Three Coupled PDE. Is my approach in the right way?












0












$begingroup$


I have the following three PDEs



begin{eqnarray}
frac{partial theta_h}{partial x} + beta_h (theta_h - theta_w) &=& 0,\
frac{partial theta_c}{partial y} + beta_c (theta_c - theta_w) &=& 0,\
lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - frac{partial theta_h}{partial x} - Vfrac{partial theta_c}{partial y} &=& 0
end{eqnarray}



From the first and second equation i expressed $frac{partial theta_h}{partial x}$ and $frac{partial theta_c}{partial y}$ in terms of $theta_w$. Then i substituted these in the third equation to yield



$$lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - (beta_h+Vbeta_c)theta_w+(beta_htheta_h+Vbeta_ctheta_c) = 0$$



The third PDE turns out to be a second order linear Elliptic PDE as $lambda_h$,$lambda_c$ and $V$ are all positive constants. I have reached a canonical form for this second order PDE. This PDE is defined on a rectangle with Neumann conditions. I plan to do the following next;




  1. Calculate $theta_w(x,y)$ from the second order PDE.

  2. Plug them in
    the first two to obtain $theta_h$ and $theta_c$


Am i following a correct approach or is there any subtlety i am over-looking ?



Attempt



The boundary conditions for the problem are as follows:



The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$.



$$frac{partial theta_w(0,y)}{partial x}=frac{partial theta_w(1,y)}{partial x}=0 $$



$$frac{partial theta_w(x,0)}{partial y}=frac{partial theta_w(x,1)}{partial y}=0 $$



$$theta_h(0,y)=1 $$$$theta_c(x,0)=0$$



After the suggestions from @Christoph here i have the following two linear third order differential equations:



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}



Both these ODEs now need to be converted to individual Boundary value problems using the BC(s). On substituting



$$theta_w(x,y) = e^{-beta_h x} F(x) e^{-beta_c y} G(y)$$ into the give BC(s), i arrive at the following



$$e^{-beta_cy}F(0)G(y)=1$$
$$e^{-beta_hx}F(x)G(0)=0$$
$$e^{-beta_cy}G(y)[F'(0)-beta_hF(0)]=0$$
$$e^{-beta_cy}e^{-beta_h}G(y)[F'(1)-beta_hF(1)]=0$$
$$e^{-beta_hx}F(x)[G'(0)-beta_cG(0)]=0$$
$$e^{-beta_hx}e^{-beta_c}F(x)[G'(1)-beta_cG(1)]=0$$



Following this (keeping in mind that exponential cannot attain a 0 value), i arrive at the following simplifications:



$$G(0)=0$$
$$G'(0)=0$$
$$frac{G'(1)}{G(1)}=beta_c$$
$$frac{F'(0)}{F(0)}=beta_h$$
$$frac{F'(1)}{F(1)}=beta_h$$



The ODEs are of third order and although i have six BC(s), on decoupling the BC(s) i get just 5. Am i misunderstanding something or is there some other way ?



Attempt 2



As @Christoph advised I made the following changes:
$$bar{{theta_h}}(x,y):=theta_h(x,y)-1$$
and the ansatz
$$theta_w(x,y)=e^{-beta_hx}f(x)e^{-beta_cy}g(y)$$ such that $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$



The third order linear DEs we arrive at still remain the same.
For figuring out the b.c.(s), the ansatz became:
$$theta_w(x,y)=e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$



But the boundary conditions now take the following form



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



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



Now i have three b.c. (s) for each boundary value problem viz. $F$ and $G$.
Each BVP (one each of $F$ and $G$) now involve one Dirichlet and two Robin type b.c.










share|cite|improve this question











$endgroup$












  • $begingroup$
    But you still have the unknown functions $theta_h$ and $theta_c$ in your PDE after this substitution?!
    $endgroup$
    – Christoph
    Jan 8 at 16:38










  • $begingroup$
    Also, there should be a factor $V$ in front of $frac{partial theta_c}{partial y}$.
    $endgroup$
    – Christoph
    Jan 8 at 16:43










  • $begingroup$
    @Christoph Made the edit, you were right about the $V$. Can't $(beta_htheta_h+Vbeta_ctheta_c)$ be together considered a variable say $k$ which would be a function of $x$ and $y$ . And a second order linear PDE requires the coefficients to be the functions of the independent variables $x$ and $y$.
    $endgroup$
    – Indrasis Mitra
    Jan 8 at 16:48












  • $begingroup$
    @Christoph. Any suggestion on how tot tackle the problem if what i was thinking is wrong
    $endgroup$
    – Indrasis Mitra
    Jan 8 at 16:50










  • $begingroup$
    It is correct to have 5 boundary conditions. As the scale of $F$ and $G$ is free for a homogeneous equation, fixing the scales adds two additional equations, making 7 conditions for a state of 6 function values and derivatives plus one parameter $mu$. // In the second approach either one of the cited BC is redundant or there is some other error, as 8 equations for 7 variables is usually not solvable.
    $endgroup$
    – LutzL
    yesterday
















0












$begingroup$


I have the following three PDEs



begin{eqnarray}
frac{partial theta_h}{partial x} + beta_h (theta_h - theta_w) &=& 0,\
frac{partial theta_c}{partial y} + beta_c (theta_c - theta_w) &=& 0,\
lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - frac{partial theta_h}{partial x} - Vfrac{partial theta_c}{partial y} &=& 0
end{eqnarray}



From the first and second equation i expressed $frac{partial theta_h}{partial x}$ and $frac{partial theta_c}{partial y}$ in terms of $theta_w$. Then i substituted these in the third equation to yield



$$lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - (beta_h+Vbeta_c)theta_w+(beta_htheta_h+Vbeta_ctheta_c) = 0$$



The third PDE turns out to be a second order linear Elliptic PDE as $lambda_h$,$lambda_c$ and $V$ are all positive constants. I have reached a canonical form for this second order PDE. This PDE is defined on a rectangle with Neumann conditions. I plan to do the following next;




  1. Calculate $theta_w(x,y)$ from the second order PDE.

  2. Plug them in
    the first two to obtain $theta_h$ and $theta_c$


Am i following a correct approach or is there any subtlety i am over-looking ?



