Eigen values of a Third Order Linear Homogenous ODE












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$begingroup$


I have two third order linear ODE which have been arrived after applying separation of variables to a system of PDEs



begin{eqnarray}
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}



$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are



For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$



For $G$:
$$G(0)=0$$
$$frac{G''(0)}{G'(0)}=beta_c$$
$$frac{G''(1)}{G'(1)}=beta_c$$



$lambda_h$, $lambda_c$, $beta_h$ and $beta_c$ are all constants $>0$.



$mu$ is the separation constant .



I need to determine eigenvalues for each BVP involving $F$ and $G$. So for finding out eigenvalues i know i need to consider all the three cases
$mu>0$, $mu<0$ and $mu=0$ and then look for non-trivial solutions by applying the specific set of b.c. Although i am acquainted with the procedure to determine eigenvalues for a second order DE, the third order of the DE(s)is something i am not familiar with.



Any recommendations on how should i go about tackling this ?



Attempt
As per @Cesareo recommendations, I arrive at the following
linear equations



$$C_1+C_2+C_3=0$$



$$frac{F''(0)}{F'(0)}=frac{{C_1}{delta_1(mu)}^2+{C_2}{delta_2(mu)}^2+{C_3}{delta_3(mu)}^2}{-{C_1}{delta_1(mu)}-{C_2}{delta_2(mu)}-{C_3}{delta_3(mu)}}=beta_h$$



$$frac{F''(1)}{F'(1)}=frac{{C_1e^{-delta_1(mu)}}{delta_1(mu)}^2+{C_2e^{-delta_2(mu)}}{delta_2(mu)}^2+{C_3e^{-delta_3(mu)}}{delta_3(mu)}^2}{-{C_1e^{-delta_1(mu)}}{delta_1(mu)}-{C_2e^{-delta_2(mu)}}{delta_2(mu)}-{C_3e^{-delta_3(mu)}}{delta_3(mu)}}=beta_h$$



I reach the following form of $M(mu).C=0$



begin{vmatrix}
1 & 1 & 1 \
{delta_1(mu)}^2+beta_hdelta_1(mu) & {delta_2(mu)}^2+beta_hdelta_2(mu) & {delta_3(mu)}^2+beta_hdelta_3(mu) \
e^{-delta_1(mu)}({delta_1(mu)}^2+beta_hdelta_1(mu)) & e^{-delta_2(mu)}({delta_2(mu)}^2+beta_hdelta_2(mu)) & e^{-delta_3(mu)}({delta_3(mu)}^2+beta_hdelta_3(mu)) \
end{vmatrix}
$=0$



Solving this determinant is supposed to give me the eigen values $mu_n$ and consequently the eigen functions. The determinant can be reduced to two $0$ in the first row by coloumn manipulation, but I do not find any way to handle the consequent equation that comes out of it which is something like this:



$$[(delta_1(mu)-delta_2(mu))(delta_1(mu)+delta_2(mu)+beta_h)[(e^{-delta_2(mu)}{delta_2(mu)}^2-e^{-delta_3(mu)}{delta_3(mu)}^2)+beta_h(e^{-delta_2(mu)}{delta_2(mu)}-e^{-delta_3(mu)}{delta_3(mu)})]]-[(delta_2(mu)-delta_3(mu))(delta_2(mu)+delta_3(mu)+beta_h)[(e^{-delta_1(mu)}{delta_1(mu)}^2-e^{-delta_2(mu)}{delta_2(mu)}^2)+beta_h(e^{-delta_1(mu)}{delta_1(mu)}-e^{-delta_2(mu)}{delta_2(mu)})]]=0$$



After this step i fail to proceed further to find the eigenvalues using this $mathbb{det}M=0$ equation










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    $begingroup$


    I have two third order linear ODE which have been arrived after applying separation of variables to a system of PDEs



    begin{eqnarray}
    lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
    V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
    end{eqnarray}



    $F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are



    For $F$:
    $$F(0)=0$$
    $$frac{F''(0)}{F'(0)}=beta_h$$
    $$frac{F''(1)}{F'(1)}=beta_h$$



    For $G$:
    $$G(0)=0$$
    $$frac{G''(0)}{G'(0)}=beta_c$$
    $$frac{G''(1)}{G'(1)}=beta_c$$



    $lambda_h$, $lambda_c$, $beta_h$ and $beta_c$ are all constants $>0$.



    $mu$ is the separation constant .



