Need explanation-hyperbolic system of equations

[mathmari]

Hey!

I have the following in my notes:

$$u_t+A(x,t,u)u_x=b(x,t,u) \ \ \ \ \ \ \ \ \ \ (1)$$
$$u=(u_1, \dots, u_n), b=(b_1, \dots, b_n)$$
$$A=[a_{ij}], i,j = 1, \dots, n$$

We set the question if there are characteristic directions at the path of which the PDE system $(1)$ is reduced to an ODE system.

So we take linear combinations of the above equations.

We use the vector $\gamma=(\gamma_1, \dots, \gamma_n) \ \ \ \ \ \ \ \ \ \ (2)$.

We take $\gamma^T(u_t+Au_x)=\gamma^Tb \ \ \ \ \ \ \ \ \ \ (3)$.

We want to conclude to a form of total derivative of the linear combination of $u$.

That means:(we consider that $u_j=u_j(x(t),t)$)

$$\frac{d}{dt}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)=\frac{\partial}{\partial{t}}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)+\frac{dx}{dt} \frac{\partial}{\partial{x}}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)$$

and we define $\frac{dx}{dt}=\lambda$

The relation above means:
$$m^T(\frac{\partial{u}}{\partial{t}}+\lambda\frac{\partial{u}}{\partial{x}})=\gamma^Tb \ \ \ \ \ \ \ \ \ \ (4)$$

From the relation $(3)$ and $(4)$, we see that $\gamma=m=(\gamma_1, \dots, \gamma_n)$ and that $\gamma^TA=\lambda \gamma^T$

$\lambda:$ eigenvalue of the matrix $A$

$\gamma^T:$ left eigenvector of $A$

We define the characteristic directions:

$\frac{dx}{dt}=\lambda$, so $\gamma^T(\frac{\partial{u}}{\partial{t}}+\frac{dx}{dt}
\frac{\partial{u}}{\partial{x}})=\gamma^Tb$

$\gamma^T \frac{du}{dt}=\gamma^Tb$

(To can reduce the system to a PDE system, the $\frac{dx}{dt}=\lambda$ must exist. To be able to apply this method, the matrix $A$ should have $n$ real values.)

We conclude that to reduce the PDE system to an ODE system, the matrix $A$ should have $n$ discrete real eigenvalues.

In this case we say that the system is hyperbolic.
Could you explain me what we have done here? I got stuck right now... (Wondering)

 

[mathmari]

Hey!

I have the following in my notes:

$$u_t+A(x,t,u)u_x=b(x,t,u) \ \ \ \ \ \ \ \ \ \ (1)$$
$$u=(u_1, \dots, u_n), b=(b_1, \dots, b_n)$$
$$A=[a_{ij}], i,j = 1, \dots, n$$

We set the question if there are characteristic directions at the path of which the PDE system $(1)$ is reduced to an ODE system.

So we take linear combinations of the above equations.

We use the vector $\gamma=(\gamma_1, \dots, \gamma_n) \ \ \ \ \ \ \ \ \ \ (2)$.

We take $\gamma^T(u_t+Au_x)=\gamma^Tb \ \ \ \ \ \ \ \ \ \ (3)$.

We want to conclude to a form of total derivative of the linear combination of $u$.

That means:(we consider that $u_j=u_j(x(t),t)$)

$$\frac{d}{dt}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)=\frac{\partial}{\partial{t}}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)+\frac{dx}{dt} \frac{\partial}{\partial{x}}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)$$

and we define $\frac{dx}{dt}=\lambda$

The relation above means:
$$m^T(\frac{\partial{u}}{\partial{t}}+\lambda\frac{\partial{u}}{\partial{x}})=\gamma^Tb \ \ \ \ \ \ \ \ \ \ (4)$$

From the relation $(3)$ and $(4)$, we see that $\gamma=m=(\gamma_1, \dots, \gamma_n)$ and that $\gamma^TA=\lambda \gamma^T$

$\lambda:$ eigenvalue of the matrix $A$

$\gamma^T:$ left eigenvector of $A$

We define the characteristic directions:

$\frac{dx}{dt}=\lambda$, so $\gamma^T(\frac{\partial{u}}{\partial{t}}+\frac{dx}{dt}
\frac{\partial{u}}{\partial{x}})=\gamma^Tb$

$\gamma^T \frac{du}{dt}=\gamma^Tb$

(To can reduce the system to a PDE system, the $\frac{dx}{dt}=\lambda$ must exist. To be able to apply this method, the matrix $A$ should have $n$ real values.)

