Mixing
Uniqueness: linear first order pde with constant coefficients
$begingroup$
Let us say I find the characteristic lines of some easy PDE $a U_x + b U_y = 0$ to be $bx-ay=c$, where $b, a, c$ are constants. Now, we say the solution must be constant along those lines, so it HAS to be a function of $(bx-ay)$. We can write $U(x,y)=f(bx-ay)$ then.
Here's my question. If I have another function $u(x,y)$ that is a solution and is constant along those same lines, but does not have the same fixed value along each line as $f(bx-ay)$, what says that it HAS to be able to be written in the form $u(x,y)=f(bx-ay)$?
pde characteristics linear-pde
edited Jan 24 at 10:35
Harry49
7,53431341
asked Jan 24 at 2:01
childishsadbinochildishsadbino
1148
$endgroup$
add a comment |
$begingroup$
Let us say I find the characteristic lines of some easy PDE $a U_x + b U_y = 0$ to be $bx-ay=c$, where $b, a, c$ are constants. Now, we say the solution must be constant along those lines, so it HAS to be a function of $(bx-ay)$. We can write $U(x,y)=f(bx-ay)$ then.
Here's my question. If I have another function $u(x,y)$ that is a solution and is constant along those same lines, but does not have the same fixed value along each line as $f(bx-ay)$, what says that it HAS to be able to be written in the form $u(x,y)=f(bx-ay)$?
pde characteristics linear-pde
edited Jan 24 at 10:35
Harry49
7,53431341
asked Jan 24 at 2:01
childishsadbinochildishsadbino
1148
$endgroup$
1
$begingroup$
The second function need not be the same as the first, $u(x,y)=g(bx-ay)$. It only shares the characteristic lines.
$endgroup$
– Chrystomath
Jan 24 at 10:06
add a comment |
$begingroup$
Let us say I find the characteristic lines of some easy PDE $a U_x + b U_y = 0$ to be $bx-ay=c$, where $b, a, c$ are constants. Now, we say the solution must be constant along those lines, so it HAS to be a function of $(bx-ay)$. We can write $U(x,y)=f(bx-ay)$ then.
Here's my question. If I have another function $u(x,y)$ that is a solution and is constant along those same lines, but does not have the same fixed value along each line as $f(bx-ay)$, what says that it HAS to be able to be written in the form $u(x,y)=f(bx-ay)$?
pde characteristics linear-pde
edited Jan 24 at 10:35
Harry49
7,53431341
asked Jan 24 at 2:01
childishsadbinochildishsadbino
1148
$endgroup$
Let us say I find the characteristic lines of some easy PDE $a U_x + b U_y = 0$ to be $bx-ay=c$, where $b, a, c$ are constants. Now, we say the solution must be constant along those lines, so it HAS to be a function of $(bx-ay)$. We can write $U(x,y)=f(bx-ay)$ then.
Here's my question. If I have another function $u(x,y)$ that is a solution and is constant along those same lines, but does not have the same fixed value along each line as $f(bx-ay)$, what says that it HAS to be able to be written in the form $u(x,y)=f(bx-ay)$?
pde characteristics linear-pde
pde characteristics linear-pde
edited Jan 24 at 10:35
Harry49
7,53431341
asked Jan 24 at 2:01
childishsadbinochildishsadbino
1148
edited Jan 24 at 10:35
Harry49
7,53431341
asked Jan 24 at 2:01
childishsadbinochildishsadbino
1148
edited Jan 24 at 10:35
Harry49
7,53431341
edited Jan 24 at 10:35
Harry49
7,53431341
edited Jan 24 at 10:35
Harry49
7,53431341
7,53431341
asked Jan 24 at 2:01
childishsadbinochildishsadbino
1148
asked Jan 24 at 2:01
childishsadbinochildishsadbino
1148
asked Jan 24 at 2:01
childishsadbinochildishsadbino
1148
1148
1
$begingroup$
The second function need not be the same as the first, $u(x,y)=g(bx-ay)$. It only shares the characteristic lines.
$endgroup$
– Chrystomath
Jan 24 at 10:06
add a comment |
1
$begingroup$
The second function need not be the same as the first, $u(x,y)=g(bx-ay)$. It only shares the characteristic lines.
$endgroup$
– Chrystomath
Jan 24 at 10:06
1
1
$begingroup$
The second function need not be the same as the first, $u(x,y)=g(bx-ay)$. It only shares the characteristic lines.
$endgroup$
– Chrystomath
Jan 24 at 10:06
$begingroup$
The second function need not be the same as the first, $u(x,y)=g(bx-ay)$. It only shares the characteristic lines.
$endgroup$
– Chrystomath
Jan 24 at 10:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The fact that $U$ must be constant along the lines $bx-ay = c$ is only a consequence of the method of characteristics. To determine the values $U = f(c)$ along those lines in a unique way, we need to specify some boundary conditions (or boundary data).
For example, let us assume that $U$ is known along the line $Ax+By=C$. For any point $(x,y)$, we construct the intersection $(x_0,y_0)$ of the line $Ax+By=C$ with the characteristic curve:
begin{aligned}
bx_0-ay_0 &= c = bx-ay\
Ax_0+By_0 &= C
end{aligned}
The determinant of this system is $Delta = aA + bB$, and $Deltaneq 0$ leads to
$$
x_0 = frac{aC + Bc}{aA + bB} ,quad y_0 = frac{bC - Ac}{aA + bB}
$$
Existence and uniqueness is only guaranteed if $Delta neq 0$, i.e. if the characteristic lines are not parallel to the line $Ax+By=C$ where $U$ is known. The function $f$ defining the solution as $U(x,y) = f(bx-ay)$ is related to the boundary data. Another set of boundary data would provide another such function $f$.
answered Jan 24 at 10:02
Harry49Harry49
7,53431341
$endgroup$
add a comment |
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$begingroup$
The fact that $U$ must be constant along the lines $bx-ay = c$ is only a consequence of the method of characteristics. To determine the values $U = f(c)$ along those lines in a unique way, we need to specify some boundary conditions (or boundary data).
