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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 105 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-105/3/a/solution