Help me solve a system of three linear ODEs. I seem to have grasped the $2\times 2$ systems but $3\times 3$ ones give me problems. I solved one $3\times 3$ using the method of elimination. I'm postponing the matrix method for a while.

Here's the system: $$ x' = 3x + y - z $$ $$ y' = x + 2y - z $$ $$ z' = 3x + 3y -z $$

What I do is I turn it into operator notation: $$ (D-3)x - y + z = 0 $$ $$ -x + (D-2)y + z = 0 $$ $$ -3x - 3y + (D+1)z = 0 $$ Then into $2\times 2$ system by eliminating $z$ - first two equations to get $$ (D-2)x + (-D+1)y = 0 $$ and then last two to get: $$ -(D-2)x + (D^2-D+1)y = 0 $$ Now, if I eliminate $x$ from the $2\times 2$ system easily by adding them together I get $$ (D^2-2D+2)y = 0 $$ which gives a complex solution $$ r_1 = 1 + i $$ and $$ r_2 = 1 - i $$ which gives as the solution for $y$: $$ y = C_1 \mathrm{e}^t \cos(t) + C_2 \mathrm{e}^t \sin(t) $$ So, only two constants - $C_1$ and $C_2$. But, it should have three constants since this is $3 \times 3$ system. At least, that's how it was in another solution of $3\times 3$ system that I solved. How do I solve this using elimination method?

  • $\begingroup$ "...and then last two to get..." what do you do there? $\endgroup$ – Pedro Tamaroff Feb 18 '12 at 18:03
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    $\begingroup$ @Peter: seems that he multiplies by $(D+1)$: differentiate the second equation and add its original version in order to obtain $(D+1)z$ $\endgroup$ – Ilya Feb 18 '12 at 18:10
  • $\begingroup$ @Ilya is correct. $\endgroup$ – aarnes Feb 18 '12 at 18:21

I have never used such method but direct elimination gives the same result (which is expected since the method you have used is just symbolically different from the direct elimination).

In order to have existence and uniqueness (since the right-hand side of the system is clearly nice enough) you should have 3 constants in the answer and it will indeed happen. After you have found the solution for $y$ let us call it $f(t,c_1,c_2)$. There are only two constants involved, but let us see what is happening when we finish the solution of the system.

What do you have further: $$ f' = x+2f-z\Leftrightarrow z = x+2f-f' $$ where both $x,z$ are still unknown. We can substitute it in the first equation and obtain: $$ x' = 3x+f-z = 2x-f+f' $$ There is only one unknown function so you can solve it and obtain the 3rd constant.

To find $z$ you write: $z =x+2f-f'$ - so the 4th constant will not appear since you won't have to solve an ODE.

Edited: let me show how the equation on $x$ should be solved: $$ x' = 2x-f+f'\Leftrightarrow x'-2x = f'-f. $$ This is not an equation with separated variables, so it cannot be integrated directly as you tried in your comment. This is also not a linear homogeneous ODE since the right-hand side is not zero but $f'-f$. So, we use the method of variation of parameters.

  1. We first solve the homogeneous version of an ODE: $$ x_h' - 2x_h = 0 $$ and obtain that the general solution has the form $x_h(t) = c\mathrm e^{2t}$.

  2. Clearly, $x_h$ does not solve the original equation - but it gives us a guess how to make a proper change of variables. We assume that it is easy to find $x$ in the form $$ x(t) = c(t)\mathrm e^{2t}. $$ where now $c$ is an unknown function. Note that with this assumption we won't lose any solutions and in the worst case will stay with an unpleasant equation on $c(t)$. If you make a substitution though, you will see that some things cancel: $$ x'-2x=f'-f $$ turns into $$ c' = \mathrm e^{-2t}(f'-f) $$ since $$ x'(t)-2x(t) = c'(t)\mathrm e^{2t}+2c(t)\mathrm e^{2t}-2c(t)\mathrm e^{2t} = c'(t)\mathrm e^{2t}. $$

As a result we obtain $$ x(t) = \mathrm e^{2t}\left(\int \mathrm e^{-2t}(f'(t,c_1,c_2)-f(t,c_1,c_2))dt+c_3\right) $$ and that is where the third constant appears.

