Initial Value Problem for a System of Linear Differential Equations

[mathmari]

Hey!

We have the following initial value problem:

$$x' = \frac{1}{2}(45 − x) + \frac{1}{4}(y − x) \\
y' = \frac{1}{4}(x − y) + \frac{1}{2}(35 − y) + \frac{1}{2}(z − y) + 20 \\
z' = \frac{1}{2}(y − z) + \frac{1}{2}(35 − z)$$

This can be written as follows:

$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}'=\begin{pmatrix}
-\frac{3}{4} & \frac{1}{4} & 0 \\
\frac{1}{4} & -\frac{5}{4} & \frac{1}{2} \\
0 & \frac{1}{2} & -1
\end{pmatrix}\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}+\begin{pmatrix}
\frac{45}{2}\\
\frac{75}{2}\\
\frac{35}{2}
\end{pmatrix}$$

right?? (Wondering)

To solve this problem do we have to solve first the homogengeous part and then the inhomogeneous one?? (Wondering)

 

[I like Serena]

Hi! (Blush)

I think it's best to first diagonalize the matrix. (Thinking)

 

[I like Serena]

Hmm, I think the bottom right entry in your matrix is not correct. (Worried)

Anyway, we can write your set of equations:
$$X'=AX+C \tag 1$$
as:
$$X' = BDB^{-1}X+C \tag 2$$
where $D$ is a diagonal matrix (or a Jordan normal form).

When we substitute:
$$X=BY \tag 3$$
we'll get a set of decoupled linear inhomogeneous differential equations. (Thinking)

 

[mathmari]

Hmm, I think the bottom right entry in your matrix is not correct. (Worried)

I edited my initial post... I wrote a wrong coefficient at the last equation.

Anyway, we can write your set of equations:
$$X'=AX+C \tag 1$$
as:
$$X' = BDB^{-1}X+C \tag 2$$
where $D$ is a diagonal matrix (or a Jordan normal form).

I found the following:

$$D=\begin{pmatrix}
-1.683 & 0 & 0 \\
0 & -0.817 & 0 \\
0 & 0 & -0.5
\end{pmatrix}$$

$$B=\begin{pmatrix}
-0.2113 & -0.7887 & -0.5774 \\
0.7887 & 0.2113 & -0.5774 \\
-0.5774 & 0.5774 & -0.5774
\end{pmatrix}$$

$$B^{-1}=\begin{pmatrix}
-0.2113 & 0.7887 & -0.5774 \\
-0.7887 & 0.2113 & 0.5774 \\
-0.5774 & -0.5774 & -0.5774
\end{pmatrix}$$

When we substitute:
$$X=BY \tag 3$$
we'll get a set of decoupled linear inhomogeneous differential equations. (Thinking)

What is $Y$ ?? (Wondering)

 

[I like Serena]

$$D=\begin{pmatrix}
-1.683 & 0 & 0 \\
0 & -0.817 & 0 \\
0 & 0 & -0.5
\end{pmatrix}$$

$$B=\begin{pmatrix}
-0.2113 & -0.7887 & -0.5774 \\
0.7887 & 0.2113 & -0.5774 \\
-0.5774 & 0.5774 & -0.5774
\end{pmatrix}$$

$$B^{-1}=\begin{pmatrix}
-0.2113 & 0.7887 & -0.5774 \\
-0.7887 & 0.2113 & 0.5774 \\
-0.5774 & -0.5774 & -0.5774
\end{pmatrix}$$

Good! (Happy)

What is $Y$ ?? (Wondering)

An arbitrary vector of 3 variables.
Let's say:
$$Y=\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}$$

And $Y'$ is its derivative:
$$Y'=\begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}$$
(Wasntme)

 

[mathmari]

An arbitrary vector of 3 variables.
Let's say:
$$Y=\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}$$

And $Y'$ is its derivative:
$$Y'=\begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}$$
(Wasntme)

I haven't understood why we should do this substitution... (Wondering)

 

[I like Serena]

I haven't understood why we should do this substitution... (Wondering)

To decouple the differential equations.

In your problem you have 3 coupled equations, meaning each equation contains a combination of $x,y,z$.
As such we cannot solve it.

After the substitution, we will have 3 equations that are decoupled, meaning that e.g. the first equation will contain only $y_1$ and $y_1'$.
Now we can solve each equation independent from the others. (Wasntme)

 

[mathmari]

To decouple the differential equations.

In your problem you have 3 coupled equations, meaning each equation contains a combination of $x,y,z$.
As such we cannot solve it.

