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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 on $\mathcal{H}$. With the initial condition $x(0)=x_0$, 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 have thought that the following sub-questions 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?

1 Answers 1

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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.