Weak solution

Remarks

The identity is what integration by parts gives when $u$ is of class $C^m$ and $P u = f$ holds pointwise. So every classical solution is a weak solution. Conversely, a weak solution of class $C^m$ satisfies $P u = f$ almost everywhere, and everywhere if $f$ is continuous. If $u$ has weak derivatives up to order $m$, the condition says that $\sum a_\alpha D^\alpha u = f$ almost everywhere.

Example: on $\R^2$ with variables $(x, t)$, let $u(x, t) = 1$ for $x > t$ and $u(x, t) = 0$ for $x \le t$, and let $P = \partial_t + \partial_x$, so $P^t \varphi = -(\varphi_t + \varphi_x)$. Put $\psi(y, t) = \varphi(y + t, t)$; then $\partial_t \psi = \varphi_x + \varphi_t$ at the point $(y + t, t)$. Hence $\int\!\!\int u \, (\varphi_t + \varphi_x) \, dx \, dt = \int_0^\infty \int_{\R} \partial_t \psi(y, t) \, dt \, dy = 0$, because $\psi(y, \cdot)$ has compact support. So $u$ is a weak solution of $u_t + u_x = 0$, though it is discontinuous.

Non-example: the step $w(x, t) = 1$ for $x > 0$ and $0$ for $x \le 0$ is not a weak solution of the same equation. Here $\int\!\!\int w \, (\varphi_t + \varphi_x) \, dx \, dt = -\int_{\R} \varphi(0, t) \, dt$, which is not zero for a test function with $\varphi(0, t) > 0$ near some $t$.

For second-order equations in divergence form, such as $-\sum_{i,j} \partial_j (a_{ij} \partial_i u) = f$ with bounded coefficients, one often asks instead that $u$ lie in a Sobolev space $H^1_0(U)$ and that $\int_U \sum_{i,j} a_{ij} \, \partial_i u \, \partial_j v \, dx = \int_U f v \, dx$ for all $v \in H^1_0(U)$. This is the variational form of the definition.