After the substitution, we will have 3 equations that are decoupled, meaning that e.g. the first equation will contain only $y_1$ and $y_1'$.
Now we can solve each equation independent from the others. (Wasntme)

We get the following:

$$X'=AX+C \Rightarrow X'=BDB^{-1}+C$$

We set $X=BY$ and we get $$BY'=BDB^{-1}BY+C \Rightarrow BY'=BDY+C \Rightarrow B^{-1}BY'=B^{-1}DY+B^{-1}C \Rightarrow Y'=Y+B^{-1}C$$

right?? (Wondering)

So, we have to solve the following problems:

$$y_1'=y_1+14.7169 \\ y_2'=y_2+0.2831 \\ y_3'=y_3-44.7446$$

The solutions are the following:

$$y_1=c_1e^t-14.7169 \ \ , \ \ y_2=c_2e^y-0.2831 \ \ , \ \ y_3=c_3e^t+44.7466$$

Now we solve for $x, y, z$ by replacing $Y$ at $X=BY$, right?? (Wondering)

 

[mathmari]

Having the initial values $$X(0)=\begin{pmatrix}
45\\
35\\
35
\end{pmatrix}$$ we get the following inital values for $Y$ $$Y(0)=\begin{pmatrix}
14.7169\\
0.2831\\
-44.7446
\end{pmatrix}$$

So, we have the folowing solutions $$y_1=29.4338e^t-14.7169 \ \ , \ \ y_2=0.5662e^t-0.2831 \ \ , \ \ y_3=-89.4892e^t+44.7446$$

Therefore, the solution of the initial problem is the following:

$$X=BY=\begin{pmatrix}
-0.2113 & -0.7887 & -0.5774 \\
0.7887 & 0.2113 & -0.5774 \\
-0.5774 & 0.5774 & -0.5774
\end{pmatrix} \begin{pmatrix}
29.4338e^t-14.7169\\
0.5662e^t-0.2831\\
-89.4892e^t+44.7446
\end{pmatrix} \Rightarrow X=\begin{pmatrix}
45.0051e^t-22.5026\\
75.0051e^t-37.5026\\
35.0029e^t-17.5015
\end{pmatrix}
$$

Is this correct?? (Wondering)

 

[I like Serena]

We get the following:

$$X'=AX+C \Rightarrow X'=BDB^{-1}+C$$

We set $X=BY$ and we get $$BY'=BDB^{-1}BY+C \Rightarrow BY'=BDY+C \Rightarrow B^{-1}BY'=B^{-1}DY+B^{-1}C \Rightarrow Y'=Y+B^{-1}C$$

right?? (Wondering)

Hold on! (Wait)
I think we have lost the diagonal matrix $D$. (Worried)

Therefore, the solution of the initial problem is the following:

$$X=BY=\begin{pmatrix}
-0.2113 & -0.7887 & -0.5774 \\
0.7887 & 0.2113 & -0.5774 \\
-0.5774 & 0.5774 & -0.5774
\end{pmatrix} \begin{pmatrix}
29.4338e^t-14.7169\\
0.5662e^t-0.2831\\
-89.4892e^t+44.7446
\end{pmatrix} \Rightarrow X=\begin{pmatrix}
45.0051e^t-22.5026\\
75.0051e^t-37.5026\\
35.0029e^t-17.5015
\end{pmatrix}
$$

Is this correct?? (Wondering)

When I substitute it in the first equation, it doesn't fit. (Doh)

It does look weird that $e^t$ wouldn't have a coefficient in its power. (Wasntme)

 

[mathmari]

I tried it again...

We get the following:

$$X'=AX+C \Rightarrow X'=BDB^{-1}+C$$

We set $X=BY$ and we get $$BY'=BDB^{-1}BY+C \\ \Rightarrow BY'=BDY+C \\ \Rightarrow B^{-1}BY'=B^{-1}BDY+B^{-1}C \\ \Rightarrow Y'=DY+B^{-1}C$$

So, we have to solve the following problems:

$$y_1'=-1.683y_1+14.7169 \\ y_2'=-0.817y_2+0.2831 \\ y_3'=-0.5y_3-44.7446$$

The solutions are the following:

$$y_1=c_1e^{-1.683t}+8.74444 \ \ , \ \ y_2=c_2e^{-0.817t}+0.346512 \ \ , \ \ y_3=c_3e^{-0.5t}-89.4892$$

The initial values are $$Y(0)=B^{-1}X(0)=\begin{pmatrix}
-2.1132\\
-7.8868\\
-66.3953
\end{pmatrix}$$

So, we get the following solution:

$$Y=\begin{pmatrix}
-10.8576e^{-1.683t}+8.74444\\
-8.23331e^{-0.817t}+0.346512\\
23.0939e^{-0.5t}-89.4892
\end{pmatrix}$$