    I need to determine eigenvalues for each BVP involving $F$ and $G$. So for finding out eigenvalues i know i need to consider all the three cases
    $mu>0$, $mu<0$ and $mu=0$ and then look for non-trivial solutions by applying the specific set of b.c. Although i am acquainted with the procedure to determine eigenvalues for a second order DE, the third order of the DE(s)is something i am not familiar with.



    Any recommendations on how should i go about tackling this ?



    Attempt
    As per @Cesareo recommendations, I arrive at the following
    linear equations



    $$C_1+C_2+C_3=0$$



    $$frac{F''(0)}{F'(0)}=frac{{C_1}{delta_1(mu)}^2+{C_2}{delta_2(mu)}^2+{C_3}{delta_3(mu)}^2}{-{C_1}{delta_1(mu)}-{C_2}{delta_2(mu)}-{C_3}{delta_3(mu)}}=beta_h$$



    $$frac{F''(1)}{F'(1)}=frac{{C_1e^{-delta_1(mu)}}{delta_1(mu)}^2+{C_2e^{-delta_2(mu)}}{delta_2(mu)}^2+{C_3e^{-delta_3(mu)}}{delta_3(mu)}^2}{-{C_1e^{-delta_1(mu)}}{delta_1(mu)}-{C_2e^{-delta_2(mu)}}{delta_2(mu)}-{C_3e^{-delta_3(mu)}}{delta_3(mu)}}=beta_h$$



    I reach the following form of $M(mu).C=0$



    begin{vmatrix}
    1 & 1 & 1 \
    {delta_1(mu)}^2+beta_hdelta_1(mu) & {delta_2(mu)}^2+beta_hdelta_2(mu) & {delta_3(mu)}^2+beta_hdelta_3(mu) \
    e^{-delta_1(mu)}({delta_1(mu)}^2+beta_hdelta_1(mu)) & e^{-delta_2(mu)}({delta_2(mu)}^2+beta_hdelta_2(mu)) & e^{-delta_3(mu)}({delta_3(mu)}^2+beta_hdelta_3(mu)) \
    end{vmatrix}
    $=0$



    Solving this determinant is supposed to give me the eigen values $mu_n$ and consequently the eigen functions. The determinant can be reduced to two $0$ in the first row by coloumn manipulation, but I do not find any way to handle the consequent equation that comes out of it which is something like this:



    $$[(delta_1(mu)-delta_2(mu))(delta_1(mu)+delta_2(mu)+beta_h)[(e^{-delta_2(mu)}{delta_2(mu)}^2-e^{-delta_3(mu)}{delta_3(mu)}^2)+beta_h(e^{-delta_2(mu)}{delta_2(mu)}-e^{-delta_3(mu)}{delta_3(mu)})]]-[(delta_2(mu)-delta_3(mu))(delta_2(mu)+delta_3(mu)+beta_h)[(e^{-delta_1(mu)}{delta_1(mu)}^2-e^{-delta_2(mu)}{delta_2(mu)}^2)+beta_h(e^{-delta_1(mu)}{delta_1(mu)}-e^{-delta_2(mu)}{delta_2(mu)})]]=0$$



    After this step i fail to proceed further to find the eigenvalues using this $mathbb{det}M=0$ equation










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have two third order linear ODE which have been arrived after applying separation of variables to a system of PDEs



      begin{eqnarray}
      lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
      V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
      end{eqnarray}



      $F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are



      For $F$:
      $$F(0)=0$$
      $$frac{F''(0)}{F'(0)}=beta_h$$
      $$frac{F''(1)}{F'(1)}=beta_h$$



      For $G$:
      $$G(0)=0$$
      $$frac{G''(0)}{G'(0)}=beta_c$$
      $$frac{G''(1)}{G'(1)}=beta_c$$



      $lambda_h$, $lambda_c$, $beta_h$ and $beta_c$ are all constants $>0$.



      $mu$ is the separation constant .



      I need to determine eigenvalues for each BVP involving $F$ and $G$. So for finding out eigenvalues i know i need to consider all the three cases
      $mu>0$, $mu<0$ and $mu=0$ and then look for non-trivial solutions by applying the specific set of b.c. Although i am acquainted with the procedure to determine eigenvalues for a second order DE, the third order of the DE(s)is something i am not familiar with.



      Any recommendations on how should i go about tackling this ?



