is just perturbation and Taylor expansion. Setting $x = y = z = 0$ (since we are seeking solution around the origin), we have the "base solution" $$ a^2 + 2 a^2 \ln a = 0 $$ Now put $z$ back and treat it as a small parameter, Taylor expand all terms $$ z^2 + 2 a z + a^2 - 4 a z + 2 z^2 \left( \ln a + \frac{z}{a} - \frac{z^2}{2a^2} + \frac{z^3}{3a^3}\right) + x = 0 $$ It is very interesting that we have to expand up to $O(z^3)$ to have nonzero terms - all lower order terms cancel out. Use the base solution and do the cancellation, we are left with $$ x + \frac{2z^3}{3a} = 0 $$

I am not sure what the 3rd question is asking for. To me the only effect of $a$ is to control the slope of the dividing line between the elliptic and hyperbolic region. With $a = 0$ the dividing line is just the $y$-axis. Please point out if I missed something important.

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