Help in building final solution after solving the separated Eigenvalue problems












0












$begingroup$


I (with help from a MSE user) used the following substitution to seperate variables in a second order linear PDE



$$theta_w = e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$



The following two ODEs (Eigenvalue problems) are a result of applying variable seperation to a system of three coupled 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}

with some separation constant $mu in mathbb{R}$.



BC(s)
$$F(0) = 0, frac{F''(0)}{F'(0)}=beta_h, frac{F''(1)}{F'(1)}=beta_h$$
Similarly,
$$G(0) = 0, frac{G''(0)}{G'(0)}=beta_c, frac{G''(1)}{G'(1)}=beta_c$$



Everything below is for $lambda_h=lambda_c=0.02$, $beta_h=beta_c=10$ and $V=1$. The next step i did was solve the Eigen BVP for $G$ using chebfun in MATLAB which gave me



14.364332916201686
17.484587457962977
20.888494184298537
24.587309921467451
28.600217347815317
32.946305486743015,....

and infinitely many towards positive as the eigenvalues. The general solution form for this ODE is:



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

the three constants i can determine by the three linear equations in $C_1,C_2,C_3$ using the boundary conditions.



My questions are:




  1. What should be my next step, substitute the same EV in the $F$ equation to find its constants?

  2. Or, the EVs of $F$ are to be found separately (which i already tried, they are of the same magnitude but opposite in sign) /

  3. How many EVs should i need to consider to build my solution (I tried the above mentioned process for one EV and my final $theta_w$ function gives results of order $O(9)$ where it should not go above 1)


  4. In general what should be my path forward? I have no clue. Most of the texts mention second order eigenvalue problems which are not helping. Have been pestering a lot over this. Any advice will help.



NOTE
When we solve the $$u_{xx}+u_{yy}=0$$ on $0<x<1$ and $0<y<1$ with bc(s) as $u(0,y)=0,u(1,y)=0,u(x,0)=0,u(x,1)=g_2(x)$ they assume a solution of the form
$$u(x,y)=X(x)Y(y)$$ which results in
$$X''+lambda X=0$$ and $$Y''-lambda Y=0$$ with the separated bc as $X(0)=0,X(1)=0,Y(0)=0,Y(1)=?$.



Solving the first eigenvalue problem gives EVs as



9.8696044010876
39.4784176043502
88.8264396097948
157.9136704174691
246.7401100272539
355.3057584392049
which is $lambda_n={{npi}^2}$ where $n$ attains positive integer values. Then we substitute this $lambda_n$ expression into the $Y$ problem to get a $Y_n$ expression.



This step makes it sure that $X$ and $Y$ are using the same EV values.



[From PDE:Methods,Applications and Theories; Harumi, Hattori]



When i solve the $Y$ expression separately for EVs with an assumed $Y(1)=0$, bc it gives me results as $lambda_n=-{(npi)}^2$ viz. same magnitude and opposite sign which is something i observe in my problem too.Then since there are no intersecting $lambda$ , this problem too should not have had a solution.



Hence i was contemplating of calculating the $F$ EVs and then substituting in $G$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I think the first issue is that you need to find eigenpairs $(mu,F,G)$ for $F$ and $G$ simultaneously. It is possible and not uncommon for boundary value problems to have no solution for certain combinations of parameters.
    $endgroup$
    – Christoph
    Jan 27 at 8:52










  • $begingroup$
    @Christoph If i understand you correctly , you mean that i need to look for values of $mu$ that satisfy both the $F$ and $G$ equation together ? I would emphasize here that when i solved $F$ and $G$ as separate Eigenvalue problems using chebfun in MATLAB, for $lambda_h$ and $beta_h$ values as mentioned in my post, they had no intersecting EVs and the found EVs had same magnitude but were opposite in sign.
    $endgroup$
    – Indrasis Mitra
    Jan 27 at 10:22










  • $begingroup$
    Yes, there must be values of $mu$ for which both problems have a solution. If there is no such $mu$, then I believe your original BVP (with the PDE) has no solution for this combination of parameters.
    $endgroup$
    – Christoph
    Jan 27 at 10:26










  • $begingroup$
    @Christoph have added a note to my OP keeping in mind your points.It is something similar i find. Have a look.
    $endgroup$
    – Indrasis Mitra
    Jan 27 at 12:24






  • 1




    $begingroup$
    But for the Laplace equation example, we have $X_n(x) = sin(n pi x)$ and $Y_n(y) = C_n sinh(n pi y)$. These functions satisfy $X_n'' + lambda_n X_n = 0$ and $Y_n'' - lambda_n Y_n = 0$ with the same values $lambda_n = (npi)^2$, $n in mathbb{N}$.
    $endgroup$
    – Christoph
    Jan 28 at 7:57


















0












$begingroup$


I (with help from a MSE user) used the following substitution to seperate variables in a second order linear PDE



$$theta_w = e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$



The following two ODEs (Eigenvalue problems) are a result of applying variable seperation to a system of three coupled 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}

with some separation constant $mu in mathbb{R}$.



