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6 linear PDE for only 3 unknowns?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to solve this degenerate parabolic euqationProving the uniqueness of a PDE's solutionHow to solve this PDE $partial_t u - Delta u + cu + dcdot nabla u = f$ with Dirichlet boundary conditions?Crank-Nicolson for quadratic PDEMatrix for discretized PDEUsing Fourier sine transform to solve PDEAn estimate for a 1d hyperbolic PDESplitting a PDE into two subdomains: Regularity conditions at the interface point(s)Two PDE for one unknown?Two PDE for one matrix-valued unknown?
$begingroup$
Let $x in (0,L)$, $t in (0,T)$, and let $u_0 = u_0(x) in mathbbR^3$, $g=g(t) in mathbbR^3$, $P = P(x,t) in mathbbR^3$ and $Q = Q(x,t) in mathbbR^3$ be continuously differentiable functions.
Denote by $hatP, hatQ$ the matrices
beginalign*
hatP = beginbmatrix
0 & -P_3 & P_2 \
Q_3 & 0 & -P_1 \
-P_2 & P_1 & 0
endbmatrix, qquad
hatQ = beginbmatrix
0 & -Q_3 & Q_2 \
Q_3 & 0 & -Q_1 \
-Q_2 & Q_1 & 0
endbmatrix.
endalign*
My question is:
Assuming that the following four conditions hold:
Condition 1:
beginalign*
partial_x P - partial_t Q = hatPQ qquad textin (0,L) times (0,T),
endalign*
Condition 2:
beginalign*
u_0(0) = g(0),
endalign*
Condition 3:
beginalign*
u_0' = - hatQu_0 qquad textfor x in (0,L),
endalign*
Condition 4:
beginalign*
g' = - hatPg qquad textfor t in (0,T),
endalign*
is there a solution $u = u(x,t) in mathbbR^3$ to
beginalign*
begincases
partial_t u = -hatPu &textin (0,L)times(0,T) \
partial_x u = - hatQu &textin (0,L)times(0,T) \
u(x,0) = u_0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T)
endcases
endalign*
or not?
In the above, $f'$, $partial_t f$a and $partial_x f$ denote the derivative, partial time derivative and partial space derivative respectively; and $P_k$, $Q_k$, $u_k$, $u_0k$ and $g_k$ denote the components of $P$, $Q$, $u$, $u_0$ and $g$ respectively.
My question with all terms developed rewrites as follows. Assuming that the following four conditions hold:
Condition 1:
beginalign*
begincases
partial_x P_1 - partial_t Q_1 = P_2 Q_3 - P_3 Q_2\
partial_x P_2 - partial_t Q_2 = P_3 Q_1 - P_1 Q_3\
partial_x P_3 - partial_t Q_3 = P_1 Q_2 - P_2 Q_1
endcases qquad textin (0,L) times (0,T),
endalign*
Condition 2:
beginalign*
u_0(0) = g(0),
endalign*
Condition 3:
beginalign*
begincases
u_01' = Q_3(cdot,0) u_02 - Q_2(cdot,0) u_03 \
u_02' = -Q_3(cdot,0) u_01 + Q_1(cdot,0) u_03 \
u_03' = Q_2(cdot,0) u_01 - Q_1(cdot,0) u_02
endcases qquad textfor x in (0,L),
endalign*
Condition 4:
beginalign*
begincases
g_1' &= P_3(0,cdot) g_2 - P_2(0,cdot) g_3 \
g_2' &= -P_3(0,cdot) g_1 + P_1(0,cdot) g_3 \
g_3' &= P_2(0,cdot) g_1 - P_1(0,cdot) g_2
endcases qquad textfor t in (0,T),
endalign*
is there a solution $u = u(x,t) in mathbbR^3$ to the problem:
beginalign*
begincases
partial_t u_1 = P_3 u_2 - P_2 u_3 &textin (0,L)times(0,T) \
partial_t u_2 = -P_3 u_1 + P_1 u_3 &textin (0,L)times(0,T) \
partial_t u_3 = P_2 u_1 - P_1 u_2 & textin (0,L)times(0,T) \
partial_x u_1 = Q_3 u_2 - Q_2 u_3 & textin (0,L)times(0,T) \
partial_x u_2 = -Q_3 u_1 + Q_1 u_3 &textin (0,L)times(0,T) \
partial_x u_3 = Q_2 u_1 - Q_1 u_2 & textin (0,L)times(0,T) \
u(x,0) = u_0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T).