So, the solution of the initial problem is $$X=BY=\begin{pmatrix}
-0.2113 & -0.7887 & -0.5774 \\
0.7887 & 0.2113 & -0.5774 \\
-0.5774 & 0.5774 & -0.5774
\end{pmatrix} \begin{pmatrix}
-10.8576e^{-1.683t}+8.74444\\
-8.23331e^{-0.817t}+0.346512\\
23.0939e^{-0.5t}-89.4892
\end{pmatrix} = \begin{pmatrix}
49.5501+2.29421 e^{-1.683 t}+6.49361 e^{-0.817 t}-13.3344 e^{-0.5 t}\\
58.641-8.56339 e^{-1.683 t}-1.7397 e^{-0.817 t}-13.3344 e^{-0.5 t}\\
46.8221+6.26918 e^{-1.683 t}-4.75391 e^{-0.817 t}-13.3344 e^{-0.5 t}
\end{pmatrix}$$

Is this correct?? (Wondering)

 

[I like Serena]

It looks correct, which is good! (Nod)

But at my first attempt to substitute the solution in the original set of equations, I found a mismatch, so there may be a mistake somewhere. (Worried)

What do you get if you substitute the solution in the original set of equations? (Wondering)

 

[mathmari]

$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}'=\begin{pmatrix}
-\frac{3}{4} & \frac{1}{4} & 0 \\
\frac{1}{4} & -\frac{5}{4} & \frac{1}{2} \\
0 & \frac{1}{2} & -1
\end{pmatrix}\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}+\begin{pmatrix}
\frac{45}{2}\\
\frac{75}{2}\\
\frac{35}{2}
\end{pmatrix}
$$

So, it is of the form $X'=AX+C$.

The eigenvalues of the matrix $A$ are $-\frac{1}{2}, -\frac{5}{4}-\frac{\sqrt{3}}{4}, \frac{\sqrt{3}}{4}-\frac{5}{4}$.

We want to write the matrix $A$ in the form $BDB^{-1}$.

The diagonal matrix $D$ is the following:

$$D=\begin{pmatrix}
-\frac{1}{2} & 0 & 0 \\
0 & -\frac{5}{4}-\frac{\sqrt{3}}{4} & 0 \\
0 & 0 & \frac{\sqrt{3}}{4}-\frac{5}{4}
\end{pmatrix}$$

Then using the eigenvectors we get the following matrix $B$:

$$B=\begin{pmatrix}
1 & 1 & 1 \\
1 & -2-\sqrt{3} & -2+\sqrt{3} \\
1 & 1+\sqrt{3} & 1-\sqrt{3}
\end{pmatrix}$$

of which the inverse is the following:

$$B^{-1}=\begin{pmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
-\frac{\sqrt{3}}{6}+\frac{1}{3} & -\frac{1}{6} & \frac{\sqrt{3}-1}{6} \\
\frac{\sqrt{3}}{6}+\frac{1}{3} & -\frac{1}{6} & \frac{-\sqrt{3}-1}{6}
\end{pmatrix}
$$

So, we have the following:

$$X'AX+C \Rightarrow X'=BDB^{-1}X+C$$

Setting $X=BY$ we get the following:

$$BY'=BDB^{-1}BY+C \Rightarrow BY'=BDY+C \Rightarrow B^{-1}BY'=B^{-1}BDY+B^{-1}C \\ \Rightarrow Y'=DY+B^{-1}C$$

where $$B^{-1}C=\begin{pmatrix}
\frac{155}{6}-\frac{10}{3}\sqrt{3}\\
-\frac{5}{3}\\
-\frac{5}{3}+\frac{55}{6}\sqrt{3}
\end{pmatrix}$$

So, we have to solve the following equations:

$$y_1'=-\frac{1}{2}y_1+\frac{155}{6}-\frac{10}{3}\sqrt{3} \\ y_2'=\left (-\frac{5}{4}-\frac{\sqrt{3}}{4}\right )y_2-\frac{5}{3} \\ y_3'=\left (\frac{\sqrt{3}}{4}-\frac{5}{4}\right )y_3-\frac{5}{3}+\frac{55}{6}\sqrt{3}$$

with the initial values $$Y(0)=B^{-1}X(0)=\begin{pmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
-\frac{\sqrt{3}}{6}+\frac{1}{3} & -\frac{1}{6} & \frac{\sqrt{3}-1}{6} \\
\frac{\sqrt{3}}{6}+\frac{1}{3} & -\frac{1}{6} & \frac{-\sqrt{3}-1}{6}
\end{pmatrix}\begin{pmatrix}
45\\
35\\
35
\end{pmatrix}=\begin{pmatrix}
\frac{115}{3}\\
\frac{10-5\sqrt{3}}{3}\\
\frac{10+5\sqrt{3}}{3}
\end{pmatrix}$$