Attempt



The boundary conditions for the problem are as follows:



The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$.



$$frac{partial theta_w(0,y)}{partial x}=frac{partial theta_w(1,y)}{partial x}=0 $$



$$frac{partial theta_w(x,0)}{partial y}=frac{partial theta_w(x,1)}{partial y}=0 $$



$$theta_h(0,y)=1 $$$$theta_c(x,0)=0$$



After the suggestions from @Christoph here i have the following two linear third order differential equations:



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}



Both these ODEs now need to be converted to individual Boundary value problems using the BC(s). On substituting



$$theta_w(x,y) = e^{-beta_h x} F(x) e^{-beta_c y} G(y)$$ into the give BC(s), i arrive at the following



$$e^{-beta_cy}F(0)G(y)=1$$
$$e^{-beta_hx}F(x)G(0)=0$$
$$e^{-beta_cy}G(y)[F'(0)-beta_hF(0)]=0$$
$$e^{-beta_cy}e^{-beta_h}G(y)[F'(1)-beta_hF(1)]=0$$
$$e^{-beta_hx}F(x)[G'(0)-beta_cG(0)]=0$$
$$e^{-beta_hx}e^{-beta_c}F(x)[G'(1)-beta_cG(1)]=0$$



Following this (keeping in mind that exponential cannot attain a 0 value), i arrive at the following simplifications:



$$G(0)=0$$
$$G'(0)=0$$
$$frac{G'(1)}{G(1)}=beta_c$$
$$frac{F'(0)}{F(0)}=beta_h$$
$$frac{F'(1)}{F(1)}=beta_h$$



The ODEs are of third order and although i have six BC(s), on decoupling the BC(s) i get just 5. Am i misunderstanding something or is there some other way ?



Attempt 2



As @Christoph advised I made the following changes:
$$bar{{theta_h}}(x,y):=theta_h(x,y)-1$$
and the ansatz
$$theta_w(x,y)=e^{-beta_hx}f(x)e^{-beta_cy}g(y)$$ such that $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$



The third order linear DEs we arrive at still remain the same.
For figuring out the b.c.(s), the ansatz became:
$$theta_w(x,y)=e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$



But the boundary conditions now take the following form



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



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



Now i have three b.c. (s) for each boundary value problem viz. $F$ and $G$.
Each BVP (one each of $F$ and $G$) now involve one Dirichlet and two Robin type b.c.










share|cite|improve this question











$endgroup$












  • $begingroup$
    But you still have the unknown functions $theta_h$ and $theta_c$ in your PDE after this substitution?!
    $endgroup$
    – Christoph
    Jan 8 at 16:38










  • $begingroup$
    Also, there should be a factor $V$ in front of $frac{partial theta_c}{partial y}$.
    $endgroup$
    – Christoph
    Jan 8 at 16:43










  • $begingroup$
    @Christoph Made the edit, you were right about the $V$. Can't $(beta_htheta_h+Vbeta_ctheta_c)$ be together considered a variable say $k$ which would be a function of $x$ and $y$ . And a second order linear PDE requires the coefficients to be the functions of the independent variables $x$ and $y$.
    $endgroup$
    – Indrasis Mitra
    Jan 8 at 16:48












  • $begingroup$
    @Christoph. Any suggestion on how tot tackle the problem if what i was thinking is wrong
    $endgroup$
    – Indrasis Mitra
    Jan 8 at 16:50










  • $begingroup$
    It is correct to have 5 boundary conditions. As the scale of $F$ and $G$ is free for a homogeneous equation, fixing the scales adds two additional equations, making 7 conditions for a state of 6 function values and derivatives plus one parameter $mu$. // In the second approach either one of the cited BC is redundant or there is some other error, as 8 equations for 7 variables is usually not solvable.
    $endgroup$
    – LutzL
    yesterday














0












0








0


1



$begingroup$


I have the following three PDEs



begin{eqnarray}
frac{partial theta_h}{partial x} + beta_h (theta_h - theta_w) &=& 0,\
frac{partial theta_c}{partial y} + beta_c (theta_c - theta_w) &=& 0,\
lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - frac{partial theta_h}{partial x} - Vfrac{partial theta_c}{partial y} &=& 0
end{eqnarray}



From the first and second equation i expressed $frac{partial theta_h}{partial x}$ and $frac{partial theta_c}{partial y}$ in terms of $theta_w$. Then i substituted these in the third equation to yield



$$lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - (beta_h+Vbeta_c)theta_w+(beta_htheta_h+Vbeta_ctheta_c) = 0$$



The third PDE turns out to be a second order linear Elliptic PDE as $lambda_h$,$lambda_c$ and $V$ are all positive constants. I have reached a canonical form for this second order PDE. This PDE is defined on a rectangle with Neumann conditions. I plan to do the following next;




  1. Calculate $theta_w(x,y)$ from the second order PDE.

  2. Plug them in
    the first two to obtain $theta_h$ and $theta_c$


Am i following a correct approach or is there any subtlety i am over-looking ?



