Mixing
Classifying linear first-order PDE system (elliptic, hyperbolic, or parabolic)
$begingroup$
- Consider the constants
$$begin{aligned}
& (text i.); a_1 = b_1 = a_2 = b_2 = 1 \
& (text {ii}.); a_1 = b_2 = 1, quad b_1 = 0, quad a_2 = -1 \
& (text {iii}.); a_1 = b_1 = b_2 = 1, quad a_2 = -1
end{aligned} tag{3}$$
classify the following system for the constants given above
$$begin{aligned}
& a_1 u_x + a_2 v_y = g_1 \
& b_1 v_x + b_2 u_y = g_2
end{aligned} tag{4}$$
where $g_1$ and $g_2$ are +ive constants.
I have already put the system in the form
$$
Aq_x + Bq_y = C,
$$
with
$$
A =
begin{bmatrix}
a_1 & 0\
0 &b_1
end{bmatrix},quad B=
begin{bmatrix}
0 & a_2\
b_2 & 0
end{bmatrix},quad q=
begin{pmatrix}
u\
v
end{pmatrix}.
$$
But then, I do not know how to continue from here on to find out the eigenvalues due to the term
$$
C =
begin{pmatrix}
g_1\
g_2
end{pmatrix}
$$
that is confusing me.
If $C$ were $0$, then I'd know how to proceed, but unfortunately this is not the case.
Any help would be appreciated.
pde linear-pde
edited Jan 22 at 15:08
Harry49
7,50431341
asked Jan 21 at 21:03
Matias SalgoMatias Salgo
113
$endgroup$
add a comment |
$begingroup$
- Consider the constants
$$begin{aligned}
& (text i.); a_1 = b_1 = a_2 = b_2 = 1 \
& (text {ii}.); a_1 = b_2 = 1, quad b_1 = 0, quad a_2 = -1 \
& (text {iii}.); a_1 = b_1 = b_2 = 1, quad a_2 = -1
end{aligned} tag{3}$$
classify the following system for the constants given above
$$begin{aligned}
& a_1 u_x + a_2 v_y = g_1 \
& b_1 v_x + b_2 u_y = g_2
end{aligned} tag{4}$$
where $g_1$ and $g_2$ are +ive constants.
I have already put the system in the form
$$
Aq_x + Bq_y = C,
$$
with
$$
A =
begin{bmatrix}
a_1 & 0\
0 &b_1
end{bmatrix},quad B=
begin{bmatrix}
0 & a_2\
b_2 & 0
end{bmatrix},quad q=
begin{pmatrix}
u\
v
end{pmatrix}.
$$
But then, I do not know how to continue from here on to find out the eigenvalues due to the term
$$
C =
begin{pmatrix}
g_1\
g_2
end{pmatrix}
$$
that is confusing me.
If $C$ were $0$, then I'd know how to proceed, but unfortunately this is not the case.
Any help would be appreciated.
pde linear-pde
edited Jan 22 at 15:08
Harry49
7,50431341
asked Jan 21 at 21:03
Matias SalgoMatias Salgo
113
$endgroup$
add a comment |
$begingroup$
- Consider the constants
$$begin{aligned}
& (text i.); a_1 = b_1 = a_2 = b_2 = 1 \
& (text {ii}.); a_1 = b_2 = 1, quad b_1 = 0, quad a_2 = -1 \
& (text {iii}.); a_1 = b_1 = b_2 = 1, quad a_2 = -1
end{aligned} tag{3}$$
classify the following system for the constants given above
$$begin{aligned}
& a_1 u_x + a_2 v_y = g_1 \
& b_1 v_x + b_2 u_y = g_2
end{aligned} tag{4}$$
where $g_1$ and $g_2$ are +ive constants.
I have already put the system in the form
$$
Aq_x + Bq_y = C,
$$
with
$$
A =
begin{bmatrix}
a_1 & 0\
0 &b_1
end{bmatrix},quad B=
begin{bmatrix}
0 & a_2\
b_2 & 0
end{bmatrix},quad q=
begin{pmatrix}
u\
v
end{pmatrix}.
$$
But then, I do not know how to continue from here on to find out the eigenvalues due to the term
$$
C =
begin{pmatrix}
g_1\
g_2
end{pmatrix}
$$
that is confusing me.
If $C$ were $0$, then I'd know how to proceed, but unfortunately this is not the case.
