Second order linear differential operator

Ready2GoXtr
Messages
74
Reaction score
0

Homework Statement


Suppose that L is a second order linear differential operator over the interval J, that f is a function defined on J, and that the function v has the property that

Lv = f on J

(a) Show that if y = u + v and that Lu = 0 on J, then Ly = f on J
(b) Show that if Ly = f on J, then y = u + v for some u such that Lu = 0 on J

Homework Equations


None that apply

The Attempt at a Solution


Well unfortunately this one I was unable to attempt because i am not even sure what it is asking, this professor I have tends to deviate from the book.
 
Physics news on Phys.org
Ready2GoXtr said:

Homework Statement


Suppose that L is a second order linear differential operator over the interval J, that f is a function defined on J, and that the function v has the property that

Lv = f on J

(a) Show that if y = u + v and that Lu = 0 on J, then Ly = f on J
(b) Show that if Ly = f on J, then y = u + v for some u such that Lu = 0 on J

Homework Equations


None that apply

The Attempt at a Solution


Well unfortunately this one I was unable to attempt because i am not even sure what it is asking, this professor I have tends to deviate from the book.

(a) If L is linear, then L(u+v)=Lu+Lv. So if we have Lu=0, then

Ly=L(u+v)=Lu+Lv=0+Lv=Lv=f.

(b) Since L is linear, then there is a zero vector u with Lu=0 and 0+y=y. Choose u=0 and v=y. Then u+v=0+y=y.
 
Thanks man I would never figure that out!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top