Qualitative dependence of solution to second-order matrix differential equation on eigenvalues
$begingroup$
Suppose we have a matrix differential equation in $vec{x}(t)=left(begin{smallmatrix}x_{1}(t) \ vdots \ x_{n}(t)end{smallmatrix}right)$, such that:
$$frac{mathrm{d}^{2}vec{x}}{mathrm{d}t^{2}}=-mathbf{M}vec{x}$$
And $mathbf{M}$ is a real, symmetric matrix, which therefore diagonalizes to $mathbf{M}=mathbf{P}mathbf{Lambda}mathbf{P}^{-1}$, with $mathbf{Lambda}=operatorname{diag}(lambda_{1},dots,lambda_{n})$. We can perform a change of basis such that $vec{x}(t)=mathbf{P}vec{eta}(t)$, and therefore we have the following:
$$frac{mathrm{d}^{2}vec{eta}}{mathrm{d}t^{2}}=-mathbf{Lambda}vec{eta} implies frac{mathrm{d}^{2}eta_{i}}{mathrm{d}t^{2}}=-lambda_{i}eta_{i}$$
This is easy to solve and gives us:
$$eta_{i}=begin{cases}A_{i}cos(2sqrt{lambda_{i}}t)+B_{i}sin(2sqrt{lambda_{i}}t) & lambda_{i} > 0 \
A_{i}t+B_{i} & lambda_{i} = 0 \
A_{i}e^{2sqrt{lambda_{i}}t}+B_{i}e^{-2sqrt{lambda_{i}}t} & lambda_{i} < 0end{cases}$$
We can therefore solve for $vec{x}$ by computing $mathbf{P}vec{eta}$. However, I am then asked how the qualitative behaviour of $vec{x}$ depends upon the eigenvalues of $mathbf{M}$. I am having trouble understanding what is required of me; we have the qualitative behaviour of $eta_{i}$ and it's dependence upon $lambda_{i}$, but without more explicit knowledge of the change of basis matrix $mathbf{P}$ I'm not sure what we can say about $vec{x}$?
The question then requires us to focus on the 2-dimensional coupled system of differential equations:
begin{align}
frac{mathrm{d}^{2}x_{1}}{mathrm{d}t^{2}}&=-kx_{1}-lx_{2} \
frac{mathrm{d}^{2}x_{2}}{mathrm{d}t^{2}}&= -lx_{1} - kx_{2}
end{align}
We can therefore construct $mathbf{M}=left(begin{smallmatrix}k & l \ l & kend{smallmatrix}right)$, which diagonalizes to give $mathbf{Lambda}=operatorname{diag}(k-l,k+l)$ and $mathbf{P}=frac{1}{sqrt{2}}left(begin{smallmatrix}-1 & 1 \ 1 & 1end{smallmatrix}right)$. Now clearly the solution can be worked out explicitly using the method above and then the qualitative behaviour can be explored, but is there anything we can say without explicit calculation?
calculus linear-algebra ordinary-differential-equations systems-of-equations
asked Dec 28 '13 at 17:05
Thomas RussellThomas Russell
7,81632551
$endgroup$
add a comment |
$begingroup$
Suppose we have a matrix differential equation in $vec{x}(t)=left(begin{smallmatrix}x_{1}(t) \ vdots \ x_{n}(t)end{smallmatrix}right)$, such that:
$$frac{mathrm{d}^{2}vec{x}}{mathrm{d}t^{2}}=-mathbf{M}vec{x}$$
And $mathbf{M}$ is a real, symmetric matrix, which therefore diagonalizes to $mathbf{M}=mathbf{P}mathbf{Lambda}mathbf{P}^{-1}$, with $mathbf{Lambda}=operatorname{diag}(lambda_{1},dots,lambda_{n})$. We can perform a change of basis such that $vec{x}(t)=mathbf{P}vec{eta}(t)$, and therefore we have the following:
$$frac{mathrm{d}^{2}vec{eta}}{mathrm{d}t^{2}}=-mathbf{Lambda}vec{eta} implies frac{mathrm{d}^{2}eta_{i}}{mathrm{d}t^{2}}=-lambda_{i}eta_{i}$$
This is easy to solve and gives us:
$$eta_{i}=begin{cases}A_{i}cos(2sqrt{lambda_{i}}t)+B_{i}sin(2sqrt{lambda_{i}}t) & lambda_{i} > 0 \
A_{i}t+B_{i} & lambda_{i} = 0 \
A_{i}e^{2sqrt{lambda_{i}}t}+B_{i}e^{-2sqrt{lambda_{i}}t} & lambda_{i} < 0end{cases}$$
We can therefore solve for $vec{x}$ by computing $mathbf{P}vec{eta}$. However, I am then asked how the qualitative behaviour of $vec{x}$ depends upon the eigenvalues of $mathbf{M}$. I am having trouble understanding what is required of me; we have the qualitative behaviour of $eta_{i}$ and it's dependence upon $lambda_{i}$, but without more explicit knowledge of the change of basis matrix $mathbf{P}$ I'm not sure what we can say about $vec{x}$?
