I am trying to teach myself how to solve first-order PDEs using the method of characteristics, and I would like to solve the following first-order PDE for $u = u \left(x, y, t \right)$:
\begin{align*} a \frac{\partial u}{\partial t} + y \frac{\partial u}{\partial x} + b u &= f \left(x, t \right) \tag{1} \end{align*}
where $a$ and $b$ are constants. From the method of characteristics we get:
\begin{alignat*}{2} & \frac{dt}{ds} &&= a &&\Rightarrow &&t \left(s \right) &&= as + c_{1} \tag{2}\\ & \frac{d y}{ds} &&= 0 &&\Rightarrow &&y \left(s \right) &&= c_{2} \tag{3}\\ & \frac{d x}{ds} &&= y &&\Rightarrow &&x \left(s \right) &&= y s + c_{3} \tag{4}\\ & \frac{d z}{ds} &&= f \left(x,t \right) - bz &&\Rightarrow &&z \left(s \right) &&= e^{-bs} \left[ \int{e^{bs} f \left(x \left(s \right), t \left(s \right) \right) ds} + c_{4} \right] \tag{5} \end{alignat*}
However, I am unsure of how to proceed from here to obtain $u \left(x, y, t \right)$. In particular, I am confused regarding how to express the integral in the last equality in terms of $x$ and $t$. Any help on this would be greatly appreciated.
As an aside, I inputted $(1)$ into WolframAlpha, which gave the following:
\begin{align*} u \left(x,y,t \right) &= e^{\frac{-bx}{y}} \left[\int_{1}^{x}{\frac{e^{\frac{b \xi}{y}} f \left(\xi, t + \frac{a \left(\xi - x \right)}{y} \right)}{y} d \xi} + g \left(y \right) \left(t - \frac{a x}{y} \right) \right] \tag{6} \end{align*}
Is this correct? I am slightly confused since from $(1)$, $f$ should depend on $x$ and $t$ only; however, from $(6)$ it appears that $f$ depends on $x$, $t$ and $y$. Apologies in advance if these questions sound silly.
Kind regards, Nathan
The equations $(5)$ and $(6)$ are both correct, however contradictory they may seem.
This is because Eq.$(5)$ is valid only on the characteristic curve but not valid everywhere. Eq.$(6)$ is valid everywhere (any $x,y,t$).
An equivalent but different presentation may be helpful to understand.
$$a \frac{\partial u}{\partial t} + y \frac{\partial u}{\partial x} =-bu+ f \left(x, t \right) $$ Your system of ODEs is correct. Written on an equivalent form (Charpit-Lagrange) : $$ds=\frac{dt}{a}=\frac{dx}{y}=\frac{dy}{0}=\frac{du}{-bu+ f \left(x, t \right)}$$ You corectly found the characteristic equation $\quad y=c_2$.
A second characteristic equation comes from solving $\quad \frac{dt}{a}=\frac{dx}{y}=\frac{dx}{c_2}$: $$t-\frac{ax}{c_2}=c_3.$$ A third characteristic equation comes from solving $\quad \frac{dx}{c_2}=\frac{du}{-bu+ f \left(x, t \right)}$ $$c_2\frac{du}{dx}=-bu+f(x,t)$$ This is a linear first order ODE wrt the unknown function $u(x)$. The solution is: $$u=e^{-b\frac{x}{c_2}}\int e^{b\frac{x}{c_2}}f(x,t)dx$$ Especially this must be valid on the characteristic curves $y=c_2$ and $t=c_3+\frac{ax}{c_2}$ : $$u=e^{-b\frac{x}{y}}\int e^{b\frac{x}{y}}f\left(x\:, \left(c_3+\frac{ax}{y} \right)\right)dx$$ With definite integral : $$u=e^{-b\frac{x}{y}}\left[\int^{\xi=x} e^{b\frac{\xi}{y}}f\left(\xi\:,\left(c_3+\frac{a\xi}{y} \right)\right)d\xi+C \left(c_2,c_3 \right)\right]$$ $$u=e^{-b\frac{x}{y}}\left[\int^x e^{b\frac{\xi}{y}}f\left(\xi\:, \left(t-\frac{ax}{y}+\frac{a\xi}{y} \right)\right)d\xi+C \left(y\:,t-\frac{ax}{y} \right)\right]$$
$$u(x,y,t)=e^{-b\frac{x}{y}}\left[\int^x e^{b\frac{\xi}{y}}f\left(\xi\:, t + \frac{a(\xi-x)}{y}\right)d\xi+C \left(y\:,t-\frac{ax}{y} \right)\right]$$ $C$ is an arbitray function of two variables.
I don't know why WolframAlpha chose $1$ as lower bound for the integral instead $0$ or any other value. So, I let the choice open.