Mixing
System of linear and nonlinear diff equations [duplicate]
$begingroup$
This question already has an answer here:
System of differential equations - linear system, shooting method
2 answers
I am looking for a help with solving this chemical-engineering problem. Is there anybody, who can help?
The task is about this:
Concentration profiles $cA,dA$ during extraction of A-component from solvent C into solvent B in countercurrent colony with axial mixing of continual phase of solvent B are described with following diff equations which I have to solve:
begin{align}
dA'' &= -(uB/E), dA'-kL ,(cA-dA/psi)\
cA' &= -(kL/uC), (cA-dA/psi),
end{align}
where $cA,dA$ represent concentrations of A-component in original and new solvent, $kL$,$uB$,$uC$,$E$ are constants, "psi" is partition coefficient. Then we have following initial and boundary conditions - z=0 m is inlet of feedstock, z=2 m is inlet of extraction agent.
$cA(0) = 100$
$uB* ,dA(2)+E,* dA'(2) = 0$
$dA'(0) = 0$
Firstly, I have to solve it for $psi = 100$ (at this moment system of diff equations is linear) and find value for $cA(2)$ in raffinate then. Then I have to calculate using shooting method and then use proper Runge-Kutta approximation (using $psi = 10*sqrt(dA)$, for non-physical option $dA$ $=<0$ I have to use $dA/psi = 0$).
Firstly I want understand the problem in general - the final numbers are not important. I want to try it using your advices (or some similar problems which I can use in my solution) and how to use the substition for the diff equation with rate 2, then how to calculate three diff equations with rate 1 with these written conditions for both cases - linear and nonlinear.
Thank you, J.
ordinary-differential-equations
asked Jan 14 at 19:33
user634771user634771
1
$endgroup$
marked as duplicate by Did, Nosrati, Cesareo, Shailesh, Leucippus Jan 21 at 3:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 4 more comments
$begingroup$
This question already has an answer here:
System of differential equations - linear system, shooting method
2 answers
I am looking for a help with solving this chemical-engineering problem. Is there anybody, who can help?
The task is about this:
Concentration profiles $cA,dA$ during extraction of A-component from solvent C into solvent B in countercurrent colony with axial mixing of continual phase of solvent B are described with following diff equations which I have to solve:
begin{align}
dA'' &= -(uB/E), dA'-kL ,(cA-dA/psi)\
cA' &= -(kL/uC), (cA-dA/psi),
end{align}
where $cA,dA$ represent concentrations of A-component in original and new solvent, $kL$,$uB$,$uC$,$E$ are constants, "psi" is partition coefficient. Then we have following initial and boundary conditions - z=0 m is inlet of feedstock, z=2 m is inlet of extraction agent.
$cA(0) = 100$
$uB* ,dA(2)+E,* dA'(2) = 0$
$dA'(0) = 0$
Firstly, I have to solve it for $psi = 100$ (at this moment system of diff equations is linear) and find value for $cA(2)$ in raffinate then. Then I have to calculate using shooting method and then use proper Runge-Kutta approximation (using $psi = 10*sqrt(dA)$, for non-physical option $dA$ $=<0$ I have to use $dA/psi = 0$).
Firstly I want understand the problem in general - the final numbers are not important. I want to try it using your advices (or some similar problems which I can use in my solution) and how to use the substition for the diff equation with rate 2, then how to calculate three diff equations with rate 1 with these written conditions for both cases - linear and nonlinear.
Thank you, J.
ordinary-differential-equations
asked Jan 14 at 19:33
user634771user634771
1
$endgroup$
marked as duplicate by Did, Nosrati, Cesareo, Shailesh, Leucippus Jan 21 at 3:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
This, System of differential equations - linear system, shooting method, looks similar. Is this a duplicate?
$endgroup$
– LutzL
Jan 14 at 20:09
$begingroup$
Yes, its the same, but this version should be more clear for understanding. And I have not understand the comments below the task on that link..
$endgroup$
– user634771
Jan 14 at 20:12
$begingroup$
The comments were about a typographical problem. You should be able to understand the answers. What tools do you have available? You need a numerical ODE solver and a routine to find roots of scalar functions.
$endgroup$
– LutzL
Jan 14 at 20:27
$begingroup$
Maple. But for the analytic solution, I should do it on paper, without any software.
