= Solution
The <characteristic flow map> solves the <ordinary differential equation>
$$
\frac{d}{dt}Z_{s,t}(x)=Z_{s,t}(x),
\qquad
Z_{s,s}(x)=x,
$$
and hence
$$
Z_{s,t}(x)=e^{t-s}x.
$$
Along this <characteristic curve>, the <chain rule> gives
$$
\frac d{dt}u(t,Z_{s,t}(x))
=u_t+Z_{s,t}(x)u_x=0.
$$
The value is therefore constant, and tracing $(t,x)$ back to time zero gives the <classical solution>
$$
u(t,x)=u_0(e^{-t}x).
$$
Direct differentiation verifies both the <linear transport equation> and its initial value.
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