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How can I solve the second order nonlinear PDE's?
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I have three PDE's and I tried to solve it through Finite Fourier Transform (FFT) method but could not get the required result due to non-linear term in Eq. (1). How can I solve it? Is there any other method to solve these equations?
beginalign*
fracpartialtheta_cpartial X-fracA_4epsilon(Y^2A_1+YA_2+A_3),,fracpartial^2theta_cpartial Y^2&=0 qquad (1) \
fracpartial^2theta_spartial Y^2-Bi(theta_s-theta_f)&=0 qquad (2) \
kappa,fracpartial^2theta_fpartial Y^2+Bi(theta_s-theta_f)-kappa,U_p,fracpartialtheta_fpartial X&=0 qquad (3)
endalign*
Boundary Conditions:
beginalign*
textatquad Y&=0,qquad theta_s=theta_f,qquad 1=-kappa,fracpartialtheta_fpartial Y-fracpartialtheta_spartial Y\
textatquad Y&=Y_p, qquad theta_s=theta_f=theta_c,qquad gamma=-frackappaepsilon,fracpartialtheta_cpartial Y=-kappa,fracpartialtheta_fpartial Y-fracpartialtheta_spartial Y \
textatquad Y&=1,qquad fracpartialtheta_cpartial Y=0 \
textatquad X&=0,qquad theta_s=theta_f=theta_c=0.
endalign*
Note that $epsilon, kappa, Bi, U_p, A_1, A_2, A_3, A_4$ are constants.
I want to solve for $theta_s,;theta_f,$ and $theta_c.$
pde fourier-transform nonlinear-system parabolic-pde
edited Aug 1 at 14:11
Adrian Keister
3,49321533
asked Aug 1 at 13:51
MIRZA FARRUKH BAIG
64
 |Â
show 10 more comments
up vote
1
down vote
favorite
I have three PDE's and I tried to solve it through Finite Fourier Transform (FFT) method but could not get the required result due to non-linear term in Eq. (1). How can I solve it? Is there any other method to solve these equations?
beginalign*
fracpartialtheta_cpartial X-fracA_4epsilon(Y^2A_1+YA_2+A_3),,fracpartial^2theta_cpartial Y^2&=0 qquad (1) \
fracpartial^2theta_spartial Y^2-Bi(theta_s-theta_f)&=0 qquad (2) \
kappa,fracpartial^2theta_fpartial Y^2+Bi(theta_s-theta_f)-kappa,U_p,fracpartialtheta_fpartial X&=0 qquad (3)
endalign*
Boundary Conditions:
beginalign*
textatquad Y&=0,qquad theta_s=theta_f,qquad 1=-kappa,fracpartialtheta_fpartial Y-fracpartialtheta_spartial Y\
textatquad Y&=Y_p, qquad theta_s=theta_f=theta_c,qquad gamma=-frackappaepsilon,fracpartialtheta_cpartial Y=-kappa,fracpartialtheta_fpartial Y-fracpartialtheta_spartial Y \
textatquad Y&=1,qquad fracpartialtheta_cpartial Y=0 \
textatquad X&=0,qquad theta_s=theta_f=theta_c=0.
endalign*
Note that $epsilon, kappa, Bi, U_p, A_1, A_2, A_3, A_4$ are constants.
I want to solve for $theta_s,;theta_f,$ and $theta_c.$
pde fourier-transform nonlinear-system parabolic-pde
edited Aug 1 at 14:11
Adrian Keister
3,49321533
asked Aug 1 at 13:51
MIRZA FARRUKH BAIG
64
Just to clarify: the pde's are linear in the dependent variables $theta_s, theta_c,$ and $theta_f$.
– Adrian Keister
Aug 1 at 14:02
Actually Eq. 1 is nonlinear PDE and the rest are the linear PDE @AdrianKeister
– MIRZA FARRUKH BAIG
Aug 2 at 4:19
All three pde's are linear in the $theta$ variables. Even your boundary conditions are linear in the $theta$ variables. Unless you are reporting one of the equations incorrectly, they are all linear. If you think the first equation is nonlinear, please point out the nonlinearity.