$endgroup$
– user634771
Jan 14 at 20:29
1
$begingroup$
The order-1 system is also nicely explained in the answer of Christoph in the other post.
$endgroup$
– LutzL
Jan 14 at 20:47
|
show 4 more comments
$begingroup$
This question already has an answer here:
System of differential equations - linear system, shooting method
2 answers
I am looking for a help with solving this chemical-engineering problem. Is there anybody, who can help?
The task is about this:
Concentration profiles $cA,dA$ during extraction of A-component from solvent C into solvent B in countercurrent colony with axial mixing of continual phase of solvent B are described with following diff equations which I have to solve:
begin{align}
dA'' &= -(uB/E), dA'-kL ,(cA-dA/psi)\
cA' &= -(kL/uC), (cA-dA/psi),
end{align}
where $cA,dA$ represent concentrations of A-component in original and new solvent, $kL$,$uB$,$uC$,$E$ are constants, "psi" is partition coefficient. Then we have following initial and boundary conditions - z=0 m is inlet of feedstock, z=2 m is inlet of extraction agent.
$cA(0) = 100$
$uB* ,dA(2)+E,* dA'(2) = 0$
$dA'(0) = 0$
Firstly, I have to solve it for $psi = 100$ (at this moment system of diff equations is linear) and find value for $cA(2)$ in raffinate then. Then I have to calculate using shooting method and then use proper Runge-Kutta approximation (using $psi = 10*sqrt(dA)$, for non-physical option $dA$ $=<0$ I have to use $dA/psi = 0$).
Firstly I want understand the problem in general - the final numbers are not important. I want to try it using your advices (or some similar problems which I can use in my solution) and how to use the substition for the diff equation with rate 2, then how to calculate three diff equations with rate 1 with these written conditions for both cases - linear and nonlinear.
Thank you, J.
ordinary-differential-equations
asked Jan 14 at 19:33
user634771user634771
1
$endgroup$
This question already has an answer here:
System of differential equations - linear system, shooting method
2 answers
I am looking for a help with solving this chemical-engineering problem. Is there anybody, who can help?
The task is about this:
Concentration profiles $cA,dA$ during extraction of A-component from solvent C into solvent B in countercurrent colony with axial mixing of continual phase of solvent B are described with following diff equations which I have to solve:
begin{align}
dA'' &= -(uB/E), dA'-kL ,(cA-dA/psi)\
cA' &= -(kL/uC), (cA-dA/psi),
end{align}
where $cA,dA$ represent concentrations of A-component in original and new solvent, $kL$,$uB$,$uC$,$E$ are constants, "psi" is partition coefficient. Then we have following initial and boundary conditions - z=0 m is inlet of feedstock, z=2 m is inlet of extraction agent.
$cA(0) = 100$
$uB* ,dA(2)+E,* dA'(2) = 0$
$dA'(0) = 0$
Firstly, I have to solve it for $psi = 100$ (at this moment system of diff equations is linear) and find value for $cA(2)$ in raffinate then. Then I have to calculate using shooting method and then use proper Runge-Kutta approximation (using $psi = 10*sqrt(dA)$, for non-physical option $dA$ $=<0$ I have to use $dA/psi = 0$).
Firstly I want understand the problem in general - the final numbers are not important. I want to try it using your advices (or some similar problems which I can use in my solution) and how to use the substition for the diff equation with rate 2, then how to calculate three diff equations with rate 1 with these written conditions for both cases - linear and nonlinear.
Thank you, J.
This question already has an answer here:
System of differential equations - linear system, shooting method
2 answers
ordinary-differential-equations
ordinary-differential-equations
asked Jan 14 at 19:33
user634771user634771
1
asked Jan 14 at 19:33
user634771user634771
1
asked Jan 14 at 19:33
user634771user634771
1
asked Jan 14 at 19:33
user634771user634771
1
asked Jan 14 at 19:33
user634771user634771
1
1
marked as duplicate by Did, Nosrati, Cesareo, Shailesh, Leucippus Jan 21 at 3:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Did, Nosrati, Cesareo, Shailesh, Leucippus Jan 21 at 3:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
This, System of differential equations - linear system, shooting method, looks similar. Is this a duplicate?
