let : $$\frac{\partial f}{\partial x}=f _{x}\,,\qquad\frac{\partial f}{\partial t}=f _{t}\,,\qquad\frac{\partial}{\partial t}\frac{\partial f}{\partial x}=f_{tx}\,,\qquad\frac{\partial}{\partial x}\frac{\partial f}{\partial x}=f_{xx} $$ And: $\,f=f(x,t)$

We have the PDE : $$\left(1-2\hspace{0.3ex}f_{x}^{2}\right)f_{t}\hspace{0.5ex}f_{xx}+2\hspace{0.3ex}f_{x}\hspace{0.4ex}f_{tx}\left(1+f_{x}^{2}\right)=0$$ Which can be written as : $$\big(2\hspace{0.3ex}f_{x}\hspace{0.3ex}f_{tx}+f_{t}\hspace{0.4ex}f_{xx}\big) \big(1+f_{x}^{2}\big)=3\hspace{0.3ex}f_{x}^{2\hspace{0.2ex}}f_{t}\hspace{0.4ex}f_{xx}$$ And we have the Dirchlet boundary data : $$ f(x,t_{i})=g_{i}(x)\;,\;\;f(x,t_{j})=g_{j}(x)$$

I am having a hard time trying to solve the problem. Any help is highly appreciated.

  • $\begingroup$ Does it mean you solve the Dirichlet problem in a strip $t_i<t<t_j$, $x\in \mathbb R$, for some numbers $t_i$, $t_j$? $\endgroup$ – Andrew Jun 27 '16 at 17:15
  • $\begingroup$ yes. the idea is to solve the problem for $t_{i}<t<t_{j}$ $\endgroup$ – Mohammad Al Jamal Jun 27 '16 at 17:32
  • 3
    $\begingroup$ Here are some observations from the group analysis (which doesn't help to solve the problem though). The equation doesn't have much symmetry but there is still some. If $f(x,t)$ is a solution of the equation then 1) $u(x,t)=k^{-1}f(kx+C_1,t)+C_2$; 2) $u(x,t)=f(x,h(t))$, where $h$ is an arbitrary (smooth) function; 3) combinations of the above transforms; are also solutions. From 2) it follows that it's enough to solve the Dirichlet problem in the strip $0<t<1$ and then get a solution $f$ in an arbitrary strip $t_i<t<t_j$ by shift and stretch wrt to $t$, say $f(x,t)=u(x,(t-t_i)/(t_j-t_i))$. $\endgroup$ – Andrew Jun 27 '16 at 17:53
  • $\begingroup$ that's somewhat assuring, since the solution is supposed to generate a homotopy that contains $g_{i}(x)$ and $g_{j}(x)$ with an additional constraint, that for an infinitesimal increment of $t$, $(x,y[=f(x,t)])$ should map to $(x+dx,y+dy)$ such that the distance between the two points is minimum . $\endgroup$ – Mohammad Al Jamal Jun 27 '16 at 18:16

$$(1-2f_x^2)f_tf_{xx}+2f_xf_{tx}(1+f_x^2)=0$$ This is a second order non-linear PDE.

Below, it is shown how to reduce it to the first order. Of course this is far to be the full solving, but I hope that it will help.

$$\frac{(1-2f_x^2)f_{xx}}{(1+f_x^2)f_x}+2\frac{f_{tx}}{f_t}=0$$ $\begin{cases} \frac{(1-2f_x^2)f_{xx}}{(1+f_x^2)f_x}=\frac{\partial}{\partial x}\left(\ln|f_x|-\frac{3}{2}\ln|1+f_x^2| \right)\\ \frac{f_{tx}}{f_t}=\frac{\partial}{\partial x}\left(\ln|f_t| \right) \end{cases} \quad\to\quad \frac{\partial}{\partial x}\left(\ln|f_x|-\frac{3}{2}\ln|1+f_x^2| +2\ln|f_t|\right)=0$

$$\frac{\partial}{\partial x}\left(\ln\left(\frac{f_x^2f_t^4}{(1+f_x^2)^3} \right)\right)=0$$ $\frac{f_x^2f_t^4}{(1+f_x^2)^3}=$ function of $t$ only.

$$(1+f_x^2)^3-f_x^2f_t^4 \varphi(t)=0 \quad \text{any function }\varphi(t)$$ This is a first order non-linear PDE.

