What is the dimension of the subspace $W=\{ F(x) \mid (D^2+1)F(x)=0 \wedge F(x)=0\}$

Question

What is the dimension of the subspace $W=\{ F(x) \mid (D^2+1)F(x)=0 \wedge F(x)=0\}$

I solved this and found $F(x)=a \cos(x)+b \sin(x)$, but due to the second condition I think it is the null space. Am I right or wrong?

Answer 1

As the assignment is currently written, my guess would be that the answer is yes. Though you may want to check if the second condition was meant to be $F(x_0)=0$ (as in $W=W_{x_0}$ being a set which depends on the parameter $x_0$).

Answer 2

I think that $W$ is the space of functions $y : \mathbb R \to \mathbb R$, which are solutions of the initial value problem

$y''+y=0$, $y(x_0)=0$,

where $x_0$ is a given point in $ \mathbb R$.

The general solution of the differential equation is given by $y(x)=a \cos(x)+b \sin(x)$.

From $0=y(x_0)$ we get $a \cos(x_0)+b \sin(x_0)=0$.

Case 1: $\cos (x_0) \ne 0$- Then $a= -\frac{b \sin(x_0)}{\cos(x_0)}$, hence

$$y(x)=b[ \sin(x)-\frac{\sin(x_0)}{\cos(x_0)} \cos(x)].$$

This gives $ \dim W=1$.

Case 2: $\sin(x_0) \ne 0$. Similar considerations (which are your turn) give again $ \dim W=1$.

Remark: the case $ \cos(x_0)=0= \sin(x_0)$ is not possible. Why ?