0

I have a PDE of the form

$$\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 = f(x,y)$$

and I am looking for a way to solve it. I tried the Method of Characteristics for the same but the Lagrange-Charpit equations turn out to be equally complicated. I am primarily looking for an analytical method to solve it. Any help or suggestions would be appreciated. Thanks

EditPiAf
  • 21,328
quirkyquark
  • 156

2 Answers2

2

As proposed here, let us use the fact that $u_y = \pm\sqrt{f-u_x^2}\,$ to compute $$ u_{yx} = \frac{f_x-2u_xu_{xx}}{\pm 2\sqrt{f-u_x^2}}\, . $$ This step amounts to differentiating the original PDE w.r.t. $x$. If we substitute $v=u_x$, then we have $$ vv_{x} \pm\sqrt{f-v^2}\,v_{y} = \tfrac12 f_x $$ for sufficiently smooth solutions. For this first-order quasi-linear PDE, the Lagrange-Charpit equations read $$ \frac{\text d x}{v} = \frac{\text d y}{\pm\sqrt{f-v^2}} = \frac{\text d v}{\tfrac12 f_x} $$ from which we obtain the following characteristic families:

  • $f_x\,\text d x = 2 v\,\text d v$ gives $f-v^2 = c_1^2$ with $c_1 = u_y$
  • $f_x\,\text d y = 2 c_1\,\text d v$ gives $\int^y f_x(x,\eta)\,\text d\eta - 2 c_1 v = c_2$

Thus, we can write \begin{aligned} |u_y| &= F\big(\textstyle\int^y f_x(x,\eta)\,\text d\eta - 2 u_y u_x\big) \\ |u_x| &= \sqrt{f - u_y^2} \end{aligned} for some $F$. Once $u_x$, $u_y$ are determined, it remains to integrate to find $u$. Fully explicit analytical solutions may be obtained for some particular functions $f$, and for some particular boundary conditions. Note that this is an eikonal equation, for which some particular solutions are known (see e.g. this post, this post and linked ones).

Note: Here, only real unknowns $u$ were considered. There might be complex solutions too.

EditPiAf
  • 21,328
  • 1
    I think in the equation for the second characteristic family, the $dx$ should be $dv$. – quirkyquark Apr 05 '20 at 23:24
  • 1
    thanks for the answer but I think my $f(x,y)$ is way complicated to be solved by this method. It is of the form $\frac{1}{((x-d)^2 + y^2))^2} + \frac{1}{((x+d)^2 + y^2))^2}$ where $d$ is a constant. At most I can set $d=0$ but I still have to solve it for $d \neq 0$ case. – quirkyquark Apr 05 '20 at 23:49
  • 1
    @NamanAgarwal For $d=0$, you may transform to polar coordinates and look for radial solutions (see linked posts in answer). For $d\neq 0$, the answer provides implicit equations that can be solved numerically (quasi-analytical solution) – EditPiAf Apr 06 '20 at 00:06
  • For $d=0$ case, the polar transformation worked well and gave me a solution. Thanks for the help! – quirkyquark Apr 06 '20 at 00:21
2

Here's a way to transform this nonlinear PDE to a linear PDE:

First note that the PDE can be written as $|\nabla u|^2 = f.$ This implies $|\nabla u| = \sqrt{f}$ and there exists some unit vector field $\vec{n}(x,y)$ such that $\nabla u = \sqrt{f}\,\vec{n}.$

Now, the rotation of a gradient vanishes, so $$ \vec{0} = \nabla\times\nabla u = \nabla\times(\sqrt{f}\,\vec{n}) = (\nabla\sqrt{f})\times\vec{n} + \sqrt{f}\,(\nabla\times\vec{n}), $$ or, when $f\neq 0,$ $$ \nabla\times\vec{n} = -\frac{\nabla\sqrt{f}}{\sqrt{f}}\times\vec{n}. $$ Setting $\vec{\varphi} = -\frac{\nabla\sqrt{f}}{\sqrt{f}}$ we arrive at $$ \nabla\times\vec{n} = \vec{\varphi}\times\vec{n}. $$

If we can solve this for $\vec{n}$ then we have $\nabla u = \sqrt{f}\,\vec{n}$ and can just set $u(x,y) = \int_\gamma \sqrt{f}\,\vec{n} \cdot d\vec{r}$ for some path $\gamma$ from $(0,0)$ to $(x,y).$

md2perpe
  • 30,042