For example, let us assume that $U$ is known along the line $Ax+By=C$. For any point $(x,y)$, we construct the intersection $(x_0,y_0)$ of the line $Ax+By=C$ with the characteristic curve:
begin{aligned}
bx_0-ay_0 &= c = bx-ay\
Ax_0+By_0 &= C
end{aligned}
The determinant of this system is $Delta = aA + bB$, and $Deltaneq 0$ leads to
$$
x_0 = frac{aC + Bc}{aA + bB} ,quad y_0 = frac{bC - Ac}{aA + bB}
$$
Existence and uniqueness is only guaranteed if $Delta neq 0$, i.e. if the characteristic lines are not parallel to the line $Ax+By=C$ where $U$ is known. The function $f$ defining the solution as $U(x,y) = f(bx-ay)$ is related to the boundary data. Another set of boundary data would provide another such function $f$.
answered Jan 24 at 10:02
Harry49Harry49
7,53431341
$endgroup$
add a comment |
$begingroup$
The fact that $U$ must be constant along the lines $bx-ay = c$ is only a consequence of the method of characteristics. To determine the values $U = f(c)$ along those lines in a unique way, we need to specify some boundary conditions (or boundary data).
For example, let us assume that $U$ is known along the line $Ax+By=C$. For any point $(x,y)$, we construct the intersection $(x_0,y_0)$ of the line $Ax+By=C$ with the characteristic curve:
begin{aligned}
bx_0-ay_0 &= c = bx-ay\
Ax_0+By_0 &= C
end{aligned}
The determinant of this system is $Delta = aA + bB$, and $Deltaneq 0$ leads to
$$
x_0 = frac{aC + Bc}{aA + bB} ,quad y_0 = frac{bC - Ac}{aA + bB}
$$
Existence and uniqueness is only guaranteed if $Delta neq 0$, i.e. if the characteristic lines are not parallel to the line $Ax+By=C$ where $U$ is known. The function $f$ defining the solution as $U(x,y) = f(bx-ay)$ is related to the boundary data. Another set of boundary data would provide another such function $f$.
answered Jan 24 at 10:02
Harry49Harry49
7,53431341
$endgroup$
add a comment |
$begingroup$
The fact that $U$ must be constant along the lines $bx-ay = c$ is only a consequence of the method of characteristics. To determine the values $U = f(c)$ along those lines in a unique way, we need to specify some boundary conditions (or boundary data).
For example, let us assume that $U$ is known along the line $Ax+By=C$. For any point $(x,y)$, we construct the intersection $(x_0,y_0)$ of the line $Ax+By=C$ with the characteristic curve:
begin{aligned}
bx_0-ay_0 &= c = bx-ay\
Ax_0+By_0 &= C
end{aligned}
The determinant of this system is $Delta = aA + bB$, and $Deltaneq 0$ leads to
$$
x_0 = frac{aC + Bc}{aA + bB} ,quad y_0 = frac{bC - Ac}{aA + bB}
$$
Existence and uniqueness is only guaranteed if $Delta neq 0$, i.e. if the characteristic lines are not parallel to the line $Ax+By=C$ where $U$ is known. The function $f$ defining the solution as $U(x,y) = f(bx-ay)$ is related to the boundary data. Another set of boundary data would provide another such function $f$.
answered Jan 24 at 10:02
Harry49Harry49
7,53431341
$endgroup$
The fact that $U$ must be constant along the lines $bx-ay = c$ is only a consequence of the method of characteristics. To determine the values $U = f(c)$ along those lines in a unique way, we need to specify some boundary conditions (or boundary data).
For example, let us assume that $U$ is known along the line $Ax+By=C$. For any point $(x,y)$, we construct the intersection $(x_0,y_0)$ of the line $Ax+By=C$ with the characteristic curve:
begin{aligned}
bx_0-ay_0 &= c = bx-ay\
Ax_0+By_0 &= C
end{aligned}
The determinant of this system is $Delta = aA + bB$, and $Deltaneq 0$ leads to
$$
x_0 = frac{aC + Bc}{aA + bB} ,quad y_0 = frac{bC - Ac}{aA + bB}
$$
Existence and uniqueness is only guaranteed if $Delta neq 0$, i.e. if the characteristic lines are not parallel to the line $Ax+By=C$ where $U$ is known. The function $f$ defining the solution as $U(x,y) = f(bx-ay)$ is related to the boundary data. Another set of boundary data would provide another such function $f$.
answered Jan 24 at 10:02
Harry49Harry49
7,53431341
edited Jan 28 at 9:47
answered Jan 24 at 10:02
Harry49Harry49
7,53431341
answered Jan 24 at 10:02
Harry49Harry49
7,53431341
answered Jan 24 at 10:02
Harry49Harry49
7,53431341
7,53431341
add a comment |
add a comment |
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$begingroup$
The second function need not be the same as the first, $u(x,y)=g(bx-ay)$. It only shares the characteristic lines.
$endgroup$
– Chrystomath
Jan 24 at 10:06