Solving the canonical form of an elliptic PDE [HEAT EQUATION]
$begingroup$
I have the following PDE
$$
lambda_h frac{partial^2 theta}{partial x^2} + lambda_c V frac{partial^2 theta}{partial y^2} + theta (k_1) + k_2 = 0
$$
This turns out to be an Elliptic second order linear PDE. On converting to the canonical form it takes the following shape
$$frac{partial^2 theta}{partial alpha^2}+frac{partial^2 theta}{partial beta^2}=q_1theta+q_2
$$
where
$q_1=-k_1/(lambda_cV),q_2=-k_2/(lambda_cV), beta=-px, alpha=y$ and $p^2=frac{lambda_cV}{lambda_h}$
What should be the approach now to move forward with this canonical form ?
Should I introduce a Laplace transform or is there a standard method for such a PDE ?
[The Laplace equation would have a zero on the RHS with no $theta$ term and the Poisson equation a function of $x$ and $y$ on RHS with no $theta$ term. So this canonical form doesn't correspond to any of them ?]
The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$. The boundary conditions are of Neumann type.
$$frac{partial theta(0,y)}{partial x}=frac{partial theta(1,y)}{partial x}=0 $$
$$frac{partial theta(x,0)}{partial y}=frac{partial theta(x,1)}{partial y}=0 $$
Attempt
As suggested i tried as following using seperation of variables:
$$theta_{alphaalpha}+theta_{betabeta}-q_1theta=0$$
let $$theta=X(alpha)Y(beta)$$
This gives: $$X^{''}Y+XY^{''}-q_1XY=0$$
$$frac{X^{''}}{X}+frac{Y^{''}}{Y}-q_1=0$$
This is not turning towards a usual variable separation problem due to the variable $q_1$. Any further suggestions ?
ordinary-differential-equations pde
asked Jan 6 at 6:28
Indrasis MitraIndrasis Mitra
80111
$endgroup$
|
show 4 more comments
$begingroup$
I have the following PDE
$$
lambda_h frac{partial^2 theta}{partial x^2} + lambda_c V frac{partial^2 theta}{partial y^2} + theta (k_1) + k_2 = 0
$$
This turns out to be an Elliptic second order linear PDE. On converting to the canonical form it takes the following shape
$$frac{partial^2 theta}{partial alpha^2}+frac{partial^2 theta}{partial beta^2}=q_1theta+q_2
$$
where
$q_1=-k_1/(lambda_cV),q_2=-k_2/(lambda_cV), beta=-px, alpha=y$ and $p^2=frac{lambda_cV}{lambda_h}$
What should be the approach now to move forward with this canonical form ?
Should I introduce a Laplace transform or is there a standard method for such a PDE ?
[The Laplace equation would have a zero on the RHS with no $theta$ term and the Poisson equation a function of $x$ and $y$ on RHS with no $theta$ term. So this canonical form doesn't correspond to any of them ?]
The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$. The boundary conditions are of Neumann type.
$$frac{partial theta(0,y)}{partial x}=frac{partial theta(1,y)}{partial x}=0 $$
$$frac{partial theta(x,0)}{partial y}=frac{partial theta(x,1)}{partial y}=0 $$
Attempt
As suggested i tried as following using seperation of variables:
$$theta_{alphaalpha}+theta_{betabeta}-q_1theta=0$$
let $$theta=X(alpha)Y(beta)$$
This gives: $$X^{''}Y+XY^{''}-q_1XY=0$$
$$frac{X^{''}}{X}+frac{Y^{''}}{Y}-q_1=0$$
This is not turning towards a usual variable separation problem due to the variable $q_1$. Any further suggestions ?
ordinary-differential-equations pde
asked Jan 6 at 6:28
Indrasis MitraIndrasis Mitra
80111
$endgroup$
$begingroup$
What is the domain you're trying to solve your PDE on? Do you have boundary conditions?
$endgroup$
– Dmoreno
Jan 6 at 11:22
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@Dmoreno I have edited the question to add the boundary conditions.
$endgroup$
– Indrasis Mitra
Jan 6 at 12:19
$begingroup$
Have you tried using separation of variables?
$endgroup$
– Dmoreno
Jan 6 at 15:34
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@Dmoreno No , i encountered this problem while analysing a heat exchanger system. I did go through some material for equations of similar type from standard PDE textbooks but encountered homogenous type and with no $u$ term. They all do use seperation of variables. Should i try the standard u= XY seperation route ?
$endgroup$
– Indrasis Mitra
Jan 6 at 17:28
$begingroup$
Yes, but first notice that your PDE is non-homogenous because of the $q_2$ term. I would first solve for the eigenmodes satisfying the same equation and BCs but for $q_2 = 0$ and then expand.
