We have the differential equation $x'(t)=ax+b(t)$ such that $b(T+t)=b(t)$ $\forall t \in R$ i.e. $b$ is periodic with $t$.

I need to show that if a is non-zero, then there is one and only one periodic solution $x(t)$ with period $T$ i.e $x(t)=x(t+T)$ $\forall t \in R$.

I have partially solved the equation with initial condition $x(0)=c$ first.

  • $\begingroup$ You might want to use Picard, but you can do it without it too, if you use the integrating factor method. $\endgroup$ – Julien Feb 14 '13 at 14:01
  • $\begingroup$ One possible approach: Suppose $x_1(t)$ and $x_2(t)$ are two period solutions of the ODE with period T, verify $\int_0^T |x_1(t)-x_2(t)|^2 dt = 0$. $\endgroup$ – achille hui Feb 14 '13 at 14:03

Suppose $x_1(t)$ and $x_2(t)$ are two periodic solutions for the ODE $$\frac{d}{dt}x(t) = a(t)x(t) + b(t)$$ with period T and $a(t)$ never change sign. Let $y(t) = x_1(t) - x_2(t)$, we have:

$$y'(t) = a(t)y(t) \,\,\text{ and }\,\, y(t) = y(t+T)$$

Multiply both side by $y(t)$ and integrate, we get:

$$\int_{0}^{T} a(t)y(t)^2 dt = \int_{0}^{T} y(t)y'(t) dt = \frac12 \int_{0}^{T} \frac{d}{dt} y(t)^2 dt = \frac12 \left[ y(t)^2 \right]_0^T = 0$$

Since $a(t)$ never change sign, this is only possible when $y(t) \equiv 0$ over $[0,T]$. Since $y(t)$ is periodic, we get $x_1(t) = x_2(t)$ for all $t \ge 0$.

The condition that $a(t)$ is non-zero is important. For example, when $a(t) \equiv 0$ and $x_1(t)$ is a period solution, so does any $x_1(t) + const$.


Since I'm doing this, let us prove this again using the integrating factor.

Let $x(t)$ be any period solution of the ODE with period $T$. Let $$\phi(t) = e^{-\int_0^t a(s)ds} \implies \frac{d}{dt}\phi(t) = -a(t) \phi(t)$$ We have: $$\begin{align} &\frac{d}{dt} [x(t)\phi(t)] = x'(t)\phi(t) - a(t)x(t)\phi(t) = b(t)\phi(t)\\ \implies & x(T)\phi(T) - x(0)\phi(0) = \int_{0}^{T} \frac{d}{dt}[ x(t)\phi(t) ]dt = \int_{0}^{T} b(t)\phi(t)dt\\ \implies & (\phi(T) - 1) x(0) = \int_{0}^{T} b(t)\phi(t)dt\tag{*} \end{align}$$ If $a(t)$ never changes sign, $\int_{0}^T a(t)dt \ne 0$ and hence $\phi(T) \ne 1$, then $(*)$ uniquely fixes the initial value $x(0)$. By the fundamental theorem of ODE, there is only one solution for the ODE with this initial value $x(0)$. So the period solution $x(t)$ is unique.

  • $\begingroup$ Thank you:), but how does y′(t)=a(t)y(t)? $\endgroup$ – Seany Adams Feb 14 '13 at 15:21
  • $\begingroup$ @seanyAdams your ODE is linear in $x$! $$y'(t) = x_1'(t) - x_2'(t) = (a(t) x_1(t) + b(t) ) - (a(t)x_2(t) + b(t)) = a(t) (x_1(t) - x_2(t)) = a(t) y(t)$$ $\endgroup$ – achille hui Feb 14 '13 at 15:28
  • $\begingroup$ Ah yeas of course!:) $\endgroup$ – Seany Adams Feb 14 '13 at 15:50
  • $\begingroup$ You argument shows uniqueness but says nothing about existence. $\endgroup$ – Artem Oct 16 '15 at 18:28
  • $\begingroup$ @Artem, will, start from any $x_0$, $x(t) = \phi(t)^{-1}\left(x_0\phi(0) + \int_0^t b(s)\phi(s)ds\right)$ provides a solution to the ODE with initial condition $x(0) = x_0$. If you choose $x_0$ to be the one that satisfies $(*)$, then you get a periodic solution. $\endgroup$ – achille hui Oct 16 '15 at 18:35

Solve by integrating factor method. There is a unique solution as this is linear regardless of any of constraints mentioned. $a$ being nonzero does not seem relevant.

$x'-ax=b$. multiply with $m=e^{\int -a(t) dt}$. Rewrite as $(mx)'=bm$. Then $mx=\int bm +K$ and $x=(\int bm +K)/m$. If $a$ and $b$ have the same period or $a$ is constant and $b$ periodic then $x$ will be periodic.

  • $\begingroup$ Apparently $a$ is a constant, so it is fine. $\endgroup$ – Julien Feb 14 '13 at 14:47

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