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I am studying PDEs and have the following (seemingly simple) problem:

Find a surface that passes through the curve $$x^2+y^2=z=1$$ and is orthogonal to the family of surfaces $$z(x+y)=c(3z+1)\qquad(c\in\Bbb R)$$

After writing down the orthogonality condition (assuming my calculations are correct $(*)$), this yields the following equation: $$u(3u+1)(u_x+u_y)-x-y=0$$

We usually solve such equations by using the method of characteristics, which tells us (using assumption $(*)$ again) to solve the following characteristic system:

$$\begin{align}\dot x=&u(3u+1)\\\dot y=&u(3u+1)\\\dot u =&x+y\end{align}$$

Differentiating the last equation of this system with respect to $t$ gives us $\ddot u=\dot x+\dot y$, which using the first two equations gives us $$\ddot u = 2u(3u+1)$$

After staring at this equation for some time, I decided to ask Wolfram|Alpha. The result seems pretty ugly, so the following questions arise:

Did I make a mistake/am I missing something? Is my approach correct? How do I proceed?

Thanks.

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Dejan, since you're studying PDEs too, you may wanna be interested in this question I posted: math.stackexchange.com/questions/154515/… –  Weltschmerz Jun 8 '12 at 7:00

1 Answer 1

up vote 2 down vote accepted

Here's what I do (I'm probably wrong, but we can compare our procedures. I'm not using the comment box because there's not enough space in it).

The intersection of the surfaces would be the set of points $(x,y,z)$ such that \begin{cases}u(x,y)-z=0\\z(x+y)-3cz-c=0,\end{cases} that is, those that lie both on the solution surface and the given surface (determined by $c$, and this must hold for every $c$).

The surfaces are orthogonal means (I'm guessing) their respective gradients at the intersection points must be orthogonal. The gradient of the solution surface is $(u_x,u_y,-1)$, and the gradient of the given surface(s) is $(z,z,x+y-3c)$. So the condition is $zu_x+zu_y-x-y+3c=0$. Since this occurs when $z=u$ we can write it as $$uu_x+uu_y=x+y-3c.$$ We can clear $c$ from the intersection equations: $c=\frac{(x+y)u}{1+3u}$, and plugging it in the last equation gives us $$uu_x+uu_y=(x+y)(\frac{1}{1+3u}).$$

So we agree there. As for what follows, the method of characteristics, I think you proceeded correctly.

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Thanks for your thoughts. –  Dejan Govc Jun 8 '12 at 10:00

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