Any help would be appreciated.
pde linear-pde
edited Jan 22 at 15:08
Harry49
7,50431341
asked Jan 21 at 21:03
Matias SalgoMatias Salgo
113
$endgroup$
- Consider the constants
$$begin{aligned}
& (text i.); a_1 = b_1 = a_2 = b_2 = 1 \
& (text {ii}.); a_1 = b_2 = 1, quad b_1 = 0, quad a_2 = -1 \
& (text {iii}.); a_1 = b_1 = b_2 = 1, quad a_2 = -1
end{aligned} tag{3}$$
classify the following system for the constants given above
$$begin{aligned}
& a_1 u_x + a_2 v_y = g_1 \
& b_1 v_x + b_2 u_y = g_2
end{aligned} tag{4}$$
where $g_1$ and $g_2$ are +ive constants.
I have already put the system in the form
$$
Aq_x + Bq_y = C,
$$
with
$$
A =
begin{bmatrix}
a_1 & 0\
0 &b_1
end{bmatrix},quad B=
begin{bmatrix}
0 & a_2\
b_2 & 0
end{bmatrix},quad q=
begin{pmatrix}
u\
v
end{pmatrix}.
$$
But then, I do not know how to continue from here on to find out the eigenvalues due to the term
$$
C =
begin{pmatrix}
g_1\
g_2
end{pmatrix}
$$
that is confusing me.
If $C$ were $0$, then I'd know how to proceed, but unfortunately this is not the case.
Any help would be appreciated.
pde linear-pde
pde linear-pde
edited Jan 22 at 15:08
Harry49
7,50431341
asked Jan 21 at 21:03
Matias SalgoMatias Salgo
113
edited Jan 22 at 15:08
Harry49
7,50431341
asked Jan 21 at 21:03
Matias SalgoMatias Salgo
113
edited Jan 22 at 15:08
Harry49
7,50431341
edited Jan 22 at 15:08
Harry49
7,50431341
edited Jan 22 at 15:08
Harry49
7,50431341
7,50431341
asked Jan 21 at 21:03
Matias SalgoMatias Salgo
113
asked Jan 21 at 21:03
Matias SalgoMatias Salgo
113
asked Jan 21 at 21:03
Matias SalgoMatias Salgo
113
113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us differentiate the first equation w.r.t. $y$ and the second one w.r.t. $x$:
begin{aligned}
a_1 u_{xy} + a_2 v_{yy} &= 0 \
b_1 v_{xx} + b_2 u_{yx} &= 0
end{aligned}
Using the equality of mixed derivatives for $u$ and the fact that $a_1 = b_2 = 1 neq 0$, we obtain the equations
$$
u_{xy} = u_{yx} = -a_2/a_1 v_{yy} = -b_1/b_2 v_{xx}
$$
Since $a_2neq 0$, we are left with the second-order linear PDE $A v_{xx} + 2 B v_{xy} + Cv_{yy} = 0$ satisfied by $v$, with $A = frac{a_1 b_1}{a_2 b_2}$, $B = 0$ and $C = -1$.
- The case (i) is hyperbolic since $B^2 - A C > 0$.
- The case (ii) is parabolic since $B^2 - A C = 0$.
- The case (iii) is elliptic since $B^2 - A C < 0$.
An alternative derivation would consist in the diagonalization of the matrices
$$
mathcal A = begin{bmatrix}
a_1 & 0 \
0 & b_1
end{bmatrix} qquadtext{and}qquad mathcal B = begin{bmatrix}
0 & a_2 \
b_2 & 0
end{bmatrix}
$$
as $mathcal A = P D_{mathcal A} P^{-1}$ and $mathcal B = P D_{mathcal B} P^{-1}$, where $D_{mathcal A}$, $D_{mathcal B}$ are diagonal matrices.
For instance, (i) leads to
$$
P = begin{bmatrix}
-1 & 1 \
1 & 1
end{bmatrix} qquadtext{and}qquad P^{-1} = begin{bmatrix}
-1/2 & 1/2 \
1/2 & 1/2
end{bmatrix}
$$
so that
$D_{mathcal A} = text{diag}(1,1)$ and $D_{mathcal B} = text{diag}(-1,1)$.
Hence, the first-order system $mathcal A q_x + mathcal B q_y = g$ with $q = (u,v)^top$ and $g = (g_1, g_2)^top$ rewrites as $p_x + D_{mathcal B} p_y = f$ with $p = P^{-1} q$ and $f = P^{-1} g$. Since $D_{mathcal B}$ has real eigenvalues, this system is a first-order hyperbolic system.
answered Jan 22 at 14:42
Harry49Harry49
7,50431341
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let us differentiate the first equation w.r.t. $y$ and the second one w.r.t. $x$:
begin{aligned}
a_1 u_{xy} + a_2 v_{yy} &= 0 \
b_1 v_{xx} + b_2 u_{yx} &= 0
end{aligned}
Using the equality of mixed derivatives for $u$ and the fact that $a_1 = b_2 = 1 neq 0$, we obtain the equations
$$
u_{xy} = u_{yx} = -a_2/a_1 v_{yy} = -b_1/b_2 v_{xx}
$$
Since $a_2neq 0$, we are left with the second-order linear PDE $A v_{xx} + 2 B v_{xy} + Cv_{yy} = 0$ satisfied by $v$, with $A = frac{a_1 b_1}{a_2 b_2}$, $B = 0$ and $C = -1$.
