Difference between initial state problem and boundary-value problem
$begingroup$
I would like to check the aspect that is not covered in the the original question.
They both seem to have the same nature, yet, I feel like the difference lies in that the initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically (I mean simulation in a single run, not trial and error until your solution is found). That is, known initial value means that you can find next value in your recursion, next value and so you can find the whole sequence by simulation. The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence. Is it right?
boundary-value-problem
edited Dec 13 '18 at 9:58
dmtri
1,5082521
asked Jul 26 '14 at 17:23
ValVal
1
$endgroup$
add a comment |
$begingroup$
I would like to check the aspect that is not covered in the the original question.
They both seem to have the same nature, yet, I feel like the difference lies in that the initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically (I mean simulation in a single run, not trial and error until your solution is found). That is, known initial value means that you can find next value in your recursion, next value and so you can find the whole sequence by simulation. The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence. Is it right?
boundary-value-problem
edited Dec 13 '18 at 9:58
dmtri
1,5082521
asked Jul 26 '14 at 17:23
ValVal
1
$endgroup$
$begingroup$
Do you mean a BVP in one dimension, or in at least two?
$endgroup$
– Semiclassical
Jul 26 '14 at 17:28
$begingroup$
@Semiclassical I do not care. The answer may tell how this aspect is important. It may change all the answer.
$endgroup$
– Val
Jul 26 '14 at 17:39
add a comment |
$begingroup$
I would like to check the aspect that is not covered in the the original question.
They both seem to have the same nature, yet, I feel like the difference lies in that the initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically (I mean simulation in a single run, not trial and error until your solution is found). That is, known initial value means that you can find next value in your recursion, next value and so you can find the whole sequence by simulation. The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence. Is it right?
boundary-value-problem
edited Dec 13 '18 at 9:58
dmtri
1,5082521
asked Jul 26 '14 at 17:23
ValVal
1
$endgroup$
I would like to check the aspect that is not covered in the the original question.
They both seem to have the same nature, yet, I feel like the difference lies in that the initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically (I mean simulation in a single run, not trial and error until your solution is found). That is, known initial value means that you can find next value in your recursion, next value and so you can find the whole sequence by simulation. The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence. Is it right?
boundary-value-problem
boundary-value-problem
edited Dec 13 '18 at 9:58
dmtri
1,5082521
asked Jul 26 '14 at 17:23
ValVal
1
edited Dec 13 '18 at 9:58
dmtri
1,5082521
asked Jul 26 '14 at 17:23
ValVal
1
edited Dec 13 '18 at 9:58
dmtri
1,5082521
edited Dec 13 '18 at 9:58
dmtri
1,5082521
edited Dec 13 '18 at 9:58
dmtri
1,5082521
1,5082521
asked Jul 26 '14 at 17:23
ValVal
1
asked Jul 26 '14 at 17:23
ValVal
1
asked Jul 26 '14 at 17:23
ValVal
1
1
$begingroup$
Do you mean a BVP in one dimension, or in at least two?
$endgroup$
– Semiclassical
Jul 26 '14 at 17:28
$begingroup$
@Semiclassical I do not care. The answer may tell how this aspect is important. It may change all the answer.
$endgroup$
– Val
Jul 26 '14 at 17:39
add a comment |
$begingroup$
Do you mean a BVP in one dimension, or in at least two?
$endgroup$
– Semiclassical
Jul 26 '14 at 17:28
$begingroup$
@Semiclassical I do not care. The answer may tell how this aspect is important. It may change all the answer.
$endgroup$
– Val
Jul 26 '14 at 17:39
$begingroup$
Do you mean a BVP in one dimension, or in at least two?
$endgroup$
– Semiclassical
Jul 26 '14 at 17:28
$begingroup$
Do you mean a BVP in one dimension, or in at least two?
$endgroup$
– Semiclassical
Jul 26 '14 at 17:28
$begingroup$
@Semiclassical I do not care. The answer may tell how this aspect is important. It may change all the answer.
