Mild solution

Remarks

The formula is the variation of constants formula, the analogue of Duhamel's principle for linear systems of ordinary differential equations. The mild solution always exists and is unique. A classical solution, that is, a continuously differentiable $u$ with values in the domain of $A$ that satisfies the equation, is equal to the mild solution. Conversely, if $u_0$ lies in the domain of $A$ and $f$ is continuously differentiable on $[0, \tau]$, the mild solution is a classical solution.

Example: let $X = \R$, $A = a$ for a real number $a \ne 0$, so $T(t) = e^{at}$, and let $f = 1$. The mild solution is $u(t) = e^{at} u_0 + (e^{at} - 1)/a$, and it satisfies $u' = a e^{at} u_0 + e^{at} = a u + 1$ with $u(0) = u_0$.

A mild solution need not be classical. Let $X = L^2(\R)$, $A = d/dx$ with domain the Sobolev space $H^1(\R)$, so $T(t)$ is the shift $(T(t) g)(x) = g(x + t)$, and let $f = 0$. For $u_0$ the indicator function of $[0, 1]$, the mild solution $u(t) = u_0(\cdot + t)$ is not differentiable as a map into $L^2(\R)$, so the problem has no classical solution.

Non-example: for $X = \R$, $A = 0$ and $f = 0$, the function $u(t) = t$ is not a mild solution of $u' = 0$, $u(0) = 0$. Here $T(t) = 1$, so the formula forces $u(t) = 0$ for all $t$.