Help with a nonlinear partial differential equation

Question

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.

Answer 1

$$(1-2f_x^2)f_tf_{xx}+2f_xf_{tx}(1+f_x^2)=0$$ 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.

Answer 2

$$(1-2f_{x}^{2})f_{t}f_{xx} + 2f_{x}f_{tx}(1+f_{x}^{2})=0$$ 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.