Register to reply

Doubt about the dimension of a 2nd order homogeneous equation

by ashok vardhan
Tags: dimension, doubt, equation, homogeneous, order
Share this thread:
ashok vardhan
#1
Mar28-12, 10:31 AM
P: 19
My doubt is that is dimension of a 2nd order homogeneous equation of form y''+p(x)y'+q(x)=0 always 2 ??? or dimension is 2 only when p(x),q(x) are contionuos on a given interval I..??
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
ashok vardhan
#2
Mar28-12, 10:32 AM
P: 19
Can any one help me to clarify my doubt..
HallsofIvy
#3
Mar28-12, 01:08 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,323
A differential equation does NOT have a "dimension". What you are asking about is the dimension of the solution set. And you will need p and q to be continuous in order to use the "existance and uniqueness theorem".

I assume you have seen the proof that the solution set does, in fact, form a vector space. You just need to observe if f and g are solutions, then, for any a and b,
(af+ bg)''+ p(x)(af+ bg)'+ q(x)(af+ bg)= af''+ bg''+ap(x)f'+ bp(x)g'+ aq(x)f+ bq(x)g= a(f''+ pf'+ qf)+ b(g''+ pg'+ qg)= a(0)+ b(0)= 0 so af+ bg is also a in that set.0

To see that "the set of all solutions to a second order, homogeneous, linear differential equation form a vector space of dimension 2", look at the inital value problems y''+ p(x)y'+ q(x)= 0, with y(a)= 1, y'(a)= 0 and with y(a)= 0, y'(a)= 1. Since y''= -p(x)y'- q(x)y, if p and q are continuous on some interval around x= a, then there exist a unique solution to each of those problems on that interval(You also need that f(x, y)= -p(x)y'+ q(x)y be "Lischitz" in y. Since it is differentiable with respect to y, that is clear.). I will call those solutions Y1(x) and Y2(x). If AY1(x)+ BY2(x)= 0 for all x, then, in particular, AY1(a)+ BY2(a)= A= 0. We then have BY2(x)= 0 for all x, and since Y2'(a) is not 0, Y2 is not a constant, so, in particular not 0 for all x, so B= 0. That proves that the two functions, Y1 and Y2, are independent.

Now let y(x) be any solution to the differential equation. Let A= y(a), B= y'(a). Then AY1+ BY2 also satisfies the differential equation and (AY1+ BY2)(a)= AY1(a)+ BY2(a)= A(1)+ B(0)= A and (AY1+ BY2)'(a)= AY1'(a)+ BY2'(a)= A(0)+ B(1)= B. Since AY1+ BY2 satisfies the same differential equation and the same initial conditions, it follows that y(x)= AY1(x)+ BY2(x). That shows that Y1 and Y2 span the space of all solutions so since they are also indepenendent, they form a basis for that space and so it is two dimensional.

ashok vardhan
#4
Apr15-12, 11:13 AM
P: 19
Doubt about the dimension of a 2nd order homogeneous equation

sir, i dont know whether i can ask this question in this forum..if not please excuse me..my doubt is i want to learn Complex analysis in details from Basics..which is the best book??


Register to reply

Related Discussions
2nd order non homogeneous equation Calculus & Beyond Homework 2
Non-homogeneous 1st order diff equation Calculus & Beyond Homework 2
Homogeneous equation (third order) Calculus & Beyond Homework 3
First Order Homogeneous Equation Calculus & Beyond Homework 2
Second-Order Homogeneous Linear Equation Differential Equations 1