Help with x=exp(t)

1. Oct 27, 2007

cks

1. The problem statement, all variables and given/known data

$$x=exp(t)$$

2. Relevant equations

Prove that $$x^2\frac{d^2}{dx^2}=(\frac{d}{dt})(\frac{d}{dt}-1)$$

3. The attempt at a solution

Let a tested function y

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2. Oct 27, 2007

Gib Z

$$\frac{d}{dt}$$ makes no sense...

3. Oct 27, 2007

cks

why you say it doesn't make sense. Actually,

it's the method to solve the linear different equation of the form

$$b_0x^nD^n+b_1x^n^-1D^n^-^1+... +b_n)y=R(x)$$

it can be solved by letting x=exp(t)

with $$xD=D_t$$

$$x^2D^2=D_t(D_t-1)$$

but I don't know how to get it.

4. Oct 27, 2007

Gib Z

Derivative of WHAT with respect to t?

5. Oct 27, 2007

cks

It's an operator w.r.t x and we can find another equivalent operator that is w.r.t t

6. Oct 27, 2007

arildno

Well, Let G(x) be a function, F(t) another function, so that G(x=exp(t))=F(t).

Then we have:
$$\frac{dF}{dt}=\frac{dG}{dx}\frac{dx}{dt}=\frac{dG}{dx}x$$
$$\frac{d^{2}F}{dt^{2}}=\frac{d^{2}G}{dx^{2}}(\frac{dx}{dt})^{2}+\frac{dG}{dx}\frac{dx}{dt}=\frac{d^{2}G}{dx^{2}}x^{2}+\frac{dG}{dx}x=x^{2}\frac{d^{2}G}{dx^{2}}+\frac{dF}{dt}$$
Rearranging, we get:
$$x^{2}\frac{d^{2}}{dx^{2}}(G)=\frac{d}{dt}(\frac{dF}{dt}-F)=\frac{d}{dt}(\frac{d}{dt}-1)(F)$$
whereby the operator equality is shown by remembering G(x=exp(t))=F(t).

Last edited: Oct 27, 2007
7. Oct 27, 2007

meopemuk

Let us introduce an arbitrary trial function $f(x)$, which can be also written as $f(e^t)$. Then

$$\frac{dx}{dt} = x$$

$$\frac{d}{dt} f(x) = \frac{df}{dx} \frac{dx}{dt} = x \frac{df}{dx}$$

$$\frac{d^2}{dt^2} f(x) = \frac{d}{dt} (\frac{df}{dx} x) = \frac{d^2f}{dx^2} \frac{dx}{dt} x + \frac{df}{dx} \frac{dx}{dt} = \frac{d^2f}{dx^2} x^2 + \frac{df}{dx} x$$

$$x^2 \frac{d^2f}{dx^2} = \frac{d^2f}{dt^2} - x\frac{df}{dx} = \frac{d^2f}{dt^2} - \frac{df}{dt}$$

$$x^2 \frac{d^2}{dx^2} = \frac{d}{dt} ( \frac{d}{dt} - 1)$$

Eugene.

EDIT: OOps! arildno was 5 min. faster than me.

Last edited: Oct 27, 2007
8. Oct 27, 2007

Gib Z

O damn I suck :( Idiocy conceded =]

9. Oct 27, 2007

cks

10. Oct 27, 2007

Hurkyl

Staff Emeritus
x and t aren't independent. You can't just interchange the order of derivatives!

Try computing
$$\frac{d}{dx} \frac{d}{dt} x$$
and
$$\frac{d}{dt} \frac{d}{dx} x$$
for a concrete example!

11. Oct 27, 2007

arildno

Remember, cks, that x and t are related by an EQUATION; they are not independent variables.

12. Oct 27, 2007

cks

ooo, I see. Thank you very much.

13. Oct 28, 2007

Pellefant

$$x^2 \frac{d^2f}{dx^2} = \frac{d^2f}{dt^2} - x\frac{df}{dx} = \frac{d^2f}{dt^2} - \frac{df}{dt}$$

This has been learning ...

But i don't get this part ...

$$- x\frac{df}{dx} = - \frac{df}{dt}$$

Kindly Pellefant ...

14. Oct 28, 2007

cristo

Staff Emeritus
$$\frac{d}{dt}=\frac{d}{dx}\frac{dx}{dt}=x\frac{d}{dx}$$ from the given relationship between x and t.

15. Oct 28, 2007

Pellefant

Thank you cristo :)

Kindly Pellefant!

16. Oct 28, 2007

cristo

Staff Emeritus
You're welcome!