Attempt



The boundary conditions for the problem are as follows:



The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$.



$$frac{partial theta_w(0,y)}{partial x}=frac{partial theta_w(1,y)}{partial x}=0 $$



$$frac{partial theta_w(x,0)}{partial y}=frac{partial theta_w(x,1)}{partial y}=0 $$



$$theta_h(0,y)=1 $$$$theta_c(x,0)=0$$



After the suggestions from @Christoph here i have the following two linear third order differential equations:



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}



Both these ODEs now need to be converted to individual Boundary value problems using the BC(s). On substituting



$$theta_w(x,y) = e^{-beta_h x} F(x) e^{-beta_c y} G(y)$$ into the give BC(s), i arrive at the following



$$e^{-beta_cy}F(0)G(y)=1$$
$$e^{-beta_hx}F(x)G(0)=0$$
$$e^{-beta_cy}G(y)[F'(0)-beta_hF(0)]=0$$
$$e^{-beta_cy}e^{-beta_h}G(y)[F'(1)-beta_hF(1)]=0$$
$$e^{-beta_hx}F(x)[G'(0)-beta_cG(0)]=0$$
$$e^{-beta_hx}e^{-beta_c}F(x)[G'(1)-beta_cG(1)]=0$$



Following this (keeping in mind that exponential cannot attain a 0 value), i arrive at the following simplifications:



$$G(0)=0$$
$$G'(0)=0$$
$$frac{G'(1)}{G(1)}=beta_c$$
$$frac{F'(0)}{F(0)}=beta_h$$
$$frac{F'(1)}{F(1)}=beta_h$$



The ODEs are of third order and although i have six BC(s), on decoupling the BC(s) i get just 5. Am i misunderstanding something or is there some other way ?



Attempt 2



As @Christoph advised I made the following changes:
$$bar{{theta_h}}(x,y):=theta_h(x,y)-1$$
and the ansatz
$$theta_w(x,y)=e^{-beta_hx}f(x)e^{-beta_cy}g(y)$$ such that $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$



The third order linear DEs we arrive at still remain the same.
For figuring out the b.c.(s), the ansatz became:
$$theta_w(x,y)=e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$



But the boundary conditions now take the following form



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



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



Now i have three b.c. (s) for each boundary value problem viz. $F$ and $G$.
Each BVP (one each of $F$ and $G$) now involve one Dirichlet and two Robin type b.c.










share|cite|improve this question











$endgroup$




I have the following three PDEs



begin{eqnarray}
frac{partial theta_h}{partial x} + beta_h (theta_h - theta_w) &=& 0,\
frac{partial theta_c}{partial y} + beta_c (theta_c - theta_w) &=& 0,\
lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - frac{partial theta_h}{partial x} - Vfrac{partial theta_c}{partial y} &=& 0
end{eqnarray}



From the first and second equation i expressed $frac{partial theta_h}{partial x}$ and $frac{partial theta_c}{partial y}$ in terms of $theta_w$. Then i substituted these in the third equation to yield



$$lambda_h frac{partial^2 theta_w}{partial x^2} + lambda_c V frac{partial^2 theta_w}{partial y^2} - (beta_h+Vbeta_c)theta_w+(beta_htheta_h+Vbeta_ctheta_c) = 0$$



The third PDE turns out to be a second order linear Elliptic PDE as $lambda_h$,$lambda_c$ and $V$ are all positive constants. I have reached a canonical form for this second order PDE. This PDE is defined on a rectangle with Neumann conditions. I plan to do the following next;




  1. Calculate $theta_w(x,y)$ from the second order PDE.

  2. Plug them in
    the first two to obtain $theta_h$ and $theta_c$


Am i following a correct approach or is there any subtlety i am over-looking ?



Attempt



The boundary conditions for the problem are as follows:



The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$.



$$frac{partial theta_w(0,y)}{partial x}=frac{partial theta_w(1,y)}{partial x}=0 $$



$$frac{partial theta_w(x,0)}{partial y}=frac{partial theta_w(x,1)}{partial y}=0 $$



$$theta_h(0,y)=1 $$$$theta_c(x,0)=0$$



After the suggestions from @Christoph here i have the following two linear third order differential equations:



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}



Both these ODEs now need to be converted to individual Boundary value problems using the BC(s). On substituting



$$theta_w(x,y) = e^{-beta_h x} F(x) e^{-beta_c y} G(y)$$ into the give BC(s), i arrive at the following



$$e^{-beta_cy}F(0)G(y)=1$$
$$e^{-beta_hx}F(x)G(0)=0$$
$$e^{-beta_cy}G(y)[F'(0)-beta_hF(0)]=0$$
$$e^{-beta_cy}e^{-beta_h}G(y)[F'(1)-beta_hF(1)]=0$$
$$e^{-beta_hx}F(x)[G'(0)-beta_cG(0)]=0$$
$$e^{-beta_hx}e^{-beta_c}F(x)[G'(1)-beta_cG(1)]=0$$



Following this (keeping in mind that exponential cannot attain a 0 value), i arrive at the following simplifications:



$$G(0)=0$$
$$G'(0)=0$$
$$frac{G'(1)}{G(1)}=beta_c$$
$$frac{F'(0)}{F(0)}=beta_h$$
$$frac{F'(1)}{F(1)}=beta_h$$



The ODEs are of third order and although i have six BC(s), on decoupling the BC(s) i get just 5. Am i misunderstanding something or is there some other way ?