- The case (i) is hyperbolic since $B^2 - A C > 0$.
- The case (ii) is parabolic since $B^2 - A C = 0$.
- The case (iii) is elliptic since $B^2 - A C < 0$.
An alternative derivation would consist in the diagonalization of the matrices
$$
mathcal A = begin{bmatrix}
a_1 & 0 \
0 & b_1
end{bmatrix} qquadtext{and}qquad mathcal B = begin{bmatrix}
0 & a_2 \
b_2 & 0
end{bmatrix}
$$
as $mathcal A = P D_{mathcal A} P^{-1}$ and $mathcal B = P D_{mathcal B} P^{-1}$, where $D_{mathcal A}$, $D_{mathcal B}$ are diagonal matrices.
For instance, (i) leads to
$$
P = begin{bmatrix}
-1 & 1 \
1 & 1
end{bmatrix} qquadtext{and}qquad P^{-1} = begin{bmatrix}
-1/2 & 1/2 \
1/2 & 1/2
end{bmatrix}
$$
so that
$D_{mathcal A} = text{diag}(1,1)$ and $D_{mathcal B} = text{diag}(-1,1)$.
Hence, the first-order system $mathcal A q_x + mathcal B q_y = g$ with $q = (u,v)^top$ and $g = (g_1, g_2)^top$ rewrites as $p_x + D_{mathcal B} p_y = f$ with $p = P^{-1} q$ and $f = P^{-1} g$. Since $D_{mathcal B}$ has real eigenvalues, this system is a first-order hyperbolic system.
answered Jan 22 at 14:42
Harry49Harry49
7,50431341
$endgroup$
add a comment |
$begingroup$
Let us differentiate the first equation w.r.t. $y$ and the second one w.r.t. $x$:
begin{aligned}
a_1 u_{xy} + a_2 v_{yy} &= 0 \
b_1 v_{xx} + b_2 u_{yx} &= 0
end{aligned}
Using the equality of mixed derivatives for $u$ and the fact that $a_1 = b_2 = 1 neq 0$, we obtain the equations
$$
u_{xy} = u_{yx} = -a_2/a_1 v_{yy} = -b_1/b_2 v_{xx}
$$
Since $a_2neq 0$, we are left with the second-order linear PDE $A v_{xx} + 2 B v_{xy} + Cv_{yy} = 0$ satisfied by $v$, with $A = frac{a_1 b_1}{a_2 b_2}$, $B = 0$ and $C = -1$.
- The case (i) is hyperbolic since $B^2 - A C > 0$.
- The case (ii) is parabolic since $B^2 - A C = 0$.
- The case (iii) is elliptic since $B^2 - A C < 0$.
An alternative derivation would consist in the diagonalization of the matrices
$$
mathcal A = begin{bmatrix}
a_1 & 0 \
0 & b_1
end{bmatrix} qquadtext{and}qquad mathcal B = begin{bmatrix}
0 & a_2 \
b_2 & 0
end{bmatrix}
$$
as $mathcal A = P D_{mathcal A} P^{-1}$ and $mathcal B = P D_{mathcal B} P^{-1}$, where $D_{mathcal A}$, $D_{mathcal B}$ are diagonal matrices.
For instance, (i) leads to
$$
P = begin{bmatrix}
-1 & 1 \
1 & 1
end{bmatrix} qquadtext{and}qquad P^{-1} = begin{bmatrix}
-1/2 & 1/2 \
1/2 & 1/2
end{bmatrix}
$$
so that
$D_{mathcal A} = text{diag}(1,1)$ and $D_{mathcal B} = text{diag}(-1,1)$.
Hence, the first-order system $mathcal A q_x + mathcal B q_y = g$ with $q = (u,v)^top$ and $g = (g_1, g_2)^top$ rewrites as $p_x + D_{mathcal B} p_y = f$ with $p = P^{-1} q$ and $f = P^{-1} g$. Since $D_{mathcal B}$ has real eigenvalues, this system is a first-order hyperbolic system.
answered Jan 22 at 14:42
Harry49Harry49
7,50431341
$endgroup$
add a comment |
$begingroup$
Let us differentiate the first equation w.r.t. $y$ and the second one w.r.t. $x$:
begin{aligned}
a_1 u_{xy} + a_2 v_{yy} &= 0 \
b_1 v_{xx} + b_2 u_{yx} &= 0
end{aligned}
Using the equality of mixed derivatives for $u$ and the fact that $a_1 = b_2 = 1 neq 0$, we obtain the equations
$$
u_{xy} = u_{yx} = -a_2/a_1 v_{yy} = -b_1/b_2 v_{xx}
$$
Since $a_2neq 0$, we are left with the second-order linear PDE $A v_{xx} + 2 B v_{xy} + Cv_{yy} = 0$ satisfied by $v$, with $A = frac{a_1 b_1}{a_2 b_2}$, $B = 0$ and $C = -1$.
