A practice question says:

Find and sketch the characteristics for $2u_t + (1+t^2)u_x = 0$

So I found that the vector field for where $u$ is constant is $$(1 + t^2, 2)$$ and so I'm looking for a set of curves such that $$\frac{d}{d\tau}(x(\tau), t(\tau)) = (1+(t(\tau))^2, 2)$$ and I got that $$t(\tau) = 2\tau + t_0$$ and therefore $$x(\tau) = \tau + \frac{1}{6}(2\tau + t_0)^3 + x_0$$ but this seems awfully complicated to sketch?! And so I think I may have done something wrong?


Just remove $\tau$ from both equations and you get


and this is just a set of cubic curves parametrized by the two constants $t_0$ and $x_0$.

  • $\begingroup$ how can you just remove $\tau$ from both equations? $\endgroup$ – user26069 May 12 '12 at 13:17
  • $\begingroup$ Invert $t(\tau)=2\tau+t_0$ to $\tau=\frac{1}{2}(t-t_0)$, put this into $x(\tau)$, and you are done. This gives the characteristic curves. $\endgroup$ – Jon May 12 '12 at 13:20
  • $\begingroup$ In other words, the solution is the one-parameter family $(x_c)_c$ defined by $x_c(t)=\frac16t^3+\frac12t+c$. $\endgroup$ – Did May 13 '12 at 8:32

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