Attempt 2



As @Christoph advised I made the following changes:
$$bar{{theta_h}}(x,y):=theta_h(x,y)-1$$
and the ansatz
$$theta_w(x,y)=e^{-beta_hx}f(x)e^{-beta_cy}g(y)$$ such that $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$



The third order linear DEs we arrive at still remain the same.
For figuring out the b.c.(s), the ansatz became:
$$theta_w(x,y)=e^{-beta_hx}F'(x)e^{-beta_cy}G'(y)$$



But the boundary conditions now take the following form



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



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



Now i have three b.c. (s) for each boundary value problem viz. $F$ and $G$.
Each BVP (one each of $F$ and $G$) now involve one Dirichlet and two Robin type b.c.







pde eigenvalues-eigenvectors elliptic-equations linear-pde






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 3:44







Indrasis Mitra

















asked Jan 8 at 16:16









Indrasis MitraIndrasis Mitra

257




257












  • $begingroup$
    But you still have the unknown functions $theta_h$ and $theta_c$ in your PDE after this substitution?!
    $endgroup$
    – Christoph
    Jan 8 at 16:38










  • $begingroup$
    Also, there should be a factor $V$ in front of $frac{partial theta_c}{partial y}$.
    $endgroup$
    – Christoph
    Jan 8 at 16:43










  • $begingroup$
    @Christoph Made the edit, you were right about the $V$. Can't $(beta_htheta_h+Vbeta_ctheta_c)$ be together considered a variable say $k$ which would be a function of $x$ and $y$ . And a second order linear PDE requires the coefficients to be the functions of the independent variables $x$ and $y$.
    $endgroup$
    – Indrasis Mitra
    Jan 8 at 16:48












  • $begingroup$
    @Christoph. Any suggestion on how tot tackle the problem if what i was thinking is wrong
    $endgroup$
    – Indrasis Mitra
    Jan 8 at 16:50










  • $begingroup$
    It is correct to have 5 boundary conditions. As the scale of $F$ and $G$ is free for a homogeneous equation, fixing the scales adds two additional equations, making 7 conditions for a state of 6 function values and derivatives plus one parameter $mu$. // In the second approach either one of the cited BC is redundant or there is some other error, as 8 equations for 7 variables is usually not solvable.
    $endgroup$
    – LutzL
    yesterday


















  • $begingroup$
    But you still have the unknown functions $theta_h$ and $theta_c$ in your PDE after this substitution?!
    $endgroup$
    – Christoph
    Jan 8 at 16:38










  • $begingroup$
    Also, there should be a factor $V$ in front of $frac{partial theta_c}{partial y}$.
    $endgroup$
    – Christoph
    Jan 8 at 16:43










  • $begingroup$
    @Christoph Made the edit, you were right about the $V$. Can't $(beta_htheta_h+Vbeta_ctheta_c)$ be together considered a variable say $k$ which would be a function of $x$ and $y$ . And a second order linear PDE requires the coefficients to be the functions of the independent variables $x$ and $y$.
    $endgroup$
    – Indrasis Mitra
    Jan 8 at 16:48












  • $begingroup$
    @Christoph. Any suggestion on how tot tackle the problem if what i was thinking is wrong
    $endgroup$
    – Indrasis Mitra
    Jan 8 at 16:50










  • $begingroup$
    It is correct to have 5 boundary conditions. As the scale of $F$ and $G$ is free for a homogeneous equation, fixing the scales adds two additional equations, making 7 conditions for a state of 6 function values and derivatives plus one parameter $mu$. // In the second approach either one of the cited BC is redundant or there is some other error, as 8 equations for 7 variables is usually not solvable.
    $endgroup$
    – LutzL
    yesterday
















$begingroup$
But you still have the unknown functions $theta_h$ and $theta_c$ in your PDE after this substitution?!
$endgroup$
– Christoph
Jan 8 at 16:38




$begingroup$
But you still have the unknown functions $theta_h$ and $theta_c$ in your PDE after this substitution?!
$endgroup$
– Christoph
Jan 8 at 16:38












$begingroup$
Also, there should be a factor $V$ in front of $frac{partial theta_c}{partial y}$.
$endgroup$
– Christoph
Jan 8 at 16:43




$begingroup$
Also, there should be a factor $V$ in front of $frac{partial theta_c}{partial y}$.
$endgroup$
– Christoph
Jan 8 at 16:43












$begingroup$
@Christoph Made the edit, you were right about the $V$. Can't $(beta_htheta_h+Vbeta_ctheta_c)$ be together considered a variable say $k$ which would be a function of $x$ and $y$ . And a second order linear PDE requires the coefficients to be the functions of the independent variables $x$ and $y$.
$endgroup$
– Indrasis Mitra
Jan 8 at 16:48






$begingroup$
@Christoph Made the edit, you were right about the $V$. Can't $(beta_htheta_h+Vbeta_ctheta_c)$ be together considered a variable say $k$ which would be a function of $x$ and $y$ . And a second order linear PDE requires the coefficients to be the functions of the independent variables $x$ and $y$.
$endgroup$
– Indrasis Mitra
Jan 8 at 16:48














$begingroup$
@Christoph. Any suggestion on how tot tackle the problem if what i was thinking is wrong
$endgroup$
– Indrasis Mitra
Jan 8 at 16:50