BC(s)
$$F(0) = 0, frac{F''(0)}{F'(0)}=beta_h, frac{F''(1)}{F'(1)}=beta_h$$
Similarly,
$$G(0) = 0, frac{G''(0)}{G'(0)}=beta_c, frac{G''(1)}{G'(1)}=beta_c$$



Everything below is for $lambda_h=lambda_c=0.02$, $beta_h=beta_c=10$ and $V=1$. The next step i did was solve the Eigen BVP for $G$ using chebfun in MATLAB which gave me



14.364332916201686
17.484587457962977
20.888494184298537
24.587309921467451
28.600217347815317
32.946305486743015,....

and infinitely many towards positive as the eigenvalues. The general solution form for this ODE is:



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

the three constants i can determine by the three linear equations in $C_1,C_2,C_3$ using the boundary conditions.



My questions are:




  1. What should be my next step, substitute the same EV in the $F$ equation to find its constants?

  2. Or, the EVs of $F$ are to be found separately (which i already tried, they are of the same magnitude but opposite in sign) /

  3. How many EVs should i need to consider to build my solution (I tried the above mentioned process for one EV and my final $theta_w$ function gives results of order $O(9)$ where it should not go above 1)


  4. In general what should be my path forward? I have no clue. Most of the texts mention second order eigenvalue problems which are not helping. Have been pestering a lot over this. Any advice will help.



NOTE
When we solve the $$u_{xx}+u_{yy}=0$$ on $0<x<1$ and $0<y<1$ with bc(s) as $u(0,y)=0,u(1,y)=0,u(x,0)=0,u(x,1)=g_2(x)$ they assume a solution of the form
$$u(x,y)=X(x)Y(y)$$ which results in
$$X''+lambda X=0$$ and $$Y''-lambda Y=0$$ with the separated bc as $X(0)=0,X(1)=0,Y(0)=0,Y(1)=?$.



Solving the first eigenvalue problem gives EVs as



9.8696044010876
39.4784176043502
88.8264396097948
157.9136704174691
246.7401100272539
355.3057584392049
which is $lambda_n={{npi}^2}$ where $n$ attains positive integer values. Then we substitute this $lambda_n$ expression into the $Y$ problem to get a $Y_n$ expression.



This step makes it sure that $X$ and $Y$ are using the same EV values.



[From PDE:Methods,Applications and Theories; Harumi, Hattori]



When i solve the $Y$ expression separately for EVs with an assumed $Y(1)=0$, bc it gives me results as $lambda_n=-{(npi)}^2$ viz. same magnitude and opposite sign which is something i observe in my problem too.Then since there are no intersecting $lambda$ , this problem too should not have had a solution.



Hence i was contemplating of calculating the $F$ EVs and then substituting in $G$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I think the first issue is that you need to find eigenpairs $(mu,F,G)$ for $F$ and $G$ simultaneously. It is possible and not uncommon for boundary value problems to have no solution for certain combinations of parameters.
    $endgroup$
    – Christoph
    Jan 27 at 8:52










  • $begingroup$
    @Christoph If i understand you correctly , you mean that i need to look for values of $mu$ that satisfy both the $F$ and $G$ equation together ? I would emphasize here that when i solved $F$ and $G$ as separate Eigenvalue problems using chebfun in MATLAB, for $lambda_h$ and $beta_h$ values as mentioned in my post, they had no intersecting EVs and the found EVs had same magnitude but were opposite in sign.
    $endgroup$
    – Indrasis Mitra
    Jan 27 at 10:22










  • $begingroup$
    Yes, there must be values of $mu$ for which both problems have a solution. If there is no such $mu$, then I believe your original BVP (with the PDE) has no solution for this combination of parameters.
    $endgroup$
    – Christoph
    Jan 27 at 10:26










  • $begingroup$
    @Christoph have added a note to my OP keeping in mind your points.It is something similar i find. Have a look.
    $endgroup$
    – Indrasis Mitra
    Jan 27 at 12:24






