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I've got this for homework:

Find general solution for this first order PDE

$\displaystyle (\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2= \frac{k}{r} - h$

where $k$ and $h$ are constant real numbers and $ r=\sqrt{x^2+y^2}$ .

Since $\displaystyle (\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2 = \vert \nabla u\vert^2$ , it is clear, from the upper equation, that $\lvert \nabla u \lvert $ depends only on $r$ which gave me an idea to look for a radial solution $u(x,y)=g(r)$. My question would be: is this only solution and how can I know if it is?

1 Answers 1

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Hint:

$\left(\dfrac{\partial u}{\partial x}\right)^2+\left(\dfrac{\partial u}{\partial y}\right)^2=\dfrac{k}{\sqrt{x^2+y^2}}-h$

$\left(\dfrac{\partial u}{\partial x}\right)^2=\dfrac{k}{\sqrt{x^2+y^2}}-h-\left(\dfrac{\partial u}{\partial y}\right)^2$

$\dfrac{\partial u}{\partial x}=\pm\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-\left(\dfrac{\partial u}{\partial y}\right)^2}$

$\dfrac{\partial^2u}{\partial x\partial y}=\pm\dfrac{-\dfrac{ky}{(x^2+y^2)^\frac{3}{2}}-2\dfrac{\partial u}{\partial y}\dfrac{\partial^2u}{\partial y^2}}{2\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-\left(\dfrac{\partial u}{\partial y}\right)^2}}$

Let $v=\dfrac{\partial u}{\partial y}$ ,

Then $\dfrac{\partial v}{\partial x}=\pm\dfrac{-\dfrac{ky}{(x^2+y^2)^\frac{3}{2}}-2v\dfrac{\partial v}{\partial y}}{2\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-v^2}}$

$\dfrac{\partial v}{\partial x}\pm\dfrac{v}{\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-v^2}}\dfrac{\partial v}{\partial y}=\mp\dfrac{ky}{2(x^2+y^2)^\frac{3}{2}\sqrt{\dfrac{k}{\sqrt{x^2+y^2}}-h-v^2}}$