$begingroup$
@Christoph. Any suggestion on how tot tackle the problem if what i was thinking is wrong
$endgroup$
– Indrasis Mitra
Jan 8 at 16:50












$begingroup$
It is correct to have 5 boundary conditions. As the scale of $F$ and $G$ is free for a homogeneous equation, fixing the scales adds two additional equations, making 7 conditions for a state of 6 function values and derivatives plus one parameter $mu$. // In the second approach either one of the cited BC is redundant or there is some other error, as 8 equations for 7 variables is usually not solvable.
$endgroup$
– LutzL
yesterday




$begingroup$
It is correct to have 5 boundary conditions. As the scale of $F$ and $G$ is free for a homogeneous equation, fixing the scales adds two additional equations, making 7 conditions for a state of 6 function values and derivatives plus one parameter $mu$. // In the second approach either one of the cited BC is redundant or there is some other error, as 8 equations for 7 variables is usually not solvable.
$endgroup$
– LutzL
yesterday










1 Answer
1






active

oldest

votes


















1












$begingroup$

Here are a few hints:




  1. Solve the two first-order PDEs for $theta_h, theta_c$ as functions of $theta_w$:
    begin{eqnarray}
    theta_h(x,y) &=& beta_h e^{-beta_h x} int e^{beta_h x} theta_w(x,y) , mathrm{d}x,\
    theta_c(x,y) &=& beta_c e^{-beta_c y} int e^{beta_c y} theta_w(x,y) , mathrm{d}y.
    end{eqnarray}


  2. Eliminate $theta_h, theta_c$ in the second-order PDE to obtain the following equation for $theta_w$:
    begin{eqnarray}
    0 &=& e^{-beta_h x} left( lambda_h e^{beta_h x} frac{partial^2 theta_w}{partial x^2} - beta_h e^{beta_h x} theta_w + beta_h^2 int e^{beta_h x} theta_w , mathrm{d}x right) +\
    && + V e^{-beta_c y} left( lambda_c e^{beta_c y} frac{partial^2 theta_w}{partial y^2} - beta_c e^{beta_c y} theta_w + beta_c^2 int e^{beta_c y} theta_w , mathrm{d}y right).
    end{eqnarray}


  3. Use separation of variables with the ansatz $theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$. You should obtain two linear third-order ODEs with constant coefficients for $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$:
    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 $mu in mathbb{R}$.


  4. From the boundary conditions on $theta_h, theta_c, theta_w$ you should obtain conditions on $F$, $G$. Once the BVPs with the third-order ODEs are solved, compute $f equiv F'$, $g equiv G'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for this guidance. I am trying to reproduce your steps. While substituting the ansatz for $theta_w$ in the PDE of the 2nd step i am encountering a term $int F(x) mathrm{d}x$. Is this term responsible for producing the third order $F^{'''}$ term ? Is it some obvious thing i am missing ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:38












  • $begingroup$
    To rephrase my concern, Since $F$ is a guessed function how are we supposed to write $int F(x) mathrm{d}x$ ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:45










  • $begingroup$
    Yes, the last two equations in my answer were obtained by taking one more derivative in order to remove the antiderivatives of $F$ and $G$. Therefore, the third derivatives appear.
    $endgroup$
    – Christoph
    Jan 9 at 5:05












  • $begingroup$
    This was really helpful. So now the two linear third order ODEs must be solved separately to get $F$ and $G$ which will give a complete $theta_w$. And then $theta_h$ and $theta_c$ could be determined from their individual relations. My only point of doubt is now the separation variable $mu$. Will that come out from the boundary conditions on the two ODEs?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 12:33










  • $begingroup$
    Yes, that's correct. From the boundary conditions on $theta_h, theta_c, theta_w$ on the four faces of the rectangular domain you should obtain boundary conditions for $F$ and $G$. The boundary-value problems (BVPs) for $F$ and $G$ should have solutions only for a discrete set of values $mu_n$, $n in mathbb{N}$. For each of these values you might find solutions $F_n(x)$ and $G_n(y)$ of the BVPs, which you can finally add up to obtain a series representation $theta_w(x,y) = sum_n c_n e^{-beta_h x} F_n(x) e^{-beta_c y} G_n(y)$, with some coefficients $c_n in mathbb{R}$.
    $endgroup$
    – Christoph
    Jan 9 at 13:50













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066372%2fthree-coupled-pde-is-my-approach-in-the-right-way%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Here are a few hints:




  1. Solve the two first-order PDEs for $theta_h, theta_c$ as functions of $theta_w$:
    begin{eqnarray}
    theta_h(x,y) &=& beta_h e^{-beta_h x} int e^{beta_h x} theta_w(x,y) , mathrm{d}x,\
    theta_c(x,y) &=& beta_c e^{-beta_c y} int e^{beta_c y} theta_w(x,y) , mathrm{d}y.
    end{eqnarray}


  2. Eliminate $theta_h, theta_c$ in the second-order PDE to obtain the following equation for $theta_w$:
    begin{eqnarray}
    0 &=& e^{-beta_h x} left( lambda_h e^{beta_h x} frac{partial^2 theta_w}{partial x^2} - beta_h e^{beta_h x} theta_w + beta_h^2 int e^{beta_h x} theta_w , mathrm{d}x right) +\
    && + V e^{-beta_c y} left( lambda_c e^{beta_c y} frac{partial^2 theta_w}{partial y^2} - beta_c e^{beta_c y} theta_w + beta_c^2 int e^{beta_c y} theta_w , mathrm{d}y right).
    end{eqnarray}


  3. Use separation of variables with the ansatz $theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$. You should obtain two linear third-order ODEs with constant coefficients for $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$:
    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 $mu in mathbb{R}$.


