1. ## Linear Map Question

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

$\displaystyle L(u)=a\frac{d^2}{dx^2}u+b\frac{d}{dx}u+cu$

Suppose that $\displaystyle u_{1},u_{2}$ is a basis for the solution space of $\displaystyle L(u)=0$. Find a basis for the solution space of the fourth order differential equation $\displaystyle L(L(u))=0$. What can you say about the kernals of $\displaystyle L$ and $\displaystyle L^2$?

This is my working out so far:

Since $\displaystyle u_{1},u_{2}$ is a basis for the solution space of $\displaystyle L(u)=0$, then $\displaystyle u_{1},u_{2}$ is a basis for the kernal of $\displaystyle L$.

Therefore we can write

$\displaystyle u=\lambda u_{1}+\mu u_{2} ,\ \ \ \ \ \ \ \forall u\ \epsilon \ Ker(L), \ \lambda,\mu \ \epsilon \ R$

So $\displaystyle Nullity(L)=2$ since $\displaystyle Ker(L)$ is spanned by two linearly independent vectors.

Now, consider $\displaystyle L(L(u))=0$. Since every solution of $\displaystyle L(u)=0$ can be represented as $\displaystyle \lambda u_{1}+\mu u_{2}$,
the equation $\displaystyle L(L(u))=0$ becomes

$\displaystyle L(u)=\lambda u_{1}+\mu u_{2}$, which then becomes

$\displaystyle a\frac{d^2}{dx^2}u+b\frac{d}{dx}u+cu=\lambda u_{1}+\mu u_{2}$.

Let $\displaystyle Y_{H }, Y_{P }$ denote the homogenous solution and the particular solution respectively.

Obviously $\displaystyle Y_{H }=\lambda u_{1}+\mu u_{2}$ as $\displaystyle u_{1},u_{2}$ is a basis for the solution space of $\displaystyle L(u)=0$.

Now this is where I got stuck. I did’t know what the particular solution is. I let $\displaystyle Y_{P}=mxu_{1}+nxu_{2},\ \ \ \ \ \ m,n\ \epsilon \ R$

and tried to solve for the constants $\displaystyle m,n$. It was a mess and I don’t think I was on the right track.

I was thinking that if I can find $\displaystyle Y_{P}$, then$\displaystyle Y_{H }, Y_{P }$ would be the basis of the fourth order differential equation $\displaystyle L(L(u))=0$. Am I correct?

Any help will be appreciated.

Thanks.

2. I think you are "overworking" this. For any linear transformation, L, L(0)= 0 so that, if u is in the kernel of L, Lu= 0, the $\displaystyle L^2u= L(Lu)= L(0)= 0$. That is, for any linear transformation, L, the kernel of L is a subspace of the kernel of $\displaystyle L^2$.

3. Thanks. Yep I understand what you mean by $\displaystyle 'Ker(L)$ is a subspace of $\displaystyle Ker(L^2)'$'.

I think $\displaystyle Nullity(L^2)=4$ because I have read somewhere that the solution space of an nth order homogeneous differential equation forms an nth dimensional vector space. Is $\displaystyle L(L(u))=0$ a fourth order homogeneous equation? How would you find the basis?

4. Bump. Anyone know how to find the basis for $\displaystyle L(L(u))$? Many thanks.