$endgroup$
– Val
Jul 26 '14 at 17:39
$begingroup$
@Semiclassical I do not care. The answer may tell how this aspect is important. It may change all the answer.
$endgroup$
– Val
Jul 26 '14 at 17:39
add a comment |
1 Answer
1
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$begingroup$
initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically
Not true as stated. There are lots of numerical boundary value problem solvers out there, one of better known is in Matlab PDE toolbox. The whole industry of Finite Element Method begs to disagree with your "needs to solved analytically" statement.
But you are right to say that
The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence.
Initial value problems (for ODE and PDE) can be solved by marching forward in time. For example, a simple solver for the diffusion equation would take three consecutive values of $u$ at time step $k$, say $u^{k}_{i-1}$, $u^k_i$, $u^k_{i+1}$, and calculate
$$u^{k+1}_i = bu^k_{i-1}+(1-2b)u^k_i+bu^k_{i+1}$$
where $0<b<1/2$ comes from the diffusion coefficient and the sizes of time step and space step.
Not so for boundary value problems. As you remarked, for ODE there is the shooting method, which attempts to solve a BVP by means of several IVPs. But the more systematic way, which works for both ODE and PDE, is to express the discrete form of PDE as a large (but sparse) linear system, and hit it with a specialized linear solver. For example, if I'm solving a boundary value problem for the Laplace equation, my linear system might consist of equations
$$u_{i j}=frac14(u_{i-1 j}+u_{i+1 j} + u_{i j-1}+u_{i j+1})$$
which together with boundary conditions form a linear system with unique solution.
One practical consequence of this difference is that solution of initial value problems for PDE can be adapted to parallel computing easier than solution of boundary value problems.
$endgroup$
add a comment |
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$begingroup$
initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically
Not true as stated. There are lots of numerical boundary value problem solvers out there, one of better known is in Matlab PDE toolbox. The whole industry of Finite Element Method begs to disagree with your "needs to solved analytically" statement.
But you are right to say that
The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence.
Initial value problems (for ODE and PDE) can be solved by marching forward in time. For example, a simple solver for the diffusion equation would take three consecutive values of $u$ at time step $k$, say $u^{k}_{i-1}$, $u^k_i$, $u^k_{i+1}$, and calculate
$$u^{k+1}_i = bu^k_{i-1}+(1-2b)u^k_i+bu^k_{i+1}$$
where $0<b<1/2$ comes from the diffusion coefficient and the sizes of time step and space step.
Not so for boundary value problems. As you remarked, for ODE there is the shooting method, which attempts to solve a BVP by means of several IVPs. But the more systematic way, which works for both ODE and PDE, is to express the discrete form of PDE as a large (but sparse) linear system, and hit it with a specialized linear solver. For example, if I'm solving a boundary value problem for the Laplace equation, my linear system might consist of equations
$$u_{i j}=frac14(u_{i-1 j}+u_{i+1 j} + u_{i j-1}+u_{i j+1})$$
which together with boundary conditions form a linear system with unique solution.
One practical consequence of this difference is that solution of initial value problems for PDE can be adapted to parallel computing easier than solution of boundary value problems.
$endgroup$
add a comment |
$begingroup$
initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically
Not true as stated. There are lots of numerical boundary value problem solvers out there, one of better known is in Matlab PDE toolbox. The whole industry of Finite Element Method begs to disagree with your "needs to solved analytically" statement.
But you are right to say that
The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence.
Initial value problems (for ODE and PDE) can be solved by marching forward in time. For example, a simple solver for the diffusion equation would take three consecutive values of $u$ at time step $k$, say $u^{k}_{i-1}$, $u^k_i$, $u^k_{i+1}$, and calculate
$$u^{k+1}_i = bu^k_{i-1}+(1-2b)u^k_i+bu^k_{i+1}$$
where $0<b<1/2$ comes from the diffusion coefficient and the sizes of time step and space step.
