Nonhomogeneous one-dimensional transport equation with boundary condition on $(x,t):x+t = 0$ instead of $t=0$

Consider the transport equation (see e.g. Evans p19) The solution of the initial value problem $u_t + b \cdot Du = f$ in $\mathbb R^n \times (0,\infty)$ with initial condition $u = g$ on $\mathbb R^n \times \{t=0\}$. is given by $u(x,t) = g(x-tb) + \int^t_0 f(x+(s-t)b,s) ds \quad (x \in \mathbb R^n, t \geq 0)$

Now, what if we have the following ($n=1$) problem: $u_t+u_x = 1$ ($f \equiv 1$ constant) in the domain $\{(x,t) \in \mathbb R^2: x+t > 0\}$ with the condition $u(r,-r) = \sin(r)$ (which is the boundary of the domain)

Can we, to solve this, apply the solution formula of the initial value problem and if so, how? (do we need to perform a transformation?) Is this still called initial value problem or is it a boundary problem?

• Regards @Mekanik. If I may, in your case, I don't think it is an initial value problem, since at $t=0$ we only know $u(0,0)=\sin(0)=0$. What is the PDE when $(x,t)$ is not in the domain? What is $f$ outside the domain in particular. – Arief Anbiya May 17 '17 at 17:51
• I dont really understand your question. We look for a function $u$ which is defined only in this open domain and there it should solve the given transport equation. – Mekanik May 17 '17 at 19:37

Initial value problems are a particular case of boundary value problems, where the solution is known at $t=0$. Here, we have a boundary value problem, where the solution is known on the line $t=-x$.
For the non-homogeneous advection equation $u_t + u_x = 1$, the method of characteristics gives the Lagrange-Charpit equations $$\frac{\text{d}t}{1} = \frac{\text{d}x}{1} = \frac{\text{d}u}{1} \, .$$ Therefore, the characteristic curves in the $x$-$t$ plane are parallel straight lines with equation $x(t) = x_0 + t-t_0$, on which $u$ is not constant: \begin{aligned} u(x(t),t) &= u(x_0,t_0) + t-t_0\, . \end{aligned} Now, since $u$ is known on the line $t=-x$, we select characteristics starting on this line, i.e., $t_0 = -x_0 = x_0 - x(t) + t$. Thus, $x_0 = \frac{1}{2}(x(t)-t)$, and \begin{aligned} u(x,t) &= u\left(\tfrac{1}{2}(x-t),\tfrac{1}{2}(t-x)\right) + t + \tfrac{1}{2}(x-t) \\ &= \sin\left(\tfrac{1}{2}(x-t)\right) + \tfrac{1}{2}(x+t) \, . \end{aligned}