The solutions are the following:

$$y_1=\frac{5}{3} (31-4 \sqrt{3}+4 (-2+\sqrt{3}) e^{-\frac{t}{2}}) \\ y_2=-\frac{5 e^{-\frac{1}{4} (5+\sqrt{3}) t} (-11+3 \sqrt{3})+20}{3 (5+\sqrt{3})} \\ y_3= \frac{5 (4-22 \sqrt{3}+(-11+19 \sqrt{3}) e^{\frac{1}{4} (-5+\sqrt{3}) t})}{3 (-5+\sqrt{3})}$$

So, the solution of the initial problem is $$\begin{pmatrix}
1 & 1 & 1 \\
1 & -2-\sqrt{3} & -2+\sqrt{3} \\
1 & 1+\sqrt{3} & 1-\sqrt{3}
\end{pmatrix} \begin{pmatrix}
\frac{5}{3} (31-4 \sqrt{3}+4 (-2+\sqrt{3}) e^{-\frac{t}{2}}) \\ -\frac{5 e^{-\frac{1}{4} (5+\sqrt{3}) t} (-11+3 \sqrt{3})+20}{3 (5+\sqrt{3})} \\ \frac{5 (4-22 \sqrt{3}+(-11+19 \sqrt{3}) e^{\frac{1}{4} (-5+\sqrt{3}) t})}{3 (-5+\sqrt{3})}
\end{pmatrix}=\begin{pmatrix}
\frac{590}{11}+\frac{5}{\sqrt{3}}+\frac{20}{3}(-2+\sqrt{3})e^{-\frac{t}{2}}-\frac{5(-11+3\sqrt{3})}{3(5+\sqrt{3})}e^{-\frac{(5+\sqrt{3})t}{4}}+\frac{5(-11+19\sqrt{3})}{3(-5+\sqrt{3})}e^{\frac{(-5+\sqrt{3})t}{4}}\\
\frac{780}{11}-\frac{55}{\sqrt{3}}+\frac{20}{3}(-2+\sqrt{3})e^{-\frac{t}{2}}+\frac{(2+\sqrt{3})5(-11+3\sqrt{3})}{3(5+\sqrt{3})}e^{-\frac{(5+\sqrt{3})t}{4}}+\frac{5(-11+19\sqrt{3})(-2+\sqrt{3})}{3(-5+\sqrt{3})}e^{\frac{(-5+\sqrt{3})t}{4}}\\
\frac{335}{11}-\frac{10}{\sqrt{3}}+\frac{20}{3}(-2+\sqrt{3})e^{-\frac{t}{2}}-\frac{(-1+\sqrt{3})5(-11+3\sqrt{3})}{3(5+\sqrt{3})}e^{-\frac{(5+\sqrt{3})t}{4}}+\frac{5(-11+19\sqrt{3})(1-\sqrt{3})}{3(-5+\sqrt{3})}e^{\frac{(-5+\sqrt{3})t}{4}}
\end{pmatrix}$$

Is this correct?? (Wondering)

 

[I like Serena]

So, the solution of the initial problem is $$\begin{pmatrix}
1 & 1 & 1 \\
1 & -2-\sqrt{3} & -2+\sqrt{3} \\
1 & 1+\sqrt{3} & 1-\sqrt{3}
\end{pmatrix} \begin{pmatrix}
\frac{5}{3} (31-4 \sqrt{3}+4 (-2+\sqrt{3}) e^{-\frac{t}{2}}) \\ -\frac{5 e^{-\frac{1}{4} (5+\sqrt{3}) t} (-11+3 \sqrt{3})+20}{3 (5+\sqrt{3})} \\ \frac{5 (4-22 \sqrt{3}+(-11+19 \sqrt{3}) e^{\frac{1}{4} (-5+\sqrt{3}) t})}{3 (-5+\sqrt{3})}
\end{pmatrix}=\begin{pmatrix}
\frac{590}{11}+\frac{5}{\sqrt{3}}+\frac{20}{3}(-2+\sqrt{3})e^{-\frac{t}{2}}-\frac{5(-11+3\sqrt{3})}{3(5+\sqrt{3})}e^{-\frac{(5+\sqrt{3})t}{4}}+\frac{5(-11+19\sqrt{3})}{3(-5+\sqrt{3})}e^{\frac{(-5+\sqrt{3})t}{4}}\\
\frac{780}{11}-\frac{55}{\sqrt{3}}+\frac{20}{3}(-2+\sqrt{3})e^{-\frac{t}{2}}+\frac{(2+\sqrt{3})5(-11+3\sqrt{3})}{3(5+\sqrt{3})}e^{-\frac{(5+\sqrt{3})t}{4}}+\frac{5(-11+19\sqrt{3})(-2+\sqrt{3})}{3(-5+\sqrt{3})}e^{\frac{(-5+\sqrt{3})t}{4}}\\
\frac{335}{11}-\frac{10}{\sqrt{3}}+\frac{20}{3}(-2+\sqrt{3})e^{-\frac{t}{2}}-\frac{(-1+\sqrt{3})5(-11+3\sqrt{3})}{3(5+\sqrt{3})}e^{-\frac{(5+\sqrt{3})t}{4}}+\frac{5(-11+19\sqrt{3})(1-\sqrt{3})}{3(-5+\sqrt{3})}e^{\frac{(-5+\sqrt{3})t}{4}}
\end{pmatrix}$$