We conclude that to reduce the PDE system to an ODE system, the matrix $A$ should have $n$ discrete real eigenvalues.

In this case we say that the system is hyperbolic.
Could you explain me what we have done here? I got stuck right now... (Wondering)

What are characteristic directions at the path of which the PDE system $(1)$ is reduced to an ODE system? (Wondering)

 

[I like Serena]

Hey! (Blush)

We set the question if there are characteristic directions at the path of which the PDE system $(1)$ is reduced to an ODE system.

So we take linear combinations of the above equations.

We use the vector $\gamma=(\gamma_1, \dots, \gamma_n) \ \ \ \ \ \ \ \ \ \ (2)$.

We take $\gamma^T(u_t+Au_x)=\gamma^Tb \ \ \ \ \ \ \ \ \ \ (3)$.

We want to conclude to a form of total derivative of the linear combination of $u$.

Can it be that it should be: "linear combination of the total derivative of $u$ with respect to $t$"?

That means:(we consider that $u_j=u_j(x(t),t)$)

$$\frac{d}{dt}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)=\frac{\partial}{\partial{t}}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)+\frac{dx}{dt} \frac{\partial}{\partial{x}}(\gamma_1u_1+\gamma_2u_2 +\dots+\gamma_n u_n)$$

and we define $\frac{dx}{dt}=\lambda$

The relation above means:
$$m^T(\frac{\partial{u}}{\partial{t}}+\lambda\frac{\partial{u}}{\partial{x}})=\gamma^Tb \ \ \ \ \ \ \ \ \ \ (4)$$

It seems the assumption is made that (3) can be written as an ODE that is a linear combination of $\frac{du_j}{dt}$. (Wondering)

That would imply there is some vector $m$ such that:
$$m^T \frac{du}{dt} = g(t)$$
where $g(t) = \gamma^T b(t)$.

From the relation $(3)$ and $(4)$, we see that $\gamma=m=(\gamma_1, \dots, \gamma_n)$ and that $\gamma^TA=\lambda \gamma^T$

$\lambda:$ eigenvalue of the matrix $A$

$\gamma^T:$ left eigenvector of $A$

I think I'm seeing another assumption here.
That from $\gamma^T x_t(t) u_x(t) = \gamma^T A(t) u_x(t)$ we can assume that $\gamma^T x_t(t) = \gamma^T A(t)$. (Wondering)

If that holds true, we get that the scalar $x_t(t)$ is an eigenvalue of $A(t)$.

We define the characteristic directions:

$\frac{dx}{dt}=\lambda$, so $\gamma^T(\frac{\partial{u}}{\partial{t}}+\frac{dx}{dt}
\frac{\partial{u}}{\partial{x}})=\gamma^Tb$

$\gamma^T \frac{du}{dt}=\gamma^Tb$

(To can reduce the system to a PDE system, the $\frac{dx}{dt}=\lambda$ must exist. To be able to apply this method, the matrix $A$ should have $n$ real values.)

We conclude that to reduce the PDE system to an ODE system, the matrix $A$ should have $n$ discrete real eigenvalues.

In this case we say that the system is hyperbolic.

If the assumptions hold and $A(t)$ is diagonalizable, we should be able to apply a change of variables and reduce the system to something like:
$$\delta^T(v_t + D(t) v_x) = \delta^T c(t)$$
where $\delta$ is some constant vector just like $\gamma$.