The question then requires us to focus on the 2-dimensional coupled system of differential equations:
begin{align}
frac{mathrm{d}^{2}x_{1}}{mathrm{d}t^{2}}&=-kx_{1}-lx_{2} \
frac{mathrm{d}^{2}x_{2}}{mathrm{d}t^{2}}&= -lx_{1} - kx_{2}
end{align}
We can therefore construct $mathbf{M}=left(begin{smallmatrix}k & l \ l & kend{smallmatrix}right)$, which diagonalizes to give $mathbf{Lambda}=operatorname{diag}(k-l,k+l)$ and $mathbf{P}=frac{1}{sqrt{2}}left(begin{smallmatrix}-1 & 1 \ 1 & 1end{smallmatrix}right)$. Now clearly the solution can be worked out explicitly using the method above and then the qualitative behaviour can be explored, but is there anything we can say without explicit calculation?
calculus linear-algebra ordinary-differential-equations systems-of-equations
asked Dec 28 '13 at 17:05
Thomas RussellThomas Russell
7,81632551
$endgroup$
1
$begingroup$
Are you sure there is more to do than you already did? You have that, depending on the occurrence of positive, negative or zero eigenvalues, the solutions are bounded for all times or exponentially resp. linearly growing in at least one time direction.
$endgroup$
– LutzL
Dec 28 '13 at 20:36
add a comment |
$begingroup$
Suppose we have a matrix differential equation in $vec{x}(t)=left(begin{smallmatrix}x_{1}(t) \ vdots \ x_{n}(t)end{smallmatrix}right)$, such that:
$$frac{mathrm{d}^{2}vec{x}}{mathrm{d}t^{2}}=-mathbf{M}vec{x}$$
And $mathbf{M}$ is a real, symmetric matrix, which therefore diagonalizes to $mathbf{M}=mathbf{P}mathbf{Lambda}mathbf{P}^{-1}$, with $mathbf{Lambda}=operatorname{diag}(lambda_{1},dots,lambda_{n})$. We can perform a change of basis such that $vec{x}(t)=mathbf{P}vec{eta}(t)$, and therefore we have the following:
$$frac{mathrm{d}^{2}vec{eta}}{mathrm{d}t^{2}}=-mathbf{Lambda}vec{eta} implies frac{mathrm{d}^{2}eta_{i}}{mathrm{d}t^{2}}=-lambda_{i}eta_{i}$$
This is easy to solve and gives us:
$$eta_{i}=begin{cases}A_{i}cos(2sqrt{lambda_{i}}t)+B_{i}sin(2sqrt{lambda_{i}}t) & lambda_{i} > 0 \
A_{i}t+B_{i} & lambda_{i} = 0 \
A_{i}e^{2sqrt{lambda_{i}}t}+B_{i}e^{-2sqrt{lambda_{i}}t} & lambda_{i} < 0end{cases}$$
We can therefore solve for $vec{x}$ by computing $mathbf{P}vec{eta}$. However, I am then asked how the qualitative behaviour of $vec{x}$ depends upon the eigenvalues of $mathbf{M}$. I am having trouble understanding what is required of me; we have the qualitative behaviour of $eta_{i}$ and it's dependence upon $lambda_{i}$, but without more explicit knowledge of the change of basis matrix $mathbf{P}$ I'm not sure what we can say about $vec{x}$?
The question then requires us to focus on the 2-dimensional coupled system of differential equations:
begin{align}
frac{mathrm{d}^{2}x_{1}}{mathrm{d}t^{2}}&=-kx_{1}-lx_{2} \
frac{mathrm{d}^{2}x_{2}}{mathrm{d}t^{2}}&= -lx_{1} - kx_{2}
end{align}
We can therefore construct $mathbf{M}=left(begin{smallmatrix}k & l \ l & kend{smallmatrix}right)$, which diagonalizes to give $mathbf{Lambda}=operatorname{diag}(k-l,k+l)$ and $mathbf{P}=frac{1}{sqrt{2}}left(begin{smallmatrix}-1 & 1 \ 1 & 1end{smallmatrix}right)$. Now clearly the solution can be worked out explicitly using the method above and then the qualitative behaviour can be explored, but is there anything we can say without explicit calculation?