Another simplification can be done :

Change of variable $\theta=\int \varphi(t)^{-1/4}dt \quad\to\quad dt=\varphi(t)^{1/4}d\theta \quad\to\quad f_\theta=f_t \: \varphi(t)^{1/4}$ $$(1+f_x^2)^3-f_x^2f_\theta^4 =0$$ where the unknown function is $f(x,\theta)$

Since $\varphi(t)$ is any function of $t$ then $\theta(t)$ is any function of $t$. So, one can forget the intermediate notation $\varphi$.

$f(x,t)=f\left(x,\theta(t)\right) \quad$ any function $\theta(t)$. This is consistent with the judicious comment of Andrew.

  • $\begingroup$ there is a minor error in the way you posted the problem. the factor $(1-f_{x}^{2})$ should be $(1-2f_{x}^{2})$. $\endgroup$ – Mohammad Al Jamal Jun 30 '16 at 16:51
  • $\begingroup$ You are right. It's a typo in coping from the draft, only in the two first equations. Nothing changes after. Correction done. Thanks for the remark. Also, see a further simplification at end of my main answer. $\endgroup$ – JJacquelin Jun 30 '16 at 17:21
  • $\begingroup$ feeding $(1+f_x^2)^3-f_x^2f_\theta^4 =0$ into mathematica, seems like $f(x,\theta)=c_{1}\theta+c_{2}x$ $\endgroup$ – Mohammad Al Jamal Jun 30 '16 at 20:21
  • $\begingroup$ I also found this particular solution at first time. But I didn't mention it because it isn't convenient according to the boundary conditions. The problem is to find a more general solution with fonctions consistent with the boundary conditions. $\endgroup$ – JJacquelin Jul 1 '16 at 4:55

$$(1-2f_{x}^{2})f_{t}f_{xx} + 2f_{x}f_{tx}(1+f_{x}^{2})=0$$ Try Separation of variables: $$f_{t}f_{xx} - 2f_{t}f_{xx}f_{x}^{2} + 2f_{x}f_{tx} + 2 f_{tx} f_{x}^{3}=0$$

write $f = X(x)T(t)$ Then you have:

$$X\dot{T}\ddot{X} T - 2X\dot{T}\ddot{X} \dot{X}^2 T^3 + 2\dot{T} \dot{X}^2 T + 2\dot{T} \dot{X}^4 T^3=0$$ $$(X\ddot{X} + 2 \dot{X}^2)\dot{T}T = (2X\ddot{X} \dot{X}^2 - 2\dot{X}^4 )\dot{T}T^3$$ $$(X\ddot{X} + 2 \dot{X}^2) = 2(X\ddot{X} - \dot{X}^2 )\dot{X}^2T^2$$ $$\frac{X\ddot{X} + 2 \dot{X}^2}{2(X\ddot{X} - \dot{X}^2 )\dot{X}^2} = T^2$$ Since the left hand side depends entirely on $x$ and the right on $t$, I think you have that they must be constant (say $k$). This implies that the Time dependence is trivial $$T=\sqrt{k}$$ So in fact your function is independent of time? Not sure if this is in the right direction, maybe I made some algebraic or other error. But if this is the case then your problem got a lot easier.

  • $\begingroup$ The solution i am looking for isn't of the form $f(x,t)=X(x)T(t)$. While separation of variables might produce a solution, it doesn't have to be unique, and it's not the form i am looking for. $\endgroup$ – Mohammad Al Jamal Jun 27 '16 at 19:54

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