  4. From the boundary conditions on $theta_h, theta_c, theta_w$ you should obtain conditions on $F$, $G$. Once the BVPs with the third-order ODEs are solved, compute $f equiv F'$, $g equiv G'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for this guidance. I am trying to reproduce your steps. While substituting the ansatz for $theta_w$ in the PDE of the 2nd step i am encountering a term $int F(x) mathrm{d}x$. Is this term responsible for producing the third order $F^{'''}$ term ? Is it some obvious thing i am missing ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:38












  • $begingroup$
    To rephrase my concern, Since $F$ is a guessed function how are we supposed to write $int F(x) mathrm{d}x$ ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:45










  • $begingroup$
    Yes, the last two equations in my answer were obtained by taking one more derivative in order to remove the antiderivatives of $F$ and $G$. Therefore, the third derivatives appear.
    $endgroup$
    – Christoph
    Jan 9 at 5:05












  • $begingroup$
    This was really helpful. So now the two linear third order ODEs must be solved separately to get $F$ and $G$ which will give a complete $theta_w$. And then $theta_h$ and $theta_c$ could be determined from their individual relations. My only point of doubt is now the separation variable $mu$. Will that come out from the boundary conditions on the two ODEs?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 12:33










  • $begingroup$
    Yes, that's correct. From the boundary conditions on $theta_h, theta_c, theta_w$ on the four faces of the rectangular domain you should obtain boundary conditions for $F$ and $G$. The boundary-value problems (BVPs) for $F$ and $G$ should have solutions only for a discrete set of values $mu_n$, $n in mathbb{N}$. For each of these values you might find solutions $F_n(x)$ and $G_n(y)$ of the BVPs, which you can finally add up to obtain a series representation $theta_w(x,y) = sum_n c_n e^{-beta_h x} F_n(x) e^{-beta_c y} G_n(y)$, with some coefficients $c_n in mathbb{R}$.
    $endgroup$
    – Christoph
    Jan 9 at 13:50


















1












$begingroup$

Here are a few hints:




  1. Solve the two first-order PDEs for $theta_h, theta_c$ as functions of $theta_w$:
    begin{eqnarray}
    theta_h(x,y) &=& beta_h e^{-beta_h x} int e^{beta_h x} theta_w(x,y) , mathrm{d}x,\
    theta_c(x,y) &=& beta_c e^{-beta_c y} int e^{beta_c y} theta_w(x,y) , mathrm{d}y.
    end{eqnarray}


  2. Eliminate $theta_h, theta_c$ in the second-order PDE to obtain the following equation for $theta_w$:
    begin{eqnarray}
    0 &=& e^{-beta_h x} left( lambda_h e^{beta_h x} frac{partial^2 theta_w}{partial x^2} - beta_h e^{beta_h x} theta_w + beta_h^2 int e^{beta_h x} theta_w , mathrm{d}x right) +\
    && + V e^{-beta_c y} left( lambda_c e^{beta_c y} frac{partial^2 theta_w}{partial y^2} - beta_c e^{beta_c y} theta_w + beta_c^2 int e^{beta_c y} theta_w , mathrm{d}y right).
    end{eqnarray}


  3. Use separation of variables with the ansatz $theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$. You should obtain two linear third-order ODEs with constant coefficients for $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$:
    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 $mu in mathbb{R}$.


  4. From the boundary conditions on $theta_h, theta_c, theta_w$ you should obtain conditions on $F$, $G$. Once the BVPs with the third-order ODEs are solved, compute $f equiv F'$, $g equiv G'$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for this guidance. I am trying to reproduce your steps. While substituting the ansatz for $theta_w$ in the PDE of the 2nd step i am encountering a term $int F(x) mathrm{d}x$. Is this term responsible for producing the third order $F^{'''}$ term ? Is it some obvious thing i am missing ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:38












  • $begingroup$
    To rephrase my concern, Since $F$ is a guessed function how are we supposed to write $int F(x) mathrm{d}x$ ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:45










  • $begingroup$
    Yes, the last two equations in my answer were obtained by taking one more derivative in order to remove the antiderivatives of $F$ and $G$. Therefore, the third derivatives appear.
    $endgroup$
    – Christoph
    Jan 9 at 5:05












  • $begingroup$
    This was really helpful. So now the two linear third order ODEs must be solved separately to get $F$ and $G$ which will give a complete $theta_w$. And then $theta_h$ and $theta_c$ could be determined from their individual relations. My only point of doubt is now the separation variable $mu$. Will that come out from the boundary conditions on the two ODEs?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 12:33










  • $begingroup$
    Yes, that's correct. From the boundary conditions on $theta_h, theta_c, theta_w$ on the four faces of the rectangular domain you should obtain boundary conditions for $F$ and $G$. The boundary-value problems (BVPs) for $F$ and $G$ should have solutions only for a discrete set of values $mu_n$, $n in mathbb{N}$. For each of these values you might find solutions $F_n(x)$ and $G_n(y)$ of the BVPs, which you can finally add up to obtain a series representation $theta_w(x,y) = sum_n c_n e^{-beta_h x} F_n(x) e^{-beta_c y} G_n(y)$, with some coefficients $c_n in mathbb{R}$.
    $endgroup$
    – Christoph
    Jan 9 at 13:50
















