System of differential equations

**AmberLamps** · May 29th 2010, 03:09 PM

Hey guys,

I have derived the system of differential equations below, and am coming unstuck on the later parts of the question. This is in an econ course so I have changed some of the symbols so i dont offend anyone in here

(we use $\Pi$ for inflation rate!).

$\begin{pmatrix} \dot {x} \\ \dot {y} \end{pmatrix}$ = $\left[ \begin {array}{cc} [\O(1+b\lambda/a) - \lambda/a]&-\O(1+b\lambda/a)\\ \noalign{\medskip} \O&-\O \end {array} \right] $ $\begin{pmatrix} x \\ y \end{pmatrix}$ + $\begin{pmatrix} u\lambda/a - g\lambda\\ 0 \end{pmatrix}$

also note that $x = \dot {G}/G$ and $y = \dot {P}/P$

the question is: examine the system above for stationary points and develop stability conditions for any such points.

The stationary point is when $\dot {x} = \dot {y}= 0$ . The answer to this is apparently

$x = y = u - ag$

Now I can kind of understand where this has come from if the fact that x and y are a function of a differential equation means that they are also 0 at a stationary point. Is this correct? Nevertheless I wouldn't mind a proof if possible as my answer is still missing its LHS

For the stability conditions, we use something called Hartman's Lemma which is pretty obscure but it basically involves finding the eigenvalues of the coefficient matrix, and using them to test for stability. What I cannot do, however, is find the eigenvalues.

Again I had scribbled down the characteristic equation as below, but I just can't see how to get to this from my coefficient matrix - and I'm sure its something to do with my poor technical math.

$Z^2 + Z\lambda(1-b\O)/a + \O\lambda/a$

Any help would be much appreciated, it is no way near as long winded as it looks!

**AllanCuz** · May 29th 2010, 04:27 PM

Originally Posted by AmberLamps

Hey guys,

I have derived the system of differential equations below, and am coming unstuck on the later parts of the question. This is in an econ course so I have changed some of the symbols so i dont offend anyone in here

(we use $\Pi$ for inflation rate!).

$\begin{pmatrix} \dot {x} \\ \dot {y} \end{pmatrix}$ = $\left[ \begin {array}{cc} [\O(1+b\lambda/a) - \lambda/a]&-\O(1+b\lambda/a)\\ \noalign{\medskip} \O&-\O \end {array} \right] $ $\begin{pmatrix} x \\ y \end{pmatrix}$ + $\begin{pmatrix} u\lambda/a - g\lambda\\ 0 \end{pmatrix}$

also note that $x = \dot {G}/G$ and $y = \dot {P}/P$

the question is: examine the system above for stationary points and develop stability conditions for any such points.

The stationary point is when $\dot {x} = \dot {y}= 0$ . The answer to this is apparently

$x = y = u - ag$

Now I can kind of understand where this has come from if the fact that x and y are a function of a differential equation means that they are also 0 at a stationary point. Is this correct? Nevertheless I wouldn't mind a proof if possible as my answer is still missing its LHS

For the stability conditions, we use something called Hartman's Lemma which is pretty obscure but it basically involves finding the eigenvalues of the coefficient matrix, and using them to test for stability. What I cannot do, however, is find the eigenvalues.

Again I had scribbled down the characteristic equation as below, but I just can't see how to get to this from my coefficient matrix - and I'm sure its something to do with my poor technical math.

$Z^2 + Z\lambda(1-b\O)/a + \O\lambda/a$

Any help would be much appreciated, it is no way near as long winded as it looks!

$\dot x =[\O(1+b\lambda/a) - \lambda/a]x -[ \O(1+b\lambda/a) ]y + u\lambda/a - g\lambda$

$\dot y = \O x - \O y$

Thus, $x = y$ and

$0 =[\O(1+b\lambda/a) - \lambda/a]x -[ \O(1+b\lambda/a) ]x + u\lambda/a - g\lambda$

$0 = - \frac{ \lambda }{a} x + u \frac{ \lambda }{a} - g \lambda$

$0 = - \frac{1}{a}x + \frac{1}{a} u - g$

$x = u - ga$

It's just multiplying out the matrix and equating both equations to 0. Nothing overly tricky!

**AmberLamps** · May 29th 2010, 04:38 PM

thanks. I just hadn't set x=y

**AllanCuz** · May 29th 2010, 04:41 PM

Originally Posted by AmberLamps

thanks. I just hadn't set x=y

Well, you don't set this. It follows from,

$\bar y = \O x - \O y$ but $\bar y = 0$

So,

$\O x = \O y$ $\to x = y$

But yeah, then you use that result to compute the rest. Otherwise this problem would really suck :P

Edit- What application does this have in economics? I really don't see how...

**AmberLamps** · May 29th 2010, 05:18 PM

Originally Posted by AllanCuz

Well, you don't set this. It follows from,

$\bar y = \O x - \O y$ but $\bar y = 0$

So,

$\O x = \O y$ $\to x = y$

But yeah, then you use that result to compute the rest. Otherwise this problem would really suck :P

Edit- What application does this have in economics? I really don't see how...

yeye thats how i understood it... i think

x is actual inflation rate, y is expected inflation rate, u is monetary growth and g is change real income trend. As displayed it doesn't actually show that much, other than the relationship between these terms. But from it you can - with stability - show the response of the inflation rate, demand for real money balances etc to an increase in monetary growth. you can also look at what happens if the monetary authority tries to control one of the variables (i.e. inflation rate) by solving as a function of monetary growth. At this point though, it is going a bit past what is required of us. thanks for the help.

Thread: System of differential equations

LinkBack

Thread Tools

Display

System of differential equations

Similar Math Help Forum Discussions

How do i solve this system of differential equations?

system of differential equations

System of differential equations

system of differential equations

System of Differential Equations

Search Tags