Consider the differential equation:-

$a \phi + (bD^3 - cD)w =0$, where $a, b$ and $c$ are constants, $D$ denotes the differential operator $\dfrac{d}{dx}$, and $w$ is a function of $x$.

I'm defining $w = Lw'$ and $x=Lx'$, where $L$ is a constant.

I'm trying to obtain $\phi$ in terms of $x'$. But I've two questions that pop into my mind immediately:-

$1.$ How do I change the differential operator from $\dfrac{d}{dx}$ to $\dfrac{d}{dx'}$, so that I can obtain $\phi$ correctly in terms of $x'$?

$2.$ Let's say I'm keeping the differential operator as such, and differentiating $w$ with respect to $x$. After the differentiation, if I substitute $x$ with $Lx'$, is $\phi$ the same as the one obtained by changing the differential operator?


Here is a start. First make the change of the dependent variable $w=Lz$ (I used z instead of w' to avoid confusion with derivative)

$$ w=Lz \implies D^n w= L D^n z,\quad D=\frac{d}{dx}, $$

so, the differential equation becomes

$$ a \phi(x) + L(bD^3 - cD)z =0 \longrightarrow (1).$$

Now, we use the other change of variables $x=Lt$ (again I let $t=x'$ avoiding the confusion) in $(1)$ as

$$ \frac{dz}{dx} = \frac{dz}{dt}\frac{dt}{dx} = \frac{1}{L} \frac{dz}{dt} $$

$$ \implies D^2 z = \frac{d^2z}{dx^2} = \frac{d}{dx}\left(\frac{1}{L}\frac{dz}{dt}\right)\frac{dt}{dx} = \frac{d}{dt}\left(\frac{1}{L}\frac{dz}{dt}\right)\frac{dt}{dx}=\frac{1}{L^2}\frac{d^2z}{dt^2}$$

$$ \implies D^3 z = \frac{1}{L^3}\frac{d^3z}{dt^3}. $$

Now, just go back and make substitutions in $(1)$. I think you can do that.

  • 1
    $\begingroup$ Thank you so much! I understand it now. $\endgroup$ – Train Heartnet Jan 27 '14 at 14:23
  • $\begingroup$ @downvoter: Is there a reason for the downvote? $\endgroup$ – Mhenni Benghorbal Feb 12 '14 at 18:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.