      Attempt
      As per @Cesareo recommendations, I arrive at the following
      linear equations



      $$C_1+C_2+C_3=0$$



      $$frac{F''(0)}{F'(0)}=frac{{C_1}{delta_1(mu)}^2+{C_2}{delta_2(mu)}^2+{C_3}{delta_3(mu)}^2}{-{C_1}{delta_1(mu)}-{C_2}{delta_2(mu)}-{C_3}{delta_3(mu)}}=beta_h$$



      $$frac{F''(1)}{F'(1)}=frac{{C_1e^{-delta_1(mu)}}{delta_1(mu)}^2+{C_2e^{-delta_2(mu)}}{delta_2(mu)}^2+{C_3e^{-delta_3(mu)}}{delta_3(mu)}^2}{-{C_1e^{-delta_1(mu)}}{delta_1(mu)}-{C_2e^{-delta_2(mu)}}{delta_2(mu)}-{C_3e^{-delta_3(mu)}}{delta_3(mu)}}=beta_h$$



      I reach the following form of $M(mu).C=0$



      begin{vmatrix}
      1 & 1 & 1 \
      {delta_1(mu)}^2+beta_hdelta_1(mu) & {delta_2(mu)}^2+beta_hdelta_2(mu) & {delta_3(mu)}^2+beta_hdelta_3(mu) \
      e^{-delta_1(mu)}({delta_1(mu)}^2+beta_hdelta_1(mu)) & e^{-delta_2(mu)}({delta_2(mu)}^2+beta_hdelta_2(mu)) & e^{-delta_3(mu)}({delta_3(mu)}^2+beta_hdelta_3(mu)) \
      end{vmatrix}
      $=0$



      Solving this determinant is supposed to give me the eigen values $mu_n$ and consequently the eigen functions. The determinant can be reduced to two $0$ in the first row by coloumn manipulation, but I do not find any way to handle the consequent equation that comes out of it which is something like this:



      $$[(delta_1(mu)-delta_2(mu))(delta_1(mu)+delta_2(mu)+beta_h)[(e^{-delta_2(mu)}{delta_2(mu)}^2-e^{-delta_3(mu)}{delta_3(mu)}^2)+beta_h(e^{-delta_2(mu)}{delta_2(mu)}-e^{-delta_3(mu)}{delta_3(mu)})]]-[(delta_2(mu)-delta_3(mu))(delta_2(mu)+delta_3(mu)+beta_h)[(e^{-delta_1(mu)}{delta_1(mu)}^2-e^{-delta_2(mu)}{delta_2(mu)}^2)+beta_h(e^{-delta_1(mu)}{delta_1(mu)}-e^{-delta_2(mu)}{delta_2(mu)})]]=0$$



      After this step i fail to proceed further to find the eigenvalues using this $mathbb{det}M=0$ equation










      share|cite|improve this question











      $endgroup$




      I have two third order linear ODE which have been arrived after applying separation of variables to a system of PDEs



      begin{eqnarray}
      lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
      V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
      end{eqnarray}



      $F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are



      For $F$:
      $$F(0)=0$$
      $$frac{F''(0)}{F'(0)}=beta_h$$
      $$frac{F''(1)}{F'(1)}=beta_h$$



      For $G$:
      $$G(0)=0$$
      $$frac{G''(0)}{G'(0)}=beta_c$$
      $$frac{G''(1)}{G'(1)}=beta_c$$



      $lambda_h$, $lambda_c$, $beta_h$ and $beta_c$ are all constants $>0$.



      $mu$ is the separation constant .



      I need to determine eigenvalues for each BVP involving $F$ and $G$. So for finding out eigenvalues i know i need to consider all the three cases
      $mu>0$, $mu<0$ and $mu=0$ and then look for non-trivial solutions by applying the specific set of b.c. Although i am acquainted with the procedure to determine eigenvalues for a second order DE, the third order of the DE(s)is something i am not familiar with.



      Any recommendations on how should i go about tackling this ?



      Attempt
      As per @Cesareo recommendations, I arrive at the following
      linear equations



      $$C_1+C_2+C_3=0$$



      $$frac{F''(0)}{F'(0)}=frac{{C_1}{delta_1(mu)}^2+{C_2}{delta_2(mu)}^2+{C_3}{delta_3(mu)}^2}{-{C_1}{delta_1(mu)}-{C_2}{delta_2(mu)}-{C_3}{delta_3(mu)}}=beta_h$$



      $$frac{F''(1)}{F'(1)}=frac{{C_1e^{-delta_1(mu)}}{delta_1(mu)}^2+{C_2e^{-delta_2(mu)}}{delta_2(mu)}^2+{C_3e^{-delta_3(mu)}}{delta_3(mu)}^2}{-{C_1e^{-delta_1(mu)}}{delta_1(mu)}-{C_2e^{-delta_2(mu)}}{delta_2(mu)}-{C_3e^{-delta_3(mu)}}{delta_3(mu)}}=beta_h$$