1












1








1





$begingroup$

Here are a few hints:




  1. Solve the two first-order PDEs for $theta_h, theta_c$ as functions of $theta_w$:
    begin{eqnarray}
    theta_h(x,y) &=& beta_h e^{-beta_h x} int e^{beta_h x} theta_w(x,y) , mathrm{d}x,\
    theta_c(x,y) &=& beta_c e^{-beta_c y} int e^{beta_c y} theta_w(x,y) , mathrm{d}y.
    end{eqnarray}


  2. Eliminate $theta_h, theta_c$ in the second-order PDE to obtain the following equation for $theta_w$:
    begin{eqnarray}
    0 &=& e^{-beta_h x} left( lambda_h e^{beta_h x} frac{partial^2 theta_w}{partial x^2} - beta_h e^{beta_h x} theta_w + beta_h^2 int e^{beta_h x} theta_w , mathrm{d}x right) +\
    && + V e^{-beta_c y} left( lambda_c e^{beta_c y} frac{partial^2 theta_w}{partial y^2} - beta_c e^{beta_c y} theta_w + beta_c^2 int e^{beta_c y} theta_w , mathrm{d}y right).
    end{eqnarray}


  3. Use separation of variables with the ansatz $theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$. You should obtain two linear third-order ODEs with constant coefficients for $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$:
    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 $mu in mathbb{R}$.


  4. From the boundary conditions on $theta_h, theta_c, theta_w$ you should obtain conditions on $F$, $G$. Once the BVPs with the third-order ODEs are solved, compute $f equiv F'$, $g equiv G'$.







share|cite|improve this answer











$endgroup$



Here are a few hints:




  1. Solve the two first-order PDEs for $theta_h, theta_c$ as functions of $theta_w$:
    begin{eqnarray}
    theta_h(x,y) &=& beta_h e^{-beta_h x} int e^{beta_h x} theta_w(x,y) , mathrm{d}x,\
    theta_c(x,y) &=& beta_c e^{-beta_c y} int e^{beta_c y} theta_w(x,y) , mathrm{d}y.
    end{eqnarray}


  2. Eliminate $theta_h, theta_c$ in the second-order PDE to obtain the following equation for $theta_w$:
    begin{eqnarray}
    0 &=& e^{-beta_h x} left( lambda_h e^{beta_h x} frac{partial^2 theta_w}{partial x^2} - beta_h e^{beta_h x} theta_w + beta_h^2 int e^{beta_h x} theta_w , mathrm{d}x right) +\
    && + V e^{-beta_c y} left( lambda_c e^{beta_c y} frac{partial^2 theta_w}{partial y^2} - beta_c e^{beta_c y} theta_w + beta_c^2 int e^{beta_c y} theta_w , mathrm{d}y right).
    end{eqnarray}


  3. Use separation of variables with the ansatz $theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$. You should obtain two linear third-order ODEs with constant coefficients for $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$:
    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 $mu in mathbb{R}$.


  4. From the boundary conditions on $theta_h, theta_c, theta_w$ you should obtain conditions on $F$, $G$. Once the BVPs with the third-order ODEs are solved, compute $f equiv F'$, $g equiv G'$.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 16:02

























answered Jan 8 at 22:22









ChristophChristoph

4616




4616












  • $begingroup$
    Thanks a lot for this guidance. I am trying to reproduce your steps. While substituting the ansatz for $theta_w$ in the PDE of the 2nd step i am encountering a term $int F(x) mathrm{d}x$. Is this term responsible for producing the third order $F^{'''}$ term ? Is it some obvious thing i am missing ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:38












  • $begingroup$
    To rephrase my concern, Since $F$ is a guessed function how are we supposed to write $int F(x) mathrm{d}x$ ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:45










  • $begingroup$
    Yes, the last two equations in my answer were obtained by taking one more derivative in order to remove the antiderivatives of $F$ and $G$. Therefore, the third derivatives appear.
    $endgroup$
    – Christoph
    Jan 9 at 5:05












  • $begingroup$
    This was really helpful. So now the two linear third order ODEs must be solved separately to get $F$ and $G$ which will give a complete $theta_w$. And then $theta_h$ and $theta_c$ could be determined from their individual relations. My only point of doubt is now the separation variable $mu$. Will that come out from the boundary conditions on the two ODEs?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 12:33










  • $begingroup$
    Yes, that's correct. From the boundary conditions on $theta_h, theta_c, theta_w$ on the four faces of the rectangular domain you should obtain boundary conditions for $F$ and $G$. The boundary-value problems (BVPs) for $F$ and $G$ should have solutions only for a discrete set of values $mu_n$, $n in mathbb{N}$. For each of these values you might find solutions $F_n(x)$ and $G_n(y)$ of the BVPs, which you can finally add up to obtain a series representation $theta_w(x,y) = sum_n c_n e^{-beta_h x} F_n(x) e^{-beta_c y} G_n(y)$, with some coefficients $c_n in mathbb{R}$.
    $endgroup$
    – Christoph
    Jan 9 at 13:50




















  • $begingroup$
    Thanks a lot for this guidance. I am trying to reproduce your steps. While substituting the ansatz for $theta_w$ in the PDE of the 2nd step i am encountering a term $int F(x) mathrm{d}x$. Is this term responsible for producing the third order $F^{'''}$ term ? Is it some obvious thing i am missing ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:38












  • $begingroup$
    To rephrase my concern, Since $F$ is a guessed function how are we supposed to write $int F(x) mathrm{d}x$ ?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 4:45










