2
$\begingroup$

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 $u$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $$ u=c_1e^{t}+c_2e^{-2t}+\frac{1}{3}e^{t}\int e^{-t}f(t)dt-\frac{1}{3}e^{-2t}\int e^tf(t)dt. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.