Let $$y' = a(x) y + b(x)$$ be a linear differential equation with continuous, periodic coefficients $a, b: \mathbb{R} \to \mathbb{R}$ that both have a period of $T > 0$. Also, we assume that $\int_0^T a(t) dt ≠ 0$.

Based on this, I now want to show that the given differential equation has exactly one $T$-periodic solution.

Thanks in advance. My thought so far was that we might somehow transfer the periodicity onto $y$ by maybe considering a function like $z(x) = y(x + T)$, and then find one solution that fits both functions, and show that it's the only one. For the existence of a solution, we probably have to use the given integral, but I'm not entirely sure how.

  • $\begingroup$ Are you sure what you wrote is actually true? Where did you come across to this question? A book, assignment? $\endgroup$ – grdgfgr May 27 '15 at 8:56
  • $\begingroup$ In an assignment indeed, yes. Is there anything wrong with it, is it only true under certain circumstances (or not at all)? I wrote it down as it was given to me, so I naturally assumed that it is true. $\endgroup$ – moran May 27 '15 at 9:00
  • $\begingroup$ It is only true in some circumstances, i'll write it soon $\endgroup$ – grdgfgr May 27 '15 at 9:01
  • $\begingroup$ Let $a(x)\equiv1$, $b(x)\equiv0$. $\endgroup$ – Christian Blatter May 27 '15 at 9:17
  • $\begingroup$ @ChristianBlatter: Constant functions do not have a period $>0$. $\endgroup$ – Alex M. May 27 '15 at 9:27


$y(0) = y_0; \tag{1}$

then the unique solution to the equation

$y' = a(x)y + b(x) \tag{2}$


$y(x) = e^{\int_0^x a(s)ds}(y_0 + \int_0^x e^{-\int_0^s a(u) du} b(s) ds); \tag{3}$

formula (3) is very well known; a derivation may be found here; for $x = T$ we thus have

$y(T) = e^{\int_0^T a(s) ds}(y_0 + \int_0^T e^{-\int_0^s a(u) du} b(s) ds). \tag{4}$

A $T$-periodic solution to (2) satisfies $y(T) = y(0) = y_0$; in this case (4) yields

$y_0 = e^{\int_0^T a(s) ds}(y_0 + \int_0^T e^{-\int_0^s a(u) du} b(s) ds), \tag{5}$

which we re-write as

$(1 - e^{\int_0^T a(s) ds})y_0 = e^{\int_0^T a(s) ds}\int_0^T e^{-\int_0^s a(u) du} b(s) ds; \tag{6}$

the hypothesis

$\int_0^T a(s) ds \ne 0 \tag{7}$

guarantees that

$e^{\int_0^T a(s) ds} \ne 1; \tag{8}$

in this case we may solve (6) for $y_0$:

$y_0 = \dfrac{\int_0^T a(s) ds}{1 - e^{\int_0^T a(s) ds}}\int_0^T e^{-\int_0^s a(u) du} b(s) ds. \tag{9}$

(9) indicates that there is at most one initial condition $y_0 = y(0)$ for which the solution $y(t)$ of (2) is periodic. Thus any periodic $y(t)$ satisfying (2) under the condition (7) must be unique; it remains to establish the existence of such a $y(t)$.

To establish the existence of a periodic solution, note that we may translate any solution to (2) forward in $x$ by $T$, obtaining a function $y(x + T)$; we have

$y'(x +T) = a(x + T)y(x + T) + b(x + T) = a(x)y(x + T) + b(x), \tag{10}$

by the $T$-periodicity of $a(x)$ and $b(x)$. Furthermore, at $x = 0$ the function $y(x + T)$ takes the value

$y(0 + T) = y(T). \tag{11}$

If follows from (11) that if we can find a solution $y(x)$ such that

$y(T) = y(0) = y_0, \tag{12}$

then both $y(x)$ and $y(x + T)$ will satisfy (2) with the same initial condition $y_0$; uniqueness of solutions then allows us to conclude that

$y(x + T) = y(x) \tag{13}$

for all $x$; i.e., the solution $y(x)$ is $T$-periodic. But if we choose $y_0$ as in (9), then (6) and hence (5) evidently bind, so that (4) yields (12), and hence (13); we have thus demonstrated the existence of a periodic solution to (2), which must then be unique by the arguments given above. QED.



I am suppressing $(x)$ in $a(x)$ etc. $$y=e^{\int a\, dx}\big(\int b\,e^{-\int a\, dx}dx +C \big)$$

Let $\int_0^T a\, dx = M$ and $\int_0^T b\, dx = N$. Then we can write $\int a\, dx=A(x)+\frac MTx$ and $\int b\, dx=B(x)+\frac NTx$ where $A$ and $B$ are zero mean AND periodic.

FACT: Integral of a zero mean periodic function is periodic. Integral of a non-zero mean periodic function is NOT periodic. But we can subtract $\text{Constant}\times x$ to make it periodic.

FACT: Exponential of a periodic function is periodic.

FACT: Product of periodic functions are also periodic.

Assuming for zero mean $a$ and $b=0$, it is very easy to see that the solution above is periodic for any C value.

However, if $\int_0^T a(t) dt ≠ 0$, lets insert what I wrote above:

$$y=e^{A(x)+\frac MTx}\big(\int b\,e^{-A(x)-\frac MTx}dx +C \big)$$

Since $e^{A(x)+\frac MTx}$ is not periodic, $C$ must be $0$

Moreover, $e^{A(x)+\frac MTx} \int b\, e ^{-A(x)-\frac MTx}dx$ is not periodic at first sight. However, it will become periodic in steady state, if $M<0$, which I do not really know how I would show.

Conclusion: If $a$ is negative mean, the solution will be periodic for $0$ initial condition in the steady state for any $b$. If $a$ is positive mean, the solution will diverge in all cases. If $a$ is 0 mean, then, well, I don't know.


For $a(x)=1+\cos x$ and $b(x)=\sin x$ you get $y(x)= \mathbb e ^{x + \sin x} C + \mathbb e ^{x + \sin x} \int \limits _1 ^x \mathbb e ^{- t - \sin t} \sin t \space \mathbb d t$, with $C$ an integration constant. Then $y(2 \pi) \neq y(0)$, so the statement is not true.

(Being lazy, all the above computations have been done with Mathematica.)

  • 2
    $\begingroup$ But this isn't actually an example, because we're supposed to assume that the average of $a(t)$ is not zero. $\endgroup$ – Chappers May 27 '15 at 9:11
  • $\begingroup$ @Chappers: Thank you for signalling the mistake. Corrected. $\endgroup$ – Alex M. May 27 '15 at 9:38

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