Not so for boundary value problems. As you remarked, for ODE there is the shooting method, which attempts to solve a BVP by means of several IVPs. But the more systematic way, which works for both ODE and PDE, is to express the discrete form of PDE as a large (but sparse) linear system, and hit it with a specialized linear solver. For example, if I'm solving a boundary value problem for the Laplace equation, my linear system might consist of equations
$$u_{i j}=frac14(u_{i-1 j}+u_{i+1 j} + u_{i j-1}+u_{i j+1})$$
which together with boundary conditions form a linear system with unique solution.
One practical consequence of this difference is that solution of initial value problems for PDE can be adapted to parallel computing easier than solution of boundary value problems.
$endgroup$
add a comment |
$begingroup$
initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically
Not true as stated. There are lots of numerical boundary value problem solvers out there, one of better known is in Matlab PDE toolbox. The whole industry of Finite Element Method begs to disagree with your "needs to solved analytically" statement.
But you are right to say that
The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence.
Initial value problems (for ODE and PDE) can be solved by marching forward in time. For example, a simple solver for the diffusion equation would take three consecutive values of $u$ at time step $k$, say $u^{k}_{i-1}$, $u^k_i$, $u^k_{i+1}$, and calculate
$$u^{k+1}_i = bu^k_{i-1}+(1-2b)u^k_i+bu^k_{i+1}$$
where $0<b<1/2$ comes from the diffusion coefficient and the sizes of time step and space step.
Not so for boundary value problems. As you remarked, for ODE there is the shooting method, which attempts to solve a BVP by means of several IVPs. But the more systematic way, which works for both ODE and PDE, is to express the discrete form of PDE as a large (but sparse) linear system, and hit it with a specialized linear solver. For example, if I'm solving a boundary value problem for the Laplace equation, my linear system might consist of equations
$$u_{i j}=frac14(u_{i-1 j}+u_{i+1 j} + u_{i j-1}+u_{i j+1})$$
which together with boundary conditions form a linear system with unique solution.
One practical consequence of this difference is that solution of initial value problems for PDE can be adapted to parallel computing easier than solution of boundary value problems.
$endgroup$
initial value problem can be solved by simulation whereas boundary value problem needs to be solved analytically
Not true as stated. There are lots of numerical boundary value problem solvers out there, one of better known is in Matlab PDE toolbox. The whole industry of Finite Element Method begs to disagree with your "needs to solved analytically" statement.
But you are right to say that
The boundary value, in contrast, means that you cannot find the next value in the sequence without finding the whole sequence.
Initial value problems (for ODE and PDE) can be solved by marching forward in time. For example, a simple solver for the diffusion equation would take three consecutive values of $u$ at time step $k$, say $u^{k}_{i-1}$, $u^k_i$, $u^k_{i+1}$, and calculate
$$u^{k+1}_i = bu^k_{i-1}+(1-2b)u^k_i+bu^k_{i+1}$$
where $0<b<1/2$ comes from the diffusion coefficient and the sizes of time step and space step.
Not so for boundary value problems. As you remarked, for ODE there is the shooting method, which attempts to solve a BVP by means of several IVPs. But the more systematic way, which works for both ODE and PDE, is to express the discrete form of PDE as a large (but sparse) linear system, and hit it with a specialized linear solver. For example, if I'm solving a boundary value problem for the Laplace equation, my linear system might consist of equations
$$u_{i j}=frac14(u_{i-1 j}+u_{i+1 j} + u_{i j-1}+u_{i j+1})$$
which together with boundary conditions form a linear system with unique solution.
One practical consequence of this difference is that solution of initial value problems for PDE can be adapted to parallel computing easier than solution of boundary value problems.
answered Jul 27 '14 at 5:00
user147263
add a comment |
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$begingroup$
Do you mean a BVP in one dimension, or in at least two?
$endgroup$
– Semiclassical
Jul 26 '14 at 17:28
$begingroup$
@Semiclassical I do not care. The answer may tell how this aspect is important. It may change all the answer.
$endgroup$
– Val
Jul 26 '14 at 17:39