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Apply $v=\ln u$ and then $v(x,y)=f(x)+g(y)$ to solve the PDE $$x^2u_x^2+y^2u_y^2=(xyu)^2.$$

Attempt: From $v=\ln u$, we have $u_x=u v_x$, $u_y=uv_y$.

Then $x^2u_x^2+y^2u_y^2=(xyu)^2$ reduces to $$x^2v_x^2+y^2v_y^2=x^2y^2, \text{ i.e. }\ \ \ x^2(f'(x))^2+y^2 (g'(y))^2=x^2y^2.$$

But I am unable to proceed further as I am unable to separate the variables.

Arctic Char
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user1942348
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1 Answers1

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From the choice $v = \ln u$ we assumed that $u>0$.

Continuing from $$x^2(f'(x))^2+y^2 (g'(y))^2=x^2y^2,$$ the above holds for all $x, y$. Choose $y=0$ gives

$$ x^2 (f'(x))^2 = 0,$$

thus $f$ is constant. Similarly by plugging $x=0$, $g$ is also constant. Thus $u$ has to be a constant function. Thus $u_x = u_y = 0$. Put this back to the original equation, we find that $u = 0$..... which is contradictory to the assumption that $u>0$. So it seems that your method can't be used to solve this PDE.

Going back to the original PDE, $$\tag{1} x^2u_x^2+y^2u_y^2=(xyu)^2,$$ let's try $u(x, y) = h(xy)$ for some function $h = h(t)$. Then $$ u_x (x, y) = y h'(xy), \ \ \ \ u_y(x, y) = xh'(xy),$$ thus (1) implies $$ 2(h'(xy))^2 = h^2(xy)\Rightarrow h'(t) = \pm\frac{1}{\sqrt 2} h(t)$$ which gives $$h(t) = C e^{ \pm \frac{1}{\sqrt 2} t}$$

and thus

$$ u(x, y) = Ce^{\pm \frac{1}{\sqrt 2} xy}.$$

Arctic Char
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