$endgroup$
– LutzL
Jan 14 at 20:09
$begingroup$
Yes, its the same, but this version should be more clear for understanding. And I have not understand the comments below the task on that link..
$endgroup$
– user634771
Jan 14 at 20:12
$begingroup$
The comments were about a typographical problem. You should be able to understand the answers. What tools do you have available? You need a numerical ODE solver and a routine to find roots of scalar functions.
$endgroup$
– LutzL
Jan 14 at 20:27
$begingroup$
Maple. But for the analytic solution, I should do it on paper, without any software.
$endgroup$
– user634771
Jan 14 at 20:29
1
$begingroup$
The order-1 system is also nicely explained in the answer of Christoph in the other post.
$endgroup$
– LutzL
Jan 14 at 20:47
|
show 4 more comments
$begingroup$
This, System of differential equations - linear system, shooting method, looks similar. Is this a duplicate?
$endgroup$
– LutzL
Jan 14 at 20:09
$begingroup$
Yes, its the same, but this version should be more clear for understanding. And I have not understand the comments below the task on that link..
$endgroup$
– user634771
Jan 14 at 20:12
$begingroup$
The comments were about a typographical problem. You should be able to understand the answers. What tools do you have available? You need a numerical ODE solver and a routine to find roots of scalar functions.
$endgroup$
– LutzL
Jan 14 at 20:27
$begingroup$
Maple. But for the analytic solution, I should do it on paper, without any software.
$endgroup$
– user634771
Jan 14 at 20:29
1
$begingroup$
The order-1 system is also nicely explained in the answer of Christoph in the other post.
$endgroup$
– LutzL
Jan 14 at 20:47
$begingroup$
This, System of differential equations - linear system, shooting method, looks similar. Is this a duplicate?
$endgroup$
– LutzL
Jan 14 at 20:09
$begingroup$
This, System of differential equations - linear system, shooting method, looks similar. Is this a duplicate?
$endgroup$
– LutzL
Jan 14 at 20:09
$begingroup$
Yes, its the same, but this version should be more clear for understanding. And I have not understand the comments below the task on that link..
$endgroup$
– user634771
Jan 14 at 20:12
$begingroup$
Yes, its the same, but this version should be more clear for understanding. And I have not understand the comments below the task on that link..
$endgroup$
– user634771
Jan 14 at 20:12
$begingroup$
The comments were about a typographical problem. You should be able to understand the answers. What tools do you have available? You need a numerical ODE solver and a routine to find roots of scalar functions.
$endgroup$
– LutzL
Jan 14 at 20:27
$begingroup$
The comments were about a typographical problem. You should be able to understand the answers. What tools do you have available? You need a numerical ODE solver and a routine to find roots of scalar functions.
$endgroup$
– LutzL
Jan 14 at 20:27
$begingroup$
Maple. But for the analytic solution, I should do it on paper, without any software.
$endgroup$
– user634771
Jan 14 at 20:29
$begingroup$
Maple. But for the analytic solution, I should do it on paper, without any software.
$endgroup$
– user634771
Jan 14 at 20:29
1
1
$begingroup$
The order-1 system is also nicely explained in the answer of Christoph in the other post.
$endgroup$
– LutzL
Jan 14 at 20:47
$begingroup$
The order-1 system is also nicely explained in the answer of Christoph in the other post.
$endgroup$
– LutzL
Jan 14 at 20:47
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint.
For $psi = C^{te}$. The linear DE system has the structure
$$
dA''+a* dA'+b* cA-b *dA/psi = 0\
cA' +c* cA-c*dA/psi=0
$$
or
$$
left(partial^2+apartial-frac{b}{psi}right)dA =-b*cA\
left(partial +cright)cA=c*dA/psi
$$
or
$$
left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)dA +b*c*dA/psi = 0
$$
or
$$
left(left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)+frac{b*c}{psi}right)dA=0
$$
so
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}
$$
where $r_k, k = 1,2,3$ are the differential operator characteristic values (roots)
and then
$$
cA = -frac 1bleft(partial^2+apartial-frac{b}{psi}right)dA
$$
so now we have $dA, cA$ with three constants to determine, ($C_k$) using the boundary conditions. For the linear DE system no shooting method is needed.
