MHB Derivative of function containing absolute value

[find_the_fun]

I'm working on a ODE with initial conditions y(2)=4 and y'(2)=1/3. I solved it to be [math]y=\frac{c_1}{|x-6|^8} + c_2|x-6|^{\frac{2}{3}}[/math]. How do I apply the second initial condition? I'm stuck at taking the derivative.

 

[Ackbach]

$$\frac{d}{dx}|x|=\text{sgn}(x),$$
where sgn is the signum (or sign) function. Out of curiosity, what is your original DE?

 

[find_the_fun]

$$\frac{d}{dx}|x|=\text{sgn}(x),$$
where sgn is the signum (or sign) function. Out of curiosity, what is your original DE?

I forget but I think I had fouled up at an earlier step than shown.

 

[Ackbach]

I forget but I think I had fouled up at an earlier step than shown.

Ok...

 

[chisigma]

$$\frac{d}{dx}|x|=\text{sgn}(x),$$
where sgn is the signum (or sign) function...

This is true only if we accept the definition of derivative...

$\displaystyle f^{\ '} (x) = \lim_{\delta \rightarrow 0} \frac{f(x + \delta) - f(x - \delta)}{2\ \delta}\ (1)$

With the more 'conventional' definition...

$\displaystyle f^{\ '} (x) = \lim_{\delta \rightarrow 0} \frac{f(x + \delta) - f(x)}{\delta}\ (2)$

... the derivative of |x| in x=0 doesn't exist...

Kind regards

$\chi$ $\sigma$

 

[Ackbach]

This is true only if we accept the definition of derivative...

$\displaystyle f^{\ '} (x) = \lim_{\delta \rightarrow 0} \frac{f(x + \delta) - f(x - \delta)}{2\ \delta}\ (1)$

Not sure I agree. It is true that using the symmetric definition yields $f'(0)=0=\text{sgn}(0)$. However, to my mind, this is a meaningless statement. $|x|$ has a corner at the origin, and should not be viewed as differentiable there. I was kind of taking it as understood that $|x|$ is not differentiable at the origin. You could think of this as an "almost everywhere" equality.

 

[chisigma]

Not sure I agree. It is true that using the symmetric definition yields $f'(0)=0=\text{sgn}(0)$. However, to my mind, this is a meaningless statement. $|x|$ has a corner at the origin, and should not be viewed as differentiable there. I was kind of taking it as understood that $|x|$ is not differentiable at the origin. You could think of this as an "almost everywhere" equality.

Ok! ... some problems, however could be born if the initial conditions have not been given at x = 2 but at x = 6 ... what to do in this case? ...

Kind regards

$\chi$ $\sigma$

 

[Ackbach]

Ok! ... some problems, however could be born if the initial conditions have not been given at x = 2 but at x = 6 ... what to do in this case? ...

Kind regards

$\chi$ $\sigma$

Well, you'd certainly need $c_1=0$ if the initial conditions are finite. As for the derivative, I really don't think it's possible even to specify two-sided derivative conditions at $x=6$. You could specify one-sided derivative conditions, though. That is, you could specify what
$$\lim_{h\to 0^{+}}\frac{y(6+h)-y(6)}{h}\quad\text{and}\quad
\lim_{h\to 0^{-}}\frac{y(6+h)-y(6)}{h}$$
are. I imagine if you specify one of them, the other might be determined.