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Let $D$ be the derivative operator, and set $$T:=\sum_{n=0}^\infty D^n. $$ It is known that under certain conditions $$T=(Id-D)^{-1} .$$ Set $v=Tu$, applying $I-D$ to both sides, in this case, gives the ODE $$v-v'=u $$ which can be solved for $v$ as follows $$v=\mathrm{e}^x \left( C-\int \mathrm{e}^{-x} u(x) \mathrm{d} x \right),$$ where $C$ is a constant of integration, originating from the indefinite integral. At the same time we also have $$v=Tu=\sum_{n=0}^\infty u^{(n)}(x), $$ so we seem to get $$ \mathrm{e}^x \left( C-\int \mathrm{e}^{-x} u(x) \mathrm{d} x \right)=\sum_{n=0}^\infty u^{(n)}(x).$$ The left hand side is an infinite 1-parametric family of functions, while the right hand side is a concrete function, provided the series converges (e.g. the case where $u(x)$ is a polynomial).

My questions are: how does one settle this issue? Is the integral on the left hand side satisfying some initial/boundary condition? What is the value of $C$?

Thank you!

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  • $\begingroup$ In general for a ODE like $v' - v = -u$ you need to apply the initial conditions to fix the value of $C$. For your case integrating $[v(x)e^{-x}]' = -u(x)e^{-x}$ over $[0,x]$ gives $v(x) = v(0) e^{x} - \int_0^x u(x') e^{x-x'}{\rm d}x'$. $\endgroup$ – Winther Jun 19 '17 at 14:03
  • $\begingroup$ What's with the hyperlink? $\endgroup$ – Aweygan Jun 19 '17 at 14:41
  • $\begingroup$ @Aweygan Oops, lol. That's not the right one. I'll fix it. $\endgroup$ – user1337 Jun 19 '17 at 14:44
  • $\begingroup$ Yeah I thought something was fishy lol. Now to the matter at hand, observe the first sentence in the properties tab of the linked page: Suppose $T$ is a bounded operator acting on the normed vector space $X$. So to apply this result, we need to have a normed vector space $D$ acts on, and $D$ needs to be bounded. Without this, the Neumann series $\sum_{k=0}^\infty D^k$ and $(I-D)^{-1}$ need not be equal. Moreover, $(I-D)^{-1}$ may not make sense. $\endgroup$ – Aweygan Jun 19 '17 at 15:05

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