      I reach the following form of $M(mu).C=0$



      begin{vmatrix}
      1 & 1 & 1 \
      {delta_1(mu)}^2+beta_hdelta_1(mu) & {delta_2(mu)}^2+beta_hdelta_2(mu) & {delta_3(mu)}^2+beta_hdelta_3(mu) \
      e^{-delta_1(mu)}({delta_1(mu)}^2+beta_hdelta_1(mu)) & e^{-delta_2(mu)}({delta_2(mu)}^2+beta_hdelta_2(mu)) & e^{-delta_3(mu)}({delta_3(mu)}^2+beta_hdelta_3(mu)) \
      end{vmatrix}
      $=0$



      Solving this determinant is supposed to give me the eigen values $mu_n$ and consequently the eigen functions. The determinant can be reduced to two $0$ in the first row by coloumn manipulation, but I do not find any way to handle the consequent equation that comes out of it which is something like this:



      $$[(delta_1(mu)-delta_2(mu))(delta_1(mu)+delta_2(mu)+beta_h)[(e^{-delta_2(mu)}{delta_2(mu)}^2-e^{-delta_3(mu)}{delta_3(mu)}^2)+beta_h(e^{-delta_2(mu)}{delta_2(mu)}-e^{-delta_3(mu)}{delta_3(mu)})]]-[(delta_2(mu)-delta_3(mu))(delta_2(mu)+delta_3(mu)+beta_h)[(e^{-delta_1(mu)}{delta_1(mu)}^2-e^{-delta_2(mu)}{delta_2(mu)}^2)+beta_h(e^{-delta_1(mu)}{delta_1(mu)}-e^{-delta_2(mu)}{delta_2(mu)})]]=0$$



      After this step i fail to proceed further to find the eigenvalues using this $mathbb{det}M=0$ equation







      ordinary-differential-equations pde eigenvalues-eigenvectors eigenfunctions






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      edited Jan 13 at 3:42







      Indrasis Mitra

















      asked Jan 11 at 12:06









      Indrasis MitraIndrasis Mitra

      2517




      2517






















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          $begingroup$

          Regarding the first DE the linear differential operator



          $$
          lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
          $$



          and the three roots $delta_i(mu), i = 1,2,3$ we have that



          $$
          F(t) = sum_k C_k e^{-delta_k(mu)t}
          $$



          using now the boundary conditions



          $$
          F(0) = sum_k C_k = 0 longrightarrow (1)
          $$



          and with



          $$
          F'(0) = -sum_k C_k delta_k(mu)\
          F''(0) = sum_k C_k delta_k(mu)^2\
          $$



          giving



          $$
          -frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
          $$



          and similarly



          $$
          -frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
          $$



          then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as



          $$
          M(mu)cdot C = 0, C = (C_k)
          $$



          This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$



          The procedure for $G$ is quite similar.



          NOTE



          Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial



          $$
          s^3+20s^2+(110-mu)s+100 = 0
          $$



          with roots $delta_1(mu),delta_2(mu),delta_3(mu)$



          The determinant after simplifications reads



          $$
          det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
          _h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
          delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
          _h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
          _3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
          _3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
          left(delta _1+delta _2+delta _3right)right)
          $$



          discarding $cosh (delta_1+delta_2+delta_3)-sinh
          (delta_1+delta_2+delta_3)=0$
          we follow with



          $$
          Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
          _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
          left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
          _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
          left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
          $$



          and then after plotting we have



          enter image description here



          In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$



          Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$



          parms = {lh -> 1/4, bh -> -10};
          F[t_, n_] := Sum[
          !(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
          !(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
          !(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
          sols = Solve[
          lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
          parms // FullSimplify
          roots = s /. sols;
          M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
          {E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
          det = -Det[M] // FullSimplify
          subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
          subdet2 = det/subdet1 // FullSimplify
          subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
          Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
          PlotRange -> {-10, 10}]
          solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
          solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
          solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
          solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
          solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
          solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
          solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
          solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
          solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
          Subscript[mu, 1] = mu /. solmu1;
          Subscript[mu, 2] = mu /. solmu2;
          Subscript[mu, 3] = mu /. solmu3;
          Subscript[mu, 4] = mu /. solmu4;
          Subscript[mu, 5] = mu /. solmu5;
          Subscript[mu, 6] = mu /. solmu6;
          Subscript[mu, 7] = mu /. solmu7;
          Subscript[mu, 8] = mu /. solmu8;
          Subscript[mu, 9] = mu /. solmu9;
          exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
          F[t, 9]