endcases
endalign*
What I started. I started reasoning the following way. For $x in (0,L)$ fixed, a function of the form
beginalign*
u_1(x,t) &= u_01(x) + int_0^t P_3(x,s)u_2(x,s) - P_2(x,s)u_3(x,s)ds\
u_2(x,t) &= u_02(x) + int_0^t -P_3(x,s)u_1(x,s) + P_1(x,s)u_3(x,s)ds\
u_3(x,t) &= u_03(x) + int_0^t P_2(x,s)u_1(x,s) - P_1(x,s)u_2(x,s)ds,
endalign*
satisfies the initial value problem involving time derivatives, with $u(cdot, 0) = u_0$, while for $t in (0,T)$ fixed, a solution of the form
beginalign*
beginaligned
u_1(x,t) &= g_1(t) + int_0^x Q_3(xi,t)u_2(xi,t) - Q_2(xi, t)u_3(xi, t)dxi \
u_2(x,t) &= g_2(t) + int_0^x -Q_3(xi, t)u_1(xi, t) + Q_1(xi, t) u_3(xi, t)dxi\
u_3(x,t) &= g_3(t) + int_0^x Q_2(xi, t)u_1(xi, t) - Q_1(xi, t)u_2(xi, t)dxi.
endaligned
endalign*
satisfies the initial value problem involving space derivatives, with $u(0, cdot) = g$. I wanted to show, using the four conditions, that both expressions for the solution coincide. Using conditions 3 and 4 we can rewrite both expressions as
beginalign*
u_1(x,t) &= u_01(0) + int_0^x Q_3(xi, 0)u_02(xi) - Q_2(xi, 0)u_03(xi)dxi \
&quad + int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s)ds + int_0^t int_0^x partial_x (P_3 u_2 - P_2 u_3 )(xi, s)dxi ds\
u_2(x,t) &= u_02(0) + int_0^x -Q_3(xi, 0)u_01 + Q_1(xi, 0)u_03(xi)dxi \
&quad + int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s)ds + int_0^t int_0^x partial_x(-P_3 u_1 + P_1 u_3)(xi, s)ds\
u_3(x,t) &= u_03(0) + int_0^x Q_2(xi, 0)u_01(xi) - Q_1(xi, 0)u_02(xi)\
&quad + int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s)ds + int_0^t int_0^x partial_x (P_2 u_1 - P_1 u_2)(xi,s)dxi ds,
endalign*
and
beginalign*
u_1(x,t) &= g_1(0) + int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s) ds \
&quad + int_0^x Q_3(xi,0)u_02(xi) - Q_2(xi, 0)u_03(xi)dxi + int_0^x int_0^t partial_t ( Q_3 u_2 - Q_2 u_3)(xi, s)dsdxi\
u_2(x,t) &= g_2(0) + int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s) ds\
&quad + int_0^x -Q_3(xi, 0)u_01(xi) + Q_1(xi, 0) u_03(xi)dxi + int_0^x int_0^t partial_t (-Q_3 u_1 + Q_1 u_3)(xi,s)dsdxi\
u_3(x,t) &= g_3(0) + int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s) ds \
&quad + int_0^x Q_2(xi, 0)u_01(xi) - Q_1(xi, 0)u_02(xi)dxi + int_0^x int_0^t partial_t ( Q_2 u_1 - Q_1 u_2)(xi, s)ds dxi
endalign*
respectively. It seems that both expressions would coincide if the following equalities hold:
beginalign*
partial_x (P_3 u_2 - P_2 u_3 ) &= partial_t ( Q_3 u_2 - Q_2 u_3)\
partial_x(-P_3 u_1 + P_1 u_3) &= partial_t (-Q_3 u_1 + Q_1 u_3)\
partial_x (P_2 u_1 - P_1 u_2) &= partial_t ( Q_2 u_1 - Q_1 u_2).
endalign*
To show that these equalities hold, I would like to use condition 1. However, I do not know how to argue from there.
