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Two PDE for one matrix-valued unknown?

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2

$begingroup$

Let $x in (0,L)$, $t in (0,T)$, and let $P = P(x,t) in mathbbR^3times 3$, $Q = Q(x,t) in mathbbR^3times 3$, $R_0 = R_0(x) in mathbbR^3times 3$ and $G= G(t) in mathbbR^3 times 3$ be continuous functions.

My question is:

Can we find a function $R = R(x,t) in mathbbR^3 times 3$ that satisfies
beginequation
begincases
partial_t R(x,t) = R(x,t) P(x,t) & textin (0,L)times(0,T)\
partial_x R(x,t) = R(x,t) Q(x,t) & textin (0,L)times(0,T)\
R(x,0) = R_0(x) & textfor x in (0,L)\
R(0,t) = G(t) & textfor t in (0,T).
endcases
endequation
with the supplementary assumptions:
beginalign*
& 1. text $R$ is unitary (i.e. $RR^intercal = R^intercal R = I$, where $I$ is the identity matrix),\
& 2. text the determinant of $R$ is equal to one,\
& 3. text $P$ and $Q$ are both skew-symmetric (i.e. $Q^intercal = -Q$ and $P^intercal = -P$).\
endalign*

(Here $partial_t$ and $partial_x$ denote the partial derivative with respect to time and space respectively.)

Following ideas from the scalar case (see Two PDE for one unknown?), I first considered, for $x in (0, L)$ fixed, the solution to the initial value problem $partial_t R = R P$:
beginalign*
R(x,t) = R_0(x) expleft( int_0^t P(x,s)ds right),
endalign*
and for $t in (0,T)$ fixed, the solution to the initial value problem $partial_x R = R Q$:
beginalign*
R(x,t) = G(t) exp left( int_0^x Q(y,t) dy right).
endalign*
I was looking for conditions on the coefficients $P$, $Q$ and on the initial and boundary data, sufficient to imply the existence of a solution $R$ (e.g. $G'(t) = G(t) Q(0,t)$, and others). However, the fact that matrices (and not scalars) are involved implies that terms do not necessarily commute (e.g. $R$ and $partial_t R$ do not necessarily commute).

Any suggestion of method, reference, explanation of why it is/is not possible to have a solution, would be welcome. Thank you.

Edit: I realize now that the formula I gave (above) for the solutions of the initial value problems are probably not correct without assuming that some quantities involving $P(x,t)$ and $Q(x,t)$ commute.

functional-analysis pde matrix-equations matrix-calculus linear-pde

share|cite|improve this question

edited Mar 21 at 15:23

user344045

asked Mar 21 at 8:13

user344045user344045

807

$endgroup$

add a comment | 

2

$begingroup$

Let $x in (0,L)$, $t in (0,T)$, and let $P = P(x,t) in mathbbR^3times 3$, $Q = Q(x,t) in mathbbR^3times 3$, $R_0 = R_0(x) in mathbbR^3times 3$ and $G= G(t) in mathbbR^3 times 3$ be continuous functions.

My question is:

Can we find a function $R = R(x,t) in mathbbR^3 times 3$ that satisfies
beginequation
begincases
partial_t R(x,t) = R(x,t) P(x,t) & textin (0,L)times(0,T)\
partial_x R(x,t) = R(x,t) Q(x,t) & textin (0,L)times(0,T)\
R(x,0) = R_0(x) & textfor x in (0,L)\
R(0,t) = G(t) & textfor t in (0,T).
endcases
endequation
with the supplementary assumptions:
beginalign*
& 1. text $R$ is unitary (i.e. $RR^intercal = R^intercal R = I$, where $I$ is the identity matrix),\
& 2. text the determinant of $R$ is equal to one,\
& 3. text $P$ and $Q$ are both skew-symmetric (i.e. $Q^intercal = -Q$ and $P^intercal = -P$).\
endalign*

(Here $partial_t$ and $partial_x$ denote the partial derivative with respect to time and space respectively.)