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          $endgroup$













          • $begingroup$
            Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 3:10










          • $begingroup$
            Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 4:26










          • $begingroup$
            my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 6:40












          • $begingroup$
            I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
            $endgroup$
            – Indrasis Mitra
            Jan 13 at 3:45










          • $begingroup$
            @IndrasisMitra See attached note.
            $endgroup$
            – Cesareo
            Jan 13 at 9:19











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          1












          $begingroup$

          Regarding the first DE the linear differential operator



          $$
          lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
          $$



          and the three roots $delta_i(mu), i = 1,2,3$ we have that



          $$
          F(t) = sum_k C_k e^{-delta_k(mu)t}
          $$



          using now the boundary conditions



          $$
          F(0) = sum_k C_k = 0 longrightarrow (1)
          $$



          and with



          $$
          F'(0) = -sum_k C_k delta_k(mu)\
          F''(0) = sum_k C_k delta_k(mu)^2\
          $$



          giving



          $$
          -frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
          $$



          and similarly



          $$
          -frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
          $$



          then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as



          $$
          M(mu)cdot C = 0, C = (C_k)
          $$



          This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$



          The procedure for $G$ is quite similar.



          NOTE



          Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial



          $$
          s^3+20s^2+(110-mu)s+100 = 0
          $$



          with roots $delta_1(mu),delta_2(mu),delta_3(mu)$



          The determinant after simplifications reads



          $$
          det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
          _h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
          delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
          _h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
          _3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
          _3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
          left(delta _1+delta _2+delta _3right)right)
          $$



          discarding $cosh (delta_1+delta_2+delta_3)-sinh
          (delta_1+delta_2+delta_3)=0$
          we follow with



          $$
          Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
          _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
          left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
          _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
          left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
          $$



          and then after plotting we have



          enter image description here



          In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$



          Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$



          parms = {lh -> 1/4, bh -> -10};
          F[t_, n_] := Sum[
          !(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
          !(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
          !(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
          sols = Solve[
          lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
          parms // FullSimplify
          roots = s /. sols;
          M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
          {E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
          det = -Det[M] // FullSimplify
          subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
          subdet2 = det/subdet1 // FullSimplify
          subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
          Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
          PlotRange -> {-10, 10}]
          solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
          solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
          solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
          solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
          solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
          solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
          solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
          solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
          solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
          Subscript[mu, 1] = mu /. solmu1;
          Subscript[mu, 2] = mu /. solmu2;
          Subscript[mu, 3] = mu /. solmu3;
          Subscript[mu, 4] = mu /. solmu4;
          Subscript[mu, 5] = mu /. solmu5;
          Subscript[mu, 6] = mu /. solmu6;
          Subscript[mu, 7] = mu /. solmu7;
          Subscript[mu, 8] = mu /. solmu8;
          Subscript[mu, 9] = mu /. solmu9;
          exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
          F[t, 9]





          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 3:10










          • $begingroup$
            Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 4:26










          • $begingroup$
            my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 6:40












          • $begingroup$
            I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
            $endgroup$
            – Indrasis Mitra
            Jan 13 at 3:45










          • $begingroup$
            @IndrasisMitra See attached note.
            $endgroup$
            – Cesareo
            Jan 13 at 9:19
















          1












          $begingroup$

          Regarding the first DE the linear differential operator



          $$
          lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
          $$



          and the three roots $delta_i(mu), i = 1,2,3$ we have that



          $$
          F(t) = sum_k C_k e^{-delta_k(mu)t}
          $$



          using now the boundary conditions



          $$
          F(0) = sum_k C_k = 0 longrightarrow (1)
          $$



          and with



          $$
          F'(0) = -sum_k C_k delta_k(mu)\
          F''(0) = sum_k C_k delta_k(mu)^2\
          $$



          giving



          $$
          -frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
          $$



          and similarly



          $$
          -frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
          $$



          then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as



          $$
          M(mu)cdot C = 0, C = (C_k)
          $$



          This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$



          The procedure for $G$ is quite similar.



