[proofplan] We multiply the classical transport equation by a compactly supported [test function](/page/Test%20Function) and integrate over spacetime. Compact support removes all boundary terms, and [integration by parts](/theorems/210) transfers derivatives from $u$ to the test function. [/proofplan]

[step: Integrate the classical equation against a test function] Let $\phi\in C_c^\infty(\mathbb R^n\times(0,T))$. Since $u$ is $C^1$ and solves the equation pointwise, \begin{align*} 0=\int_0^{\,T}\int_{\mathbb R^n}\phi(x,t)\left(\partial_tu(x,t)+b\cdot\nabla u(x,t)\right)\,d\mathcal L^n(x)\,dt. \end{align*} The compact support of $\phi$ in $\mathbb R^n\times(0,T)$ ensures that there are no boundary contributions at $t=0$, at $t=T$, or at spatial infinity. [/step]

[step: Transfer the derivatives to the test function][Integration by parts](/theorems/2098) in time gives \begin{align*} \int_0^{\,T}\int_{\mathbb R^n}\phi\,\partial_tu\,d\mathcal L^n\,dt=-\int_0^{\,T}\int_{\mathbb R^n}u\,\partial_t\phi\,d\mathcal L^n\,dt. \end{align*} For the spatial term, the vector $b$ is constant, so \begin{align*} \int_0^{\,T}\int_{\mathbb R^n}\phi\,b\cdot\nabla u\,d\mathcal L^n\,dt=-\int_0^{\,T}\int_{\mathbb R^n}u\,b\cdot\nabla\phi\,d\mathcal L^n\,dt. \end{align*} Substituting these two identities into the integrated classical equation yields \begin{align*} 0=-\int_0^{\,T}\int_{\mathbb R^n}u(x,t)\left(\partial_t\phi(x,t)+b\cdot\nabla\phi(x,t)\right)\,d\mathcal L^n(x)\,dt. \end{align*} Multiplying by $-1$ gives the asserted weak identity.[/step]

[guided]Start with the pointwise equation $\partial_tu+b\cdot\nabla u=0$ and multiply it by the test function $\phi$. After integrating over spacetime, this gives \begin{align*} 0=\int_0^{\,T}\int_{\mathbb R^n}\phi(x,t)\left(\partial_tu(x,t)+b\cdot\nabla u(x,t)\right)\,d\mathcal L^n(x)\,dt. \end{align*} Because $\phi$ is compactly supported in $\mathbb R^n\times(0,T)$, integration by parts in time has no endpoint contribution, so \begin{align*} \int_0^{\,T}\int_{\mathbb R^n}\phi\,\partial_tu\,d\mathcal L^n\,dt=-\int_0^{\,T}\int_{\mathbb R^n}u\,\partial_t\phi\,d\mathcal L^n\,dt. \end{align*} The vector $b$ is constant, so spatial integration by parts similarly gives \begin{align*} \int_0^{\,T}\int_{\mathbb R^n}\phi\,b\cdot\nabla u\,d\mathcal L^n\,dt=-\int_0^{\,T}\int_{\mathbb R^n}u\,b\cdot\nabla\phi\,d\mathcal L^n\,dt. \end{align*} Combining the two transferred terms yields \begin{align*} 0=-\int_0^{\,T}\int_{\mathbb R^n}u(x,t)\left(\partial_t\phi(x,t)+b\cdot\nabla\phi(x,t)\right)\,d\mathcal L^n(x)\,dt. \end{align*} Multiplying this identity by $-1$ gives the desired weak formulation.[/guided]