# Energy estimate of the differential equation $\dot{x}=Ax$

Conside the differential equation $$\dot{x}=Ax,\qquad x(t):{\bf R}\to{\mathcal H}$$ where $\mathcal{H}$ is a Hilbert space and $A$ is a bounded linear operator. With the initial condition, one can have $x(t)=e^{At}x_0$ (Is this legal in the infinite dimension case?). With the spectral method(under the assumption that $x_0$ is an eigenvector of $A$), one has the estimate $$\|x(t)\|\leq e^{\omega t}\|x_0\|.$$

Here is my question:

• With some additional assumption, can one estimate $\|x(t)\|$ without using the spectral method, say, simply taking the inner product?

I think the following sub-question may be helpful:

• What is $\frac{d}{dt}\langle x(t),x(t)\rangle$?

• In the finite dimension case, this can be done with the product rule. What about the infinite dimension case?

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Yes, if $A$ is a bounded linear operator $e^{At}$ is quite OK. It can be defined by the Taylor series, for example, and [EDIT: for $t \ge 0$ ] we have $$\|e^{At}\| \le e^{\|A\| t}.$$ If $x_0$ is an eigenvector for $A$ with eigenvalue $\lambda$, then $x(t) = e^{\lambda t} x_0$.
As for your sub-question, it is still true that $$\frac{d}{dt} \langle x(t),\, x(t) \rangle = \langle x'(t), x(t) \rangle + \langle x(t), x'(t) \rangle = 2 \Re \langle x(t), A x(t) \rangle.$$ The proof using difference quotients is basically the same as in single-variable calculus.