Showing the uniqueness of a solution to an ordinary differential equation

Question

$$ \dot{y}(x)+y(x)=u(x) $$ $$ y_c(x) = ce^{-x} + \int_{0}^{x} u(t) e^{(t-x)} dt $$

Show that if u is periodic with a period of T, then there exists exactly one solution to the differential equation with the period T. Where here $y_c(0) = c$ holds. The use of the second equation is hinted.

I am stuck trying to demonstrate the proposed property. Could someone provide a hint on how to proceed? Your help will be greatly appreciated.

Answer 1

1. If $u$ is sufficiently well-behaved, the solution to the initial value problem $\dot{y}+y=u(x)$ and $y(0)=c$ is unique$^{(*)}$ and given by $$ y_c(x) = ce^{-x} + \int_{0}^{x} u(t) e^{t-x}\, dt. \tag{1} $$

2. If there exists $c$ such that $y_c$ is $T$-periodic, then $y_c(T)=y_c(0)$. According to $(1)$, this implies $$ ce^{-T}+\int_0^Tu(t)e^{t-T}\,dt=c \implies c=\frac{1}{1-e^{-T}}\int_0^Tu(t)e^{t-T}\,dt, \tag{2} $$ i.e., there can be at most one value of $c$ such that $y_c$ is $T$-periodic.

3. Eq. $(2)$ is a necessary condition for $y_c$ to be $T$-periodic. I will now show that, if $u$ is $T$-periodic, then it also is a sufficient condition. Indeed, according to $(1)$, \begin{align} y_c(x+T)-y_c(x)&=ce^{-x-T} + \int_{0}^{x+T} u(t) e^{t-(x+T)}\, dt -ce^{-x} - \int_{0}^{x} u(t) e^{t-x}\, dt \\ &=e^{-x}\left(ce^{-T}-c+\int_{0}^{x+T} u(t) e^{t-T}\, dt - \int_{0}^{x} u(t) e^{t}\, dt\right). \tag{3} \end{align} Since $u(t)=u(t+T)$, the last integral in $(3)$ can be rewritten as $$ \int_{0}^{x} u(t) e^{t}\, dt = \int_{0}^{x} u(\underbrace{t+T}_{=\,\tau}) e^{t}\, dt =\int_T^{x+T}u(\tau)e^{\tau-T}\,d\tau, \tag{4} $$ hence $$ \int_{0}^{x+T} u(t) e^{t-T}\, dt - \int_{0}^{x} u(t) e^{t}\, dt =\int_0^T u(t)e^{t-T}\,dt. \tag{5} $$ It follows from $(2)$, $(3)$ and $(5)$ that $y_c(x+T)=y_c(x)$.

4. In summary, if $u$ is $T$-periodic (and sufficiently well-behaved), then there exists only one solution to the differential equation $\dot{y}+y=u(x)$ that is also $T$-periodic. It is the function $y_c$ defined by Eq. $(1)$, with $c$ given by Eq. $(2)$.

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$^{(*)}$ See, for instance, https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem.