Coupled differential equations using matrices

[Marcus95]

Homework Statement

We can treat the following coupled system of differential equations as an eigenvalue
problem:
$2 \frac{dy_1}{dt} = 2f_1 - 3y_1 + y_2$
$2\frac{dy_2}{dt} = 2f_2 + y_1 -3y_2$
$\frac{dy_3}{dt} = f_3 - 4y_3$

where f1, f2 and f3 is a set of time-dependent sources, and y1, y2 and y3 is a set of
time-dependent responses.
(a) If these equations are written using matrix notation,
$\frac{d\vec{y}}{dt} + K \vec{y} = \vec{f}$

what are the elements of K? Find the eigenvalues and eigenvectors of K.

(b) In the case when the system is not excited, f = 0, find all of the solutions having
the form
$\vec{y}=\vec{y}_0 e^{-\gamma t}$

where $\gamma$ > 0 is a decay constant.

(c) If f is held constant at f0, the response vector y has the steady state value y0 (that
is, with $\frac{d\vec{y}}{dt} = 0$). Write down y₀ in terms of f₀, and find y₀ in the case where f₀ = (1, 1, 1)^T .

(d) Assume that y starts in the steady state solution y₀ given in (c) with f₀ = (1, 1, 1)^T . Now suppose the source function abruptly falls to zero, f₀ = (0, 0, 0)^T , so that the response vector y moves away from y₀. Writing y as a linear combination of the allowed solutions found in (b), derive an expression for the subsequent time evolution
of the system.

Homework Equations

Eigenvalue equation: $det(M - \lambda I ) = 0$

The Attempt at a Solution

Part a) was relatively straight forward, by just rearanging and observing I find:
K = ( c₁ c₂ c₃) with c₁ =(3/2, -1/2, 0)^T, c₂= (-1/2, 3/2, 0)^T and c₃=(0, 0, 4)^T.

I also found the eigenvalues to be λ = 1, 2, 4 with eigenvectors (1, 1, 0) , (1, -1, 0) and (0, 0, 1).

However, after this I am completely stuck. I have no idea how to apply this to the differential equations in part b). I can solve the last equation: $y_3 = y_{03} e^{-4t}$, but the first two equations are coupled and I am not sure how to deak with it (with or without matrix notation).

Thank you for any help! :)

 

[Orodruin]

How does the system evolve if it is initially in a state that is proportional to one of the eigenvectors?

Hint: Rewrite the system of equations as a system of equations for the coefficients multiplying the eigenvectors.

 

[Ray Vickson]

Homework Statement

We can treat the following coupled system of differential equations as an eigenvalue
problem:
$2 \frac{dy_1}{dt} = 2f_1 - 3y_1 + y_2$
$2\frac{dy_2}{dt} = 2f_2 + y_1 -3y_2$
$\frac{dy_3}{dt} = f_3 - 4y_3$

where f1, f2 and f3 is a set of time-dependent sources, and y1, y2 and y3 is a set of
time-dependent responses.
(a) If these equations are written using matrix notation,
$\frac{d\vec{y}}{dt} + K \vec{y} = \vec{f}$

what are the elements of K? Find the eigenvalues and eigenvectors of K.

(b) In the case when the system is not excited, f = 0, find all of the solutions having
the form
$\vec{y}=\vec{y}_0 e^{-\gamma t}$

where $\gamma$ > 0 is a decay constant.

(c) If f is held constant at f0, the response vector y has the steady state value y0 (that
is, with $\frac{d\vec{y}}{dt} = 0$). Write down y₀ in terms of f₀, and find y₀ in the case where f₀ = (1, 1, 1)^T .

(d) Assume that y starts in the steady state solution y₀ given in (c) with f₀ = (1, 1, 1)^T . Now suppose the source function abruptly falls to zero, f₀ = (0, 0, 0)^T , so that the response vector y moves away from y₀. Writing y as a linear combination of the allowed solutions found in (b), derive an expression for the subsequent time evolution
of the system.

Homework Equations

Eigenvalue equation: $det(M - \lambda I ) = 0$

The Attempt at a Solution

Part a) was relatively straight forward, by just rearanging and observing I find:
K = ( c₁ c₂ c₃) with c₁ =(3/2, -1/2, 0)^T, c₂= (-1/2, 3/2, 0)^T and c₃=(0, 0, 4)^T.

