- #1

sakodo

- 21

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## Homework Statement

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]?

## Homework Equations

## 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.