Any suggestion of how to proceed from here, any alternative way of reasoning, or explanation of why there would be no solution, would be welcome. Thank you.
This question is also posed on MathOverflow: https://mathoverflow.net/questions/326280/6-linear-pde-for-only-3-unknowns.
pde systems-of-equations linear-pde
asked Mar 24 at 14:33
user344045user344045
907
$endgroup$
add a comment |
$begingroup$
Let $x in (0,L)$, $t in (0,T)$, and let $u_0 = u_0(x) in mathbbR^3$, $g=g(t) in mathbbR^3$, $P = P(x,t) in mathbbR^3$ and $Q = Q(x,t) in mathbbR^3$ be continuously differentiable functions.
Denote by $hatP, hatQ$ the matrices
beginalign*
hatP = beginbmatrix
0 & -P_3 & P_2 \
Q_3 & 0 & -P_1 \
-P_2 & P_1 & 0
endbmatrix, qquad
hatQ = beginbmatrix
0 & -Q_3 & Q_2 \
Q_3 & 0 & -Q_1 \
-Q_2 & Q_1 & 0
endbmatrix.
endalign*
My question is:
Assuming that the following four conditions hold:
Condition 1:
beginalign*
partial_x P - partial_t Q = hatPQ qquad textin (0,L) times (0,T),
endalign*
Condition 2:
beginalign*
u_0(0) = g(0),
endalign*
Condition 3:
beginalign*
u_0' = - hatQu_0 qquad textfor x in (0,L),
endalign*
Condition 4:
beginalign*
g' = - hatPg qquad textfor t in (0,T),
endalign*
is there a solution $u = u(x,t) in mathbbR^3$ to
beginalign*
begincases
partial_t u = -hatPu &textin (0,L)times(0,T) \
partial_x u = - hatQu &textin (0,L)times(0,T) \
u(x,0) = u_0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T)
endcases
endalign*
or not?
In the above, $f'$, $partial_t f$a and $partial_x f$ denote the derivative, partial time derivative and partial space derivative respectively; and $P_k$, $Q_k$, $u_k$, $u_0k$ and $g_k$ denote the components of $P$, $Q$, $u$, $u_0$ and $g$ respectively.
My question with all terms developed rewrites as follows. Assuming that the following four conditions hold:
Condition 1:
beginalign*
begincases
partial_x P_1 - partial_t Q_1 = P_2 Q_3 - P_3 Q_2\
partial_x P_2 - partial_t Q_2 = P_3 Q_1 - P_1 Q_3\
partial_x P_3 - partial_t Q_3 = P_1 Q_2 - P_2 Q_1
endcases qquad textin (0,L) times (0,T),
endalign*
Condition 2:
beginalign*
u_0(0) = g(0),
endalign*
Condition 3:
beginalign*
begincases
u_01' = Q_3(cdot,0) u_02 - Q_2(cdot,0) u_03 \
u_02' = -Q_3(cdot,0) u_01 + Q_1(cdot,0) u_03 \
u_03' = Q_2(cdot,0) u_01 - Q_1(cdot,0) u_02
endcases qquad textfor x in (0,L),
endalign*
Condition 4:
beginalign*
begincases
g_1' &= P_3(0,cdot) g_2 - P_2(0,cdot) g_3 \
g_2' &= -P_3(0,cdot) g_1 + P_1(0,cdot) g_3 \
g_3' &= P_2(0,cdot) g_1 - P_1(0,cdot) g_2
endcases qquad textfor t in (0,T),
endalign*
is there a solution $u = u(x,t) in mathbbR^3$ to the problem:
beginalign*
begincases
partial_t u_1 = P_3 u_2 - P_2 u_3 &textin (0,L)times(0,T) \
partial_t u_2 = -P_3 u_1 + P_1 u_3 &textin (0,L)times(0,T) \
partial_t u_3 = P_2 u_1 - P_1 u_2 & textin (0,L)times(0,T) \
partial_x u_1 = Q_3 u_2 - Q_2 u_3 & textin (0,L)times(0,T) \
partial_x u_2 = -Q_3 u_1 + Q_1 u_3 &textin (0,L)times(0,T) \
partial_x u_3 = Q_2 u_1 - Q_1 u_2 & textin (0,L)times(0,T) \
u(x,0) = u_0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T).