  • $\begingroup$ So, I put my solution for y $ y=C_1e^tcos(t)+C_2e^tsin(t) $ into $ z = x + 2y - y' $? Doing that doesn't produce a third constant. $\endgroup$ – aarnes Feb 18 '12 at 21:44
  • $\begingroup$ @aarnes: it does since you have this constant in $x$ $\endgroup$ – Ilya Feb 19 '12 at 10:11
  • $\begingroup$ I think you are saying that solving for $y$ only requires $2$ constants, but to solve for all $3$ unknowns, requires the $3$ constants expected. Is that so? $\endgroup$ – robjohn Feb 19 '12 at 10:19
  • $\begingroup$ @robjohn: yes, that is what I meant $\endgroup$ – Ilya Feb 19 '12 at 10:33
  • $\begingroup$ First of, let me say that I'm sorry for responding so late but I'm probably in a different time zone and last night was late and I slept for quite a while. What I still don't understand is how do I get a solution for x then in order to get that third constant? Putting solution for y (with two constants) inside $ z = x + 2y - y' $ gives $ z = C_1e^tcos(t) + C_1e^tsin(t) + C_2e^tsin(t) - C_2e^tcos(t) + x(t) $ $\endgroup$ – aarnes Feb 19 '12 at 18:39

Note that since $$ D\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3&1&-1\\1&2&-1\\3&3&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\tag{1} $$ we have the general solution $$ \begin{bmatrix}x\\y\\z\end{bmatrix}=\exp\left(\begin{bmatrix}3&1&-1\\1&2&-1\\3&3&-1\end{bmatrix}t\right)\begin{bmatrix}x_0\\y_0\\z_0\end{bmatrix}\tag{2} $$ Computing the Jordan Normal Form, we get $$ \begin{align} & \begin{bmatrix}3&1&-1\\1&2&-1\\3&3&-1\end{bmatrix}\\ &=\begin{bmatrix}3&1&1\\0&1-i&1+i\\3&3&3\end{bmatrix} \begin{bmatrix}2&0&0\\0&1-i&0\\0&0&1+i\end{bmatrix} \begin{bmatrix}3&1&1\\0&1-i&1+i\\3&3&3\end{bmatrix}^{-1}\tag{3} \end{align} $$ So that $$ \newcommand{\cis}{\operatorname{cis}} \begin{align} & \exp\left(\begin{bmatrix}3&1&-1\\1&2&-1\\3&3&-1\end{bmatrix}t\right)\\ &=\begin{bmatrix}3&1&1\\0&1-i&1+i\\3&3&3\end{bmatrix} \begin{bmatrix}e^{2t}&0&0\\0&e^t\cis(-t)&0\\0&0&e^t\cis(t)\end{bmatrix} \begin{bmatrix}3&1&1\\0&1-i&1+i\\3&3&3\end{bmatrix}^{-1}\\[12pt] &=\small{\begin{bmatrix} 3e^{2t}+e^t(\sin(t)-\cos(t))&2e^t\sin(t)&-e^{2t}-e^t(\sin(t)-\cos(t))\\ 2e^t\sin(t)&2e^t(\sin(t)+\cos(t))&-2e^t\sin(t)\\ 3e^{2t}+3e^t(\sin(t)-\cos(t))&6e^t\sin(t)&-e^{2t}+3e^t(\sin(t)-\cos(t)) \end{bmatrix}}\tag{4} \end{align} $$ Plugging $(4)$ into $(2)$, we get $$ y=2e^t(\sin(t)(x_0+y_0-z_0)+\cos(t)y_0)\tag{5} $$ Thus, $y$ is dependent on $2$ constants, $x_0+y_0-z_0$ and $y_0$, just as in your solution.

However, consider the simpler set of equations: $$ D\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}\tag{6} $$ whose solution is $$ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}e^t&0&0\\0&e^{2t}&0\\0&0&e^{3t}\end{bmatrix}\begin{bmatrix}x_0\\y_0\\z_0\end{bmatrix}\tag{7} $$ In this case, each of $x$, $y$, and $z$ are dependent on only $1$ constant, but now I think the important point is hopefully clearer, that the solution system, $\{x,y,z\}$, is dependent on $3$ constants.

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    $\begingroup$ In other words: Just because there may be $n$ degrees of freedom in the full solution to an $n\times n$ system, doesn't mean each individual component has $n$ degrees of freedom. $\endgroup$ – anon Feb 19 '12 at 0:07
  • $\begingroup$ @anon: very concisely put! That is indeed the point I was trying to make. $\endgroup$ – robjohn Feb 19 '12 at 0:19
  • $\begingroup$ Aha, I see. I found this problem in a book entitled "Fundamentals of Differential Equations". The solution for y seems exactly like mine plus when I saw your explanation why the individual solution doesn't have to have n constants in a nxn system my mind was at ease. However, I still don't know how to get x(t) and z(t). $\endgroup$ – aarnes Feb 19 '12 at 18:49
  • $\begingroup$ @aarnes: plug $(4)$ into $(2)$ and don't throw away the $x$ and $z$ parts. :-) $$ x=e^{2t}(3x_0-z_0)+e^t(2y_0\sin(t)+(x_0-z_0)(\sin(t)-\cos(t))) $$ and $$ z=e^{2t}(3x_0-z_0)+3e^t(2y_0\sin(t)+(x_0+z_0)(\sin(t)-\cos(t))) $$ $\endgroup$ – robjohn Feb 20 '12 at 23:44

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