          NOTE



          Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial



          $$
          s^3+20s^2+(110-mu)s+100 = 0
          $$



          with roots $delta_1(mu),delta_2(mu),delta_3(mu)$



          The determinant after simplifications reads



          $$
          det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
          _h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
          delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
          _h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
          _3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
          _3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
          left(delta _1+delta _2+delta _3right)right)
          $$



          discarding $cosh (delta_1+delta_2+delta_3)-sinh
          (delta_1+delta_2+delta_3)=0$
          we follow with



          $$
          Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
          _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
          left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
          _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
          left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
          $$



          and then after plotting we have



          enter image description here



          In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$



          Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$



          parms = {lh -> 1/4, bh -> -10};
          F[t_, n_] := Sum[
          !(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
          !(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
          !(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
          sols = Solve[
          lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
          parms // FullSimplify
          roots = s /. sols;
          M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
          {E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
          det = -Det[M] // FullSimplify
          subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
          subdet2 = det/subdet1 // FullSimplify
          subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
          Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
          PlotRange -> {-10, 10}]
          solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
          solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
          solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
          solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
          solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
          solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
          solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
          solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
          solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
          Subscript[mu, 1] = mu /. solmu1;
          Subscript[mu, 2] = mu /. solmu2;
          Subscript[mu, 3] = mu /. solmu3;
          Subscript[mu, 4] = mu /. solmu4;
          Subscript[mu, 5] = mu /. solmu5;
          Subscript[mu, 6] = mu /. solmu6;
          Subscript[mu, 7] = mu /. solmu7;
          Subscript[mu, 8] = mu /. solmu8;
          Subscript[mu, 9] = mu /. solmu9;
          exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
          F[t, 9]





          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 3:10










          • $begingroup$
            Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 4:26










          • $begingroup$
            my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 6:40












          • $begingroup$
            I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
            $endgroup$
            – Indrasis Mitra
            Jan 13 at 3:45










          • $begingroup$
            @IndrasisMitra See attached note.
            $endgroup$
            – Cesareo
            Jan 13 at 9:19














          1












          1








          1





          $begingroup$

          Regarding the first DE the linear differential operator



          $$
          lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
          $$



          and the three roots $delta_i(mu), i = 1,2,3$ we have that



          $$
          F(t) = sum_k C_k e^{-delta_k(mu)t}
          $$



          using now the boundary conditions



          $$
          F(0) = sum_k C_k = 0 longrightarrow (1)
          $$



          and with



          $$
          F'(0) = -sum_k C_k delta_k(mu)\
          F''(0) = sum_k C_k delta_k(mu)^2\
          $$



          giving



          $$
          -frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
          $$



          and similarly



          $$
          -frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
          $$



          then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as



          $$
          M(mu)cdot C = 0, C = (C_k)
          $$



          This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$



          The procedure for $G$ is quite similar.



          NOTE



          Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial



          $$
          s^3+20s^2+(110-mu)s+100 = 0
          $$



          with roots $delta_1(mu),delta_2(mu),delta_3(mu)$



          The determinant after simplifications reads



          $$
          det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
          _h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
          delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
          _h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
          _3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
          _3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
          left(delta _1+delta _2+delta _3right)right)
          $$



          discarding $cosh (delta_1+delta_2+delta_3)-sinh
          (delta_1+delta_2+delta_3)=0$
          we follow with



          $$
          Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
          _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
          left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
          _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
          left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
          $$



          and then after plotting we have



          enter image description here



          In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$



          Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$



          parms = {lh -> 1/4, bh -> -10};
          F[t_, n_] := Sum[
          !(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
          !(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
          !(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
          sols = Solve[
          lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
          parms // FullSimplify
          roots = s /. sols;
          M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
          {E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
          det = -Det[M] // FullSimplify
          subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
          subdet2 = det/subdet1 // FullSimplify
          subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
          Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
          PlotRange -> {-10, 10}]
          solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
          solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
          solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
          solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
          solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
          solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
          solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
          solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
          solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
          Subscript[mu, 1] = mu /. solmu1;
          Subscript[mu, 2] = mu /. solmu2;
          Subscript[mu, 3] = mu /. solmu3;
          Subscript[mu, 4] = mu /. solmu4;
          Subscript[mu, 5] = mu /. solmu5;
          Subscript[mu, 6] = mu /. solmu6;
          Subscript[mu, 7] = mu /. solmu7;
          Subscript[mu, 8] = mu /. solmu8;
          Subscript[mu, 9] = mu /. solmu9;
          exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
          F[t, 9]