endcases
endalign*
What I started. I started reasoning the following way. For $x in (0,L)$ fixed, a function of the form
beginalign*
u_1(x,t) &= u_01(x) + int_0^t P_3(x,s)u_2(x,s) - P_2(x,s)u_3(x,s)ds\
u_2(x,t) &= u_02(x) + int_0^t -P_3(x,s)u_1(x,s) + P_1(x,s)u_3(x,s)ds\
u_3(x,t) &= u_03(x) + int_0^t P_2(x,s)u_1(x,s) - P_1(x,s)u_2(x,s)ds,
endalign*
satisfies the initial value problem involving time derivatives, with $u(cdot, 0) = u_0$, while for $t in (0,T)$ fixed, a solution of the form
beginalign*
beginaligned
u_1(x,t) &= g_1(t) + int_0^x Q_3(xi,t)u_2(xi,t) - Q_2(xi, t)u_3(xi, t)dxi \
u_2(x,t) &= g_2(t) + int_0^x -Q_3(xi, t)u_1(xi, t) + Q_1(xi, t) u_3(xi, t)dxi\
u_3(x,t) &= g_3(t) + int_0^x Q_2(xi, t)u_1(xi, t) - Q_1(xi, t)u_2(xi, t)dxi.
endaligned
endalign*
satisfies the initial value problem involving space derivatives, with $u(0, cdot) = g$. I wanted to show, using the four conditions, that both expressions for the solution coincide. Using conditions 3 and 4 we can rewrite both expressions as
beginalign*
u_1(x,t) &= u_01(0) + int_0^x Q_3(xi, 0)u_02(xi) - Q_2(xi, 0)u_03(xi)dxi \
&quad + int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s)ds + int_0^t int_0^x partial_x (P_3 u_2 - P_2 u_3 )(xi, s)dxi ds\
u_2(x,t) &= u_02(0) + int_0^x -Q_3(xi, 0)u_01 + Q_1(xi, 0)u_03(xi)dxi \
&quad + int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s)ds + int_0^t int_0^x partial_x(-P_3 u_1 + P_1 u_3)(xi, s)ds\
u_3(x,t) &= u_03(0) + int_0^x Q_2(xi, 0)u_01(xi) - Q_1(xi, 0)u_02(xi)\
&quad + int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s)ds + int_0^t int_0^x partial_x (P_2 u_1 - P_1 u_2)(xi,s)dxi ds,
endalign*
and
beginalign*
u_1(x,t) &= g_1(0) + int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s) ds \
&quad + int_0^x Q_3(xi,0)u_02(xi) - Q_2(xi, 0)u_03(xi)dxi + int_0^x int_0^t partial_t ( Q_3 u_2 - Q_2 u_3)(xi, s)dsdxi\
u_2(x,t) &= g_2(0) + int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s) ds\
&quad + int_0^x -Q_3(xi, 0)u_01(xi) + Q_1(xi, 0) u_03(xi)dxi + int_0^x int_0^t partial_t (-Q_3 u_1 + Q_1 u_3)(xi,s)dsdxi\
u_3(x,t) &= g_3(0) + int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s) ds \
&quad + int_0^x Q_2(xi, 0)u_01(xi) - Q_1(xi, 0)u_02(xi)dxi + int_0^x int_0^t partial_t ( Q_2 u_1 - Q_1 u_2)(xi, s)ds dxi
endalign*
respectively. It seems that both expressions would coincide if the following equalities hold:
beginalign*
partial_x (P_3 u_2 - P_2 u_3 ) &= partial_t ( Q_3 u_2 - Q_2 u_3)\
partial_x(-P_3 u_1 + P_1 u_3) &= partial_t (-Q_3 u_1 + Q_1 u_3)\
partial_x (P_2 u_1 - P_1 u_2) &= partial_t ( Q_2 u_1 - Q_1 u_2).
endalign*
To show that these equalities hold, I would like to use condition 1. However, I do not know how to argue from there.
Any suggestion of how to proceed from here, any alternative way of reasoning, or explanation of why there would be no solution, would be welcome. Thank you.