$endgroup$
– Dmoreno
Jan 6 at 17:47
|
show 4 more comments
$begingroup$
I have the following PDE
$$
lambda_h frac{partial^2 theta}{partial x^2} + lambda_c V frac{partial^2 theta}{partial y^2} + theta (k_1) + k_2 = 0
$$
This turns out to be an Elliptic second order linear PDE. On converting to the canonical form it takes the following shape
$$frac{partial^2 theta}{partial alpha^2}+frac{partial^2 theta}{partial beta^2}=q_1theta+q_2
$$
where
$q_1=-k_1/(lambda_cV),q_2=-k_2/(lambda_cV), beta=-px, alpha=y$ and $p^2=frac{lambda_cV}{lambda_h}$
What should be the approach now to move forward with this canonical form ?
Should I introduce a Laplace transform or is there a standard method for such a PDE ?
[The Laplace equation would have a zero on the RHS with no $theta$ term and the Poisson equation a function of $x$ and $y$ on RHS with no $theta$ term. So this canonical form doesn't correspond to any of them ?]
The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$. The boundary conditions are of Neumann type.
$$frac{partial theta(0,y)}{partial x}=frac{partial theta(1,y)}{partial x}=0 $$
$$frac{partial theta(x,0)}{partial y}=frac{partial theta(x,1)}{partial y}=0 $$
Attempt
As suggested i tried as following using seperation of variables:
$$theta_{alphaalpha}+theta_{betabeta}-q_1theta=0$$
let $$theta=X(alpha)Y(beta)$$
This gives: $$X^{''}Y+XY^{''}-q_1XY=0$$
$$frac{X^{''}}{X}+frac{Y^{''}}{Y}-q_1=0$$
This is not turning towards a usual variable separation problem due to the variable $q_1$. Any further suggestions ?
ordinary-differential-equations pde
asked Jan 6 at 6:28
Indrasis MitraIndrasis Mitra
80111
$endgroup$
I have the following PDE
$$
lambda_h frac{partial^2 theta}{partial x^2} + lambda_c V frac{partial^2 theta}{partial y^2} + theta (k_1) + k_2 = 0
$$
This turns out to be an Elliptic second order linear PDE. On converting to the canonical form it takes the following shape
$$frac{partial^2 theta}{partial alpha^2}+frac{partial^2 theta}{partial beta^2}=q_1theta+q_2
$$
where
$q_1=-k_1/(lambda_cV),q_2=-k_2/(lambda_cV), beta=-px, alpha=y$ and $p^2=frac{lambda_cV}{lambda_h}$
What should be the approach now to move forward with this canonical form ?
Should I introduce a Laplace transform or is there a standard method for such a PDE ?
[The Laplace equation would have a zero on the RHS with no $theta$ term and the Poisson equation a function of $x$ and $y$ on RHS with no $theta$ term. So this canonical form doesn't correspond to any of them ?]
The PDE needs to be solved on a rectangular region where $x$ varies between $0$ to $1$ and $y$ varies between $0$ to $1$. The boundary conditions are of Neumann type.
$$frac{partial theta(0,y)}{partial x}=frac{partial theta(1,y)}{partial x}=0 $$
$$frac{partial theta(x,0)}{partial y}=frac{partial theta(x,1)}{partial y}=0 $$
Attempt
As suggested i tried as following using seperation of variables:
$$theta_{alphaalpha}+theta_{betabeta}-q_1theta=0$$
let $$theta=X(alpha)Y(beta)$$
This gives: $$X^{''}Y+XY^{''}-q_1XY=0$$
$$frac{X^{''}}{X}+frac{Y^{''}}{Y}-q_1=0$$
This is not turning towards a usual variable separation problem due to the variable $q_1$. Any further suggestions ?
ordinary-differential-equations pde
ordinary-differential-equations pde
asked Jan 6 at 6:28
Indrasis MitraIndrasis Mitra
80111
asked Jan 6 at 6:28
Indrasis MitraIndrasis Mitra
80111
edited Jan 7 at 13:32
Indrasis Mitra
asked Jan 6 at 6:28
Indrasis MitraIndrasis Mitra
80111
asked Jan 6 at 6:28
Indrasis MitraIndrasis Mitra
80111
asked Jan 6 at 6:28
Indrasis MitraIndrasis Mitra
80111
80111
$begingroup$
What is the domain you're trying to solve your PDE on? Do you have boundary conditions?
$endgroup$
– Dmoreno
Jan 6 at 11:22
$begingroup$
@Dmoreno I have edited the question to add the boundary conditions.
$endgroup$
– Indrasis Mitra
Jan 6 at 12:19
$begingroup$
Have you tried using separation of variables?
