General equation for change of variable in a differential equation

silmaril89
Messages
85
Reaction score
0
I had a second order differential equation where \psi is the unknown function and it is a function of x. We then introduced the following change of variable x = \sqrt{\frac{\hbar}{m \omega}} \xi. When all was said and done I found that,

\frac{d^2 \psi}{d \xi^2} = \bigg(\frac{dx}{d \xi}\bigg)^2 \frac{d^2\psi}{dx^2}

My question is, given an arbitrary change of variable for x and given an arbitrary order of the differential equation will the following formula always work?

\frac{d^n \psi}{d \xi^n} = \bigg(\frac{dx}{d\xi}\bigg)^n \frac{d^n \psi}{dx^n}
 
Physics news on Phys.org
That will very not work. What you might have wanted is

\frac{d^n \psi}{d \xi^n} = \bigg(\frac{dx}{d\xi} \frac{d}{dx}\bigg)^n \psi

The trouble is that in general d/dx and dx/dxi do not commute and cannot be reordered. Try some examples like
x=exp(u)
x^2=u
and so on to see this.
 
Thanks for the response, but I don't seem to see the difference.

<br /> \frac{d^n \psi}{d \xi^n} = \bigg(\frac{dx}{d \xi} \frac{d}{dx} \bigg)^n \psi<br /> = \bigg(\bigg(\frac{dx}{d \xi} \bigg)^n \bigg(\frac{d}{dx} \bigg)^n \bigg) \psi<br /> = \bigg(\frac{dx}{d \xi} \bigg)^n \frac{d^n \psi}{dx^n}<br />

Did I make any mistakes in the above? I'm not doing any reordering.
 
You are reordering
when you write

(a b)^3=a^3 b^3
a b a b a b=a a a b b b

you are making an implicit assumption that the order is not important
your change of variable equation holds when one variable is a constant multiple of the other

suppose
d/du=x d/dx
(d/du)^n=(x d/dx)^n
(d/du)^2=(x d/dx)^2=(x d/dx)(x d/dx)=x^2 (d/dx)^2+x d/dx
this is not equal to x^2 (d/dx)^2 because x d/dx is not 0
 
I see now. Thank you for clearing this up for me.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top