Inhomogeneous linear transport equation

Question

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.

Answer 1

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.