  • 1




    $begingroup$
    But for the Laplace equation example, we have $X_n(x) = sin(n pi x)$ and $Y_n(y) = C_n sinh(n pi y)$. These functions satisfy $X_n'' + lambda_n X_n = 0$ and $Y_n'' - lambda_n Y_n = 0$ with the same values $lambda_n = (npi)^2$, $n in mathbb{N}$.
    $endgroup$
    – Christoph
    Jan 28 at 7:57
















0












0








0





$begingroup$


I (with help from a MSE user) used the following substitution to seperate variables in a second order linear PDE



$$theta_w = e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$



The following two ODEs (Eigenvalue problems) are a result of applying variable seperation to a system of three coupled 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}

with some separation constant $mu in mathbb{R}$.



BC(s)
$$F(0) = 0, frac{F''(0)}{F'(0)}=beta_h, frac{F''(1)}{F'(1)}=beta_h$$
Similarly,
$$G(0) = 0, frac{G''(0)}{G'(0)}=beta_c, frac{G''(1)}{G'(1)}=beta_c$$



Everything below is for $lambda_h=lambda_c=0.02$, $beta_h=beta_c=10$ and $V=1$. The next step i did was solve the Eigen BVP for $G$ using chebfun in MATLAB which gave me



14.364332916201686
17.484587457962977
20.888494184298537
24.587309921467451
28.600217347815317
32.946305486743015,....

and infinitely many towards positive as the eigenvalues. The general solution form for this ODE is:



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

the three constants i can determine by the three linear equations in $C_1,C_2,C_3$ using the boundary conditions.



My questions are:




  1. What should be my next step, substitute the same EV in the $F$ equation to find its constants?

  2. Or, the EVs of $F$ are to be found separately (which i already tried, they are of the same magnitude but opposite in sign) /

  3. How many EVs should i need to consider to build my solution (I tried the above mentioned process for one EV and my final $theta_w$ function gives results of order $O(9)$ where it should not go above 1)


  4. In general what should be my path forward? I have no clue. Most of the texts mention second order eigenvalue problems which are not helping. Have been pestering a lot over this. Any advice will help.



NOTE
When we solve the $$u_{xx}+u_{yy}=0$$ on $0<x<1$ and $0<y<1$ with bc(s) as $u(0,y)=0,u(1,y)=0,u(x,0)=0,u(x,1)=g_2(x)$ they assume a solution of the form
$$u(x,y)=X(x)Y(y)$$ which results in
$$X''+lambda X=0$$ and $$Y''-lambda Y=0$$ with the separated bc as $X(0)=0,X(1)=0,Y(0)=0,Y(1)=?$.



Solving the first eigenvalue problem gives EVs as



9.8696044010876
39.4784176043502
88.8264396097948
157.9136704174691
246.7401100272539
355.3057584392049
which is $lambda_n={{npi}^2}$ where $n$ attains positive integer values. Then we substitute this $lambda_n$ expression into the $Y$ problem to get a $Y_n$ expression.



This step makes it sure that $X$ and $Y$ are using the same EV values.



[From PDE:Methods,Applications and Theories; Harumi, Hattori]



When i solve the $Y$ expression separately for EVs with an assumed $Y(1)=0$, bc it gives me results as $lambda_n=-{(npi)}^2$ viz. same magnitude and opposite sign which is something i observe in my problem too.Then since there are no intersecting $lambda$ , this problem too should not have had a solution.



Hence i was contemplating of calculating the $F$ EVs and then substituting in $G$.










share|cite|improve this question











$endgroup$




I (with help from a MSE user) used the following substitution to seperate variables in a second order linear PDE



$$theta_w = e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$



The following two ODEs (Eigenvalue problems) are a result of applying variable seperation to a system of three coupled 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}

with some separation constant $mu in mathbb{R}$.



BC(s)
$$F(0) = 0, frac{F''(0)}{F'(0)}=beta_h, frac{F''(1)}{F'(1)}=beta_h$$
Similarly,
$$G(0) = 0, frac{G''(0)}{G'(0)}=beta_c, frac{G''(1)}{G'(1)}=beta_c$$



Everything below is for $lambda_h=lambda_c=0.02$, $beta_h=beta_c=10$ and $V=1$. The next step i did was solve the Eigen BVP for $G$ using chebfun in MATLAB which gave me



14.364332916201686
17.484587457962977
20.888494184298537
24.587309921467451
28.600217347815317
32.946305486743015,....

and infinitely many towards positive as the eigenvalues. The general solution form for this ODE is:



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

the three constants i can determine by the three linear equations in $C_1,C_2,C_3$ using the boundary conditions.



