Why is a boundary condition at x=0 redundant for this Cauchy problem?

[onie mti]

I have this cauchy problem
U_t(x,t)= c_0[tanhx]u_x(x,t)=0
U(x,0)= u_0(x)

I managed to prove that it has at most one solution my question is why would it be redundant to have a boundary condition at x=0

 

[Ackbach]

I have this cauchy problem
U_t(x,t)= c_0[tanhx]u_x(x,t)=0
U(x,0)= u_0(x)

I managed to prove that it has at most one solution my question is why would it be redundant to have a boundary condition at x=0

If it is, as you say, that
\begin{align*}
u_t(x,t)&=C_0 \tanh(x) \, u_x(x,t) \\
u(x,0)&=u_0(x),
\end{align*}
then the DE itself says that $u_t(0,t)=C_0 \tanh(0) \, u_x(0,t)$; but $\tanh(0)=0$. So, at the very least, $u(0,t)$ cannot change in time, since $u_t(0,t)=0$.

It follows that $u(0,0)=u_0(0)$, and hence $u(0,t)=u_0(0)$.

But what would a boundary condition have to look like?