Second order differential equation

[sakodo]

Homework Statement

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

L(u)=au''+bu'+cu

Suppose that u_{1},u_{2} 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 L^{2}?

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 u_{1},u_{2}.

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

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

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

So the characteristic equation is:

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

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

\lambda^{2}(a\lambda+b)^{2}+2c\lambda(a\lambda+b)+c^{2}=0

(\lambda(a\lambda+b)+c)^{2}=0

a\lambda^{2}+b\lambda+c=0

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

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

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.

 

[HallsofIvy]

No, the solution space to L(L(u))= 0 is not the same as the solution space to L(u)= 0. If L(u)= 0, then certainly L(L(u))= L(0)= 0 so the solution set for L(u)= 0 is a subspace of the solution set for L(u).

But further, you have already that L(u)= 0 only for vectors of the form au_1+ bu_2 for numbers a and b, so that if L(L(u))= 0 then you must have L(u)= au_1+ bu_2. Think of that as a non-homogeneous second order differential equation. Of course, u_1 and u_2 are already solutions to the homogeneous equation so you should try solutions like u= xu_1(x) and u= xu_2(x).