– Adrian Keister
Aug 2 at 10:38
Because Eq. 1 involves the Y^2 term, that's why I consider it nonlinear PDE. If it is linear, as you said, so which method is suitable to solve this PDE
– MIRZA FARRUKH BAIG
Aug 2 at 12:03
Usually the linearity of a DE is determined by whether its dependent variables are linear. In this case, that is your $theta_s, theta_f, theta_c.$ Because $Y$ is an independent variable, how it shows up in the DE has nothing to do with linearity. Personally, I would hand this system of pde's over to a computer to do numerically. I doubt you could find an analytical solution. One thing I would point out: the first equation alone has $theta_c$ in it, and it doesn't have $theta_s$ or $theta_f$. That is, $theta_c$ is de-coupled in the actual equations. It's coupled in the BC's, though.
– Adrian Keister
Aug 2 at 12:43
 |Â
show 10 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have three PDE's and I tried to solve it through Finite Fourier Transform (FFT) method but could not get the required result due to non-linear term in Eq. (1). How can I solve it? Is there any other method to solve these equations?
beginalign*
fracpartialtheta_cpartial X-fracA_4epsilon(Y^2A_1+YA_2+A_3),,fracpartial^2theta_cpartial Y^2&=0 qquad (1) \
fracpartial^2theta_spartial Y^2-Bi(theta_s-theta_f)&=0 qquad (2) \
kappa,fracpartial^2theta_fpartial Y^2+Bi(theta_s-theta_f)-kappa,U_p,fracpartialtheta_fpartial X&=0 qquad (3)
endalign*
Boundary Conditions:
beginalign*
textatquad Y&=0,qquad theta_s=theta_f,qquad 1=-kappa,fracpartialtheta_fpartial Y-fracpartialtheta_spartial Y\
textatquad Y&=Y_p, qquad theta_s=theta_f=theta_c,qquad gamma=-frackappaepsilon,fracpartialtheta_cpartial Y=-kappa,fracpartialtheta_fpartial Y-fracpartialtheta_spartial Y \
textatquad Y&=1,qquad fracpartialtheta_cpartial Y=0 \
textatquad X&=0,qquad theta_s=theta_f=theta_c=0.
endalign*
Note that $epsilon, kappa, Bi, U_p, A_1, A_2, A_3, A_4$ are constants.
I want to solve for $theta_s,;theta_f,$ and $theta_c.$
pde fourier-transform nonlinear-system parabolic-pde
edited Aug 1 at 14:11
Adrian Keister
3,49321533
asked Aug 1 at 13:51
MIRZA FARRUKH BAIG
64
I have three PDE's and I tried to solve it through Finite Fourier Transform (FFT) method but could not get the required result due to non-linear term in Eq. (1). How can I solve it? Is there any other method to solve these equations?
beginalign*
fracpartialtheta_cpartial X-fracA_4epsilon(Y^2A_1+YA_2+A_3),,fracpartial^2theta_cpartial Y^2&=0 qquad (1) \
fracpartial^2theta_spartial Y^2-Bi(theta_s-theta_f)&=0 qquad (2) \
kappa,fracpartial^2theta_fpartial Y^2+Bi(theta_s-theta_f)-kappa,U_p,fracpartialtheta_fpartial X&=0 qquad (3)
endalign*
Boundary Conditions:
beginalign*
textatquad Y&=0,qquad theta_s=theta_f,qquad 1=-kappa,fracpartialtheta_fpartial Y-fracpartialtheta_spartial Y\
textatquad Y&=Y_p, qquad theta_s=theta_f=theta_c,qquad gamma=-frackappaepsilon,fracpartialtheta_cpartial Y=-kappa,fracpartialtheta_fpartial Y-fracpartialtheta_spartial Y \
textatquad Y&=1,qquad fracpartialtheta_cpartial Y=0 \
textatquad X&=0,qquad theta_s=theta_f=theta_c=0.
endalign*
Note that $epsilon, kappa, Bi, U_p, A_1, A_2, A_3, A_4$ are constants.