$endgroup$
– Dmoreno
Jan 6 at 15:34
$begingroup$
@Dmoreno No , i encountered this problem while analysing a heat exchanger system. I did go through some material for equations of similar type from standard PDE textbooks but encountered homogenous type and with no $u$ term. They all do use seperation of variables. Should i try the standard u= XY seperation route ?
$endgroup$
– Indrasis Mitra
Jan 6 at 17:28
$begingroup$
Yes, but first notice that your PDE is non-homogenous because of the $q_2$ term. I would first solve for the eigenmodes satisfying the same equation and BCs but for $q_2 = 0$ and then expand.
$endgroup$
– Dmoreno
Jan 6 at 17:47
|
show 4 more comments
$begingroup$
What is the domain you're trying to solve your PDE on? Do you have boundary conditions?
$endgroup$
– Dmoreno
Jan 6 at 11:22
$begingroup$
@Dmoreno I have edited the question to add the boundary conditions.
$endgroup$
– Indrasis Mitra
Jan 6 at 12:19
$begingroup$
Have you tried using separation of variables?
$endgroup$
– Dmoreno
Jan 6 at 15:34
$begingroup$
@Dmoreno No , i encountered this problem while analysing a heat exchanger system. I did go through some material for equations of similar type from standard PDE textbooks but encountered homogenous type and with no $u$ term. They all do use seperation of variables. Should i try the standard u= XY seperation route ?
$endgroup$
– Indrasis Mitra
Jan 6 at 17:28
$begingroup$
Yes, but first notice that your PDE is non-homogenous because of the $q_2$ term. I would first solve for the eigenmodes satisfying the same equation and BCs but for $q_2 = 0$ and then expand.
$endgroup$
– Dmoreno
Jan 6 at 17:47
$begingroup$
What is the domain you're trying to solve your PDE on? Do you have boundary conditions?
$endgroup$
– Dmoreno
Jan 6 at 11:22
$begingroup$
What is the domain you're trying to solve your PDE on? Do you have boundary conditions?
$endgroup$
– Dmoreno
Jan 6 at 11:22
$begingroup$
@Dmoreno I have edited the question to add the boundary conditions.
$endgroup$
– Indrasis Mitra
Jan 6 at 12:19
$begingroup$
@Dmoreno I have edited the question to add the boundary conditions.
$endgroup$
– Indrasis Mitra
Jan 6 at 12:19
$begingroup$
Have you tried using separation of variables?
$endgroup$
– Dmoreno
Jan 6 at 15:34
$begingroup$
Have you tried using separation of variables?
$endgroup$
– Dmoreno
Jan 6 at 15:34
$begingroup$
@Dmoreno No , i encountered this problem while analysing a heat exchanger system. I did go through some material for equations of similar type from standard PDE textbooks but encountered homogenous type and with no $u$ term. They all do use seperation of variables. Should i try the standard u= XY seperation route ?
$endgroup$
– Indrasis Mitra
Jan 6 at 17:28
$begingroup$
@Dmoreno No , i encountered this problem while analysing a heat exchanger system. I did go through some material for equations of similar type from standard PDE textbooks but encountered homogenous type and with no $u$ term. They all do use seperation of variables. Should i try the standard u= XY seperation route ?
$endgroup$
– Indrasis Mitra
Jan 6 at 17:28
$begingroup$
Yes, but first notice that your PDE is non-homogenous because of the $q_2$ term. I would first solve for the eigenmodes satisfying the same equation and BCs but for $q_2 = 0$ and then expand.
$endgroup$
– Dmoreno
Jan 6 at 17:47
$begingroup$
Yes, but first notice that your PDE is non-homogenous because of the $q_2$ term. I would first solve for the eigenmodes satisfying the same equation and BCs but for $q_2 = 0$ and then expand.
$endgroup$
– Dmoreno
Jan 6 at 17:47
|
show 4 more comments
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$begingroup$
What is the domain you're trying to solve your PDE on? Do you have boundary conditions?
$endgroup$
– Dmoreno
Jan 6 at 11:22
$begingroup$
@Dmoreno I have edited the question to add the boundary conditions.
$endgroup$
– Indrasis Mitra
Jan 6 at 12:19
$begingroup$
Have you tried using separation of variables?
$endgroup$
– Dmoreno
Jan 6 at 15:34
$begingroup$
@Dmoreno No , i encountered this problem while analysing a heat exchanger system. I did go through some material for equations of similar type from standard PDE textbooks but encountered homogenous type and with no $u$ term. They all do use seperation of variables. Should i try the standard u= XY seperation route ?
$endgroup$
– Indrasis Mitra
Jan 6 at 17:28
$begingroup$
Yes, but first notice that your PDE is non-homogenous because of the $q_2$ term. I would first solve for the eigenmodes satisfying the same equation and BCs but for $q_2 = 0$ and then expand.
$endgroup$
– Dmoreno
Jan 6 at 17:47