This question is also posed on MathOverflow: https://mathoverflow.net/questions/326280/6-linear-pde-for-only-3-unknowns.
pde systems-of-equations linear-pde
asked Mar 24 at 14:33
user344045user344045
907
$endgroup$
add a comment |
$begingroup$
Let $x in (0,L)$, $t in (0,T)$, and let $u_0 = u_0(x) in mathbbR^3$, $g=g(t) in mathbbR^3$, $P = P(x,t) in mathbbR^3$ and $Q = Q(x,t) in mathbbR^3$ be continuously differentiable functions.
Denote by $hatP, hatQ$ the matrices
beginalign*
hatP = beginbmatrix
0 & -P_3 & P_2 \
Q_3 & 0 & -P_1 \
-P_2 & P_1 & 0
endbmatrix, qquad
hatQ = beginbmatrix
0 & -Q_3 & Q_2 \
Q_3 & 0 & -Q_1 \
-Q_2 & Q_1 & 0
endbmatrix.
endalign*
My question is:
Assuming that the following four conditions hold:
Condition 1:
beginalign*
partial_x P - partial_t Q = hatPQ qquad textin (0,L) times (0,T),
endalign*
Condition 2:
beginalign*
u_0(0) = g(0),
endalign*
Condition 3:
beginalign*
u_0' = - hatQu_0 qquad textfor x in (0,L),
endalign*
Condition 4:
beginalign*
g' = - hatPg qquad textfor t in (0,T),
endalign*
is there a solution $u = u(x,t) in mathbbR^3$ to
beginalign*
begincases
partial_t u = -hatPu &textin (0,L)times(0,T) \
partial_x u = - hatQu &textin (0,L)times(0,T) \
u(x,0) = u_0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T)
endcases
endalign*
or not?
In the above, $f'$, $partial_t f$a and $partial_x f$ denote the derivative, partial time derivative and partial space derivative respectively; and $P_k$, $Q_k$, $u_k$, $u_0k$ and $g_k$ denote the components of $P$, $Q$, $u$, $u_0$ and $g$ respectively.
My question with all terms developed rewrites as follows. Assuming that the following four conditions hold:
Condition 1:
beginalign*
begincases
partial_x P_1 - partial_t Q_1 = P_2 Q_3 - P_3 Q_2\
partial_x P_2 - partial_t Q_2 = P_3 Q_1 - P_1 Q_3\
partial_x P_3 - partial_t Q_3 = P_1 Q_2 - P_2 Q_1
endcases qquad textin (0,L) times (0,T),
endalign*
Condition 2:
beginalign*
u_0(0) = g(0),
endalign*
Condition 3:
beginalign*
begincases
u_01' = Q_3(cdot,0) u_02 - Q_2(cdot,0) u_03 \
u_02' = -Q_3(cdot,0) u_01 + Q_1(cdot,0) u_03 \
u_03' = Q_2(cdot,0) u_01 - Q_1(cdot,0) u_02
endcases qquad textfor x in (0,L),
endalign*
Condition 4:
beginalign*
begincases
g_1' &= P_3(0,cdot) g_2 - P_2(0,cdot) g_3 \
g_2' &= -P_3(0,cdot) g_1 + P_1(0,cdot) g_3 \
g_3' &= P_2(0,cdot) g_1 - P_1(0,cdot) g_2
endcases qquad textfor t in (0,T),
endalign*
is there a solution $u = u(x,t) in mathbbR^3$ to the problem:
beginalign*
begincases
partial_t u_1 = P_3 u_2 - P_2 u_3 &textin (0,L)times(0,T) \
partial_t u_2 = -P_3 u_1 + P_1 u_3 &textin (0,L)times(0,T) \
partial_t u_3 = P_2 u_1 - P_1 u_2 & textin (0,L)times(0,T) \
partial_x u_1 = Q_3 u_2 - Q_2 u_3 & textin (0,L)times(0,T) \
partial_x u_2 = -Q_3 u_1 + Q_1 u_3 &textin (0,L)times(0,T) \
partial_x u_3 = Q_2 u_1 - Q_1 u_2 & textin (0,L)times(0,T) \
u(x,0) = u_0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T).