          share|cite|improve this answer











          $endgroup$



          Regarding the first DE the linear differential operator



          $$
          lambda_h delta^3 - 2 lambda_h beta_h delta^2 + left( (lambda_h beta_h - 1) beta_h - mu right) delta + beta_h^2=0
          $$



          and the three roots $delta_i(mu), i = 1,2,3$ we have that



          $$
          F(t) = sum_k C_k e^{-delta_k(mu)t}
          $$



          using now the boundary conditions



          $$
          F(0) = sum_k C_k = 0 longrightarrow (1)
          $$



          and with



          $$
          F'(0) = -sum_k C_k delta_k(mu)\
          F''(0) = sum_k C_k delta_k(mu)^2\
          $$



          giving



          $$
          -frac{sum_k C_k delta_k(mu)^2}{sum_k C_k delta_k(mu)}=beta_hlongrightarrow (2)
          $$



          and similarly



          $$
          -frac{sum_k C_k delta_k(mu)^2e^{-delta_k(mu)}}{sum_k C_k delta_k(mu)e^{-delta_k(mu)}}=beta_hlongrightarrow (3)
          $$



          then we have three linear equations $(1,2,3)$ in $C_k$ that can be arranged as



          $$
          M(mu)cdot C = 0, C = (C_k)
          $$



          This system have nontrivial solution for $det(M(mu)) = 0$ hence the roots for this determinant equation are the eigenvalues $mu_n$ and the eigenfunctions are $e^{-delta_k(mu_n)t}$



          The procedure for $G$ is quite similar.



          NOTE



          Assuming numerical values $lambda_h = 1,beta_h = -10$ we have the operator polynomial



          $$
          s^3+20s^2+(110-mu)s+100 = 0
          $$



          with roots $delta_1(mu),delta_2(mu),delta_3(mu)$



          The determinant after simplifications reads



          $$
          det(M) = left(e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta
          _h+delta _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3}
          delta _2 left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta
          _h+delta _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta
          _3right) left(beta _h+delta _1right) left(beta _h+delta _2+delta
          _3right)right) left(cosh left(delta _1+delta _2+delta _3right)-sinh
          left(delta _1+delta _2+delta _3right)right)
          $$



          discarding $cosh (delta_1+delta_2+delta_3)-sinh
          (delta_1+delta_2+delta_3)=0$
          we follow with



          $$
          Delta(mu)=e^{delta _1+delta _2} left(delta _1-delta _2right) delta _3 left(beta _h+delta
          _1+delta _2right) left(beta _h+delta _3right)-e^{delta _1+delta _3} delta _2
          left(delta _1-delta _3right) left(beta _h+delta _2right) left(beta _h+delta
          _1+delta _3right)+e^{delta _2+delta _3} delta _1 left(delta _2-delta _3right)
          left(beta _h+delta _1right) left(beta _h+delta _2+delta _3right)=0
          $$



          and then after plotting we have



          enter image description here



          In red Re[$Delta(mu)$] and in blue Im[$Delta(mu)$]. The zeroes are the eigenvalues $mu_n$



          Attached a very basic MATHEMATICA script in order to obtain the first $mu_k$ for $lambda_h = frac 14, beta_h = -10$



