Does $\partial_t u(x,t) - \partial_x u(x,t) = 0$ have a non-trivial solution? Consider the linear PDE $$\partial_t u(x,t) - \partial_x u(x,t) = 0$$ on the domain $$\Omega=\{(x,t)\;\;|\;\;0\leq x \leq 1,\;0\leq t \leq T\}$$
with boundary conditions $$u(0,t)=u(1,t)=c$$ and the initial condition $$u(x,0)=u_0(x).$$
This is a special case of the advection-diffusion equation or the heat equation with additional terms. In trying to solve this PDE using separation of variables and specifying $u(0,t)=u(1,t)=0$, I find only the trivial solution. In trying to use Fourier methods to solve the more general problem \begin{align}\partial_{t} u(x,t)&=k\,\partial_{xx}u(x,t)+A\,\partial_x u(x,t)+B\,u(x,t)\\u(0,t)&=u(L,t)=0\\u(x,0)&=u_0(x)\end{align}
on the domain $$\Omega=\{(x,t)\;\;|\;\;0\leq x \leq L,\;0\leq t \leq T\},$$ I find the solution to be $$u(x,t)=e^{-\frac{A}{2k}x+(-\frac{A^{2}}{4k}+B)t}\sum_{n=1}^{\infty}b_{n}sin\left(\frac{n\pi x}{L}\right)e^{-n^{2}\pi^{2}kt/L^{2}}$$where$$
b_{n}=\int_{0}^{L}e^{\frac{A}{2k}\xi}f(\xi)\,sin\left(\frac{n\pi\xi}{L}\right)\,d\xi$$
and $f(x)=u(x,0)$ (see P. 52 of O'Neil, Beginning Partial Differential Equations, 3rd Edition, Wiley (2014)). However, for the case of $k=B=0$, the Fourier coefficients blow up.
Does this equation have a non-trivial solution?
 A: Given the boundary conditions, there is only the trivial solution, $u\equiv c$ if $T\ge 1$. However, for $T<1$, non-trivial solutions are possible. This can be seen by using the method of characteristics:
Here, the characteristics are easily found, they are the lines parametrized by $s\mapsto (-s, s)$, along these lines the solution of your PDE has to be constant. By ignoring the particular choice of $T$ and the boundary condition at $x=0$, we get the unique solution:
$$u(x,t) = \begin{cases}u_0(x+t),\quad &\text{for }0<x<1-t, \\
c,\quad &\text{for }0<x\ge 1-t   \end{cases}.$$
Now we need to also take the B.C. at $x = 0$ into account, which is fulfilled iff (according to the above representation of $u$) $c = u(0,t) = u_0(t)$ for $t<T$. If $u_0$ doesn't fulfill this condition, there is no solution to the problem at all.
If $T\ge 1$, this means that $u_0 \equiv c$ is necessary, and $u\equiv c$ is the only solution, which is trivial.
If $T<1$, we are free to choose $u_0(x)\neq c$ for $1>x>T$, without conflicting the left boundary condition.
A: For your first equation, the characteristics are given by $dx/dt = -1$. So on $t$-vs-$x$ plane, they "travel" backwards with slope 1.
Thus, if there exists $x \in (0,1)$ such that $u_0(x) \neq c$, then there will be a discontinuity at some $t$ and at $x = 0$. As I understand it, these are sometimes called weak solutions (my terminology might be wrong here).
If you strictly want continuous solutions, then there is only the trivial solution.
A: Let $U(x,t)$ and $V(x,t)$ be any two solutions to the PDE $u_{t} - u_{x} = 0$. If $U(x,t)$ and $V(x,t)$ have different initial and boundary conditions, then of course the solution would be different also. So assume the initial and boundary conditions are the same. 
Both are solutions, so 
$$ U_{t} - U_{x} = 0 , \:\:\:\:\: V_{t} - V_{x} = 0$$
Now let $W(x,t) = U(x,t) - V(x,t)$, we get
$$ W_{t}  - W_{x} = (U_{t} - V_{t}) - (U_{x} - V_{x}) = 0 $$
so $W(x,t)$ is also a solution. 
Now notice that $$ W(x,0) = U(x,0) - V(x,0) = 0 $$, and the same applies at the boundaries. 
Since there is no disturbances, source, or sink, then $W(x,t)$ will not have any change. So $W(x,t) = 0$ for $0<x<L$, $t>0$. Which means $U(x,t) = V(x,t)$.
But another way, you may also use your Fourier method result to prove that :
$$ W(x,0) = 0 \:\:\: \implies \:\:\: b_{n} = 0 $$
then see what this means. 
Hope this helps, thanks. 