NOTE
Applying the differential operator on $dA$ we get at
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}\
cA = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }e^{r_1 t}+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }e^{r_2 t}+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }e^{r_3 t}
$$
now with the boundary conditions we have
$$
cA(0) = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }=100\
uB dA(2)+E dA'(2) = uB left(C_1 e^{2 r_1}+C_2 e^{2 r_2}+C_3 e^{2 r_3}right)+E left(C_1 e^{2 r_1} r_1+C_2 e^{2 r_2} r_2+C_3
e^{2 r_3} r_3right) = 0\
dA'(0) = C_1r_1+C_2r_2+C_3r_3 = 0
$$
so we have a linear system to solve for $(C_1,C_2,C_3)$
answered Jan 14 at 21:20
CesareoCesareo
8,8993516
$endgroup$
$begingroup$
I understood the first part of yours, but still I have not deal with founding of the boundary condition that I need...
$endgroup$
– user634771
Jan 20 at 8:15
$begingroup$
@user634771 Please. See note attached.
$endgroup$
– Cesareo
Jan 20 at 10:52
$begingroup$
In general, I think I get it, but can you tell me what does it mean using this symbol "ϕ". Thanks
$endgroup$
– user634771
Jan 20 at 11:19
$begingroup$
@user634771 Sorry. Is is a typo now corrected.
$endgroup$
– Cesareo
Jan 20 at 13:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint.
For $psi = C^{te}$. The linear DE system has the structure
$$
dA''+a* dA'+b* cA-b *dA/psi = 0\
cA' +c* cA-c*dA/psi=0
$$
or
$$
left(partial^2+apartial-frac{b}{psi}right)dA =-b*cA\
left(partial +cright)cA=c*dA/psi
$$
or
$$
left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)dA +b*c*dA/psi = 0
$$
or
$$
left(left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)+frac{b*c}{psi}right)dA=0
$$
so
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}
$$
where $r_k, k = 1,2,3$ are the differential operator characteristic values (roots)
and then
$$
cA = -frac 1bleft(partial^2+apartial-frac{b}{psi}right)dA
$$
so now we have $dA, cA$ with three constants to determine, ($C_k$) using the boundary conditions. For the linear DE system no shooting method is needed.
NOTE
Applying the differential operator on $dA$ we get at
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}\
cA = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }e^{r_1 t}+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }e^{r_2 t}+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }e^{r_3 t}
$$
now with the boundary conditions we have
$$
cA(0) = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }=100\
uB dA(2)+E dA'(2) = uB left(C_1 e^{2 r_1}+C_2 e^{2 r_2}+C_3 e^{2 r_3}right)+E left(C_1 e^{2 r_1} r_1+C_2 e^{2 r_2} r_2+C_3
e^{2 r_3} r_3right) = 0\
dA'(0) = C_1r_1+C_2r_2+C_3r_3 = 0
$$
so we have a linear system to solve for $(C_1,C_2,C_3)$
answered Jan 14 at 21:20
CesareoCesareo
8,8993516
$endgroup$
$begingroup$
I understood the first part of yours, but still I have not deal with founding of the boundary condition that I need...
$endgroup$
– user634771
Jan 20 at 8:15
$begingroup$
@user634771 Please. See note attached.
$endgroup$
– Cesareo
Jan 20 at 10:52
$begingroup$
In general, I think I get it, but can you tell me what does it mean using this symbol "ϕ". Thanks
$endgroup$
– user634771
Jan 20 at 11:19
$begingroup$
@user634771 Sorry. Is is a typo now corrected.
$endgroup$
– Cesareo
Jan 20 at 13:12
add a comment |
$begingroup$
Hint.
For $psi = C^{te}$. The linear DE system has the structure
$$
dA''+a* dA'+b* cA-b *dA/psi = 0\
cA' +c* cA-c*dA/psi=0
$$
or
$$
left(partial^2+apartial-frac{b}{psi}right)dA =-b*cA\
left(partial +cright)cA=c*dA/psi
$$
or
$$
left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)dA +b*c*dA/psi = 0
$$
or
$$
left(left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)+frac{b*c}{psi}right)dA=0
$$
so
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}
$$
where $r_k, k = 1,2,3$ are the differential operator characteristic values (roots)
and then
$$
cA = -frac 1bleft(partial^2+apartial-frac{b}{psi}right)dA
$$
so now we have $dA, cA$ with three constants to determine, ($C_k$) using the boundary conditions. For the linear DE system no shooting method is needed.
