Consider the PDE $$u_t+tu_x=0 \\ u(0,t)=f(t)$$ We have that general solution is $u(x,t)=\phi(\frac{t^2}{2}-x)$. Notice that the solution is not unique since there is a family of characteristic curves that pass through the origin.
If $t>0, $ then there is no solution unless $\frac{t^2}{2}-x=c>0$
I don't understand why there won't be solutions unless $\frac{t^2}{2}-x=c>0$.
Why $\frac{t^2}{2}-x=c<0$ doesn't work?
Could someone explain this please?
You wrote : $u(0,t)=f(x)$ which is non-sens since $u(0,t)$ is function of $t$, not of $x$. So, I suppose that the correct equation is : $$u_t+tu_x=0 \\ u(0,t)=f(t)$$ $u(x,t)=\phi(\frac{t^2}{2}-x)$ is correct.
Condition : $u(0,t)=\phi(\frac{t^2}{2})=f(t)$
Let $X=\frac{t^2}{2}\geq 0\quad$ thus $\quad t=\sqrt{2X}\quad$ and so $\quad \phi(X)=f(\sqrt{2X})$
Now, the function $\phi(X)$ is determined. Putting it into the above general solution where $X=\frac{t^2}{2}-x$ leads to :
$$u(x,t)=f\left(\sqrt{2\left(\frac{t^2}{2}-x\right)}\right)$$ $$u(x,t)=f\left(\sqrt{t^2-2x}\right)\quad \text{in case of}\quad t^2-2x\geq 0$$