endcases
endalign*
What I started. I started reasoning the following way. For $x in (0,L)$ fixed, a function of the form
beginalign*
u_1(x,t) &= u_01(x) + int_0^t P_3(x,s)u_2(x,s) - P_2(x,s)u_3(x,s)ds\
u_2(x,t) &= u_02(x) + int_0^t -P_3(x,s)u_1(x,s) + P_1(x,s)u_3(x,s)ds\
u_3(x,t) &= u_03(x) + int_0^t P_2(x,s)u_1(x,s) - P_1(x,s)u_2(x,s)ds,
endalign*
satisfies the initial value problem involving time derivatives, with $u(cdot, 0) = u_0$, while for $t in (0,T)$ fixed, a solution of the form
beginalign*
beginaligned
u_1(x,t) &= g_1(t) + int_0^x Q_3(xi,t)u_2(xi,t) - Q_2(xi, t)u_3(xi, t)dxi \
u_2(x,t) &= g_2(t) + int_0^x -Q_3(xi, t)u_1(xi, t) + Q_1(xi, t) u_3(xi, t)dxi\
u_3(x,t) &= g_3(t) + int_0^x Q_2(xi, t)u_1(xi, t) - Q_1(xi, t)u_2(xi, t)dxi.
endaligned
endalign*
satisfies the initial value problem involving space derivatives, with $u(0, cdot) = g$. I wanted to show, using the four conditions, that both expressions for the solution coincide. Using conditions 3 and 4 we can rewrite both expressions as
beginalign*
u_1(x,t) &= u_01(0) + int_0^x Q_3(xi, 0)u_02(xi) - Q_2(xi, 0)u_03(xi)dxi \
&quad + int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s)ds + int_0^t int_0^x partial_x (P_3 u_2 - P_2 u_3 )(xi, s)dxi ds\
u_2(x,t) &= u_02(0) + int_0^x -Q_3(xi, 0)u_01 + Q_1(xi, 0)u_03(xi)dxi \
&quad + int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s)ds + int_0^t int_0^x partial_x(-P_3 u_1 + P_1 u_3)(xi, s)ds\
u_3(x,t) &= u_03(0) + int_0^x Q_2(xi, 0)u_01(xi) - Q_1(xi, 0)u_02(xi)\
&quad + int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s)ds + int_0^t int_0^x partial_x (P_2 u_1 - P_1 u_2)(xi,s)dxi ds,
endalign*
and
beginalign*
u_1(x,t) &= g_1(0) + int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s) ds \
&quad + int_0^x Q_3(xi,0)u_02(xi) - Q_2(xi, 0)u_03(xi)dxi + int_0^x int_0^t partial_t ( Q_3 u_2 - Q_2 u_3)(xi, s)dsdxi\
u_2(x,t) &= g_2(0) + int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s) ds\
&quad + int_0^x -Q_3(xi, 0)u_01(xi) + Q_1(xi, 0) u_03(xi)dxi + int_0^x int_0^t partial_t (-Q_3 u_1 + Q_1 u_3)(xi,s)dsdxi\
u_3(x,t) &= g_3(0) + int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s) ds \
&quad + int_0^x Q_2(xi, 0)u_01(xi) - Q_1(xi, 0)u_02(xi)dxi + int_0^x int_0^t partial_t ( Q_2 u_1 - Q_1 u_2)(xi, s)ds dxi
endalign*
respectively. It seems that both expressions would coincide if the following equalities hold:
beginalign*
partial_x (P_3 u_2 - P_2 u_3 ) &= partial_t ( Q_3 u_2 - Q_2 u_3)\
partial_x(-P_3 u_1 + P_1 u_3) &= partial_t (-Q_3 u_1 + Q_1 u_3)\
partial_x (P_2 u_1 - P_1 u_2) &= partial_t ( Q_2 u_1 - Q_1 u_2).
endalign*
To show that these equalities hold, I would like to use condition 1. However, I do not know how to argue from there.
Any suggestion of how to proceed from here, any alternative way of reasoning, or explanation of why there would be no solution, would be welcome. Thank you.
This question is also posed on MathOverflow: https://mathoverflow.net/questions/326280/6-linear-pde-for-only-3-unknowns.
pde systems-of-equations linear-pde
asked Mar 24 at 14:33
user344045user344045
907
$endgroup$
Let $x in (0,L)$, $t in (0,T)$, and let $u_0 = u_0(x) in mathbbR^3$, $g=g(t) in mathbbR^3$, $P = P(x,t) in mathbbR^3$ and $Q = Q(x,t) in mathbbR^3$ be continuously differentiable functions.
