0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $x_1(t) + const$.

EDIT

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

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

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.