(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let V denote a vector space of twice differentiable functions on R. Define a linear map L on V by the formula:

[tex]L(u)=au''+bu'+cu[/tex]

Suppose that [tex]u_{1},u_{2}[/tex] is a basis for the solution space of L(u)=0. Find a basis for the solution space of the fourth order equation L(L(u))=0. What can you say about the kernels of L and [tex]L^{2}[/tex]?

2. Relevant equations

3. The attempt at a solution

Since u1,u2 is a basis for the solution space of L(u)=0, then the kernel of L is a plane spanned by [tex]u_{1},u_{2}[/tex].

Now, we want to find the basis for L(L(u))=0. Expanding L(L(u)), we get:

[tex]a(au''+bu'+c)''+b(au''+bu'+c)'+c(au''+bu'+c)=0[/tex], simplifying we get:

[tex]a^{2}u''''+2abu'''+(2ac+b^{2})u''+2bau'+c^{2}u=0[/tex]

So the characteristic equation is:

[tex]a^{2}\lambda^{4}+2ab\lambda^{3}+(2ac+b^{2})\lambda^{2}+2bc\lambda+c^{2}=0[/tex]

[tex]a^{2}\lambda^{4}+2ab\lambda^{3}+b^{2}\lambda^{2}+2ac\lambda^{2}+2bc\lambda+c^{2}=0[/tex]

[tex]\lambda^{2}(a\lambda+b)^{2}+2c\lambda(a\lambda+b)+c^{2}=0[/tex]

[tex](\lambda(a\lambda+b)+c)^{2}=0[/tex]

[tex]a\lambda^{2}+b\lambda+c=0[/tex]

Now, since [tex]u_{1},u_{2}[/tex] is a basis for L(u)=0, then [tex]u_{1},u_{2}[/tex] must satisfies the characteristic equation [tex]a\lambda^{2}+b\lambda+c=0[/tex].

Thus, a basis for L(L(u))=0 is [tex]u_{1},u_{2}[/tex]. The kernel of [tex]L^{2}[/tex] is also a plane spanned by [tex]u_{1},u_{2}[/tex].

If this is true, then the solution space of L(L(u))=0 is exactly the same as L(u)=0. I don't really understand the meaning of this, as in why is it the same? I think something is wrong but I am not sure.

Any help would be appreciated.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Second order differential equation

**Physics Forums | Science Articles, Homework Help, Discussion**