Method of characteristics for linear partial differential equation

Question

I am trying to solve $$\frac{\partial Q(z,t)}{\partial t}+(a-b*z)(z-1)\frac{\partial Q(z,t)}{\partial z}=0$$ by method of characteristics. So far , I have: $\frac{\partial t}{\partial r}=1$, $\frac{\partial z}{\partial r}=(a-b*z)(z-1)$, $\frac{\partial Q}{\partial r}=0$. Then, $t=r$, $z=\frac{\exp(b(c_{1}+r))-a*\exp(a(c_{1}+r))}{\exp(b(c_{1}+r))-b*\exp(a(c_{1}+r))}$, $Q=c_{2}$. The initial conditions are $z=s$, $Q=s^{n_{0}}$. And from here, I am lost. I am not sure how to get rid off $c_{1}$ and $c_{2}$.

Answer 1

With the solution for the first equation and the one for the second, $r=t$, we have a relation between $t$ and $z$. You isolated $r$ from the solution for the first, but it seemed better to me use it in its original form (btw I think the equation you wrote has some mistakes).

$$c_1+r=\dfrac{1}{a-b}\ln\left(\dfrac{z-1}{a-bz}\right)$$

Having $t$ and $z$ the relation,

$$c_1+t=\dfrac{1}{a-b}\ln\left(\dfrac{z-1}{a-bz}\right)$$

We have too $Q=c_2$. A solution sets a functional relation between $c_1$ and $c_2$, say $c_2=f(c_1)$ ($f$ continuous and differentiable):

$$Q=f\left(\dfrac{1}{a-b}\ln\left(\dfrac{z-1}{a-bz}\right)-t\right)$$

To know $f$ we have to impose the initial conditions. So is, we need the value for $Q$ along a suitable curve. I suppose it is $z=s $ and $t=0$. Along this curve $Q=s^{n_0}$

$$s^{n_0}=f\left(\dfrac{1}{a-b}\ln\left(\dfrac{s-1}{a-bs}\right)\right)$$

Making $u=\dfrac{1}{a-b}\ln\left(\dfrac{s-1}{a-bs}\right)$ and isolating $s$ we get,

$$s=\dfrac{a\,e^{(a-b)u}+1}{b\,e^{(a-b)u}+1}$$

Giving an explicit form for $f$

$$f(u)=\left(\dfrac{a\,e^{(a-b)u}+1}{b\,e^{(a-b)u}+1}\right)^{n_0}$$

$Q(z,t)=\left(\dfrac{a\,e^{(a-b)\left(\frac{1}{a-b}\ln\left(\frac{z-1}{a-bz}\right)-t\right)}+1}{b\,e^{(a-b)\left(\frac{1}{a-b}\ln\left(\frac{z-1}{a-bz}\right)-t\right)}+1}\right)^{n_0}$

$Q(z,t)=\left(\dfrac{a(z-1)e^{(b-a)t}+a-bz}{b(z-1)e^{(b-a)t}+a-bz}\right)^{n_0}$