Following ideas from the scalar case (see Two PDE for one unknown?), I first considered, for $x in (0, L)$ fixed, the solution to the initial value problem $partial_t R = R P$:
beginalign*
R(x,t) = R_0(x) expleft( int_0^t P(x,s)ds right),
endalign*
and for $t in (0,T)$ fixed, the solution to the initial value problem $partial_x R = R Q$:
beginalign*
R(x,t) = G(t) exp left( int_0^x Q(y,t) dy right).
endalign*
I was looking for conditions on the coefficients $P$, $Q$ and on the initial and boundary data, sufficient to imply the existence of a solution $R$ (e.g. $G'(t) = G(t) Q(0,t)$, and others). However, the fact that matrices (and not scalars) are involved implies that terms do not necessarily commute (e.g. $R$ and $partial_t R$ do not necessarily commute).

Any suggestion of method, reference, explanation of why it is/is not possible to have a solution, would be welcome. Thank you.

Edit: I realize now that the formula I gave (above) for the solutions of the initial value problems are probably not correct without assuming that some quantities involving $P(x,t)$ and $Q(x,t)$ commute.

functional-analysis pde matrix-equations matrix-calculus linear-pde

share|cite|improve this question

edited Mar 21 at 15:23

user344045

asked Mar 21 at 8:13

user344045user344045

807

$endgroup$

add a comment | 

$begingroup$

Let $x in (0,L)$, $t in (0,T)$, and let $P = P(x,t) in mathbbR^3times 3$, $Q = Q(x,t) in mathbbR^3times 3$, $R_0 = R_0(x) in mathbbR^3times 3$ and $G= G(t) in mathbbR^3 times 3$ be continuous functions.

My question is:

Can we find a function $R = R(x,t) in mathbbR^3 times 3$ that satisfies
beginequation
begincases
partial_t R(x,t) = R(x,t) P(x,t) & textin (0,L)times(0,T)\
partial_x R(x,t) = R(x,t) Q(x,t) & textin (0,L)times(0,T)\
R(x,0) = R_0(x) & textfor x in (0,L)\
R(0,t) = G(t) & textfor t in (0,T).
endcases
endequation
with the supplementary assumptions:
beginalign*
& 1. text $R$ is unitary (i.e. $RR^intercal = R^intercal R = I$, where $I$ is the identity matrix),\
& 2. text the determinant of $R$ is equal to one,\
& 3. text $P$ and $Q$ are both skew-symmetric (i.e. $Q^intercal = -Q$ and $P^intercal = -P$).\
endalign*

(Here $partial_t$ and $partial_x$ denote the partial derivative with respect to time and space respectively.)

Following ideas from the scalar case (see Two PDE for one unknown?), I first considered, for $x in (0, L)$ fixed, the solution to the initial value problem $partial_t R = R P$:
beginalign*
R(x,t) = R_0(x) expleft( int_0^t P(x,s)ds right),
endalign*
and for $t in (0,T)$ fixed, the solution to the initial value problem $partial_x R = R Q$:
beginalign*
R(x,t) = G(t) exp left( int_0^x Q(y,t) dy right).
endalign*
I was looking for conditions on the coefficients $P$, $Q$ and on the initial and boundary data, sufficient to imply the existence of a solution $R$ (e.g. $G'(t) = G(t) Q(0,t)$, and others). However, the fact that matrices (and not scalars) are involved implies that terms do not necessarily commute (e.g. $R$ and $partial_t R$ do not necessarily commute).

Any suggestion of method, reference, explanation of why it is/is not possible to have a solution, would be welcome. Thank you.

Edit: I realize now that the formula I gave (above) for the solutions of the initial value problems are probably not correct without assuming that some quantities involving $P(x,t)$ and $Q(x,t)$ commute.

functional-analysis pde matrix-equations matrix-calculus linear-pde

share|cite|improve this question

edited Mar 21 at 15:23

user344045

asked Mar 21 at 8:13

user344045user344045

807

$endgroup$

share|cite|improve this question

edited Mar 21 at 15:23

user344045

asked Mar 21 at 8:13

user344045user344045

807

share|cite|improve this question

edited Mar 21 at 15:23

user344045

asked Mar 21 at 8:13

user344045user344045

807

asked Mar 21 at 8:13

user344045user344045

807

asked Mar 21 at 8:13

user344045user344045

807