Denote by $hatP, hatQ$ the matrices
beginalign*
hatP = beginbmatrix
0 & -P_3 & P_2 \
Q_3 & 0 & -P_1 \
-P_2 & P_1 & 0
endbmatrix, qquad
hatQ = beginbmatrix
0 & -Q_3 & Q_2 \
Q_3 & 0 & -Q_1 \
-Q_2 & Q_1 & 0
endbmatrix.
endalign*
My question is:
Assuming that the following four conditions hold:
Condition 1:
beginalign*
partial_x P - partial_t Q = hatPQ qquad textin (0,L) times (0,T),
endalign*
Condition 2:
beginalign*
u_0(0) = g(0),
endalign*
Condition 3:
beginalign*
u_0' = - hatQu_0 qquad textfor x in (0,L),
endalign*
Condition 4:
beginalign*
g' = - hatPg qquad textfor t in (0,T),
endalign*
is there a solution $u = u(x,t) in mathbbR^3$ to
beginalign*
begincases
partial_t u = -hatPu &textin (0,L)times(0,T) \
partial_x u = - hatQu &textin (0,L)times(0,T) \
u(x,0) = u_0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T)
endcases
endalign*
or not?
In the above, $f'$, $partial_t f$a and $partial_x f$ denote the derivative, partial time derivative and partial space derivative respectively; and $P_k$, $Q_k$, $u_k$, $u_0k$ and $g_k$ denote the components of $P$, $Q$, $u$, $u_0$ and $g$ respectively.
My question with all terms developed rewrites as follows. Assuming that the following four conditions hold:
Condition 1:
beginalign*
begincases
partial_x P_1 - partial_t Q_1 = P_2 Q_3 - P_3 Q_2\
partial_x P_2 - partial_t Q_2 = P_3 Q_1 - P_1 Q_3\
partial_x P_3 - partial_t Q_3 = P_1 Q_2 - P_2 Q_1
endcases qquad textin (0,L) times (0,T),
endalign*
Condition 2:
beginalign*
u_0(0) = g(0),
endalign*
Condition 3:
beginalign*
begincases
u_01' = Q_3(cdot,0) u_02 - Q_2(cdot,0) u_03 \
u_02' = -Q_3(cdot,0) u_01 + Q_1(cdot,0) u_03 \
u_03' = Q_2(cdot,0) u_01 - Q_1(cdot,0) u_02
endcases qquad textfor x in (0,L),
endalign*
Condition 4:
beginalign*
begincases
g_1' &= P_3(0,cdot) g_2 - P_2(0,cdot) g_3 \
g_2' &= -P_3(0,cdot) g_1 + P_1(0,cdot) g_3 \
g_3' &= P_2(0,cdot) g_1 - P_1(0,cdot) g_2
endcases qquad textfor t in (0,T),
endalign*
is there a solution $u = u(x,t) in mathbbR^3$ to the problem:
beginalign*
begincases
partial_t u_1 = P_3 u_2 - P_2 u_3 &textin (0,L)times(0,T) \
partial_t u_2 = -P_3 u_1 + P_1 u_3 &textin (0,L)times(0,T) \
partial_t u_3 = P_2 u_1 - P_1 u_2 & textin (0,L)times(0,T) \
partial_x u_1 = Q_3 u_2 - Q_2 u_3 & textin (0,L)times(0,T) \
partial_x u_2 = -Q_3 u_1 + Q_1 u_3 &textin (0,L)times(0,T) \
partial_x u_3 = Q_2 u_1 - Q_1 u_2 & textin (0,L)times(0,T) \
u(x,0) = u_0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T).