NOTE
Applying the differential operator on $dA$ we get at
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}\
cA = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }e^{r_1 t}+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }e^{r_2 t}+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }e^{r_3 t}
$$
now with the boundary conditions we have
$$
cA(0) = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }=100\
uB dA(2)+E dA'(2) = uB left(C_1 e^{2 r_1}+C_2 e^{2 r_2}+C_3 e^{2 r_3}right)+E left(C_1 e^{2 r_1} r_1+C_2 e^{2 r_2} r_2+C_3
e^{2 r_3} r_3right) = 0\
dA'(0) = C_1r_1+C_2r_2+C_3r_3 = 0
$$
so we have a linear system to solve for $(C_1,C_2,C_3)$
answered Jan 14 at 21:20
CesareoCesareo
8,8993516
$endgroup$
$begingroup$
I understood the first part of yours, but still I have not deal with founding of the boundary condition that I need...
$endgroup$
– user634771
Jan 20 at 8:15
$begingroup$
@user634771 Please. See note attached.
$endgroup$
– Cesareo
Jan 20 at 10:52
$begingroup$
In general, I think I get it, but can you tell me what does it mean using this symbol "ϕ". Thanks
$endgroup$
– user634771
Jan 20 at 11:19
$begingroup$
@user634771 Sorry. Is is a typo now corrected.
$endgroup$
– Cesareo
Jan 20 at 13:12
add a comment |
$begingroup$
Hint.
For $psi = C^{te}$. The linear DE system has the structure
$$
dA''+a* dA'+b* cA-b *dA/psi = 0\
cA' +c* cA-c*dA/psi=0
$$
or
$$
left(partial^2+apartial-frac{b}{psi}right)dA =-b*cA\
left(partial +cright)cA=c*dA/psi
$$
or
$$
left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)dA +b*c*dA/psi = 0
$$
or
$$
left(left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)+frac{b*c}{psi}right)dA=0
$$
so
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}
$$
where $r_k, k = 1,2,3$ are the differential operator characteristic values (roots)
and then
$$
cA = -frac 1bleft(partial^2+apartial-frac{b}{psi}right)dA
$$
so now we have $dA, cA$ with three constants to determine, ($C_k$) using the boundary conditions. For the linear DE system no shooting method is needed.
NOTE
Applying the differential operator on $dA$ we get at
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}\
cA = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }e^{r_1 t}+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }e^{r_2 t}+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }e^{r_3 t}
$$
now with the boundary conditions we have
$$
cA(0) = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }=100\
uB dA(2)+E dA'(2) = uB left(C_1 e^{2 r_1}+C_2 e^{2 r_2}+C_3 e^{2 r_3}right)+E left(C_1 e^{2 r_1} r_1+C_2 e^{2 r_2} r_2+C_3
e^{2 r_3} r_3right) = 0\
dA'(0) = C_1r_1+C_2r_2+C_3r_3 = 0
$$
so we have a linear system to solve for $(C_1,C_2,C_3)$
answered Jan 14 at 21:20
CesareoCesareo
8,8993516
$endgroup$
Hint.
For $psi = C^{te}$. The linear DE system has the structure
$$
dA''+a* dA'+b* cA-b *dA/psi = 0\
cA' +c* cA-c*dA/psi=0
$$
or
$$
left(partial^2+apartial-frac{b}{psi}right)dA =-b*cA\
left(partial +cright)cA=c*dA/psi
$$
or
$$
left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)dA +b*c*dA/psi = 0
$$
or
$$
left(left(partial +cright)left(partial^2+apartial-frac{b}{psi}right)+frac{b*c}{psi}right)dA=0
$$
so
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}
$$
where $r_k, k = 1,2,3$ are the differential operator characteristic values (roots)
and then
$$
cA = -frac 1bleft(partial^2+apartial-frac{b}{psi}right)dA
$$
so now we have $dA, cA$ with three constants to determine, ($C_k$) using the boundary conditions. For the linear DE system no shooting method is needed.