I also found the eigenvalues to be λ = 1, 2, 4 with eigenvectors (1, 1, 0) , (1, -1, 0) and (0, 0, 1).

However, after this I am completely stuck. I have no idea how to apply this to the differential equations in part b). I can solve the last equation: $y_3 = y_{03} e^{-4t}$, but the first two equations are coupled and I am not sure how to deak with it (with or without matrix notation).

Thank you for any help! :)

Just so you know: you can easily do matrices in LaTeX; there are two styles:
$$(1) \hspace{4mm} A = \pmatrix{3/2&-1/2&0\\-1/2&3/2& 0 \\ 0&0&4}$$
or
$$(2) \hspace{4mm} A = \begin{bmatrix}
3/2&-1/2&0\\-1/2&3/2& 0 \\ 0&0&4
\end{bmatrix}$$
Just right-click on the formulas to see the commands.

 

[Marcus95]

How does the system evolve if it is initially in a state that is proportional to one of the eigenvectors?

Well in that case we have:$\frac{d\vec{y}}{dt} + K \vec{y} =\frac{d\vec{y}}{dt} + \lambda \vec{y} = 0$

so:
$\frac{d\vec{y}}{dt} =- \lambda \vec{y}$
but this only applies to eigenvectors and is hence not of much use?

Hint: Rewrite the system of equations as a system of equations for the coefficients multiplying the eigenvectors.

Well this reminds me of diagonalisation. So if R is the eigen-vector matrix, we have:

$R\frac{d\vec{y}}{dt} = RKR^{-1} R\vec{y}$ and this should give 3 solvable ODEs?

 

[Orodruin]

but this only applies to eigenvectors and is hence not of much use?

On the contrary, is extremely useful as your differential equation is linear. Write the initial condition as a superposition of eigenvectors and profit.

 

[Marcus95]

On the contrary, is extremely useful as your differential equation is linear. Write the initial condition as a superposition of eigenvectors and profit.

I am not entierly sure how that would be done. If we for instance have $\frac{d\vec{y_i}(0)}{dx} = i$ doesn't the matrix K itself change? Or would that be if I have initial conditions for $\vec{y_i}(x)$ itself, and not for the derivative?

 

[Orodruin]

Consider your case (but restricting to two because I want to write less on my phone). The eigenvectors are $\vec y_1 = \vec e_1 + \vec e_2$, $\vec y_2 = \vec e_1 - \vec e_2$. What is the general time dependent linear combination expressed in terms of the eigenvectors? (Write it out explicitly)
What happens when you insert this into the differential equation (and note that the eigenvectors are linearly independent)?

 

[Ray Vickson]

Well in that case we have:$\frac{d\vec{y}}{dt} + K \vec{y} =\frac{d\vec{y}}{dt} + \lambda \vec{y} = 0$

so:
$\frac{d\vec{y}}{dt} =- \lambda \vec{y}$
but this only applies to eigenvectors and is hence not of much use?
Well this reminds me of diagonalisation. So if R is the eigen-vector matrix, we have:

$R\frac{d\vec{y}}{dt} = RKR^{-1} R\vec{y}$ and this should give 3 solvable ODEs?

Using eigenvalues/eigenvectors is very profitable. Suppose your three eigenvectors are $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$. These form a basis for $R^3$, so any initial state can be written in terms of them:
$$\mathbf{x}(0) = \pmatrix{x_{01}\\x_{02}\\x_{03}} = c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + c_3 \mathbf{u}_3,$$
where $c_1, c_2, c_3$ are some computable constants. Then, if the eigenvalues are $\gamma_i, i=1,2,3$ (corresponding to the $\mathbf{u}_i, i=1,2,3$) it is easy to write the evolution of the homogeneous solution:
$$ \mathbf{x}(t) = c_1 \mathbf{u}_1 e^{\gamma_1 t} + c_2 \mathbf{u}_2 e^{\gamma_2 t} + c_3 \mathbf{u}_3 e^{\gamma_3 t}. $$

You can figure out a particular non-homogeneous solution in terms of the $\mathbf{u}_i$ as well.