endcases
endalign*
What I started. I started reasoning the following way. For $x in (0,L)$ fixed, a function of the form
beginalign*
u_1(x,t) &= u_01(x) + int_0^t P_3(x,s)u_2(x,s) - P_2(x,s)u_3(x,s)ds\
u_2(x,t) &= u_02(x) + int_0^t -P_3(x,s)u_1(x,s) + P_1(x,s)u_3(x,s)ds\
u_3(x,t) &= u_03(x) + int_0^t P_2(x,s)u_1(x,s) - P_1(x,s)u_2(x,s)ds,
endalign*
satisfies the initial value problem involving time derivatives, with $u(cdot, 0) = u_0$, while for $t in (0,T)$ fixed, a solution of the form
beginalign*
beginaligned
u_1(x,t) &= g_1(t) + int_0^x Q_3(xi,t)u_2(xi,t) - Q_2(xi, t)u_3(xi, t)dxi \
u_2(x,t) &= g_2(t) + int_0^x -Q_3(xi, t)u_1(xi, t) + Q_1(xi, t) u_3(xi, t)dxi\
u_3(x,t) &= g_3(t) + int_0^x Q_2(xi, t)u_1(xi, t) - Q_1(xi, t)u_2(xi, t)dxi.
endaligned
endalign*
satisfies the initial value problem involving space derivatives, with $u(0, cdot) = g$. I wanted to show, using the four conditions, that both expressions for the solution coincide. Using conditions 3 and 4 we can rewrite both expressions as
beginalign*
u_1(x,t) &= u_01(0) + int_0^x Q_3(xi, 0)u_02(xi) - Q_2(xi, 0)u_03(xi)dxi \
&quad + int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s)ds + int_0^t int_0^x partial_x (P_3 u_2 - P_2 u_3 )(xi, s)dxi ds\
u_2(x,t) &= u_02(0) + int_0^x -Q_3(xi, 0)u_01 + Q_1(xi, 0)u_03(xi)dxi \
&quad + int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s)ds + int_0^t int_0^x partial_x(-P_3 u_1 + P_1 u_3)(xi, s)ds\
u_3(x,t) &= u_03(0) + int_0^x Q_2(xi, 0)u_01(xi) - Q_1(xi, 0)u_02(xi)\
&quad + int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s)ds + int_0^t int_0^x partial_x (P_2 u_1 - P_1 u_2)(xi,s)dxi ds,
endalign*
and
beginalign*
u_1(x,t) &= g_1(0) + int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s) ds \
&quad + int_0^x Q_3(xi,0)u_02(xi) - Q_2(xi, 0)u_03(xi)dxi + int_0^x int_0^t partial_t ( Q_3 u_2 - Q_2 u_3)(xi, s)dsdxi\
u_2(x,t) &= g_2(0) + int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s) ds\
&quad + int_0^x -Q_3(xi, 0)u_01(xi) + Q_1(xi, 0) u_03(xi)dxi + int_0^x int_0^t partial_t (-Q_3 u_1 + Q_1 u_3)(xi,s)dsdxi\
u_3(x,t) &= g_3(0) + int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s) ds \
&quad + int_0^x Q_2(xi, 0)u_01(xi) - Q_1(xi, 0)u_02(xi)dxi + int_0^x int_0^t partial_t ( Q_2 u_1 - Q_1 u_2)(xi, s)ds dxi
endalign*
respectively. It seems that both expressions would coincide if the following equalities hold:
beginalign*
partial_x (P_3 u_2 - P_2 u_3 ) &= partial_t ( Q_3 u_2 - Q_2 u_3)\
partial_x(-P_3 u_1 + P_1 u_3) &= partial_t (-Q_3 u_1 + Q_1 u_3)\
partial_x (P_2 u_1 - P_1 u_2) &= partial_t ( Q_2 u_1 - Q_1 u_2).
endalign*
To show that these equalities hold, I would like to use condition 1. However, I do not know how to argue from there.
Any suggestion of how to proceed from here, any alternative way of reasoning, or explanation of why there would be no solution, would be welcome. Thank you.
This question is also posed on MathOverflow: https://mathoverflow.net/questions/326280/6-linear-pde-for-only-3-unknowns.
pde systems-of-equations linear-pde
pde systems-of-equations linear-pde
asked Mar 24 at 14:33
user344045user344045
907
asked Mar 24 at 14:33
user344045user344045
907
edited Mar 28 at 8:10
user344045
asked Mar 24 at 14:33
user344045user344045
907
asked Mar 24 at 14:33
user344045user344045
907
asked Mar 24 at 14:33
user344045user344045
907
907
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