My questions are:




  1. What should be my next step, substitute the same EV in the $F$ equation to find its constants?

  2. Or, the EVs of $F$ are to be found separately (which i already tried, they are of the same magnitude but opposite in sign) /

  3. How many EVs should i need to consider to build my solution (I tried the above mentioned process for one EV and my final $theta_w$ function gives results of order $O(9)$ where it should not go above 1)


  4. In general what should be my path forward? I have no clue. Most of the texts mention second order eigenvalue problems which are not helping. Have been pestering a lot over this. Any advice will help.



NOTE
When we solve the $$u_{xx}+u_{yy}=0$$ on $0<x<1$ and $0<y<1$ with bc(s) as $u(0,y)=0,u(1,y)=0,u(x,0)=0,u(x,1)=g_2(x)$ they assume a solution of the form
$$u(x,y)=X(x)Y(y)$$ which results in
$$X''+lambda X=0$$ and $$Y''-lambda Y=0$$ with the separated bc as $X(0)=0,X(1)=0,Y(0)=0,Y(1)=?$.



Solving the first eigenvalue problem gives EVs as



9.8696044010876
39.4784176043502
88.8264396097948
157.9136704174691
246.7401100272539
355.3057584392049
which is $lambda_n={{npi}^2}$ where $n$ attains positive integer values. Then we substitute this $lambda_n$ expression into the $Y$ problem to get a $Y_n$ expression.



This step makes it sure that $X$ and $Y$ are using the same EV values.



[From PDE:Methods,Applications and Theories; Harumi, Hattori]



When i solve the $Y$ expression separately for EVs with an assumed $Y(1)=0$, bc it gives me results as $lambda_n=-{(npi)}^2$ viz. same magnitude and opposite sign which is something i observe in my problem too.Then since there are no intersecting $lambda$ , this problem too should not have had a solution.



Hence i was contemplating of calculating the $F$ EVs and then substituting in $G$.







calculus ordinary-differential-equations pde boundary-value-problem eigenfunctions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 12:22







Indrasis Mitra

















asked Jan 24 at 11:00









Indrasis MitraIndrasis Mitra

75111




75111












  • $begingroup$
    I think the first issue is that you need to find eigenpairs $(mu,F,G)$ for $F$ and $G$ simultaneously. It is possible and not uncommon for boundary value problems to have no solution for certain combinations of parameters.
    $endgroup$
    – Christoph
    Jan 27 at 8:52










  • $begingroup$
    @Christoph If i understand you correctly , you mean that i need to look for values of $mu$ that satisfy both the $F$ and $G$ equation together ? I would emphasize here that when i solved $F$ and $G$ as separate Eigenvalue problems using chebfun in MATLAB, for $lambda_h$ and $beta_h$ values as mentioned in my post, they had no intersecting EVs and the found EVs had same magnitude but were opposite in sign.
    $endgroup$
    – Indrasis Mitra
    Jan 27 at 10:22










  • $begingroup$
    Yes, there must be values of $mu$ for which both problems have a solution. If there is no such $mu$, then I believe your original BVP (with the PDE) has no solution for this combination of parameters.
    $endgroup$
    – Christoph
    Jan 27 at 10:26










  • $begingroup$
    @Christoph have added a note to my OP keeping in mind your points.It is something similar i find. Have a look.
    $endgroup$
    – Indrasis Mitra
    Jan 27 at 12:24






  • 1




    $begingroup$
    But for the Laplace equation example, we have $X_n(x) = sin(n pi x)$ and $Y_n(y) = C_n sinh(n pi y)$. These functions satisfy $X_n'' + lambda_n X_n = 0$ and $Y_n'' - lambda_n Y_n = 0$ with the same values $lambda_n = (npi)^2$, $n in mathbb{N}$.
    $endgroup$
    – Christoph
    Jan 28 at 7:57




















  • $begingroup$
    I think the first issue is that you need to find eigenpairs $(mu,F,G)$ for $F$ and $G$ simultaneously. It is possible and not uncommon for boundary value problems to have no solution for certain combinations of parameters.
    $endgroup$
    – Christoph
    Jan 27 at 8:52