I want to solve for $theta_s,;theta_f,$ and $theta_c.$
pde fourier-transform nonlinear-system parabolic-pde
edited Aug 1 at 14:11
Adrian Keister
3,49321533
asked Aug 1 at 13:51
MIRZA FARRUKH BAIG
64
edited Aug 1 at 14:11
Adrian Keister
3,49321533
edited Aug 1 at 14:11
Adrian Keister
3,49321533
edited Aug 1 at 14:11
Adrian Keister
3,49321533
3,49321533
asked Aug 1 at 13:51
MIRZA FARRUKH BAIG
64
asked Aug 1 at 13:51
MIRZA FARRUKH BAIG
64
asked Aug 1 at 13:51
MIRZA FARRUKH BAIG
64
64
Just to clarify: the pde's are linear in the dependent variables $theta_s, theta_c,$ and $theta_f$.
– Adrian Keister
Aug 1 at 14:02
Actually Eq. 1 is nonlinear PDE and the rest are the linear PDE @AdrianKeister
– MIRZA FARRUKH BAIG
Aug 2 at 4:19
All three pde's are linear in the $theta$ variables. Even your boundary conditions are linear in the $theta$ variables. Unless you are reporting one of the equations incorrectly, they are all linear. If you think the first equation is nonlinear, please point out the nonlinearity.
– Adrian Keister
Aug 2 at 10:38
Because Eq. 1 involves the Y^2 term, that's why I consider it nonlinear PDE. If it is linear, as you said, so which method is suitable to solve this PDE
– MIRZA FARRUKH BAIG
Aug 2 at 12:03
Usually the linearity of a DE is determined by whether its dependent variables are linear. In this case, that is your $theta_s, theta_f, theta_c.$ Because $Y$ is an independent variable, how it shows up in the DE has nothing to do with linearity. Personally, I would hand this system of pde's over to a computer to do numerically. I doubt you could find an analytical solution. One thing I would point out: the first equation alone has $theta_c$ in it, and it doesn't have $theta_s$ or $theta_f$. That is, $theta_c$ is de-coupled in the actual equations. It's coupled in the BC's, though.
– Adrian Keister
Aug 2 at 12:43
 |Â
show 10 more comments
Just to clarify: the pde's are linear in the dependent variables $theta_s, theta_c,$ and $theta_f$.
– Adrian Keister
Aug 1 at 14:02
Actually Eq. 1 is nonlinear PDE and the rest are the linear PDE @AdrianKeister
– MIRZA FARRUKH BAIG
Aug 2 at 4:19
All three pde's are linear in the $theta$ variables. Even your boundary conditions are linear in the $theta$ variables. Unless you are reporting one of the equations incorrectly, they are all linear. If you think the first equation is nonlinear, please point out the nonlinearity.
– Adrian Keister
Aug 2 at 10:38
Because Eq. 1 involves the Y^2 term, that's why I consider it nonlinear PDE. If it is linear, as you said, so which method is suitable to solve this PDE
– MIRZA FARRUKH BAIG
Aug 2 at 12:03
Usually the linearity of a DE is determined by whether its dependent variables are linear. In this case, that is your $theta_s, theta_f, theta_c.$ Because $Y$ is an independent variable, how it shows up in the DE has nothing to do with linearity. Personally, I would hand this system of pde's over to a computer to do numerically. I doubt you could find an analytical solution. One thing I would point out: the first equation alone has $theta_c$ in it, and it doesn't have $theta_s$ or $theta_f$. That is, $theta_c$ is de-coupled in the actual equations. It's coupled in the BC's, though.
– Adrian Keister
Aug 2 at 12:43
Just to clarify: the pde's are linear in the dependent variables $theta_s, theta_c,$ and $theta_f$.
– Adrian Keister
Aug 1 at 14:02
Just to clarify: the pde's are linear in the dependent variables $theta_s, theta_c,$ and $theta_f$.