Impressive! (Bow) (Bow)

Is this correct?? (Wondering)

Did you substitute it in the original set of equations? (Wondering)

 

[mathmari]

I think it is wrong again... (Sadface)

I try it again... (Wait)

 

[mathmari]

$$A=\begin{pmatrix}
-\frac{3}{4} & \frac{1}{4} & 0 \\
\frac{1}{4} & -\frac{5}{4} & \frac{1}{2} \\
0 & \frac{1}{2} & -1
\end{pmatrix}$$

The eigenvalues of the matrix $A$ are $-\frac{1}{2}, -\frac{5}{4}-\frac{\sqrt{3}}{4}, \frac{\sqrt{3}}{4}-\frac{5}{4}$, right?? (Wondering)

To find the eigenvector for the eigenvalue $-\frac{1}{2}$ we do the following:

$$\begin{pmatrix}
-\frac{3}{4}-\left (-\frac{1}{2}\right ) & \frac{1}{4} & 0 \\
\frac{1}{4} & -\frac{5}{4}-\left (-\frac{1}{2}\right ) & \frac{1}{2} \\
0 & \frac{1}{2} & -1 -\left (-\frac{1}{2}\right )
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}0\\0\\0\end{pmatrix} \\ \Rightarrow \begin{pmatrix}
-\frac{1}{4} & \frac{1}{4} & 0 \\
\frac{1}{4} & -\frac{3}{4} & \frac{1}{2} \\
0 & \frac{1}{2} & -\frac{1}{2}
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}0\\0\\0\end{pmatrix} \\ \Rightarrow x=y=z$$

Doesn't this mean that we can take the vector $\begin{pmatrix}1\\1\\1\end{pmatrix}$ ?? (Wondering)

I checked it in Matlab and for the eigenvalue $-\frac{1}{2}$ the corresponding eigenvector is $\begin{pmatrix}-0.5774\\-0.5774\\-0.5774\end{pmatrix}$.

How do we find this one?? (Wondering)

 

[I like Serena]

The eigenvalues of the matrix $A$ are $-\frac{1}{2}, -\frac{5}{4}-\frac{\sqrt{3}}{4}, \frac{\sqrt{3}}{4}-\frac{5}{4}$, right?? (Wondering)

I'm certain about the $-\frac{1}{2}$. The approximate values you had previously were also correct. (Nod)

Doesn't this mean that we can take the vector $\begin{pmatrix}1\\1\\1\end{pmatrix}$ ?? (Wondering)

I checked it in Matlab and for the eigenvalue $-\frac{1}{2}$ the corresponding eigenvector is $\begin{pmatrix}-0.5774\\-0.5774\\-0.5774\end{pmatrix}$.

How do we find this one?? (Wondering)

It's the same eigenvector.
Only the direction is relevant, the length is arbitrary.
Note that if $Av=\lambda v$, we will also have $A(2v) = \lambda (2v)$. (Wasntme)

 

[mathmari]

To find the eigenvector for the eigenvalue $-\frac{5}{4}-\frac{\sqrt{3}}{4}$ we do the following:

$$\begin{pmatrix}
-\frac{3}{4}-\left (-\frac{5}{4}-\frac{\sqrt{3}}{4}\right ) & \frac{1}{4} & 0 \\
\frac{1}{4} & -\frac{5}{4}-\left (-\frac{5}{4}-\frac{\sqrt{3}}{4}\right ) & \frac{1}{2} \\
0 & \frac{1}{2} & -1 -\left (-\frac{5}{4}-\frac{\sqrt{3}}{4}\right )
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}0\\0\\0\end{pmatrix} \\ \Rightarrow \begin{pmatrix}
\frac{1}{2}+\frac{\sqrt{3}}{4} & \frac{1}{4} & 0 \\
\frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{1}{2} \\
0 & \frac{1}{2} & \frac{1}{4}+\frac{\sqrt{3}}{4}
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}0\\0\\0\end{pmatrix} \\ \Rightarrow y=-(2+\sqrt{3})x\ \ ,\ \ z=\frac{2(2+\sqrt{3})}{1+\sqrt{3}}x$$

So, we take the vector $$\begin{pmatrix} 1\\-(2+\sqrt{3})\\\frac{2(2+\sqrt{3})}{1+\sqrt{3}}\end{pmatrix}$$ right??

