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Find all the possible values of $a$ and $b$, so that the equation: $$ \ddot{x} + a\dot{x} + bx = \sin t $$

  1. Has only bounded solutions on $\mathbb{R}$
  2. Has only periodical solutions

In general, we know that every solution to this ODE is of the form: $x_h + x_p$, where $x_h$ is the solution to homogeneous equation: $$ \ddot{x} + a\dot{x} + bx = 0 $$ and $x_p$ is a particular solution to the original equation.

We can also (quite easily, I think?) find the particular solution as a function of $a$ and $b$, for instance by assuming that $x_p$ is of the form: $$ A\cos t + B\sin t $$ Differentiating it, plugging into the original equation, and then solving as a system of equations for $A$ and $B$ (The result is undefined for some values of $a$ and $b$, though).

Now here is the thing - for the homogeneous equation, we know that to satisfy condition (1) we have to choose $a$ and $b$ so that all the roots of $x^2 + ax + b$ will have negative or zero real part, and to satisfy (2), we need need the real part to be zero.

For the particular solution, assuming that it is indeed of the form $A\cos t + B\sin t$ (when $A$ and $B$ as functions of $a$ and $b$ are defined), the choice of $a$ and $b$ doesn't really matter, because it is bounded and periodical on all $\mathbb{R}$ anyway.

But I have a strong feeling that I am missing something here. Any ideas what that might be?

Thanks!

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I think problems might arise when $\sin t$ is a solution to the homogeneous equation. In this case, the particular solution may have terms of the form $At\sin t$ and $Bt\cos t$. –  David Mitra Feb 26 '12 at 18:19
    
@DavidMitra Is there a more intelligent way than mine to approach this question? –  Hila Feb 26 '12 at 18:25
    
I think it's ok (I haven't really thought through it carefully, though); but I'd treat the special case when $\sin t$ is a solution of the homogeneous equation separately. This occurs only when the c.p. is $x^2+1=0$ (i.e, $a=0, b= 1$). –  David Mitra Feb 26 '12 at 18:32
    
@DavidMitra It is one of cases where $A$ and $B$ are undefined in my method. The other one is $a = 1$ and $b = 2$. –  Hila Feb 26 '12 at 18:40
    
You may also consider, for example, when the function $\cos( ax)+\cos( x)$ is periodic. –  David Mitra Feb 26 '12 at 18:46
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