How to Prove the Equation for x=exp(t)?

[cks]

Homework Statement

$$x=exp(t)$$

Homework Equations

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

The Attempt at a Solution

Let a tested function y

 

[Gib Z]

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

 

[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.

 

[Gib Z]

Derivative of WHAT with respect to t?

 

[cks]

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

 

[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).

 

[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.

 

[Gib Z]

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

O damn I suck :( Idiocy conceded =]

 

[cks]

I'll read your all answers later, but somehow I still couldn't detect my mistakes.

 

[Hurkyl]

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!

 

[arildno]

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

 

[cks]

ooo, I see. Thank you very much.

 

[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 ...

 

[cristo]

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

 

[Pellefant]

Thank you cristo :)

Kindly Pellefant!

 

[cristo]

You're welcome!