Variation of constants for the inhomogeneous abstract Cauchy problem

Duhamel rigidity. Let $u$ be a classical solution and fix $t\in [0,T_0]$. The curve $g(s):=T(t-s)u(s)$ on $[0,t]$ is differentiable: $g^{\mathrm{\prime }}(s)=T(t-s)u^{\mathrm{\prime }}(s)-AT(t-s)u(s)=T(t-s)(u^{\mathrm{\prime }}(s)-Au(s))=T(t-s)f(s)$, where $u(s)\in D(A)$ and [F2] was used. The classical conditions make $u\mathrm{\prime }-Au$ continuous on $[0,T_0]$, so $f$ agrees on the interior with this continuous extension. Its product with $T(t-s)$ is continuous: an increment is bounded by $K\mathrm{\parallel }f(s)-f(s_0)\mathrm{\parallel }+\mathrm{\parallel }(T(t-s)-T(t-s_0))f(s_0)\mathrm{\parallel }$, which tends to zero. Thus $g\mathrm{\prime }$ extends continuously to the endpoints, and by the fundamental theorem of calculus [F2], $u(t)=g(t)=g(0)+{\int }_0^t{g}^{\mathrm{\prime }}(s)ds=T(t)x+{\int }_0^{t}T(t-s)f(s)ds$.

The formula is continuous and well defined. By [F1] $u$ is a well-defined continuous function on $[0,T_0]$; this is the mild solution of the problem in the sense of Classical, strong and mild abstract Cauchy solutions.

First take $f(s)=x_0{\mathbf{1}}_E(s)$. Put $J_{r}y={\int }_0^{r}T(q)ydq$. The needed exchange of vector integrals is justified directly: on the compact triangle $0\le s\le r\le t$, the curve $T(r-s)x_0$ is uniformly continuous. On a fine square grid approximate it uniformly by finitely valued functions sampled at points of the intersecting triangle cells, and multiply by ${\mathbf{1}}_E(s){\mathbf{1}}_{s\le r}$. Scalar Fubini (Fubini's theorem for L^1 functions on a sigma-finite product) applies to each indicator coefficient. Both iterated integral errors are at most $t^2$ times the uniform approximation error by [F1], so exchange remains valid in the limit. Consequently ${\int }_0^{t}u(r)dr=J_{t}x+{\int }_{E\cap (0,t)}{J}_{t-s}{x}_{0}ds$. The latter integral lies in $D(A)$ and its $A$-image is ${\int }_{E\cap (0,t)}(T(t-s)x_0-x_0)ds$: the pair $(J_{t-s}{x}_0,A{J}_{t-s}{x}_0)$ is continuous by [F2], and sampled step approximations, multiplied by ${\mathbf{1}}_E$, have graph-valued integrals; closedness of $A$ retains the limiting pair. Thus $A{\int }_0^{t}u=u(t)-x-{\int }_0^{t}f$. Finite linearity proves this for every integrable simple $f_n$. For general $f$, take defining simple $f_n$ with $\int \mathrm{\parallel }f-f_n\mathrm{\parallel }\to 0$. The local bound $K$ gives ${\sup }_t\mathrm{\parallel }{u}_n(t)-u(t)\mathrm{\parallel }\le K\int \mathrm{\parallel }f-f_n\mathrm{\parallel }\to 0$, so both coordinates of the graph pair converge: ${\int }_0^t{u}_n\to {\int }_0^{t}u$ and $A{\int }_0^t{u}_n\to u(t)-x-{\int }_0^{t}f$. Closedness proves the integral-solution identity for $u$.

Uniqueness among integral solutions. Let $v$ be an integral solution of $u^{\mathrm{\prime }}=Au+f$, $u(0)=x$, and put $d:=v-u$, which is continuous with ${\int }_0^{t}d\in D(A)$ and satisfies $d(t)=A{\int }_0^{t}d(r)dr$ for all $t$. For fixed $t$ define $h(s):=T(t-s){\int }_0^{s}d(r)dr$; then $h$ is differentiable with $h^{\mathrm{\prime }}(s)=T(t-s)(d(s)-A{\int }_0^{s}d)=0$ by [F2] and the equation for $d$, so $h$ is constant and ${\int }_0^{t}d(r)dr=h(t)-h(0)=0$ (using $T(0)=I$ and the primitive's value $0$ at $s=0$). Hence ${\int }_0^{t}d=0$ for every $t$; differentiating in $t$ with the fundamental theorem of calculus gives $d(t)=0$ for all $t$, so $v=u$.

Assume $x\in D(A)$ and $f(t)=f(0)+{\int }_0^{t}g(s)ds$ with $g$ Bochner integrable; the $C^1$ case is $g=f\mathrm{\prime }$. Write $v(t)={\int }_0^{t}T(r)f(t-r)dr$ (reflecting equal partitions under $s=t-r$ gives the same sampled sums, hence the same Bochner integral, for this continuous integrand). Substituting the primitive representation and exchanging the triangle integrals gives $v(t)={\int }_0^t[T(r)f(0)+{\int }_0^{r}T(r-s)g(s)ds]dr$. This exchange follows by the same grid argument as step 1.3 for simple $g$, and by $L^1$ approximation for general $g$: both errors are at most $K{T}_0\int \mathrm{\parallel }g-g_n\mathrm{\parallel }$. The integrand is continuous by [F1] applied to $g$, so the FTC gives $v\mathrm{\prime }(t)=T(t)f(0)+{\int }_0^{t}T(t-s)g(s)ds$, continuously on $[0,T_0]$. Since $T(t)x$ is $C^1$ by [F2], $u=T(\cdot )x+v$ is $C^1$.

Since $u$ is an integral solution by [step 1.3], subtraction gives $A{\int }_t^{t+h}u(r)dr=u(t+h)-u(t)-{\int }_t^{t+h}f(r)dr$ for all $0\le t<t+h\le T_0$; at $t=T_0$ use the corresponding backward difference. Divide by $h$: on the right, $\frac{u(t+h)-u(t)}{h}\to u^{\mathrm{\prime }}(t)$ and $\frac{1}{h}{\int }_t^{t+h}f\to f(t)$ by average convergence for the continuous $f$, so the right side tends to $u^{\mathrm{\prime }}(t)-f(t)$; on the left, $\frac{1}{h}{\int }_t^{t+h}u\to u(t)$ by average convergence for the continuous $u$. Since $A$ is closed, the limit pair $(u(t),u^{\mathrm{\prime }}(t)-f(t))$ lies in the graph of $A$; hence $u(t)\in D(A)$ and $Au(t)=u^{\mathrm{\prime }}(t)-f(t)$, that is $u^{\mathrm{\prime }}(t)=Au(t)+f(t)$ pointwise, and $u$ is a classical solution.

Claims (1), (2) and (3) are [step 1.1], [steps 1.2-1.4] and [steps 2.1, 3.1]; the classical upgrade holds for the stated $C^1$ or Bochner-primitive forcing.