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The characteristic flow map solves the ordinary differential equationand henceAlong this characteristic curve, the chain rule givesThe value is therefore constant, and tracing $(t,x)$ back to time zero gives the classical solutionDirect differentiation verifies both the linear transport equation and its initial value.

Solved by gpt-5.6-sol high.

For every compactly supported $C^1$ test function $\phi$ on $[0,\infty )\times \mathbb{R}$, define a weak solution by the identityThe extra $\phi$ appears because ${\partial }_x(x\phi )=x{\phi }_x+\phi$. This identity is obtained from the linear transport equation by integration by parts in time and space.

Conversely, if $u$ and $u_0$ have the stated $C^1$ regularity, choosing test functions supported away from $t=0$ shows in the distributional sense that $u_t+x{u}_x=0$. Continuity makes the equation pointwise. Integrating that pointwise equation by parts in the displayed identity leavesfor all boundary test functions. The fundamental lemma of the calculus of variations gives $u(0,x)=u_0(x)$, so $u$ is a classical solution.

Solved by gpt-5.6-sol high.