linear differential operator proof

Question

Suppose that $L$ is a linear differential operator and $L(\cos x)$ doesn't equal to $0$. $L(\sin x)$ doesn't equal to $0$. Prove that the equation $Ly=\cos x$ has a solution of the form $A \cos x + B \sin x$.

Answer 1

I am going to assume that $L$ is a linear differential operator with constant real coefficients; if we allow the coefficients to depend on $x$, then I think all bets are off as far as the truth of this assertion is concerned. So let us posit that $L$ take the form

$L = \displaystyle \sum_0^n a_i D^i, \tag{1}$

where the $a_i \in \Bbb R$. Here

$D \equiv \dfrac{d}{dx}, \tag{2}$

the usual differentiation with respect to $x$. Note that our hypothesis

$L(\cos x) \ne 0 \ne L(\sin x) \tag{3}$

implies two facts of interest: first and most obvious, $L \ne 0$; second, the polynomial

$p_L(\lambda) = \displaystyle \sum_0^n a_i \lambda^i \in R[\lambda] \tag{4}$

cannot be divisible by $\lambda^2 + 1$. For if $\lambda^2 + 1 \mid p_L(\lambda)$, then there would exist $q(\lambda) \in \Bbb C[x]$ such that

$p_L(\lambda) = q(\lambda)(\lambda^2 +1); \tag{5}$

and then we would have

$L(\cos x) = p_L(D) (\cos x) = q(D)(D^2+ 1)(\cos x) = 0, \tag{6}$

since

$(D^2 + 1)(\cos x) = 0, \tag{7}$

as is most easily seen. Likewise we would have

$L(\sin x) = 0, \tag{8}$

again since

$(D^2 + 1)(\sin x) = 0. \tag{9}$

(6) and (8) contradict (3), our given hypothesis on $L$; thus $\lambda^2 + 1$ cannot be a factor of $p_L(\lambda)$. Since $\lambda^2 + 1$ is not a factor of $p_L(\lambda)$, it follows that $\pm i$ cannot be roots of $p_L(\lambda)$; that is

$p_L(i) \ne 0 \ne p_L(-i). \tag{10}$

We make use of (10) to find a solution to

$Ly = \cos x \tag{11}$

by considering the equations

$Lz_\pm = e^{\pm \, ix}. \tag{12}$

We search for solutions to (12) of the form $z_\pm = c_\pm e^{\pm \, ix}$, with $c_\pm \in \Bbb C$. We find

$e^{\pm \, ix} = Lz_\pm = L(c_\pm e^{\pm \, ix}) = c_\pm L(e^{\pm \, ix}) = c_\pm p_L(\pm i)e^{\pm \, ix}, \tag{13}$

using the well-known identity

$L(e^{\mu x}) = p_L(\mu)e^{\mu x}, \tag{14}$

which follows readily from $D(e^{\mu x}) = \mu e^{\mu x}$. Cancelling the factor of $e^{\pm \, ix}$ from both sides of (13) yields

$c_\pm p_L(\pm \, i) = 1, \tag{15}$

so by (10) we have

$c_\pm = (p_L(\pm i))^{-1}; \tag{16}$

thus

$L((p_L(\pm \, i))^{-1}e^{\pm \, ix}) = e^{\pm \, ix}. \tag{17}$

Now using

$\cos x = (e^{ix} + e^{-ix}) / 2 \tag{18}$

and the linearity of $L$ we have

$L(((p_L(i))^{-1} e^{ix} + (p_L(-i))^{-1} e^{-ix}) /2) = \cos x; \tag{19}$

setting

$y(x) = ((p_L(i))^{-1} e^{ix} + (p_L(-i))^{-1} e^{-ix}) /2 \tag{20}$

gives us the desired result once $y$ is re-written using the identity $e^{\pm \, ix} = \cos x \pm i \sin x$; apparently

$y(x) = (((p_L(i))^{-1} + (p_L(-i))^{-1})/2) \cos x + (i((p_L(i))^{-1} - (p_L(-i))^{-1})/2) \sin x. \tag{21}$

It should be observed that in the event that $L$ is real, i.e. the $a_i \in \Bbb R$, then $p_L(-i) = p_L(i)^*$ and the coefficients of $\cos x$ and $\sin x$ reduce to real numbers. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!