# On the existence of a particular solution for an ODE

The problem asks to find a bounded $u(\cdot) \in \mathcal{C}^2(\mathbb{R})$ such that $$u''+u'-2u=f$$ where $f$ is a bounded continuous function on the real line.

[Observations, Editted] We can reduce the 2-order ODE to first-order one simply by letting $v=(u\ u')^t\in\mathbb{R}^2$, then we find $${dv \over dt}=Av+\bar{f}$$ where $A=\begin{pmatrix} 0 & 1\\2 & -1 \end{pmatrix}$, and $\bar{f}=\begin{pmatrix} 0 \\ f \end{pmatrix}$. Since the eigenvalue of $A$ is $1,-2$, we can associate some invertible $P$ such that $A=P^{-1}\begin{pmatrix}1 & 0\\0 & -2\end{pmatrix}P$. It's an easy calculation that $$P=\begin{pmatrix} 1 & 1\\1 & -2\end{pmatrix},\quad P^{-1}={1 \over 3}\begin{pmatrix} 2 & 1\\1 & -1\end{pmatrix}$$ then we find \begin{equation*} \begin{split} v(t)&=P^{-1}\begin{pmatrix}e^t & 0\\0& e^{-2t}\end{pmatrix}Pv(0)+\int_0^t P^{-1}\begin{pmatrix}e^{t-s} & 0\\0 &e^{-2(t-s)}\end{pmatrix}P\bar{f}(s)\ ds\\ &= P^{-1}\begin{pmatrix} e^t v_1+\int_0^t e^{t-s}f(s)\ ds\\e^{-2t} v_2-2\int_0^t e^{-2(t-s)}f(s)\ ds\end{pmatrix} \end{split} \end{equation*} where $Pv(0)=(v_1\ v_2)^t$. Hence $$3u(t)=2\underline{\left(e^t v_1+\int_0^t e^{t-s}f(s)\ ds\right)}_{(1)}+\underline{\left(e^{-2t} v_2-2\int_0^t e^{-2(t-s)}f(s)\ ds\right)}_{(2)}$$ Now I want to choose some $v_1,v_2$ making $(1)$ stays bounded as $t \to +\infty$ and $(2)$ stays bounded as $t \to -\infty$. Since we are considering linear ODEs, we may assume $f\leq 0$. If $f$ is integrable on $\mathbb{R}$, then simply by letting $$v_1=\int_0^\infty e^{-s} f(s)\ ds,v_2=2\int_0^{-\infty}e^{-2s}f(s)\ ds$$ we will find the conclusion. Replacing $u$ in the ODE system by $-u$ we find the conlusion holds for non-negative integrable $f$ hence for all integrable $f$. Now we consider the truncated interval $I_n=[-n,n]$, and let $f_n=f\cdot \chi_{I_n}$, then $f_n$ is integrable on the real line, hence there must exists some global $u_n$ which satisfies the ODE system with $f$ replaced by $f_n$ and stays bounded on $\mathbb{R}$. Since the solution is determined locally, we find $u_k=u_l$ on $I_{\min{k,l}}$. This seems to complete the proof. But I am not satisfied with the proof since the assumption $f$ seems to be wield, is there any other more simpler approach that can be used to show the existence of some bounded $u$?

I don't know why you need $f\le 0$, or why $f$ itself should be integrable. For any bounded continuous $f$, the product of $f$ with a decaying exponential function is integrable, by comparison principle.
Also, I think it's simpler to use Green's function $G(t,a)$, which by definition is a solution of $$u''+u'-2u=\delta_a \tag1$$ vanishing at infinity. The delta-function singularity will come from $u''$, provided that $$u'(a+)=u'(a-)+1\tag2$$ Since the homogeneous equation has solutions $u=e^{t}$, $u=e^{-2t}$, we should have $u(t)=c_1e^{t}$ for $t<a$ and $u(t)=c_2e^{-2t}$ for $t>a$, so that $u$ vanishes at infinity. Since we don't want $u$ itself to have a discontinuity (that would create a terrible singularity of $u''$), the formula for $u$ can be simplified to $$u(t)=\begin{cases}ce^{t-a}, \quad &t\ge a \\ ce^{2a-2t}, \quad & t\le a \end{cases}\tag3$$ The value of $c$ is determined by plugging (3) into (2): $$c=-2c+1,\quad \text{hence } \ c=\frac13$$ Thus, Green's function is $$G(t,a)=\begin{cases}\frac13 e^{t-a}, \quad &t\ge a \\ \frac13e^{2a-2t}, \quad & t\le a \end{cases}\tag3$$ For any bounded continuous $f$, the desired solution is $u(t)=\int_{-\infty}^\infty f(s)G(t,s)\,ds$. Indeed, this is the convolution of $f$ with $G(t,0)$; applying the differential operator yields the convolution of $f$ with $\delta_0$, which is $f$ itself. A rigorous proof is best given for Green functions in general, rather than in every special case.
I do not know why you used matrices to solve the problem. For the second order DEs, you just simply use the formula to the solution. Suppose that $u_1,u_2$ are two linearly independet solutions of $u''+bu'+cu=0$. Then the solution of $u''+bu'+cu=f$ is $$u=c_1u_1(t)+c_2u_2(t)-u_1(t)\int\frac{u_2(t)f(t)}{W(u_1,u_2)}dt+u_2(t)\int\frac{u_1(t)f(t)}{W(u_1,u_2)}dt.$$ For your equation, $$u_1=e^t,u_2=e^{-2t}$$ and $$W(u_1,u_2)=-3e^{-t}$$ and hence the solution of your eq. is $$u=c_1e^{t}+c_2e^{-2t}+\frac{1}{3}e^{t}\int e^{-t}f(t)dt-\frac{1}{3}e^{-2t}\int e^tf(t)dt.$$