In Matlab the result is $$\begin{pmatrix} -0.2113\\
0.7887\\
-0.5774 \end{pmatrix}$$ Is this the same?? (Wondering)

 

[I like Serena]

So, we take the vector $$\begin{pmatrix} 1\\-(2+\sqrt{3})\\\frac{2(2+\sqrt{3})}{1+\sqrt{3}}\end{pmatrix}$$ right??

In Matlab the result is $$\begin{pmatrix} -0.2113\\
0.7887\\
-0.5774 \end{pmatrix}$$ Is this the same?? (Wondering)

Let's see...
$$\begin{pmatrix} 1\\-(2+\sqrt{3})\\\frac{2(2+\sqrt{3})}{1+\sqrt{3}}\end{pmatrix}
\approx \begin{pmatrix} 1\\-(2+1.7)\\\frac{2(2+1.7)}{1+1.7}\end{pmatrix}
\approx \begin{pmatrix} 1\\-3.7\\\frac{7.4}{2.7}\end{pmatrix}
\approx \begin{pmatrix} 1\\-3.7\\\frac{8}{3}\end{pmatrix}
\approx \begin{pmatrix} 1\\-3.7\\2.7\end{pmatrix}
$$
Divide by $-5$:
$$-\frac 1 5\begin{pmatrix} 1\\-(2+\sqrt{3})\\\frac{2(2+\sqrt{3})}{1+\sqrt{3}}\end{pmatrix}
\approx -\frac 1 5\begin{pmatrix} 1\\-3.7\\2.7\end{pmatrix}
\approx \begin{pmatrix} -0.2\\0.7\\-0.5\end{pmatrix}
$$

Yep. I think it is the same! (Nod)

 

[mathmari]

So, the vectors of Matlab are the normalized of the vectors that I found, right?? (Wondering)

So, the normalized vectors are
$$\frac{1}{\sqrt{3}}\begin{pmatrix}1\\1\\1\end{pmatrix}, \sqrt{\frac{3-\sqrt{3}}{6}}\begin{pmatrix}1\\-(2+\sqrt{3})\\\frac{2(2+\sqrt{3})}{1+\sqrt{3}}\end{pmatrix}, \sqrt{\frac{3+\sqrt{3}}{6}}\begin{pmatrix}1\\\sqrt{3}-2\\\frac{2(2-\sqrt{3})}{1-\sqrt{3}}\end{pmatrix}$$

So, is the array of eigenvectors the following?? (Wondering)

$$B=\begin{pmatrix}
\frac{1}{\sqrt{3}} & \sqrt{\frac{3-\sqrt{3}}{6}} & \sqrt{\frac{3+\sqrt{3}}{6}} \\
\frac{1}{\sqrt{3}} & -\sqrt{\frac{3-\sqrt{3}}{6}}(2+\sqrt{3}) & \sqrt{\frac{3+\sqrt{3}}{6}}(\sqrt{3}-2) \\
\frac{1}{\sqrt{3}} & 2\sqrt{\frac{3-\sqrt{3}}{6}}\frac{(2+\sqrt{3})}{1+\sqrt{3}} & \sqrt{\frac{3+\sqrt{3}}{3}} \frac{(2-\sqrt{3})}{1-\sqrt{3}}
\end{pmatrix}$$

 

[I like Serena]

So, the vectors of Matlab are the normalized of the vectors that I found, right?? (Wondering)

Looks that way. (Nod)

So, is the array of eigenvectors the following?? (Wondering)

$$B=\begin{pmatrix}
\frac{1}{\sqrt{3}} & \sqrt{\frac{3-\sqrt{3}}{6}} & \sqrt{\frac{3+\sqrt{3}}{6}} \\
\frac{1}{\sqrt{3}} & -\sqrt{\frac{3-\sqrt{3}}{6}}(2+\sqrt{3}) & \sqrt{\frac{3+\sqrt{3}}{6}}(\sqrt{3}-2) \\
\frac{1}{\sqrt{3}} & 2\sqrt{\frac{3-\sqrt{3}}{6}}\frac{(2+\sqrt{3})}{1+\sqrt{3}} & \sqrt{\frac{3+\sqrt{3}}{3}} \frac{(2-\sqrt{3})}{1-\sqrt{3}}
\end{pmatrix}$$

I can tell the the version with approximations that you had before, was correct.
The check to make, is that $A = BDB^{-1}$, which it was. (Wasntme)

 

[mathmari]

I tried it again...