  • $begingroup$
    Yes, the last two equations in my answer were obtained by taking one more derivative in order to remove the antiderivatives of $F$ and $G$. Therefore, the third derivatives appear.
    $endgroup$
    – Christoph
    Jan 9 at 5:05












  • $begingroup$
    This was really helpful. So now the two linear third order ODEs must be solved separately to get $F$ and $G$ which will give a complete $theta_w$. And then $theta_h$ and $theta_c$ could be determined from their individual relations. My only point of doubt is now the separation variable $mu$. Will that come out from the boundary conditions on the two ODEs?
    $endgroup$
    – Indrasis Mitra
    Jan 9 at 12:33










  • $begingroup$
    Yes, that's correct. From the boundary conditions on $theta_h, theta_c, theta_w$ on the four faces of the rectangular domain you should obtain boundary conditions for $F$ and $G$. The boundary-value problems (BVPs) for $F$ and $G$ should have solutions only for a discrete set of values $mu_n$, $n in mathbb{N}$. For each of these values you might find solutions $F_n(x)$ and $G_n(y)$ of the BVPs, which you can finally add up to obtain a series representation $theta_w(x,y) = sum_n c_n e^{-beta_h x} F_n(x) e^{-beta_c y} G_n(y)$, with some coefficients $c_n in mathbb{R}$.
    $endgroup$
    – Christoph
    Jan 9 at 13:50


















$begingroup$
Thanks a lot for this guidance. I am trying to reproduce your steps. While substituting the ansatz for $theta_w$ in the PDE of the 2nd step i am encountering a term $int F(x) mathrm{d}x$. Is this term responsible for producing the third order $F^{'''}$ term ? Is it some obvious thing i am missing ?
$endgroup$
– Indrasis Mitra
Jan 9 at 4:38






$begingroup$
Thanks a lot for this guidance. I am trying to reproduce your steps. While substituting the ansatz for $theta_w$ in the PDE of the 2nd step i am encountering a term $int F(x) mathrm{d}x$. Is this term responsible for producing the third order $F^{'''}$ term ? Is it some obvious thing i am missing ?
$endgroup$
– Indrasis Mitra
Jan 9 at 4:38














$begingroup$
To rephrase my concern, Since $F$ is a guessed function how are we supposed to write $int F(x) mathrm{d}x$ ?
$endgroup$
– Indrasis Mitra
Jan 9 at 4:45




$begingroup$
To rephrase my concern, Since $F$ is a guessed function how are we supposed to write $int F(x) mathrm{d}x$ ?
$endgroup$
– Indrasis Mitra
Jan 9 at 4:45












$begingroup$
Yes, the last two equations in my answer were obtained by taking one more derivative in order to remove the antiderivatives of $F$ and $G$. Therefore, the third derivatives appear.
$endgroup$
– Christoph
Jan 9 at 5:05






$begingroup$
Yes, the last two equations in my answer were obtained by taking one more derivative in order to remove the antiderivatives of $F$ and $G$. Therefore, the third derivatives appear.
$endgroup$
– Christoph
Jan 9 at 5:05














$begingroup$
This was really helpful. So now the two linear third order ODEs must be solved separately to get $F$ and $G$ which will give a complete $theta_w$. And then $theta_h$ and $theta_c$ could be determined from their individual relations. My only point of doubt is now the separation variable $mu$. Will that come out from the boundary conditions on the two ODEs?
$endgroup$
– Indrasis Mitra
Jan 9 at 12:33




$begingroup$
This was really helpful. So now the two linear third order ODEs must be solved separately to get $F$ and $G$ which will give a complete $theta_w$. And then $theta_h$ and $theta_c$ could be determined from their individual relations. My only point of doubt is now the separation variable $mu$. Will that come out from the boundary conditions on the two ODEs?
$endgroup$
– Indrasis Mitra
Jan 9 at 12:33












$begingroup$
Yes, that's correct. From the boundary conditions on $theta_h, theta_c, theta_w$ on the four faces of the rectangular domain you should obtain boundary conditions for $F$ and $G$. The boundary-value problems (BVPs) for $F$ and $G$ should have solutions only for a discrete set of values $mu_n$, $n in mathbb{N}$. For each of these values you might find solutions $F_n(x)$ and $G_n(y)$ of the BVPs, which you can finally add up to obtain a series representation $theta_w(x,y) = sum_n c_n e^{-beta_h x} F_n(x) e^{-beta_c y} G_n(y)$, with some coefficients $c_n in mathbb{R}$.
$endgroup$
– Christoph
Jan 9 at 13:50






$begingroup$
Yes, that's correct. From the boundary conditions on $theta_h, theta_c, theta_w$ on the four faces of the rectangular domain you should obtain boundary conditions for $F$ and $G$. The boundary-value problems (BVPs) for $F$ and $G$ should have solutions only for a discrete set of values $mu_n$, $n in mathbb{N}$. For each of these values you might find solutions $F_n(x)$ and $G_n(y)$ of the BVPs, which you can finally add up to obtain a series representation $theta_w(x,y) = sum_n c_n e^{-beta_h x} F_n(x) e^{-beta_c y} G_n(y)$, with some coefficients $c_n in mathbb{R}$.
$endgroup$
– Christoph
Jan 9 at 13:50




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066372%2fthree-coupled-pde-is-my-approach-in-the-right-way%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

1300-talet

1300-talet

Display a custom attribute below product name in the front-end Magento 1.9.3.8