Unsolvable partial differential equation I was given the following differential equation:
$$\left\{\begin{matrix}u_t+2u_x=0, \ \ x>0, \ \ t>0, \\ u(x,0)=\arctan(x), \ \ x>0, \hspace{0.15cm} { }\\ u(0,t)=\dfrac{t}{1+t^2}, \ \ t>0. \hspace{0.8cm} { }\end{matrix}\right.$$
But I think that it has no solution. If I'm right, maybe it is very obvious, but I'll give a proof.
Let's consider the reduced problem: $$\left\{\begin{matrix}u_t+2u_x=0, \ \ x>0, \ \ t>0, \\ u(x,0)=\arctan(x), \ \ x>0. \hspace{0.15cm} { }\end{matrix}\right. $$ It is not difficult to check that the function $\widetilde{u}(x,t)=\arctan(x-2t)$ is a solution to this problem. But it is a Cauchy problem, so $\widetilde{u}$ is unique. However, it is clear that the complete solution to the problem (taking $u(x,0)$ and $u(0,t)$), let's call it $u(x,t)$, is also a solution to the reduced problem. By uniqueness of the Cauchy problem, $u(x,t)=\widetilde{u}(x,t)$. But this is absurd, because $\widetilde{u}(0,t)=\arctan(-2t)\not=u(0,t)=t/(1+t^2)$. Then, the complete problem can't have solution.
Finally, am I right? If not, what can be wrong in my proof?
Thanks for the help.
Edit (30/06/22): if possible, the solution I am looking for must be a classical solution.
 A: The subtle error in the logic above lies in the fact that the solution you find in the reduced problem does not define a solution in the entirety of $(x,t)\in\mathbb{R}^2$, but only in a subset of it, defined by the equation $x>2t$. The reason why this happens is because not all characteristic curves of the PDE above pass through $\mathcal{C}_1=(x,0), x>0$. When combined with $\mathcal{C}_2=(0,t), t>0$, however, it is now the case that all characteristic curves cut through the initial surface and hence define a solution for any point in $\mathbb{R}^2$.
For a complete solution, note that the characteristic passing through an arbitrary point $(x_0,t_0)$ is given by the equation
$$x-2t=x_0-2t_0$$
When $x_0-2t_0\geq 0$ it is the case that the characteristic hits the initial value surface $\partial \Omega=\mathcal{C}_1\star\mathcal{C}_2$ at the point $(x_0-2t_0,0)$ while for $x_0-2t_0< 0$ the intersection point is $(0,t_0-x_0/2)$. Then, given the constancy of $u$ on the characteristic curves, the solution is simply given by
$$u(x_0,t_0)=\begin{Bmatrix}\arctan( x_0-2t_0),&x_0-2t_0\geq 0\\\frac{t_0-x_0/2}{1+(t_0-x_0/2)^2},&x_0-2t_0<0\end{Bmatrix}$$
Note that the solution is continuous on the break line $x=2t$, as dictated by the continuity of the initial condition on the boundary curve.
A: $$u_t+2u_x=0, \quad x>0, \quad t>0, $$
The general solution is :
$$u(x,t)=F(x-2t)$$
with arbitrary function $F$ ( as long as no condition is takent into account ).
Initial condition :
$$u(x,0)=\arctan(x), \quad x>0 \implies F(x)=\arctan(x), \quad x>0 $$
Boundary condition :
$$u(0,t)=\dfrac{t}{1+t^2}, \quad t>0 \quad\implies \quad F(-2t)=\dfrac{t}{1+t^2}, \quad t>0$$
Let $X=-2t$
$$F(X)=\frac{-X/2}{1+(-X/2)^2}\quad X<0$$
Thus the function $F$ is a piecewise function :
$$F(X)=\begin{cases}
=X\qquad X>0\\
=\frac{-X/2}{1+(-X/2)^2}\quad X<0
\end{cases}$$
Now the function $F$ is determined. We put it into the above general solution where $X=x-2t$ :
$$u(x,t)=\begin{cases}
=x-2t\qquad x>2t\\
=\frac{t-\frac{x}{2}}{1+(t-\frac{x}{2})^2}\quad x<2t
\end{cases}$$
This is the particular solution which satisfies both the PDE and the conditions.
Not forgetting that $x>0$ and $t>0$ as specified in the wording of the problem.

