Substitution

ubhik · #1 $Old$ April 30th, 2008, 01:26 PM

I have the following equation in my notes and need to follow it to do another question, can someone tell me how the following line:

$(Ly)=(\frac{d^2}{dx^2}-\frac{2}{x}\frac{d}{dx}+\frac{2}{x^2})y(x)=xln(x)=h(x)$

The kernel of $L$ is found by solving $Ly = 0$

This form suggests trying $y = x^m$ . Substituting,

(I don't understand how the following line is obtained???

)

$L(x^m) = [m(m-1)-2m+2]x^{m-2}= 0$

How does m turn into the base?

I know that:

$y=x^m$

$\frac{dy}{dx}=mx^{m-1}$

and

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

but I still don't understand, how the previous bit is obtained!

topsquark · #2 $Old$ April 30th, 2008, 01:36 PM

Quote:

Originally Posted by ubhik $View Post$

I have the following equation in my notes and need to follow it to do another question, can someone tell me how the following line:

$(Ly)=(\frac{d^2}{dx^2}-\frac{2}{x}\frac{d}{dx}+\frac{2}{x^2})y(x)=xln(x)=h(x)$

The kernel of $L$ is found by solving $Ly = 0$

This form suggests trying $y = x^m$ . Substituting,

(I don't understand how the following line is obtained???

)

$L(x^m) = [m(m-1)-2m+2]x^{m-2}= 0$

How does m turn into the base?

I know that:

$y=x^m$

$\frac{dy}{dx}=mx^{m-1}$

and

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

but I still don't understand, how the previous bit is obtained!

$L(x^m) = \frac{d^2}{dx^2} \left ( x^m \right ) - \frac{2}{x} \frac{d}{dx} \left ( x^m \right ) + \frac{2}{x^2} \cdot x^m$

$= m(m - 1)x^{m - 2} - \frac{2}{x} \cdot mx^{m - 1} + \frac{2}{x^2} \cdot x^m$

$= m(m - 1)x^{m - 2} - 2mx^{m - 2} + 2x^{m - 2}$

$= [m(m - 1) - 2m + 2]x^{m - 2}$

I'm not sure what else to give you, unless I'm completely misunderstanding your question.

-Dan

flyingsquirrel · #3 $Old$ April 30th, 2008, 01:36 PM

Hi
$L(x^m)= \frac{\mathrm{d^2}}{\mathrm{d}x^2}(x^m) -2 \frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}x}(x^m) +2\frac{x^m}{x^2}$
Using the derivatives you've evaluated :
$L(x^m)= m(m-1)x^{m-2}-2\frac{mx^{m-1}}{x}+2\frac{x^m}{x^2}=m(m-1)x^{m-2}-2mx^{m-2}+2x^{m-2}$

Factor by $x^{m-2}$ and you get $L(x^m)=(m(m-1)-2m+2)x^{m-2}$

ubhik · #4 $Old$ April 30th, 2008, 02:28 PM

Thanks for that, I understand that bit a lot more now!!!

but I need to apply it the following question, but I can't seem to get it right!!!

$Ay:=\frac{d^2y}{dx^2} - y$

If I try substiting $y = x^m$

then I get

$Ay = m(m-1)x^{m-2}-x^m$

$Ay = (m^2-m)x^{m-2}-x^m$

$Ay = m^2x^{m-2}- mx^{m-2}-x^m$

I am confused because in the previous example, I was able to find the values of m by solving a quadratic equation of m, but this example, I cannot do that... Does that mean I need to substitute something different for $y$ instead of $x^m$ , because I need to get it into such a form that I can find out what the values of m are as this is going to be my kernel for the rest of the problem... I don't really know what to use instead!!! Can you please help!!!

ubhik

flyingsquirrel · #5 $Old$ April 30th, 2008, 02:53 PM

Quote:

Originally Posted by ubhik $View Post$

Thanks for that, I understand that bit a lot more now!!!

but I need to apply it the following question, but I can't seem to get it right!!!

$Ay:=\frac{d^2y}{dx^2} - y$

If I try substiting $y = x^m$

It won't work because the solutions of this equation can't be polynomials.

If you want us to help you, you should tell us what you are asked to do and what is the link between this question and the previous one.

ubhik · #6 $Old$ April 30th, 2008, 03:14 PM

Here is the question that I need to answer:

The function $x = h(x)$ is defined piecewise by

$h(x): = 0,$
$0<= x < 1$

$h(x): = x-1,$
$x >=1$

with $h$ defined above, solve the initial value problem:

$(Ay)(x)= h(x),$

$Ay:=\frac{d^2y}{dx^2} - y$

$y(0)=y'(0)=0$

by finding the Green's function of $A$ , using the variation of parameters method.

This isthe question, and in my notes the first that has been done is that the kernel has been found, but I don't know how to do this for this question, like it was with my other question that has been explained earlier by people on the MHF

flyingsquirrel · #7 $Old$ May 1st, 2008, 01:43 AM

If I've understood what is the "Green function", you have to begin by solving $\left\{\begin{array}{ll}\frac{\mathrm{d}^2y}{\mathrm{d}x^2}-y=0 & (1)\\y(0)=y'(0)=0 & (2)\end{array}\right.$

For this, write the characteristic equation* and find its roots. Let's call them $r_1$ and $r_2$ . In the case of two different real roots, the solutions of (1) are given by $y(x)=A\exp(r_1x)+B\exp(r_2x)$ . Then, use (2) to find the values of $A$ and $B$ . Once you have done this, will go through the variation of parameters method.

*For a differential equation which as the form $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+ a_1\frac{\mathrm{d}y}{\mathrm{d}x}+ a_2y=0$ , the characteristic equation is $r^2+a_1r+a_2=0$ .