Perhaps this is considered a hyperbolic form? (Wondering)

What are characteristic directions at the path of which the PDE system $(1)$ is reduced to an ODE system? (Wondering)

I have just found Method_of_characteristics on wiki that seems to describe this method.

 

[mathmari]

Can it be that it should be: "linear combination of the total derivative of $u$ with respect to $t$"?

In my notes, it is as I wrote it:
"We want to conclude to a form of total derivative of the linear combination of $u$."

Does it mean that we are looking for the derivative of $(\gamma_1 u_1 +\gamma_2 u_2+ \dots +\gamma_n u_n)$ ??
What does the part "a form of total derivative" mean?? (Wondering)

It seems the assumption is made that (3) can be written as an ODE that is a linear combination of $\frac{du_j}{dt}$. (Wondering)

That would imply there is some vector $m$ such that:
$$m^T \frac{du}{dt} = g(t)$$
where $g(t) = \gamma^T b(t)$.

I haven't understood this part... (Doh) (Worried)

I think I'm seeing another assumption here.
That from $\gamma^T x_t(t) u_x(t) = \gamma^T A(t) u_x(t)$ we can assume that $\gamma^T x_t(t) = \gamma^T A(t)$. (Wondering)

If that holds true, we get that the scalar $x_t(t)$ is an eigenvalue of $A(t)$.

$(3): \gamma^T(u_t+Au_x)= \gamma^Tb$
$(4): m^T(u_t+\lambda u_x)=\gamma^Tb$

$\Rightarrow \gamma^T(u_t+Au_x)=m^T(u_t+\lambda u_x) \Rightarrow \gamma^T u_t+\gamma^T A u_x=m^T u_t+m^T \lambda u_x \Rightarrow \gamma^T=m^T \text{ AND } \gamma^T A= m^T \lambda \overset{\gamma^T=m^T}{\Rightarrow} \gamma^T A=\lambda \gamma^T$$\gamma^T=m^T \Rightarrow \gamma=m$

Is this correct?? (Thinking)

Does it stand that $m^T \lambda=\lambda m^T$ because $\lambda$ is a number?? (Wondering)

Also how do we conlcude that
$\lambda:$ eigenvalue of the matrix $A$
and
$\gamma^T:$ left eigenvector of $A$
??

Do we suppose that these two hold??

If the assumptions hold and $A(t)$ is diagonalizable, we should be able to apply a change of variables and reduce the system to something like:
$$\delta^T(v_t + D(t) v_x) = \delta^T c(t)$$
where $\delta$ is some constant vector just like $\gamma$.

Perhaps this is considered a hyperbolic form? (Wondering)

So that the eigenvalues exist means that the matrix has $n$ real values?? (Wondering)
And is that equivalent to that the matrix $A$ is diagonalizable??
Why?? (Sweating)

And also, why do we consider this a hyperbolic form??

 

[mathmari]

There is the following example in my notes:

$$\left.\begin{matrix}
\frac{\partial{u_1}}{\partial{t}}+a \frac{\partial{u_2}}{\partial{x}}=0\\
\frac{\partial{u_2}}{\partial{t}}+b \frac{\partial{u_1}}{\partial{x}}=0
\end{matrix}\right\}$$

$a,b>0$
$\displaystyle{u=u(u_1, u_2)}$

$$A=\begin{bmatrix}
0 & a\\
b & 0
\end{bmatrix}$$

$\displaystyle{u_t+Au_x=0}$

To check if it is hypebolic, we have to find the eigenvalues.
$\displaystyle{det(A-\lambda I)=\lambda^2-ab=0 \Rightarrow \lambda= \pm \sqrt{ab}}$

$\gamma=(\gamma_1, \gamma_2)$

$\displaystyle{\gamma_1(\frac{\partial{u_1}}{\partial{t}}+a \frac{\partial{u_2}}{\partial{x}})+\gamma_2(\frac{\partial{u_2}}{\partial{t}}+b \frac{\partial{u_1}}{\partial{x}})=0}$

