Doubt about the dimension of a 2nd order homogeneous equation

[ashok vardhan]

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..??

 

[ashok vardhan]

Can anyone help me to clarify my doubt..

 

[HallsofIvy]

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]

sir, i don't 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??

 

[blue_raver22]

The dimension of a 2nd order homogeneous equation is not always 2. It depends on the specific conditions and constraints of the equation. In general, the dimension can be determined by the number of independent variables and the order of the equation. In the case of a 2nd order homogeneous equation of the form y''+p(x)y'+q(x)=0, the dimension is 2 if p(x) and q(x) are continuous on a given interval I. However, if there are other constraints or discontinuities in the equation, the dimension may be different. It is important to carefully consider all factors when determining the dimension of an equation.