$v$ - volume, $x$ - position, $t$ - time. Bone Growth is proportional to surface area of growth sites --> $ dv/dt$ ~$\pi$ $r^2$~$(r^3)$^(2/3)~$v$^(2/3). I have the equation: $v_x + v_t $= kv ^2/3$, $k is a real number. (Which i found by taking a total derivative of volume with respect to time).Here the subscriptes denote partial derivatives. $k$ is a real valued scalar. After non-dimensionalizing this equation using scales: $x=x^*X, t=t^*T,v=v^*\hat V$, I derived the dimensionless equation:
$\hat V^.66667 $ = $\hat V_X+\hat V_T$. With initial conditions:$\hat V(X,0)$=$0 $ and $V(0,T)=1$. Now I need to use the method of characteristics to solve this equation, but I'm having trouble with it. How to you derive the characteristic curves? I know they are something like $T=X+C$ , where C is a real scalar. Also, using characteristics we have to solvesolve $V'=V^.6667$ which I know how to solve, Basically I'm just curious about how one would use method of characteristics to solve this PDE. PS: Sorry for the messy notation it's been a while since I posted.
If you want to solve $u_{t} + u_{x} = u^{2/3}$ ($u = \hat{v}$) then your characteristic equations are
\begin{align} \frac{dt}{ds} &= 1 \quad (1)\\ \frac{dx}{ds} &= 1 \quad (2)\\ \frac{du}{ds} &= u^{2/3} \quad(3) \end{align}
Using the first two ODEs to eliminate $s$, we find
\begin{align} \frac{dx}{dt} &= 1 \\ \implies x(t) &= t + x_{0} \end{align}
The solving the third
\begin{align} \frac{du}{dt} &= u^{2/3} \\ \implies 3u^{1/3} &= t + f(x_{0}) \\ &= t + f(x-t) \\ \implies u(x,t) &= \bigg(\frac{t + f(x-t)}{3} \bigg)^{3} \end{align}