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Let $f(x,y,z,w)$ be a real function with continuous partial derivatives, satisfying the following partial differential equation $$ 2\left(\frac{\partial f}{\partial x}\right)\left(\frac{\partial f}{\partial y}\right) =f\left(\frac{\partial^2 f}{\partial x\partial x}\right) $$

What is the solution of that equation?

Edit: We can assume that $\frac{\partial^2 f}{\partial x\partial x}=\frac{\partial^2 f}{\partial y\partial x}$

My attempt: I noticed that $$ (\ast)\quad \frac{\partial(1/f)}{\partial x}=-\frac{1}{f^2}\frac{\partial f}{\partial x} $$ so $$ \frac{\partial^2(1/f)}{\partial x\partial y}=\frac{2}{f^3}\left(\frac{\partial f}{\partial y}\right)\left(\frac{\partial f}{\partial x}\right)-\frac{1}{f^2}\frac{\partial^2f}{\partial x\partial y}= \frac{1}{f^3}\left[2\left(\frac{\partial f}{\partial y}\right)\left(\frac{\partial f}{\partial x}\right)-f\frac{\partial^2f}{\partial x\partial y}\right] $$ Therefore, $(\ast)$ is satisfied iff $$ \frac{\partial^2(1/f)}{\partial x\partial y}=0 $$ What next?

boaz
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1 Answers1

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Next, integrate in $x$ and then in $y$ to get $1/f = g(y,z,w)+h(x,z,w)$ for some functions $g,h$.

dfr
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  • Thanks. I was not sure if I can integrate. Is the continuous partial derivatives allow us to integrate? – boaz Nov 02 '23 at 13:59
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    Yes, but if you don’t like the idea of integrating, just think that the equation is saying that the Hessian of $1/f$ is diagonal, hence once you derive w.r.t. $x$ you must end up with a function which does not depends on $y$ and vice versa. Then the dependence on those two variables must be “split” in two separate functions. – dfr Nov 02 '23 at 14:02
  • By the way, $g$ and $h$ can not be any two function, right? For instance, if $f(x,y,z,w)=\frac{z}{x}+\frac{w}{y}$. – boaz Nov 02 '23 at 14:37