$\Rightarrow \displaystyle{\gamma_1(\frac{\partial{u_1}}{\partial{t}}+\frac{b \gamma_2}{\gamma_1} \frac{\partial{u_1}}{\partial{x}})+\gamma_2(\frac{\partial{u_2}}{\partial{t}}+\frac{a \gamma_1}{\gamma_2} \frac{\partial{u_2}}{\partial{x}})=0}$

Since it should be in the form: $\gamma^T(u_t+\lambda u_x)=\gamma^T b$, it should be: $\displaystyle{\frac{b \gamma_2}{\gamma_1}=\frac{a \gamma_1}{\gamma_2}}$

We set $\displaystyle{\gamma_1=1}$, so $\displaystyle{\gamma_2^2=\frac{a}{b} \Rightarrow \gamma_2= \pm \sqrt{\frac{a}{b}}}$

We take $\gamma_2=\sqrt{\frac{a}{b}}$:

At the case $\lambda=\sqrt{ab}$:

$\displaystyle{(\frac{\partial{u_1}}{\partial{t}}+\sqrt{ab} \frac{\partial{u_1}}{\partial{x}})+\sqrt{\frac{a}{b}}(\frac{\partial{u_2}}{\partial{t}}+\sqrt{ab} \frac{\partial{u_2}}{\partial{x}})=0}$

$\displaystyle{\frac{\partial}{\partial{t}}(u_1+\sqrt{\frac{a}{b}}u_2)+\sqrt{ab} \frac{\partial}{\partial{x}}(u_1+\sqrt{\frac{a}{b}}u_2)=0}$

We set $\displaystyle{v_+=u_1+\sqrt{\frac{a}{b}}u_2}$. So we have: $\displaystyle{\frac{\partial{v_+}}{\partial{t}}+\sqrt{ab} \frac{\partial{v_+}}{\partial{x}}=0}$

The characteristic system is:
$$\frac{dt}{1}=\frac{dx}{\sqrt{ab}}=\frac{dv_+}{0}$$

$\displaystyle{v=\text{ constant }, v=u_1+\sqrt{\frac{a}{b}}u_2}$, when $\displaystyle{\frac{dx}{dt}=\sqrt{ab}}$

$\displaystyle{x=\sqrt{ab}t+c \Rightarrow c=x- \sqrt{ab}t}$

$\displaystyle{v}$ is constant when $\displaystyle{x-\sqrt{ab}t}$ is constant.

That means that $$u_1+\sqrt{\frac{a}{b}}u_2=\frac{f(x-\sqrt{ab}t)}{2}$$At the case $\lambda=-\sqrt{ab}$:

We conclude that $\displaystyle{v_-=u_1-\sqrt{\frac{a}{b}}u_2= \text{ constant }}$ when $\displaystyle{x+\sqrt{ab}t}$ is constant.

$$u_1-\sqrt{\frac{a}{b}}u_2=\frac{g(x+\sqrt{ab}t)}{2}$$So $$u_1=f(x-\sqrt{ab}t)+g(x+\sqrt{ab}t)$$
$$u_2=\sqrt{\frac{b}{a}}[f(x-\sqrt{ab}t)-g(x+\sqrt{ab}t)]$$

Instead of $\gamma_2=\sqrt{\frac{a}{b}}$, we could also use $\gamma_2=-\sqrt{\frac{a}{b}}$ and we would get the same results, right?? (Wondering)

 

[mathmari]

It seems the assumption is made that (3) can be written as an ODE that is a linear combination of $\frac{du_j}{dt}$. (Wondering)

That would imply there is some vector $m$ such that:
$$m^T \frac{du}{dt} = g(t)$$
where $g(t) = \gamma^T b(t)$.

I still don't understand what $m$ is... (Doh)(Worried)(Wondering)

 

[I like Serena]

I still don't understand what $m$ is... (Doh)(Worried)(Wondering)

As far as I can tell $m$ is some vector of constants.
It would identify a linear combination.
That's all I've got. (Blush)