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While reading this book I came across a differential equation $$t^5\frac{d^2y}{dt^2}+2t^4\frac{dy}{dt}-y=0$$ that was then rewritten in terms of the Euler operator, $\delta=t\frac{d}{dt}$, with the resulting operator being

$$\delta^2-\delta-\frac{1}{t^3}=0.$$

I'm trying to understand why does it seem as if $$\delta^2 = t^2\frac{d^2}{dt^2} .$$ Shouldn't $$\delta^2=\delta(\delta)=t\frac{d}{dt}\left(t\frac{d}{dt}\right)= t\frac{d}{dt}+t^2\frac{d^2}{dt^2}.$$ Also, I'm looking for a possible generalization of this for $n^{th}$ order polynomials, that is, given a differential operator $$A(x)\frac{d^2}{dt^2}+B(x)\frac{d}{dt}+C(x)=0$$ where $A(x), B(x)$ and $C(x)$ are $n^{th}$ degree polynomials, how can it be expressed in terms of the Euler operator

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3 Answers 3

I think you are right about the definition of $\delta^2$, but the operator should be:

$\delta^2+\delta-1/t^3$, check this with your definition and it all cancels correctly.

Then the generalisation would be:

$\frac{A(t)}{t^2}\delta^2+(B(t)-\frac{A(t)}{t})\delta+C(t)$

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Your operator acts on sufficiently differentiable functions. Look:

Let $f$ be a sufficiently differentiable function. The meaning of $\delta^2f$ is: $$\delta^2f=\delta(\delta f).$$ Then: \begin{align*} \bigl(\delta^2f) &=t\frac{\mathrm{d}}{\mathrm{d}t}\left(t\frac{\mathrm{d}f}{\mathrm{d}t}\right)\\ &=t\left(\frac{\mathrm{d}f}{\mathrm{d}t}+t\frac{\mathrm{d}^2f}{\mathrm{d}t^2}\right)\\ &=t\frac{\mathrm{d}f}{\mathrm{d}t}+t^2\frac{\mathrm{d}^2f}{\mathrm{d}t^2} \end{align*}

Conclusion: $$\delta^2=t\frac{\mathrm{d}}{\mathrm{d}t}+t^2\frac{\mathrm{d}^2}{\mathrm{d}t^2}=\delta+t^2\frac{\mathrm{d}^2}{\mathrm{d}t^2}$$

Regarding the other part of your question: Let $A$, $B$ and $C$ be any function.

Observe that $$\frac{\mathrm{d}^2}{\mathrm{d}t^2}=\frac1{t^2}\bigl(\delta^2-\delta\bigr).$$

Then:

$$A(t)\frac{\mathrm{d}^2}{\mathrm{d}t^2}+B(t)\frac{\mathrm{d}}{\mathrm{d}t}+C(t)=\frac{A(t)}{t^2}\delta^2+\left(\frac{tB(t)-A(t)}{t^2}\right)\delta+C(t).$$

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No, the identity $$\delta^2=\delta(\delta)=t\frac{d}{dt}\left(t\frac{d}{dt}\right)= t\frac{d}{dt}+t^2\frac{d^2}{dt^2}.$$ not correct, it should be as you said first $\delta^2=t^2\frac{d^2}{dt^2}$. Since \begin{align} \delta ^2 = \delta \left( \delta \right) = t\frac{d}{{dt}}\left( {t\frac{d}{{dt}}} \right) = t\left\{ {\frac{d}{{dt}}\left( {t\frac{d}{{dt}}} \right)} \right\} &= t\left\{ {t\frac{{d^2 }}{{dt^2 }} + \frac{d}{{dt}}\underbrace {\left( {\frac{{dt}}{{dt}}} \right)}_{ = 1}} \right\} \\ &= t\left\{ {t\frac{{d^2 }}{{dt^2 }} + 0} \right\} \\ &= t^2 \frac{{d^2 }}{{dt^2 }} \end{align}

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gniourf_gniourf, you made the same mistake, it should be as $ \delta ^2 f = t\frac{{df}}{{dt}}\left( {t\frac{{df}}{{dt}}} \right) $ –  mwomath Jul 3 at 15:31
    
I'm sorry but you made a mistake. Take for example $f(t)=t$. You're claiming that $\delta^2f=0$. Yet, $\delta f=f$ so $\delta^2f=f\neq0$. –  gniourf_gniourf Jul 3 at 15:44
    
for the last comment you are right, sorry! –  mwomath Jul 3 at 15:49
    
Since $\delta$ is an operator, the only sensible definition for $\delta^2$ is $\delta\circ\delta$, hence the one I gave. –  gniourf_gniourf Jul 3 at 15:53
    
opps, i didn't mean what i wrote. not focus –  mwomath Jul 5 at 4:04

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