# Inhomogeneous linear transport equation

Let $$(a,b)$$ be a subintervall of $$(0,1)$$. We consider the nonhomogeneous transport equation \eqalign{ & {y_t}(t,x) + c{y_x}(t,x) = {1_{(a,b)}}(x)f(t){\text{ }}{\text{, }}\left( {{\text{t}}{\text{,x}}} \right) \in (0,\infty ) \times (0,1), \cr & y(0,x) = {y_0}(t,x) \cr} where $$c>0$$. By the method of characterestic we find a explicit solution \eqalign{ & y(t,x) = {y_0}(x - ct),{\text{ if x}} \notin {\text{(a}}{\text{,b)}} \cr & y(t,x) = {y_0}(x - ct) + \int_0^t {{1_{(a,b)}}(x - ct + cs)f(s)ds} ,{\text{ if x}} \in {\text{(a}}{\text{,b)}} \cr} I want to find $$t$$ so that $$\int_0^t {{1_{(a,b)}}(x - ct + cs)f(s)ds} \ne 0,{\text{ if x}} \in {\text{(a}}{\text{,b)}}.$$ firstable, we know that $$t$$ must satisfies $$x - ct \in (0,1)$$, consequentely $$t\in(\dfrac{x-1}{c},\dfrac{x}{c})$$ with $$x\in(a,b)$$, I find this so strange and I'm sure that it is not true. Any ideas?. Thanks.

Let us assume an infinite domain (otherwise, boundary conditions must be taken into consideration). The method of characteristics gives $$x'(t)=c$$. Letting $$x(0)=x_0$$, we know $$x=x_0+ct$$. Moreover, since $$c>0$$, we have \begin{aligned} y'(t) &= \Bbb{I}_{[a,b]}(x(t))\, f(t)\\ &= \Bbb{I}_{[a,b]}(x_0+ct)\, f(t)\\ &= \Bbb{I}_{[a-x_0,b-x_0]/c}(t)\, f(t) \, . \end{aligned} Letting $$y(0)=y_0(x_0)$$, we know $$y(t) = y_0(x_0) + \int_0^t \Bbb{I}_{[a-x_0,b-x_0]/c}(\tau)\, f(\tau)\,\text d\tau \, .$$ If we denote by $$F$$ an antiderivative of $$f$$, we have $$y(t) = y_0(x_0) + F\big(\max\lbrace\min\lbrace t, \tfrac{b-x_0}{c}\rbrace, 0\rbrace\big) - F\big(\min\lbrace\max \lbrace 0, \tfrac{a-x_0}{c}\rbrace, t\rbrace\big) \, ,$$ which can be rewritten in terms of $$(x,t)$$ by using the expression $$x_0=x-ct$$ of the characteristic lines.