calculus linear-algebra ordinary-differential-equations systems-of-equations
asked Dec 28 '13 at 17:05
Thomas RussellThomas Russell
7,81632551
$endgroup$
Suppose we have a matrix differential equation in $vec{x}(t)=left(begin{smallmatrix}x_{1}(t) \ vdots \ x_{n}(t)end{smallmatrix}right)$, such that:
$$frac{mathrm{d}^{2}vec{x}}{mathrm{d}t^{2}}=-mathbf{M}vec{x}$$
And $mathbf{M}$ is a real, symmetric matrix, which therefore diagonalizes to $mathbf{M}=mathbf{P}mathbf{Lambda}mathbf{P}^{-1}$, with $mathbf{Lambda}=operatorname{diag}(lambda_{1},dots,lambda_{n})$. We can perform a change of basis such that $vec{x}(t)=mathbf{P}vec{eta}(t)$, and therefore we have the following:
$$frac{mathrm{d}^{2}vec{eta}}{mathrm{d}t^{2}}=-mathbf{Lambda}vec{eta} implies frac{mathrm{d}^{2}eta_{i}}{mathrm{d}t^{2}}=-lambda_{i}eta_{i}$$
This is easy to solve and gives us:
$$eta_{i}=begin{cases}A_{i}cos(2sqrt{lambda_{i}}t)+B_{i}sin(2sqrt{lambda_{i}}t) & lambda_{i} > 0 \
A_{i}t+B_{i} & lambda_{i} = 0 \
A_{i}e^{2sqrt{lambda_{i}}t}+B_{i}e^{-2sqrt{lambda_{i}}t} & lambda_{i} < 0end{cases}$$
We can therefore solve for $vec{x}$ by computing $mathbf{P}vec{eta}$. However, I am then asked how the qualitative behaviour of $vec{x}$ depends upon the eigenvalues of $mathbf{M}$. I am having trouble understanding what is required of me; we have the qualitative behaviour of $eta_{i}$ and it's dependence upon $lambda_{i}$, but without more explicit knowledge of the change of basis matrix $mathbf{P}$ I'm not sure what we can say about $vec{x}$?
The question then requires us to focus on the 2-dimensional coupled system of differential equations:
begin{align}
frac{mathrm{d}^{2}x_{1}}{mathrm{d}t^{2}}&=-kx_{1}-lx_{2} \
frac{mathrm{d}^{2}x_{2}}{mathrm{d}t^{2}}&= -lx_{1} - kx_{2}
end{align}
We can therefore construct $mathbf{M}=left(begin{smallmatrix}k & l \ l & kend{smallmatrix}right)$, which diagonalizes to give $mathbf{Lambda}=operatorname{diag}(k-l,k+l)$ and $mathbf{P}=frac{1}{sqrt{2}}left(begin{smallmatrix}-1 & 1 \ 1 & 1end{smallmatrix}right)$. Now clearly the solution can be worked out explicitly using the method above and then the qualitative behaviour can be explored, but is there anything we can say without explicit calculation?
calculus linear-algebra ordinary-differential-equations systems-of-equations
calculus linear-algebra ordinary-differential-equations systems-of-equations
asked Dec 28 '13 at 17:05
Thomas RussellThomas Russell
7,81632551
asked Dec 28 '13 at 17:05
Thomas RussellThomas Russell
7,81632551
edited Feb 12 '14 at 16:46
Thomas Russell
asked Dec 28 '13 at 17:05
Thomas RussellThomas Russell
7,81632551
asked Dec 28 '13 at 17:05
Thomas RussellThomas Russell
7,81632551
asked Dec 28 '13 at 17:05
Thomas RussellThomas Russell
7,81632551
7,81632551
1
$begingroup$
Are you sure there is more to do than you already did? You have that, depending on the occurrence of positive, negative or zero eigenvalues, the solutions are bounded for all times or exponentially resp. linearly growing in at least one time direction.
$endgroup$
– LutzL
Dec 28 '13 at 20:36
add a comment |
1
$begingroup$
Are you sure there is more to do than you already did? You have that, depending on the occurrence of positive, negative or zero eigenvalues, the solutions are bounded for all times or exponentially resp. linearly growing in at least one time direction.
$endgroup$
– LutzL
Dec 28 '13 at 20:36
1
1
$begingroup$
Are you sure there is more to do than you already did? You have that, depending on the occurrence of positive, negative or zero eigenvalues, the solutions are bounded for all times or exponentially resp. linearly growing in at least one time direction.
$endgroup$
– LutzL
Dec 28 '13 at 20:36
$begingroup$
Are you sure there is more to do than you already did? You have that, depending on the occurrence of positive, negative or zero eigenvalues, the solutions are bounded for all times or exponentially resp. linearly growing in at least one time direction.
$endgroup$
– LutzL
Dec 28 '13 at 20:36
add a comment |
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$begingroup$
Are you sure there is more to do than you already did? You have that, depending on the occurrence of positive, negative or zero eigenvalues, the solutions are bounded for all times or exponentially resp. linearly growing in at least one time direction.
$endgroup$
– LutzL
Dec 28 '13 at 20:36