0
$\begingroup$

Every resource I have seen on undetermined coefficients for inhomogeneous systems uses only examples in which the components of the forcing term have a common replication under differentiation. Thus the solution techniques suggest guesses of forms such as $\vec{x_p} = \vec{a}e^{2t}$ or $\vec{x_p} = \vec{a}t^2 + \vec{b}t + \vec{c}$, where each component of one of the undetermined vectors plays the same role (i.e., $b_1$ and $b_2$ are the coefficients of the $t$ terms).

Can undetermined coefficients also be used to solve systems with forcing terms in which each component has a different replication under differentiation, by making the usual guesses component-wise? For example, if the forcing term is $\begin{bmatrix} .5t \\ \sin(3t) \end{bmatrix}$, the guess would be $\vec{x_p} = \begin{bmatrix} at + b \\ c\cos(3t) + d\sin(3t) \end{bmatrix}$. For each component, the guess is made which would be made if that component were the forcing term in a second-or-higher-order single ODE. Is this a legitimate way to generalize undetermined coefficients for systems, or does the technique only work if all components share a common replication under differentiation?

$\endgroup$
1
  • $\begingroup$ @Moo So does what I suggested work as a shortcut, or do you have to solve the two systems separately? $\endgroup$
    – user10478
    Feb 24 '20 at 5:09
1
$\begingroup$

The first step is to come as close as possible to decoupling the variables. For example, consider $$\begin{bmatrix} \dot{x_1}\\ \dot{x_2}\end{bmatrix}=\begin{bmatrix}5&4\\-2&1 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \end{bmatrix} +\begin{bmatrix} f(t)\\g(t)\end{bmatrix} $$, The matrix $\begin{bmatrix} 5&4\\-2&1\end{bmatrix} $ has eigenvalues $3\pm 2i$. There exists a real non-singular matrix $P$ such that $$P^{-1}\begin{bmatrix} 5&4\\-2&1\end{bmatrix}P=\begin{bmatrix}3&-2\\2&3 \end{bmatrix}. $$ Make the change of variable $$\begin{bmatrix} x_1\\x_2\end{bmatrix} =P\begin{bmatrix} u_1\\u_2\end{bmatrix} $$ Then $$\begin{bmatrix}\dot{u_1}\\ \dot{u_2}\end{bmatrix}=\begin{bmatrix}3&-2\\2&3\end{bmatrix} \begin{bmatrix}u_1\\u_2\end{bmatrix} +P^{-1}\begin{bmatrix}f(t)\\g(t)\end{bmatrix} $$.Let us suppose, for the sake of illustration, that $$P^{-1}\begin{bmatrix}f(t)\\g(t)\end{bmatrix}=\begin{bmatrix}2t+1\\5\sin(2t)\end{bmatrix} $$ which gives us $$\begin{bmatrix}\dot{u_1}\\ \dot{u_2}\end{bmatrix}=\begin{bmatrix}3&-2\\2&3\end{bmatrix} \begin{bmatrix}u_1\\u_2\end{bmatrix} +\begin{bmatrix}2t+1\\5\sin(2t)\end{bmatrix} $$. Our second step is to use the method of undetermined coefficients to find particular solutions $u_{1p}$ and $u_{2p}$ of these equations. Your question is what expressions to use. Both for $u_1$ and $u_2$ we take any expression that appears in either one of the equations. So we set $$u_{1p}=at+b+r\cos(2t)+s\sin(2t), $$ $$ u_{2p}=a’t+b’+r’\cos(2t)+s’\sin(2t). $$. Substitute these expressions for $u_{1p}$ and $u_{2p}$ into $\dot{u_1}-(3u_1-2u_2)$and $\dot{u_2}-(2u_1+3u_2)$ respectively and obtain expressions $(…)t+(…)+(…)\cos(2t)+(…)\sin(2t)$. Set each (…) equal to 0, giving 8 linear equations in 8 unknowns. Solve these. The complementary solutions, i.e. the solutions of the homogeneous equations for $u_1$ and $u_2$ are $$u_{1c}=Ce^{3t}\cos(2t)+De^{3t}\sin(2t), u_{2c}=-De^{3t}\cos(2t)+Ce^{3t}\sin(2t) $$ The general solution for $u_1$ and $u_2$ is $$u_1=u_{1c}+u_{1p}, u_2=u_{2c}+u_{2p}$$. From initial conditions on $x_1$ and $x_2$ , we multiply by $P^{-1}$ to obtain initial conditions on $u_1$ and $u_2$, which we use to find $C$ and $D$. Having completely found $u_1$ and $u_2$,we multiply by $P$ to find $x_1$ and $x_2$.

$\endgroup$
2
  • $\begingroup$ What is the point of partially decoupling the ODE? I thought decoupling was only used because systems with diagonal matrices are trivial. Could this step be omitted? $\endgroup$
    – user10478
    Feb 27 '20 at 19:42
  • $\begingroup$ The point of partial decoupling, which is the best we can do if we wish to stay inside the real numbers, is to obtain a block-diaonal form in which the blocks have a very special form, i.e. a rotation matrix, which allows us to write the solution of the corresponding homogeneous equation, the so-called "complementary" solution, in a particularly nice form. $\endgroup$ Feb 27 '20 at 20:48

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.