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8. Variation of parameters

In the forced problem \(\mathbf{x}'=\mathbf{A}\mathbf{x}+\bff(t)\), the first step is to find a homogeneous solution \(\mathbf{x}_h\) in the form \(\mathbf{X}(t)\mathbf{c}\) for a fundamental matrix \(\mathbf{X}(t)\) and constant vector \(\mathbf{c}\). To finish the problem, we have to find a particular solution of the forced equation. We can use the same technique of variation of parameters as in ../first_linear/variation_parameters.

We look for a solution in the form

\[ \mathbf{x}_p(t) = \mathbf{X}(t)\mathbf{k}(t) \]

and substitute into the ODE to get

\[ \mathbf{X}'\mathbf{k} + \mathbf{X}\mathbf{k}' = \mathbf{A} (\mathbf{X}\mathbf{k}) + \mathbf{f}. \]

Since \(\mathbf{X}\) is a fundamental matrix, it satisfies \(\mathbf{X}'=\mathbf{A}\mathbf{X}\). So the equation above becomes simply \(\mathbf{X}\mathbf{k}' = \mathbf{f}\), or

\[ \mathbf{k}(t) = \int [\mathbf{X}(t)]^{-1} \mathbf{f}(t)\, dt. \]

That is,

\[ \mathbf{x}_p(t) = \mathbf{X}(t) \int [\mathbf{X}(t)]^{-1} \mathbf{f}(t)\, dt. \]

Example

The matrix

\[\mathbf{A} = \twomat{1}{-1}{5}{-3}\]

has eigenvalues \(-1 \pm i\) and eigenvectors \(\twovec{1}{2\mp i}\). Find a particular solution of \(\mathbf{x}'=\mathbf{A}\mathbf{x}+\twovec{0}{1}\).

Solution

In the preceding section we computed

\[ e^{t \mathbf{A}} = e^{-t} \,\twomat{\cos(t)+2\sin(t)}{-\sin(t)}{5\sin(t)}{\cos(t)-2\sin(t)}. \]

This is a fundamental matrix whose inverse is easy to get:

\[ e^{-t \mathbf{A}} = e^{t}\, \twomat{\cos(t)-2\sin(t)}{\sin(t)}{-5\sin(t)}{\cos(t)+2\sin(t)}. \]

For the VoP formula, we need to integrate \(\mathbf{X}(t)^{-1} \bfx_0\), or

\[ e^{-t \mathbf{A}} \twovec{0}{1} = \twovec{e^{t}\sin(t)}{e^{t}\cos(t)+2e^{t}\sin(t)}. \]

Thus,

\[\begin{align*} \bfx_p(t) &= e^{-t} \,\twomat{\cos(t)+2\sin(t)}{-\sin(t)}{5\sin(t)}{\cos(t)-2\sin(t)} \twovec{\int e^{t}\sin(t)\, dt}{\int [e^{t}\cos(t)+2e^{t}\sin(t) ]\, dt}. \end{align*}\]

Let’s not punish ourselves with the rest of the algebra.

8.1. IVP formula

It is interesting to update the IVP formulas we derived in the scalar linear case. For the general case of

\[ \bfx' = \bfA(t)\bfx + \bff(t), \quad \bfx(0)=\bfx_0, \]

we need any fundamental matrix \(\bfX(t)\). A particular solution that is zero at \(t=0\) is given by the definite integral

\[ \bfx_p(t) = \bfX(t) \int_0^t \bfX(s)^{-1} \bff(s)\, d s. \]

Then we add homogeneous and particular solutions to get

\[ \bfx(t) = \bfX(t) \bfX(0)^{-1} \mathbf{x}_0 + \bfX(t) \int_0^t \bfX(s)^{-1} \bff(s)\, d s. \]

For a constant coefficient matrix, the matrix exponential is the most convenient fundamental matrix. In that case,

\[ \bfx(t) = e^{t\bfA} \mathbf{x}_0 + \int_0^t e^{(t-s)\bfA} \bff(s)\, d s. \]

For all its complications, this yields to the same interpretation as the scalar case: a combination of the homogeneous (i.e. free) response to the initial value, plus all the contributions obtained by propagation of the forcing function at each earlier moment.

7. The matrix exponential 1. Nonlinear ODEs