I found the matrix $$B=\begin{pmatrix}
\frac{1}{\sqrt{3}} & \frac{1}{3+\sqrt{3}} & \frac{1}{3-\sqrt{3}} \\
\frac{1}{\sqrt{3}} & -\frac{3+\sqrt{3}}{6} & \frac{\sqrt{3}-3}{6} \\
\frac{1}{\sqrt{3}} & \frac{\sqrt{3}}{3} & -\frac{\sqrt{3}}{6}
\end{pmatrix}=\begin{pmatrix} 0.5774 & 0.2113 & 0.7887 \\
0.5774 & -0.7887 & -0.2113 \\
0.5774 & 0.5774 & -0.2887\end{pmatrix}$$

In Matlab with the command [m][V,D]=eig(A)[/m] I get the following:

$$V =\begin{pmatrix}
-0.2113 & -0.7887 & -0.5774 \\
0.7887 & 0.2113 & -0.5774 \\
-0.5774 & 0.5774 & -0.5774\end{pmatrix}$$$$D =\begin{pmatrix}
-1.6830 & 0 & 0 \\
0 & -0.8170 & 0 \\
0 & 0 & -0.5000\end{pmatrix}$$So, the matrix $V$ is the same as I found, $B$, but the vectors are not in the same order and it has a different sign... Does it make a difference?? (Wondering)

 

[I like Serena]

I tried it again...

I found the matrix $$B=\begin{pmatrix}
\frac{1}{\sqrt{3}} & \frac{1}{3+\sqrt{3}} & \frac{1}{3-\sqrt{3}} \\
\frac{1}{\sqrt{3}} & -\frac{3+\sqrt{3}}{6} & \frac{\sqrt{3}-3}{6} \\
\frac{1}{\sqrt{3}} & \frac{\sqrt{3}}{3} & -\frac{\sqrt{3}}{6}
\end{pmatrix}=\begin{pmatrix} 0.5774 & 0.2113 & 0.7887 \\
0.5774 & -0.7887 & -0.2113 \\
0.5774 & 0.5774 & -0.2887\end{pmatrix}$$

In Matlab with the command [m][V,D]=eig(A)[/m] I get the following:

$$V =\begin{pmatrix}
-0.2113 & -0.7887 & -0.5774 \\
0.7887 & 0.2113 & -0.5774 \\
-0.5774 & 0.5774 & -0.5774\end{pmatrix}$$$$D =\begin{pmatrix}
-1.6830 & 0 & 0 \\
0 & -0.8170 & 0 \\
0 & 0 & -0.5000\end{pmatrix}$$So, the matrix $V$ is the same as I found, $B$, but the vectors are not in the same order and it has a different sign... Does it make a difference?? (Wondering)

Good! (Smile)

And no, it doesn't make a difference. (Shake)
It does mean that a different diagonal matrix belongs to your $B$ with for instance $-0.5$ at the left top.
And it also means that when you continue, the intermediate $y_i$ will be swapped around.

More specifically, the matrix $D$ contains the eigenvalues, which are unique, but you are free to order them.
Matlab will use a standard ordering, probably putting them in order of descending absolute value.
The matrix $V$ contains the corresponding eigenvectors in the same order.
Each of those eigenvectors has an arbitrary length and sign.
Matlab will normalize them to unit length, and may ensure that the determinant is positive. (Nerd)

 

[mathmari]

For an orthonormal matrix $P$ it stands that $P^{T}=P^{-1}$, right?? (Wondering)

In Matlab I calculated the inverse and the transpose matrix of $V$ and the result was the same matrix...

When I calculated in Matlab the inverse and the transpose of $B$ I didn't get the same result... Why?? (Wondering)

 

[I like Serena]

For an orthonormal matrix $P$ it stands that $P^{T}=P^{-1}$, right?? (Wondering)

In Matlab I calculated the inverse and the transpose matrix of $V$ and the result was the same matrix...

When I calculated in Matlab the inverse and the transpose of $B$ I didn't get the same result... Why?? (Wondering)

A matrix is orthogonal iff all column vectors have unit length and are orthogonal to each other.

Turns out that your eigenvectors happen to be orthogonal.
As a result your $B$ and $V$ will be orthogonal matrices if they contain eigenvectors with unit length. (Thinking)

It appears your matrix $B$ contains a mistake though. (Worried)
I think the bottom rightmost entry should be $-\frac{\sqrt 3}{3} \approx -0.5774$, which is what you have for the corresponding eigenvector in $V$.

 

[mathmari]

You are right!

Finally I found the correct solution of the system! (Dance)

I am asked to find the values of $x$, $y$ and $z$ when the system is in steady state..

I found that the values are $$\begin{pmatrix} \frac{545}{11} \\ \frac{645}{11} \\ \frac{515}{11}\end{pmatrix}$$

After that I am asked to find in how many hours the system reaches the steady state... How do we find that?? (Wondering)

 

[I like Serena]

You are right!

Finally I found the correct solution of the system! (Dance)

Congratulations! (Party)

I am asked to find the values of $x$, $y$ and $z$ when the system is in steady state..

