Solve a system of nonhomogeneous DEs

[skrat]

Homework Statement

Solve
$$\ddot {\vec a}=A\vec a+B\dot{ \vec a}+\vec F$$

if A and B are known matrices and F is a constant vector.

Homework Equations

The Attempt at a Solution

My plan was to define $\vec b=\dot{\vec a}$ to move from second order to first order system $\dot{\vec b}=A\vec a+B\vec b+\vec F$

But now... well, I have got absolutely no idea how to continue! I wanted to use theorem 23 on page 580 in chapter 11.7 in this work: http://www.math.utah.edu/~gustafso/2250systems-de.pdf BUT I have got a $A\vec a$ term... which... wel... I don't know what to do with it!

 

[Ray Vickson]

Homework Statement

Solve
$$\ddot {\vec a}=A\vec a+B\dot{ \vec a}+\vec F$$

if A and B are known matrices and F is a constant vector.

Homework Equations

The Attempt at a Solution

My plan was to define $\vec b=\dot{\vec a}$ to move from second order to first order system $\dot{\vec b}=A\vec a+B\vec b+\vec F$

But now... well, I have got absolutely no idea how to continue! I wanted to use theorem 23 on page 580 in chapter 11.7 in this work: http://www.math.utah.edu/~gustafso/2250systems-de.pdf BUT I have got a $A\vec a$ term... which... wel... I don't know what to do with it!

If $A$ is invertible there is no reason to appeal to Theorem 23; that is just "overkill". Think first how you would deal with the scalar DE $x'' = A x + B x' + c$ when $A, B, c$ are constants and $A \neq 0$.

 

[skrat]

In case of a scalar DE i would rewrite your equation to $${x}''-B{x}'-Ax=c$$
Than the solution of a complementary part are roots of this characteristic equation $$D^2-BD-A=0$$ meaning $$D_{1,2}=\frac B 2 \pm \frac 1 2 \sqrt{B^2+4A}$$ therefore $$x_{hom}(t)=C_1e^{(\frac B 2 + \frac 1 2 \sqrt{B^2+4A})t}+C_2e^{(\frac B 2 - \frac 1 2 \sqrt{B^2+4A})t}.$$
To get the particular solution we could guess it. Or not? Let's say $$x_{part}=C_0$$
Right?

 

[Ray Vickson]

In case of a scalar DE i would rewrite your equation to $${x}''-B{x}'-Ax=c$$
Than the solution of a complementary part are roots of this characteristic equation $$D^2-BD-A=0$$ meaning $$D_{1,2}=\frac B 2 \pm \frac 1 2 \sqrt{B^2+4A}$$ therefore $$x_{hom}(t)=C_1e^{(\frac B 2 + \frac 1 2 \sqrt{B^2+4A})t}+C_2e^{(\frac B 2 - \frac 1 2 \sqrt{B^2+4A})t}.$$
To get the particular solution we could guess it. Or not? Let's say $$x_{part}=C_0$$
Right?

If $A,B,C$ are constants and $A \neq 0$ the DE $x'' = Ax + Bx' + C$ is the same as the homogeneous DE
$$(x+C/A)'' = A(x+C/A) + B(x+C/A)' .$$

In your case it is not completely clear what to do in the case that $A$ is not invertible and the linear system $Ar = F$ has no solution.

BTW: my answer assumes that you know how to deal with the homogeneous system, and are just confused what to do about the non-homogeneous system (but with a constant "forcing" term).

 

[skrat]

If I understand you correctly, than in my case of system of equations and assuming A is invertible $${(x-A^{-1}C)}''=A(x-A^{-1}C)+B{(x-A^{-1}C)}'$$ with $x$ being a vector $\vec x=(x(t),y(t))$ or did I not understand you correctly?

Your assumption is wrong, to be honest. I am sure I used to know how to do this with matrices, yet I can't seem to remember, nor can I find any good paper with a nice explanation and an example.

However, if I define $\vec y=(\vec x-A^{-1}C)$ than the equation above can be written $${\vec y}''-B{\vec y}'-A\vec y=0.$$

Now all I was able to figure out from several papers is that it makes sense to define $\vec u ={\vec y}'$ to reduce the order of the differential equation. Doing so brings me to $$ { \vec u}'-B\vec u-A\vec y=0.$$

Now you can safely assume that the $-A\vec y$ term is confusing me. Without it I would find the eigenvalues and eigenvectors of a matrix B. With that term I have no idea!

 

[Ray Vickson]

If I understand you correctly, than in my case of system of equations and assuming A is invertible $${(x-A^{-1}C)}''=A(x-A^{-1}C)+B{(x-A^{-1}C)}'$$ with $x$ being a vector $\vec x=(x(t),y(t))$ or did I not understand you correctly?

**********************
You can check this for yourself: do you get back the same DE?

**********************

Your assumption is wrong, to be honest. I am sure I used to know how to do this with matrices, yet I can't seem to remember, nor can I find any good paper with a nice explanation and an example.

***********************
The pdf file you cited has lots of examples in it.

************************However, if I define $\vec y=(\vec x-A^{-1}C)$ than the equation above can be written $${\vec y}''-B{\vec y}'-A\vec y=0.$$

Now all I was able to figure out from several papers is that it makes sense to define $\vec u ={\vec y}'$ to reduce the order of the differential equation. Doing so brings me to $$ { \vec u}'-B\vec u-A\vec y=0.$$

Now you can safely assume that the $-A\vec y$ term is confusing me. Without it I would find the eigenvalues and eigenvectors of a matrix B. With that term I have no idea!

The standard method is to re-write your system as
$$\vec{y}' = \vec{u}, \; \; \vec{u}' = A \vec{y} + B \vec{u}$$
or
$$\frac{d}{dt}\pmatrix{y \\ u} = \pmatrix{0 & I \\ A & B} \pmatrix{y\\u}$$
This is of the form $\vec{z}' = {\mathbf M} \vec{z}$.