  • $begingroup$
    @Christoph If i understand you correctly , you mean that i need to look for values of $mu$ that satisfy both the $F$ and $G$ equation together ? I would emphasize here that when i solved $F$ and $G$ as separate Eigenvalue problems using chebfun in MATLAB, for $lambda_h$ and $beta_h$ values as mentioned in my post, they had no intersecting EVs and the found EVs had same magnitude but were opposite in sign.
    $endgroup$
    – Indrasis Mitra
    Jan 27 at 10:22










  • $begingroup$
    Yes, there must be values of $mu$ for which both problems have a solution. If there is no such $mu$, then I believe your original BVP (with the PDE) has no solution for this combination of parameters.
    $endgroup$
    – Christoph
    Jan 27 at 10:26










  • $begingroup$
    @Christoph have added a note to my OP keeping in mind your points.It is something similar i find. Have a look.
    $endgroup$
    – Indrasis Mitra
    Jan 27 at 12:24






  • 1




    $begingroup$
    But for the Laplace equation example, we have $X_n(x) = sin(n pi x)$ and $Y_n(y) = C_n sinh(n pi y)$. These functions satisfy $X_n'' + lambda_n X_n = 0$ and $Y_n'' - lambda_n Y_n = 0$ with the same values $lambda_n = (npi)^2$, $n in mathbb{N}$.
    $endgroup$
    – Christoph
    Jan 28 at 7:57


















$begingroup$
I think the first issue is that you need to find eigenpairs $(mu,F,G)$ for $F$ and $G$ simultaneously. It is possible and not uncommon for boundary value problems to have no solution for certain combinations of parameters.
$endgroup$
– Christoph
Jan 27 at 8:52




$begingroup$
I think the first issue is that you need to find eigenpairs $(mu,F,G)$ for $F$ and $G$ simultaneously. It is possible and not uncommon for boundary value problems to have no solution for certain combinations of parameters.
$endgroup$
– Christoph
Jan 27 at 8:52












$begingroup$
@Christoph If i understand you correctly , you mean that i need to look for values of $mu$ that satisfy both the $F$ and $G$ equation together ? I would emphasize here that when i solved $F$ and $G$ as separate Eigenvalue problems using chebfun in MATLAB, for $lambda_h$ and $beta_h$ values as mentioned in my post, they had no intersecting EVs and the found EVs had same magnitude but were opposite in sign.
$endgroup$
– Indrasis Mitra
Jan 27 at 10:22




$begingroup$
@Christoph If i understand you correctly , you mean that i need to look for values of $mu$ that satisfy both the $F$ and $G$ equation together ? I would emphasize here that when i solved $F$ and $G$ as separate Eigenvalue problems using chebfun in MATLAB, for $lambda_h$ and $beta_h$ values as mentioned in my post, they had no intersecting EVs and the found EVs had same magnitude but were opposite in sign.
$endgroup$
– Indrasis Mitra
Jan 27 at 10:22












$begingroup$
Yes, there must be values of $mu$ for which both problems have a solution. If there is no such $mu$, then I believe your original BVP (with the PDE) has no solution for this combination of parameters.
$endgroup$
– Christoph
Jan 27 at 10:26




$begingroup$
Yes, there must be values of $mu$ for which both problems have a solution. If there is no such $mu$, then I believe your original BVP (with the PDE) has no solution for this combination of parameters.
$endgroup$
– Christoph
Jan 27 at 10:26












$begingroup$
@Christoph have added a note to my OP keeping in mind your points.It is something similar i find. Have a look.
$endgroup$
– Indrasis Mitra
Jan 27 at 12:24




$begingroup$
@Christoph have added a note to my OP keeping in mind your points.It is something similar i find. Have a look.
$endgroup$
– Indrasis Mitra
Jan 27 at 12:24




1




1




$begingroup$
But for the Laplace equation example, we have $X_n(x) = sin(n pi x)$ and $Y_n(y) = C_n sinh(n pi y)$. These functions satisfy $X_n'' + lambda_n X_n = 0$ and $Y_n'' - lambda_n Y_n = 0$ with the same values $lambda_n = (npi)^2$, $n in mathbb{N}$.
$endgroup$
– Christoph
Jan 28 at 7:57






$begingroup$
But for the Laplace equation example, we have $X_n(x) = sin(n pi x)$ and $Y_n(y) = C_n sinh(n pi y)$. These functions satisfy $X_n'' + lambda_n X_n = 0$ and $Y_n'' - lambda_n Y_n = 0$ with the same values $lambda_n = (npi)^2$, $n in mathbb{N}$.
$endgroup$
– Christoph
Jan 28 at 7:57












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