          parms = {lh -> 1/4, bh -> -10};
          F[t_, n_] := Sum[
          !(*SubscriptBox[(c), ({1, j})]) Exp[exps[[j]][[1]] t] +
          !(*SubscriptBox[(c), ({2, j})]) Exp[exps[[j]][[2]] t] +
          !(*SubscriptBox[(c), ({3, j})]) Exp[exps[[j]][[3]] t], {j, 1,n}]
          sols = Solve[
          lh s^3 - 2 lh bh s^2 + ((lh bh - 1) bh - mu) s + bh^2 == 0, s] /.
          parms // FullSimplify
          roots = s /. sols;
          M = {{1, 1, 1}, {r1^2 + bh r1, r2^2 + bh r2, r3^2 + bh r3},
          {E^(-r1) (r1^2 + bh r1), E^(-r2) (r2^2 + bh r2), E^(-r3) (r3^2 + bh r3)}};
          det = -Det[M] // FullSimplify
          subdet1 = (Cosh[r1 + r2 + r3] - Sinh[r1 + r2 + r3])
          subdet2 = det/subdet1 // FullSimplify
          subdet20 = subdet2 /. Thread[{r1, r2, r3} -> roots] /. parms;
          Plot[Im[subdet20], {mu, -100, 0}, PlotStyle -> {Thick, Black},
          PlotRange -> {-10, 10}]
          solmu1 = FindRoot[Im[subdet20] == 0, {mu, 0}]
          solmu2 = FindRoot[Im[subdet20] == 0, {mu, -7}]
          solmu3 = FindRoot[Im[subdet20] == 0, {mu, -20}]
          solmu4 = FindRoot[Im[subdet20] == 0, {mu, -40}]
          solmu5 = FindRoot[Im[subdet20] == 0, {mu, -60}]
          solmu6 = FindRoot[Im[subdet20] == 0, {mu, -80}]
          solmu7 = FindRoot[Im[subdet20] == 0, {mu, -110}]
          solmu8 = FindRoot[Im[subdet20] == 0, {mu, -150}]
          solmu9 = FindRoot[Im[subdet20] == 0, {mu, -190}]
          Subscript[mu, 1] = mu /. solmu1;
          Subscript[mu, 2] = mu /. solmu2;
          Subscript[mu, 3] = mu /. solmu3;
          Subscript[mu, 4] = mu /. solmu4;
          Subscript[mu, 5] = mu /. solmu5;
          Subscript[mu, 6] = mu /. solmu6;
          Subscript[mu, 7] = mu /. solmu7;
          Subscript[mu, 8] = mu /. solmu8;
          Subscript[mu, 9] = mu /. solmu9;
          exps = Table[roots /. parms /. {mu -> Subscript[mu, k]}, {k, 1, 9}]
          F[t, 9]






          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 19 at 18:38

























          answered Jan 11 at 21:58









          CesareoCesareo

          8,6793516




          8,6793516












          • $begingroup$
            Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 3:10










          • $begingroup$
            Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 4:26










          • $begingroup$
            my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 6:40












          • $begingroup$
            I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
            $endgroup$
            – Indrasis Mitra
            Jan 13 at 3:45










          • $begingroup$
            @IndrasisMitra See attached note.
            $endgroup$
            – Cesareo
            Jan 13 at 9:19


















          • $begingroup$
            Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 3:10










          • $begingroup$
            Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 4:26










          • $begingroup$
            my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
            $endgroup$
            – Indrasis Mitra
            Jan 12 at 6:40












          • $begingroup$
            I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
            $endgroup$
            – Indrasis Mitra
            Jan 13 at 3:45










          • $begingroup$
            @IndrasisMitra See attached note.
            $endgroup$
            – Cesareo
            Jan 13 at 9:19
















          $begingroup$
          Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
          $endgroup$
          – Indrasis Mitra
          Jan 12 at 3:10




          $begingroup$
          Thanks for the explanation. I guess you used $F'/F''$ while writing $(2)$ and $(3)$ while the BC are actually $F''/F'$
          $endgroup$
          – Indrasis Mitra
          Jan 12 at 3:10












          $begingroup$
          Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
          $endgroup$
          – Indrasis Mitra
          Jan 12 at 4:26




          $begingroup$
          Also will solving for $det(M(mu))=0$ take care of all the possible values of $mu$ that needs to be considered i.e. $mu>0$,$mu=0$ and $mu<0$ ? for finding the eigenvalues ?
          $endgroup$
          – Indrasis Mitra
          Jan 12 at 4:26












          $begingroup$
          my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
          $endgroup$
          – Indrasis Mitra
          Jan 12 at 6:40






          $begingroup$
          my last comment is pretty ignorant. So actually $F(t)=C_1e^{-delta_1(mu)t}+C_2e^{-delta_2(mu)t}+C_3e^{-delta_3(mu)t}$, for the three roots of the characteristic equation. Am i right ?
          $endgroup$
          – Indrasis Mitra
          Jan 12 at 6:40














          $begingroup$
          I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
          $endgroup$
          – Indrasis Mitra
          Jan 13 at 3:45




          $begingroup$
          I followed the steps you suggested to atrrive at $mathbb{det}(M(mu))=0$. I find that it has $delta_1(mu)$,$delta_2(mu)$ and $delta_3(mu)$. I have edited the original question to reflect my attempt. I cannot figure out how to proceed further, am i doing something wrong ?
          $endgroup$
          – Indrasis Mitra
          Jan 13 at 3:45












          $begingroup$
          @IndrasisMitra See attached note.
          $endgroup$
          – Cesareo
          Jan 13 at 9:19




          $begingroup$
          @IndrasisMitra See attached note.
          $endgroup$
          – Cesareo
          Jan 13 at 9:19


















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