NOTE
Applying the differential operator on $dA$ we get at
$$
dA = C_1 e^{r_1 t}+C_2 e^{r_2 t}+C_3 e^{r_3 t}\
cA = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }e^{r_1 t}+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }e^{r_2 t}+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }e^{r_3 t}
$$
now with the boundary conditions we have
$$
cA(0) = frac{left(b-psi r_1 left(a+r_1right)right)C_1}{b psi }+frac{left(b-psi r_2 left(a+r_2right)right)C_2}{b psi }+frac{left(b-psi r_3 left(a+r_3right)right)C_3}{b psi }=100\
uB dA(2)+E dA'(2) = uB left(C_1 e^{2 r_1}+C_2 e^{2 r_2}+C_3 e^{2 r_3}right)+E left(C_1 e^{2 r_1} r_1+C_2 e^{2 r_2} r_2+C_3
e^{2 r_3} r_3right) = 0\
dA'(0) = C_1r_1+C_2r_2+C_3r_3 = 0
$$
so we have a linear system to solve for $(C_1,C_2,C_3)$
answered Jan 14 at 21:20
CesareoCesareo
8,8993516
edited Jan 20 at 13:11
answered Jan 14 at 21:20
CesareoCesareo
8,8993516
answered Jan 14 at 21:20
CesareoCesareo
8,8993516
answered Jan 14 at 21:20
CesareoCesareo
8,8993516
8,8993516
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I understood the first part of yours, but still I have not deal with founding of the boundary condition that I need...
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– user634771
Jan 20 at 8:15
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@user634771 Please. See note attached.
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– Cesareo
Jan 20 at 10:52
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In general, I think I get it, but can you tell me what does it mean using this symbol "ϕ". Thanks
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– user634771
Jan 20 at 11:19
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@user634771 Sorry. Is is a typo now corrected.
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– Cesareo
Jan 20 at 13:12
add a comment |
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I understood the first part of yours, but still I have not deal with founding of the boundary condition that I need...
$endgroup$
– user634771
Jan 20 at 8:15
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@user634771 Please. See note attached.
$endgroup$
– Cesareo
Jan 20 at 10:52
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In general, I think I get it, but can you tell me what does it mean using this symbol "ϕ". Thanks
$endgroup$
– user634771
Jan 20 at 11:19
$begingroup$
@user634771 Sorry. Is is a typo now corrected.
$endgroup$
– Cesareo
Jan 20 at 13:12
$begingroup$
I understood the first part of yours, but still I have not deal with founding of the boundary condition that I need...
$endgroup$
– user634771
Jan 20 at 8:15
$begingroup$
I understood the first part of yours, but still I have not deal with founding of the boundary condition that I need...
$endgroup$
– user634771
Jan 20 at 8:15
$begingroup$
@user634771 Please. See note attached.
$endgroup$
– Cesareo
Jan 20 at 10:52
$begingroup$
@user634771 Please. See note attached.
$endgroup$
– Cesareo
Jan 20 at 10:52
$begingroup$
In general, I think I get it, but can you tell me what does it mean using this symbol "ϕ". Thanks
$endgroup$
– user634771
Jan 20 at 11:19
$begingroup$
In general, I think I get it, but can you tell me what does it mean using this symbol "ϕ". Thanks
$endgroup$
– user634771
Jan 20 at 11:19
$begingroup$
@user634771 Sorry. Is is a typo now corrected.
$endgroup$
– Cesareo
Jan 20 at 13:12
$begingroup$
@user634771 Sorry. Is is a typo now corrected.
$endgroup$
– Cesareo
Jan 20 at 13:12
add a comment |
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This, System of differential equations - linear system, shooting method, looks similar. Is this a duplicate?
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– LutzL
Jan 14 at 20:09
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Yes, its the same, but this version should be more clear for understanding. And I have not understand the comments below the task on that link..
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– user634771
Jan 14 at 20:12
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The comments were about a typographical problem. You should be able to understand the answers. What tools do you have available? You need a numerical ODE solver and a routine to find roots of scalar functions.
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– LutzL
Jan 14 at 20:27
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Maple. But for the analytic solution, I should do it on paper, without any software.
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– user634771
Jan 14 at 20:29
1
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The order-1 system is also nicely explained in the answer of Christoph in the other post.
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– LutzL
Jan 14 at 20:47