I found that the values are $$\begin{pmatrix} \frac{545}{11} \\ \frac{645}{11} \\ \frac{515}{11}\end{pmatrix}$$

After that I am asked to find in how many hours the system reaches the steady state... How do we find that?? (Wondering)

Steady state is when $X'=0$, or equivalently, when $Y'=BX'=0$.
Since $Y'=DY+B^{-1}C$, it follows that in steady state we have $DY+B^{-1}C = 0 \Rightarrow Y=D^{-1}B^{-1}C$.
Since you've solved $Y$, how about filling in its solution? (Thinking)

 

[mathmari]

To find the values of $(x,y,z)$ of my previous post I did the following:

$X'=0 \Rightarrow AX+C=0 \Rightarrow AX=-C \Rightarrow X=A^{-1}(-C)$.

This is the same way as you said, without setting $X=BY$, right?? (Wondering)

Since these values are the constant term of the solution of the system, that means that the terms $e^{\text{ constant }t}$ should be $0$, right?? (Wondering)

 

[mathmari]

The initial value problem is:

$$x' = m_0(45 − x) + m_1(y − x) \\
y' = m_1(x − y) + m_2(35 − y) + m_3(z − y) + 20 \\
z' = m_3(y − z) + m_4(35 − z)$$

where $m_0=\frac{1}{2}, m_1=\frac{1}{4}, m_2=\frac{1}{2}, m_3=\frac{1}{2}, m_4=\frac{1}{2}$. I am also asked to find the values of $m_1$ so that the value of $z$ is greater that $x$ in the steady state... How do we find these values?? (Wondering)

 

[I like Serena]

To find the values of $(x,y,z)$ of my previous post I did the following:

$X'=0 \Rightarrow AX+C=0 \Rightarrow AX=-C \Rightarrow X=A^{-1}(-C)$.

This is the same way as you said, without setting $X=BY$, right?? (Wondering)

Since these values are the constant term of the solution of the system, that means that the terms $e^{\text{ constant }t}$ should be $0$, right?? (Wondering)

I believe you're supposed to find the solution for $X = X(t)$, which contains factors $e^{\lambda t}$.
Those terms with $e^{\lambda t}$ will not and cannot be zero. (Wasntme)

The initial value problem is:

$$x' = m_0(45 − x) + m_1(y − x) \\
y' = m_1(x − y) + m_2(35 − y) + m_3(z − y) + 20 \\
z' = m_3(y − z) + m_4(35 − z)$$

where $m_0=\frac{1}{2}, m_1=\frac{1}{4}, m_2=\frac{1}{2}, m_3=\frac{1}{2}, m_4=\frac{1}{2}$.

I am also asked to find the values of $m_1$ so that the value of $z$ is greater that $x$ in the steady state... How do we find these values?? (Wondering)

This is again a problem of the form $X'=AX+C$.
Which steady state will it have? (Wondering)

 

[mathmari]

We have

$$A=\begin{pmatrix}-(m_0+m_1) & m_1 & 0 \\ m_1 & -(m_1+m_2+m_3) & m_3 \\ 0 & m_3 & -(m_3+m_4)\end{pmatrix}$$

$$C=\begin{pmatrix}45m_0\\20\\35m_4\end{pmatrix}$$

So, we have to solve $X=-A^{-1}C$ subject to $z>x$, right?? (Wondering)

- - - Updated - - -

I believe you're supposed to find the solution for $X = X(t)$, which contains factors $e^{\lambda t}$.
Those terms with $e^{\lambda t}$ will not and cannot be zero. (Wasntme)

We will get three relations of the form $e^{\lambda t}=0$, right?? What does this mean?? (Wondering)

 

[I like Serena]

We have

$$A=\begin{pmatrix}-(m_0+m_1) & m_1 & 0 \\ m_1 & -(m_1+m_2+m_3) & m_3 \\ 0 & m_3 & -(m_3+m_4)\end{pmatrix}$$

$$C=\begin{pmatrix}45m_0\\20\\35m_4\end{pmatrix}$$

So, we have to solve $X=-A^{-1}C$ subject to $z>x$, right?? (Wondering)

Yep. (Smile)

- - - Updated - - -

We will get three relations of the form $e^{\lambda t}=0$, right?? What does this mean?? (Wondering)

I'm not so sure that's what we will get. How did you get it? (Wondering)
If we would, there would be no solution. (Wasntme)

 

[mathmari]

Yep. (Smile)

How can we find the inverse of $A$?? I got stuck right now...

 

[mathmari]

I'm not so sure that's what we will get. How did you get it? (Wondering)
If we would, there would be no solution. (Wasntme)

$\lambda$ is negative... So, that means that $t \rightarrow +\infty$, or not?? (Wondering)

 

[I like Serena]

$\lambda$ is negative... So, that means that $t \rightarrow +\infty$, or not?? (Wondering)

If $e^{\lambda t} = 0$. What is the reason you think it is? (Wondering)