Classifying linear first-order PDE system (elliptic, hyperbolic, or parabolic)












1












$begingroup$




  1. 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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$




    1. 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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$




      1. 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







      share|cite|improve this question











      $endgroup$






      1. 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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 22 at 15:08









      Harry49

      7,50431341




      7,50431341










      asked Jan 21 at 21:03









      Matias SalgoMatias Salgo

      113




      113






















          1 Answer
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          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













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            0












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$



                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 22 at 16:43

























                answered Jan 22 at 14:42









                Harry49Harry49

                7,50431341




                7,50431341






























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