Section 9.4 Solutions to Abstract ODEs
We consider the abstract ODE on a Banach space \(X\) with unbounded operator \(A\) as
\begin{equation}
u_t = Au + f(t), \qquad u(0)= u_0 \in X,\label{abstractODE}\tag{9.4.1}
\end{equation}
where we assume \(u_0\in D(A)\text{,}\) \(f\in C([0,T], X) \text{.}\) For \(T>0\) we are interested in a classical solution
\begin{equation*}
u\in C^1([0,T], X) \cap C([0,T], D(A)).
\end{equation*}
Theorem 9.4.1.
Assume \(A\) is infinitesimal generator of a \(C^0\)-semigroup \(\{T(t)\}\text{.}\) If \(u\) is a classical solution of ((9.4.1)), then
\begin{equation}
u(t) = T(t) u_0 +\int_0^t T(t-s) f(s) ds.\tag{9.4.2}
\end{equation}
Proof.
\begin{align*}
\frac{dg}{ds} \amp=\amp - A T(t-s) u(s) + T(t-s) \dot u(s)\\
\amp=\amp - A T(t-s) u(s) +T(t-s)(A u(s) + f(s))\\
\amp=\amp T(t-s) f(s).
\end{align*}
Hence
\begin{equation*}
g(t) - g(0)= u(t) - T(t) u_0 = \int_0^t T(t-s) f(s) ds .
\end{equation*}
Definition 9.4.2.
Given
\begin{equation*}
\dot u = Au + f(t), \qquad \mbox{in }\quad X,
\end{equation*}
and \(f\in L^1((0,T), X)\text{.}\) Then \(u\) is called a mild solution if it satisfies the variation of constant formula (Theorem 9.4.1).Remark 9.4.3.
Note that a mild solution does not need any differentiation.Remark 9.4.4.
Note that a classical solution is automatically a mild solution. However, a mild solution is not necessarily a weak or classical solution. But mild solutions can be used to find those.Theorem 9.4.5.
Assume \(f\in C([0,T], X)\) and \(u_0\in D(A)\text{.}\) Assume that \(f\) satisfies at least one of the following conditions: either \(f\in W^{1,1}([0,T], X)\) or \(f\in L^1(0,T, D(A))\text{.}\) Then a mild solution \(u\) is also a classical solution.Proof.
\begin{equation}
v(t) := \int_0^t T(t-s) f(s) ds .\label{defv}\tag{9.4.3}
\end{equation}
We need to show that
\begin{equation*}
v\in \underbrace{C^1([0,T], X)}_{(1)}\cap \underbrace{C([0,T], D(A))}_{(2)}.
\end{equation*}
We compute
\begin{align*}
\underbrace{\frac{T(h) - I}{h}}_{(3)} v \amp=\amp \frac{1}{h} \left( \int_0^t T(h+t-s) f(s) ds - \int_0^t T(t-s) f(s) ds \right)\\
\amp=\amp \frac{1}{h} \left( \underbrace{\int_0^{t+h} T(h+t-s) f(s) ds}_{v(t+h)} - \int_t^{t+h} T(h+t-s) f(s) ds - \underbrace{\int_0^t T(t-s) f(s) ds}_{v(t)} \right)\\
\amp=\amp \underbrace{\frac{1}{h} (v(t+h)-v(t))}_{(4)} - \frac{1}{h} \int_t^{t+h} T(h+t-s) f(s) ds.
\end{align*}
From Theorem (Theorem 9.3.3) we evaluate the limit of the last term
\begin{equation*}
\lim_{h\to 0} \frac{1}{h} \int_t^{t+h} T(h+t-s) f(s) ds = f(t).
\end{equation*}
The limits of (3) and (4) need some more attention. In fact, if (1) is true, then (4) converges
\begin{equation*}
\lim_{h\to 0} \frac{1}{h}(v(t+h)-v(t)) = \dot v(t).
\end{equation*}
In this case (3) converges to \(Av\) and (2) is true, i.e (1) implies (2). If (2) is true then (3) converges
\begin{equation*}
\lim_{h\to 0} \frac{T(h) - I}{h} v(t) = A v(t) .
\end{equation*}
Moreover, if (3) converges, then also (4), which implies (1). Hence (1) implies (2). Also, if (4) converges, then also (3). We find that (1) implies (2) and vice versa. We break this circle conclusion by the additional conditions on \(f\text{.}\) If \(f\in L^1(0,T, D(A))\text{,}\) then (3) converges by the definition of \(v(t)\) in ((9.4.3)). If \(f\in W^{1,1}([0,T], X)\) we use the convolution identity
\begin{equation*}
v(t) = \int_0^t T(s) f(t-s) ds
\end{equation*}
and differentiate
\begin{equation*}
\dot v(t) = T(t) f(0) + \int_0^t T(s) \dot f(t-s) ds ,
\end{equation*}
which is well defined. Hence (1) holds. Altogether we have a classical solution of \(\dot u = Au + f(t)\text{,}\) \(u(0)=u_0\text{.}\)