A while back, I asked how to solve a very ugly little duckling of a PDE, and if it could be done. No response. This is a better way repeating the question - if I have a PDE, where inside the equation itself, I have a term that is fixed in one parameter, but not the other, is the PDE solvable, and if so, how? Say I have, $$ G_{t}(t,x) = a(x)G_{x}(t,0)+b(x)G_{x}(t,1)+c(x)G_{x}(t,x);\\ G(\infty,x)=0; \quad G(0,x)=f(x), $$ where $f$ is some polynomial. Is there any way to solve this, or is it a lost cause?
Now, I understand one may be tempted to say $$\frac{\text{d}u}{\text{d}x}(1,t)=g(t),$$ for some arbitrary function, $f$, but what then? It is certainly the case that $f(t)$ is in no way arbitrary, so how could one justify treating it as such?
EDIT: Upon having it pointed out that a PDE without boundary conditions can be nearly anything, I have decided to write a simplification of the problem that spawned this, to which I know the conditions constraining it. There aren't many.
In the original wording of the question was : $$u_t(x,t)=ax\frac{\partial u}{\partial x}(x,t)+bx^2\frac{\partial u}{\partial x}(1,t), \tag 1$$ Let $\quad\frac{\partial u}{\partial x}(1,t)=f(t).\quad$ Of course, at this point $f(t)$ is an unknown function which has to be determined later.
Rearranging : $$-ax\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t}=bx^2f(t)\tag 2$$ The method of characteristic consists in solving the system of ODEs : $$\frac{dx}{-ax}=\frac{dt}{1}=\frac{du}{bx^2f(t)}$$ A first family of characteristic curves comes from $\quad \frac{dx}{-ax}=\frac{dt}{1}.\quad$ Solving it leads to : $$xe^{at}=c_1$$ A second family of characteristic curves comes from $\quad \frac{dt}{1}=\frac{du}{bx^2f(t)}$.
$du=bx^2f(t)dt=b\left(c_1e^{-at} \right)^2f(t)dt$ $$u-bc_1^2 \int_0^t e^{-2a\tau}f(\tau)d\tau =c_2$$ The general solution of Eq.$(2)$ comes from $\quad c_2=F(c_1)$
where $F$ is an arbitrary function to be determined according to some boundary or initial condition. $$u-bc_1^2 \int_0^t e^{-2a\tau}f(\tau)d\tau =F\left(xe^{at}\right)$$ $$u(x,t)=bx^2e^{2at} \int_0^t e^{-2a\tau}f(\tau)d\tau +F\left(xe^{at}\right)$$ $\frac{\partial u}{\partial x}=2bxe^{2at} \int_0^t e^{-2a\tau}f(\tau)d\tau +e^{at}F'\left(xe^{at}\right)$
$f(t)=\left(\frac{\partial u}{\partial x}\right)_{x=1}=2be^{2at} \int_0^t e^{-2a\tau}f(\tau)d\tau +e^{at}F'\left(e^{at}\right)$
So, we have to solve for $f(t)$ the integral equation : $$f(t)=2be^{2at} \int_0^t e^{-2a\tau}f(\tau)d\tau +e^{at}F'\left(e^{at}\right)$$ Thus $f(t)$ depends on the arbitrary function $F$. This makes difficult the solving in the general case. That is why if a boundary or initial condition is specified, the problem becomes simpler.
For example, if the initial condition is : $$u(x,0)=u_0(x),$$ we have $u(x,0)=bx^2e^{2at} \int_0^0 e^{-2a\tau}f(\tau)d\tau +F\left(xe^{0}\right)=F(x).$
The function $F$ is determined : $\quad F(x)=u_0(x)\quad$ which is a given function.
$$u(x,t)=bx^2e^{2at} \int_0^t e^{-2a\tau}f(\tau)d\tau +u_0\left(xe^{at}\right)$$ and the integral equation becomes : $$f(t)=2be^{2at} \int_0^t e^{-2a\tau}f(\tau)d\tau +e^{at}u_0'\left(e^{at}\right)$$ $e^{-2at}f(t)= \int_0^t e^{-2a\tau}f(\tau)d\tau +e^{-at}u_0'\left(e^{at}\right)$
Differentiating leads to :
$-2ae^{-2at}f(t)+e^{-2at}f'(t)=e^{-2at}f(t)-ae^{-at}u_0'\left(e^{at}\right)+au_0''\left(e^{at}\right)$
Thus, $f(t)$ is solution of the differential equation : $$f'(t)-(1+2a)f(t) =-ae^{at}u_0'\left(e^{at}\right)+ae^{2at}u_0''\left(e^{at}\right)$$ This is a first order linear ODE. Since $u_0(t)$ is a known function, this ODE can be solved. The solution includes an arbitrary constant. Putting back the solution into the integral equation allows to determine the constant.
So, the function $f(t)$ is known now. With this, the solution of the PDE according to the initial condition is : $$u(x,t)=bx^2e^{2at} \int_0^t e^{-2a\tau}f(\tau)d\tau +u_0\left(xe^{at}\right)$$
Meanwhile, the wording of the question has change which makes it much more complicated because it involves some non-explicit functions $a(x)$ , $b(x)$ and $c(x)$. The integrals of those functions will appear into the calculus thanks to the method of characteristics. Probably an explicit solution will no longer be obtained for all kind of functions $a(x)$ , $b(x)$ and $c(x)$. Nevertheless, the simpler case above shows the method to tackle the problem.