- The case (i) is hyperbolic since $B^2 - A C > 0$.
- The case (ii) is parabolic since $B^2 - A C = 0$.
- The case (iii) is elliptic since $B^2 - A C < 0$.
An alternative derivation would consist in the diagonalization of the matrices
$$
mathcal A = begin{bmatrix}
a_1 & 0 \
0 & b_1
end{bmatrix} qquadtext{and}qquad mathcal B = begin{bmatrix}
0 & a_2 \
b_2 & 0
end{bmatrix}
$$
as $mathcal A = P D_{mathcal A} P^{-1}$ and $mathcal B = P D_{mathcal B} P^{-1}$, where $D_{mathcal A}$, $D_{mathcal B}$ are diagonal matrices.
For instance, (i) leads to
$$
P = begin{bmatrix}
-1 & 1 \
1 & 1
end{bmatrix} qquadtext{and}qquad P^{-1} = begin{bmatrix}
-1/2 & 1/2 \
1/2 & 1/2
end{bmatrix}
$$
so that
$D_{mathcal A} = text{diag}(1,1)$ and $D_{mathcal B} = text{diag}(-1,1)$.
Hence, the first-order system $mathcal A q_x + mathcal B q_y = g$ with $q = (u,v)^top$ and $g = (g_1, g_2)^top$ rewrites as $p_x + D_{mathcal B} p_y = f$ with $p = P^{-1} q$ and $f = P^{-1} g$. Since $D_{mathcal B}$ has real eigenvalues, this system is a first-order hyperbolic system.
answered Jan 22 at 14:42
Harry49Harry49
7,50431341
$endgroup$
Let us differentiate the first equation w.r.t. $y$ and the second one w.r.t. $x$:
begin{aligned}
a_1 u_{xy} + a_2 v_{yy} &= 0 \
b_1 v_{xx} + b_2 u_{yx} &= 0
end{aligned}
Using the equality of mixed derivatives for $u$ and the fact that $a_1 = b_2 = 1 neq 0$, we obtain the equations
$$
u_{xy} = u_{yx} = -a_2/a_1 v_{yy} = -b_1/b_2 v_{xx}
$$
Since $a_2neq 0$, we are left with the second-order linear PDE $A v_{xx} + 2 B v_{xy} + Cv_{yy} = 0$ satisfied by $v$, with $A = frac{a_1 b_1}{a_2 b_2}$, $B = 0$ and $C = -1$.
- The case (i) is hyperbolic since $B^2 - A C > 0$.
- The case (ii) is parabolic since $B^2 - A C = 0$.
- The case (iii) is elliptic since $B^2 - A C < 0$.
An alternative derivation would consist in the diagonalization of the matrices
$$
mathcal A = begin{bmatrix}
a_1 & 0 \
0 & b_1
end{bmatrix} qquadtext{and}qquad mathcal B = begin{bmatrix}
0 & a_2 \
b_2 & 0
end{bmatrix}
$$
as $mathcal A = P D_{mathcal A} P^{-1}$ and $mathcal B = P D_{mathcal B} P^{-1}$, where $D_{mathcal A}$, $D_{mathcal B}$ are diagonal matrices.
For instance, (i) leads to
$$
P = begin{bmatrix}
-1 & 1 \
1 & 1
end{bmatrix} qquadtext{and}qquad P^{-1} = begin{bmatrix}
-1/2 & 1/2 \
1/2 & 1/2
end{bmatrix}
$$
so that
$D_{mathcal A} = text{diag}(1,1)$ and $D_{mathcal B} = text{diag}(-1,1)$.
Hence, the first-order system $mathcal A q_x + mathcal B q_y = g$ with $q = (u,v)^top$ and $g = (g_1, g_2)^top$ rewrites as $p_x + D_{mathcal B} p_y = f$ with $p = P^{-1} q$ and $f = P^{-1} g$. Since $D_{mathcal B}$ has real eigenvalues, this system is a first-order hyperbolic system.
answered Jan 22 at 14:42
Harry49Harry49
7,50431341
edited Jan 22 at 16:43
answered Jan 22 at 14:42
Harry49Harry49
7,50431341
answered Jan 22 at 14:42
Harry49Harry49
7,50431341
answered Jan 22 at 14:42
Harry49Harry49
7,50431341
7,50431341
add a comment |
add a comment |
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