I am solving the initial value problem:
$$u_t +u_x = -\sigma(x)u+m(x), \quad u(x,0)=\phi(x), \quad u(0,t)=\gamma(t)$$
I need to get solution like this:
\begin{equation*} u(x,t)=\phi(x-t) e^{-\int_{x-t}^{x} -\sigma(s)ds}+\int_{x-t}^{x} m(\tau)e^{-\int_{\tau}^{x} -\sigma(s)ds}d\tau, \quad t<x \end{equation*}
\begin{equation*} u(x,t)=\gamma(t-x) e^{-\int_{0}^{x} -\sigma(s)ds}+\int_{0}^{x} m(\tau)e^{-\int_{\tau}^{x} -\sigma(s)ds}d\tau, \quad x<t \end{equation*}
I have tried to use characteristic line G: $$x(\tau)=x+\tau, \quad t(\tau)=t+\tau$$ And i also know, that on this line: $$\frac{du(x+\tau,t+\tau)}{d\tau} = -\sigma(x+\tau)u(x+\tau,t+\tau) + m(x+\tau)$$
I can solve it firstly like homogeneous problem: $$\frac{du(x+\tau,t+\tau)}{u(x+\tau,t+\tau)} = -\sigma(x+\tau)d\tau$$ By integrating it $\tau$ from -t to zero for first case x>t.
And i got $$u(x,t)=u(x-t,0)e^{-\int_{-t}^{0} -\sigma(x+\tau) d\tau}$$
So I don't know what I should to do next. Because I don't have some const to use method of constant variance. Please help me complete the task.
$$u_t +u_x = -\sigma(x)u+m(x), \quad u(x,0)=\phi(x), \quad u(0,t)=\gamma(t)$$
Charpit-Lagrange characteristic ODEs : $$\frac{dt}{1}=\frac{dx}{1}=\frac{du}{-\sigma(x)u+m(x)}$$ A first characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{1}$ : $$x-t=c_1$$ A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{-\sigma(x)u+m(x)}\quad\implies\quad\frac{du}{dx}=-\sigma(x)u+m(x)$
This is a first order linear ODE which is solved thanks to classical method : $$u=c_2\exp\left(-\int_{0}^x \sigma(\xi)d\xi \right) +\exp\left(-\int_{0}^x \sigma(\xi)d\xi \right)\int_{0}^x \exp\left(\int_{0}^\zeta\sigma(\xi)d\xi \right)m(\zeta)d\zeta $$ The genaral solution of the PDE from implicit form $c_2=F(c_1)$ is : $$\boxed{u(x,t)=F(x-t)\exp\left(-\int_{0}^x \sigma(\xi)d\xi \right) +\exp\left(-\int_{0}^x \sigma(\xi)d\xi \right)\int_{0}^x \exp\left(\int_{0}^\zeta\sigma(\xi)d\xi \right)m(\zeta)d\zeta} $$ $F$ is an arbitrary function (to be determined according to the specified conditions).
Conditions : $u(x,0)=\phi(x), \quad u(0,t)=\gamma(t)$
One observes that the two conditions are not necessarily compatible at $(0,0)$. Compatibility would implies $u(0,0)=\phi(0)=\gamma(0)$. For any $\phi(x)$ and $\gamma(t)$ the solution will be a piecewise function. $u(x,t)$ should have two different forms in the two areas separated by the boundary $x-t=0$ : The functions $F$ will be not the same.
In order to determine the two functions $F$ one have successively to put the conditions into the above general solution. After a function $F(x)$ and a different $F(t)$ are determined one have to put them back into the general solution with the argument $(x-t)$ instead of $x$ or $t$. Certainly some transformations (possibly arduous) will be necessary to go to the expected forms of equations.