– Adrian Keister
Aug 1 at 14:02
Actually Eq. 1 is nonlinear PDE and the rest are the linear PDE @AdrianKeister
– MIRZA FARRUKH BAIG
Aug 2 at 4:19
Actually Eq. 1 is nonlinear PDE and the rest are the linear PDE @AdrianKeister
– MIRZA FARRUKH BAIG
Aug 2 at 4:19
All three pde's are linear in the $theta$ variables. Even your boundary conditions are linear in the $theta$ variables. Unless you are reporting one of the equations incorrectly, they are all linear. If you think the first equation is nonlinear, please point out the nonlinearity.
– Adrian Keister
Aug 2 at 10:38
All three pde's are linear in the $theta$ variables. Even your boundary conditions are linear in the $theta$ variables. Unless you are reporting one of the equations incorrectly, they are all linear. If you think the first equation is nonlinear, please point out the nonlinearity.
– Adrian Keister
Aug 2 at 10:38
Because Eq. 1 involves the Y^2 term, that's why I consider it nonlinear PDE. If it is linear, as you said, so which method is suitable to solve this PDE
– MIRZA FARRUKH BAIG
Aug 2 at 12:03
Because Eq. 1 involves the Y^2 term, that's why I consider it nonlinear PDE. If it is linear, as you said, so which method is suitable to solve this PDE
– MIRZA FARRUKH BAIG
Aug 2 at 12:03
Usually the linearity of a DE is determined by whether its dependent variables are linear. In this case, that is your $theta_s, theta_f, theta_c.$ Because $Y$ is an independent variable, how it shows up in the DE has nothing to do with linearity. Personally, I would hand this system of pde's over to a computer to do numerically. I doubt you could find an analytical solution. One thing I would point out: the first equation alone has $theta_c$ in it, and it doesn't have $theta_s$ or $theta_f$. That is, $theta_c$ is de-coupled in the actual equations. It's coupled in the BC's, though.
– Adrian Keister
Aug 2 at 12:43
Usually the linearity of a DE is determined by whether its dependent variables are linear. In this case, that is your $theta_s, theta_f, theta_c.$ Because $Y$ is an independent variable, how it shows up in the DE has nothing to do with linearity. Personally, I would hand this system of pde's over to a computer to do numerically. I doubt you could find an analytical solution. One thing I would point out: the first equation alone has $theta_c$ in it, and it doesn't have $theta_s$ or $theta_f$. That is, $theta_c$ is de-coupled in the actual equations. It's coupled in the BC's, though.
– Adrian Keister
Aug 2 at 12:43
 |Â
show 10 more comments
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Just to clarify: the pde's are linear in the dependent variables $theta_s, theta_c,$ and $theta_f$.
– Adrian Keister
Aug 1 at 14:02
Actually Eq. 1 is nonlinear PDE and the rest are the linear PDE @AdrianKeister
– MIRZA FARRUKH BAIG
Aug 2 at 4:19
All three pde's are linear in the $theta$ variables. Even your boundary conditions are linear in the $theta$ variables. Unless you are reporting one of the equations incorrectly, they are all linear. If you think the first equation is nonlinear, please point out the nonlinearity.
– Adrian Keister
Aug 2 at 10:38
Because Eq. 1 involves the Y^2 term, that's why I consider it nonlinear PDE. If it is linear, as you said, so which method is suitable to solve this PDE
– MIRZA FARRUKH BAIG
Aug 2 at 12:03
Usually the linearity of a DE is determined by whether its dependent variables are linear. In this case, that is your $theta_s, theta_f, theta_c.$ Because $Y$ is an independent variable, how it shows up in the DE has nothing to do with linearity. Personally, I would hand this system of pde's over to a computer to do numerically. I doubt you could find an analytical solution. One thing I would point out: the first equation alone has $theta_c$ in it, and it doesn't have $theta_s$ or $theta_f$. That is, $theta_c$ is de-coupled in the actual equations. It's coupled in the BC's, though.
– Adrian Keister
Aug 2 at 12:43