In some of my clases, they teach me some function which describe infiltration in the soil. The equation is known as Kostiakov's equation. We denote the Kostiakov's equation as $I(t)$, which maps $I:\Bbb R_{\geq 0}\mapsto \Bbb R_{\leq 0}$ and is related by the formula:
$$I(t)=C(t)^m\quad \forall C<0 \quad \forall m\in (0,1)$$
I was trying to define a new equation with the same "form" as the previous equation but also for each value of $x$, which phisically means the position in the soil surface; and impose as condition a bijective function of $t$, such that $f:\Bbb R_{\geq 0}\mapsto \Bbb R_{\geq 0}$, where $f(t)$ phisically means the position of the "front" of the water moving in the soil surface. I intended to make an equation that would describe the total infiltration for each position $x$.
Then I define a new function $\hat I(t,x)$, $\hat I:\Bbb R_{\geq 0}^2\mapsto \Bbb R_{\leq 0}$ and the first thing I notice is that it should have the condition that $\hat I(t,f(t))=0 \quad \forall x> f(t) $, meaning that the function is $0$ when the "front" has not passed some position $x$. Second, it should have the condition that for a pair $(a,b)\in \Bbb R_{\geq 0}^2,\ \hat I(a,b)=I(a-f^{-1}(b))\quad \forall f^{-1}(b)\leq a$. And a third condition, which is $\hat I(t,0)=I(t)\quad \forall t\in\Bbb R_{\geq 0}$.
Just by observing the second condition, I guess that:
$$\hat I(t,x)=\hat I\left(t-f^{-1}(x)\right)=C\left(t-f^{-1}(x)\right)^m $$
and checked that fulfill the third condition. The first condition is fulfill if $\hat I$ is define as a piecewise function:
$$\hat I(t,x)= \begin{cases} C\left(t-f^{-1}(x)\right)^m & ,x\leq f(t)\\[2ex] 0 &,x>f(t) \end{cases} $$
Then I realize that if $f(t)=k\cdot t$ for a constant $k\in\Bbb R_{\geq 0}$ and taking the double partial derivatives with respect to $t$ and $x$ of $\hat I$, it satisfied the hyperbolic PDE:
$${\partial^2\hat I\over\partial x^2} - {1\over k^2}{\partial^2\hat I\over\partial t^2}=0$$
and so the solutions of that PDE are any function of the form $H_1\left(t-{x\over k}\right)+H_2\left(t+{x\over k}\right)$
Now if we use the general case for any $f(t)$ and taking partial derivatives it gives the follow equations:
1. $${\partial \hat I\over \partial x}={\partial\hat I\over\partial t}\left(-{dt\over dx}\right)$$
$$\Rightarrow {\partial \hat I\over \partial x}\dot{x}+{\partial\hat I\over\partial t}=0$$
2. $${\partial^2 \hat I\over \partial x^2}= {\partial^2\hat I\over\partial t^2}\left(dt\over dx\right)^2-{\partial\hat I\over\partial t}\left(d^2 t\over dx^2 \right)$$
$$={\partial^2\hat I\over\partial t^2} \left(1\over \dot{x}\right)^2+{\partial\hat I\over\partial t}{\ddot x\over{\left(\dot x\right)^3}} $$
$$={\partial^2\hat I\over\partial t^2} \left(1\over \dot{x}\right)^2+{\partial\hat I\over\partial x}{\ddot x\over{\left(\dot x\right)^2}} $$
$$\Rightarrow {\partial^2 \hat I\over \partial x^2}\left(\dot x\right)^2 -{\partial\hat I\over\partial x}{\ddot x} - {\partial^2\hat I\over\partial t^2} = 0$$
So my question is how one can resolve the first equation, and maybe the second. The last one looks very hard so I dont think there's a known method for that. Also, for what I read, maybe the function is the solution of a Delay PDE, but I don't understand that type of PDE so I don't know how to write the Delay PDE which has for a particular solution of the function I made. For last, if $\ddot x=0$, then $\dot x=c, c\in\Bbb R$, so the equation become the normal wave equation. And as can be interpreted, the solution could be related to traveling waves with non-constant velocity.