Question : What is the solution of the Cauchy Problem pde
\[u_t-uu_x=0\ ,\ \ x\in R,\ t>0\] \[u\left(x,0\right)=x,\ \ x\in R\]
Solution :
For the given Cauchy Problem pde
\[u_t-uu_x=0\ ,\ \ x\in R,\ t>0\]
\[u\left(x,0\right)=x,\ \ x\in R\]Substituting x=x, t=y, u=z
The characterstics equations are
\[\frac{dx}{dt}=-z,\ \ \frac{dy}{dt}=1,\ \ \frac{dz}{dt}=0\]
with parametrization \[\gamma:\ \left(f\left(s\right),\ g\left(s\right),\ h(s)\right)\] where initialy \[f\left(s\right)=s,\ g\left(s\right)=0,\ h\left(s\right)=s\]
integrating
\[\frac{dy}{dt}=1\]\[y=t+c_1\]
Initialy
\[t=0,\ y=g\left(s\right)=0=>c_1=0\]\[=>y=t\]Now integrating
\[\frac{dz}{dt}=0\]\[=>z=c_2\]Initialy \[z=h(s)=s=> c2=s\]Now integrating
\[\frac{dx}{dt}=-z=-s\]
\[=>x=-st+c_3\]Initialy x=s, t=0 then
\[=>c_3=s\]
\[=>x=-st+s\]\[=>s=\frac{x}{1-t}\]
Since
\[u=h\left(s\right)=s=\frac{x}{1-t}\]\[=>u=\frac{x}{1-t}\]This is the required solution of the given Cauchy Problem.