Uniqueness: linear first order pde with constant coefficients












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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







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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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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edited Jan 24 at 10:35









Harry49

7,53431341




7,53431341










asked Jan 24 at 2:01









childishsadbinochildishsadbino

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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














  • 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










1 Answer
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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 $f$.






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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 $f$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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 $f$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $f$.






        share|cite|improve this answer











        $endgroup$



        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 $f$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 28 at 9:47

























        answered Jan 24 at 10:02









        Harry49Harry49

        7,53431341




        7,53431341






























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