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I apologize for the title being slightly unclear (at least to me it seems so), so if anyone has a better suggestion feel free to change it.

Anyways, for example, when dealing with a second order differential equation of the form: $$ ay'' + by' + cy = f(x) $$ with solutions $y_1(x)$ and $y_2(x)$ we can say that there is resonance in the system provided that $f(x)$ is linearly dependant on $y_1(x)$ and/or $y_2(x)$.

Now consider the linear system $$\textbf{x}' (t) + \textbf{P}\textbf{x}(t) = \textbf{z}(t)$$ where $\textbf{x}(t)$, $\textbf{z}(t)$ are n-vectors, $\textbf{P}$ is a n-by-n matrix and $\textbf{z}(t)$ is a forcing term of the system. As the solutions are then of the form $$\textbf{x}_i(t)=a_i \textbf{v}_i e^{\lambda_i t} $$ where the $\lambda_i$ and $\textbf{v}_i$ are eigenvalues and eigenvectors of $\textbf{P}$ respectively (assuming the eigenvectors are orthogonal), can we say that resonance is occuring when the $\textbf{z}(t)$ is linearly dependant of the solutions $\textbf{x}_i(t)$?

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Suppose $\mathbf{z}(t)=\sum_i a_i\mathbf{v}_ie^{\lambda_i t}$ (for fixed $a_i$, not all zero), and based on that, we look for a particular solution to the nonhomogeneous problem $$\textbf{x}' (t) + \textbf{P}\,\textbf{x}(t) = \textbf{z}(t),\tag{N}$$ of the form $$\mathbf{x}_p(t)=\sum_{i} b_i\mathbf{v}_ie^{\lambda_i t},$$ (where the $b_i$ are to be determined). Then we are destined to fail: upon substituting $\mathbf{x}_p$ back into (N), the left-hand side becomes zero (due to the fact that (N) is linear and $\mathbf{x}_p$ is a linear combination of solutions to the homogeneous problem), whereas the right-hand side is not.

Thus, the method of undetermined coefficients---whether vector or scalar form---requires multiplying the original "guess" for $\mathbf{x}_p$ by the smallest power of $t$ for which the resulting function is not in the span of the fundamental set of solutions (to the homogeneous problem), $\{\mathbf{v}_ie^{\lambda_i t}\}$.

(I presume this bold part above is what you mean by resonance.)

Hope that helps.

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The bold part is indeed what I mean by resonance. This is what I thought should happen - what happens in the case when we have $\textbf{u}_i \neq \textbf{v}_i$ in $\textbf{z}(t)$ instead? –  Andrew D Jan 8 '13 at 23:34
    
It depends on what the $\mathbf{u}_i$ are. It could be that the method of undetermined coefficients (MUC) simply doesn't produce a particular solution to (N), but this happened for scalar problems too. However, for the class of forcing functions for which MUC does work, either $\mathbf{z}$ can be written as $\mathbf{z}(t)=\sum_i a_i\mathbf{v}_ie^{\lambda_i t}$ or it can't. If it can, proceed as above. If it can't, go with another technique, such as variation of parameters, Laplace transforms, etc. –  JohnD Jan 8 '13 at 23:48
    
Ah, okay, I see. I take it that it's probably better to use something like variation of parameters to figure out the particular integral when formulating a system using vectors/matrices, rather than applying MUC, which while works fine for ODE's, seems to be quite awkward for problems such as these. Anyways, thanks! –  Andrew D Jan 8 '13 at 23:53
    
Well, be careful though, Andrew. Whether we are solving a scalar or vector version of a nonhomogeneous problem, there is no "one size fits all". Variation of parameters always "works" in theory, but in practice you can be left with intractable integrals. Undetermined coefficients "works" and is pretty straightforward modulo the algebraic tedium, but it doesn't apply to all possible forcing functions. Similarly, transform methods only help when the forcing function is transformable and results in a "doable" problem in the transform domain. This is why we learn multiple techniques! –  JohnD Jan 9 '13 at 0:05
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