Non-linear partial differential equation with conditions

Question

I have a pde with conditions, for which I'm looking for an analytical solution :

$\partial_t f(t,x)+f(t,x)\partial_x f(t,x)=0$.

$f(0,x)=0 \, , \, f(t,0)=0 $.

$f(t,x)$ is defined over : $\mathbb{R}^+ \times \mathbb{R}$.

Edit

• A solution of the equation is $f(t,x)=cst$. Obviously, it doesn't obey the conditions, except if $f(t,x)=0$, but it's not intersting.

Very similarly, if $f(t,x)=f(x-u t)=f(X)$.

$-u \partial_X f +f \partial_X f=0 \Rightarrow \partial_X f( f - u) =0$ which is leading to $f=cst$.

• A more interesting solution was proposed , if $f(x,t)=F(x)G(t)$. We obtain :

$\frac{dG}{dt}=C G(t)^2$ and $F(x)=-C$

So the solution is of the form : $f(x)=\frac{C x + D}{C t +E}$. But this is not compatible with the conditions.

Edit 2

Just to explain where my conditions come from : In physics, in linear elasticity, when you elongate or compress one can write the velocity in the transverse direction $t$: $\partial_t v_t+ v_t \nabla \underline{v} =0 $, one can easily show that it's tantamount to $\partial_t v_t+ (2-1/\nu)v_t \partial_t v_t =0 $ with $\nu$ a constant (Poisson modulus) which is tantamount to solve an equation $\partial_t f(t,x)+f(t,x)\partial_x f(t,x)=0$. Given that before you exert a force on the material there is no displacement $f(0,x)=0$ and given that there should be a symmetry in the item $v(t,0)=0$.

So I deleted one condition at $+\infty$.

Answer 1

$$\partial_t f(t,x)+f(t,x)\partial_x f(t,x)=0$$

$$f(0,x)=0 \, , \, f(t,0)=0 $$An obvious solution is the trivial $$f(t,x)=0$$ which satisfies the PDE and the conditions.

The question is : Are there other solutions ?

The Lagrange-Charpit equations are $$\frac{dt}{1}=\frac{dx}{f}=\frac{df}{0}$$ A first family of characteristic curves comes from $\frac{df}{0}\neq 0$ $\implies\:df=0$. $$f=c_1$$ A second family of characteristic curves comes from $\frac{dt}{1}=\frac{dx}{c_1}\quad;\quad c_1t-x=c_2$ $$tf-x=c_2$$ The general solution of the PDE $f_t+f_xf=0$ is : $$f=\Phi(tf-x)$$ where $\Phi$ is an arbitrary function to be determined according to boundary conditions.

$f(0,x)=0\implies 0=\Phi(0f-x)=\Phi(-x)$ , any $x$. Thus the function $\Phi$ is constant$=0$.

$f(t,0)=0\implies 0=\Phi(t0-0)=\Phi(0)$ is consistant with the preceding result.

Now, the function $\Phi$ is determined and equal to $0$. Puting it into the genertal solution $f=\Phi(tf-x)$ leads to $f=0$ any $x$ and $t$.

So, the solution exists and is unique : $$f(t,x)=0$$

If this analytic solution is not consistent with what is expected from the real behahiour in physics, this means that the modeling is not perfect. There might be something missing or not well defined in the PDE and/or in the conditions.

Answer 2

$\frac{\partial f}{\partial t}+f\frac{\partial f}{\partial x}=0$ $f(0,x)=0, f(+\infty,x)=0, f(t,0)=0$

$f=A(x)B(t)$

$A\frac{dB}{dt}+AB^2\frac{dA}{dx}=0$

$A\frac{dB}{dt}+AB^2\frac{dA}{dx}=0$

$\frac{1}{B^2}\frac{dB}{dt}+\frac{dA}{dx}=0$

$\frac{dA}{dx}=\lambda,\frac{1}{B^2}\frac{dB}{dt}=-\lambda$

$A=\lambda x+c_1$

$\frac{-1}{B}=-\lambda t+c_2$

$f=\frac{(\lambda x +c_1)}{\lambda t+c_2}$

$c_1=0$ by last equation.

$f=\frac{(x/t)}{1+c_2/(\lambda t)}$

$f=\frac{x}{t}(1-c_2/\lambda t+c_2^2/\lambda^2t^2-...)$

Not promising. Perhaps it can be proven that the initial condition can't be zero. Amc curious about that. Will play with it some more.