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Find the surface which is orthogonal to the system $z=cxy(x^2+y^2)$ and which passes through the hyperbola $(x^2-y^2)=a^2$.

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We can consider $c$ as function of $x,y,z$.

$G(x,y,z)=c=\dfrac{z}{xy(x^2+y^2)}$ Its gradient is a vector orthogonal to each of the surfaces $G(x,y,z)=c$ constant.

$\nabla G=-\dfrac{z(y^2+3x^2)}{yx^2(x^2+y^2)^2}\hat i-\dfrac{z(x^2+3y^2)}{y^2x(x^2+y^2)^2}\hat j+\dfrac{1}{xy(x^2+y^2)}\hat k$

We have to find a function $z(x,y)$ orthogonal in each of its points to one of the surfaces with $G(x,y,z)=c$ constant. In each point of $z(x,y)$, the vectors, the orthogonal to $z\to(z_x,z_y,-1)$ and the orthogonal to $G\to\nabla G$, have to be orthogonals, so their dot product is zero.

For convenience we can use some other vector proportional to $\nabla G$

$W=\dfrac{(y^2+3x^2)}{x}\hat i+\dfrac{(x^2+3y^2)}{y}\hat j-\dfrac{(x^2+y^2)}{z}\hat k=a\hat i+b\hat j+c\hat k$

But still is valid: $(a,b,c)·(z_x,z_y,-1)=0$ Or, making $u=z$,

$\dfrac{y^2+3x^2}{x}u_x+\dfrac{x^2+3y^2}{y}u_y=\dfrac{-(x^2+y^2)}{u}$

It's a first order semilinear pde. To solve it, we use the method of characteristics. We write,

$\dfrac{xdx}{y^2+3x^2}=\dfrac{ydy}{x^2+3y^2}=\dfrac{-udu}{y^2+x^2}$

First surface for the characteristics: Adding the first and second ratios and equating to the third $\dfrac{xdx+ydy}{4(x^2+y^2)}=\dfrac{-udu}{y^2+x^2}\implies 2xdx+2ydy=-8udu\implies x^2+y^2=-4u^2+c_1$

Second surface: Substracting the second ratio to the first and equating to the third, $\dfrac{xdx-ydy}{2(x^2-y^2)}=\dfrac{-udu}{x^2+y^2}\implies\dfrac{2xdx-2ydy}{x^2-y^2}=\dfrac{-4udu}{-4u^2+c_1}\implies$

$\implies\ln\vert x^2-y^2\vert=(1/2)\ln\vert -4u^2+c_1\vert+k\implies (x^2-y^2)^{2}=(x^2+y^2)c_2$

Or $\dfrac{(x^2-y^2)^2}{x^2+y^2}=c_2$

For the general solution we have to set $c_1=f(c_2)$ for some $f$ single variable, differentiable function.

$$4u^2=f\left(\dfrac{(x^2-y^2)^2}{x^2+y^2}\right)-(x^2+y^2)$$

Now we have to impose the boundary conditions, so is, $u(x,y)=0$ for the hyperbola $x^2-y^2=a^2$ ($y^2=x^2-a^2$)

$$0=f\left(\frac{a^4}{2x^2-a^2}\right)-(2x^2-a^2)$$

Let $v=\dfrac{a^4}{2x^2-a^2}$, then $2x^2-a^2=\dfrac{a^4}{v}$, so is,

$f(v)=\dfrac{a^4}{v}$ Leading to the two particular solutions,

$$u(x,y)=\pm\sqrt{\dfrac{a^4(x^2+y^2)}{4(x^2-y^2)^2}-\dfrac{x^2+y^2}{4}}$$

The solution is not unique, but both solutions, understood as surfaces, form the surface solution.

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