NDSolve:PDE system, initial-boundary value problem:warning:NDSolve::mconly: For the method NDSolve`IDA, only...
I tried to NDSolve the PDE system
$$partial_t w =xcdot wquadquadpartial_z x=w$$
for
$$(t,z)in[0,1]times[0,pi]$$
with boundary conditions
$$x(t,0)=w(t,0)=w(t,pi)=0$$
and initial conditions
$$w(0,z)=sin zquadquad x(0,z)=1-cos z$$
Here's my code:
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == Sin[z], x[0, z] == 1 - Cos[z],
w[t, 0] == 0, w[t, π] == 0, x[t, 0] == 0}, {w , x}, {t, 0,
1}, {z, 0, π}]
Mathematica displays the following warning:
"NDSolve::mconly: For the method NDSolve`IDA, only machine real code
is available. Unable to continue with complex values or beyond
floating-point exceptions."
I would appreciate any help on how to overcome this error or solve numerically this kind of PDE system anyway.
differential-equations numerical-integration error boundary-conditions
add a comment |
I tried to NDSolve the PDE system
$$partial_t w =xcdot wquadquadpartial_z x=w$$
for
$$(t,z)in[0,1]times[0,pi]$$
with boundary conditions
$$x(t,0)=w(t,0)=w(t,pi)=0$$
and initial conditions
$$w(0,z)=sin zquadquad x(0,z)=1-cos z$$
Here's my code:
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == Sin[z], x[0, z] == 1 - Cos[z],
w[t, 0] == 0, w[t, π] == 0, x[t, 0] == 0}, {w , x}, {t, 0,
1}, {z, 0, π}]
Mathematica displays the following warning:
"NDSolve::mconly: For the method NDSolve`IDA, only machine real code
is available. Unable to continue with complex values or beyond
floating-point exceptions."
I would appreciate any help on how to overcome this error or solve numerically this kind of PDE system anyway.
differential-equations numerical-integration error boundary-conditions
add a comment |
I tried to NDSolve the PDE system
$$partial_t w =xcdot wquadquadpartial_z x=w$$
for
$$(t,z)in[0,1]times[0,pi]$$
with boundary conditions
$$x(t,0)=w(t,0)=w(t,pi)=0$$
and initial conditions
$$w(0,z)=sin zquadquad x(0,z)=1-cos z$$
Here's my code:
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == Sin[z], x[0, z] == 1 - Cos[z],
w[t, 0] == 0, w[t, π] == 0, x[t, 0] == 0}, {w , x}, {t, 0,
1}, {z, 0, π}]
Mathematica displays the following warning:
"NDSolve::mconly: For the method NDSolve`IDA, only machine real code
is available. Unable to continue with complex values or beyond
floating-point exceptions."
I would appreciate any help on how to overcome this error or solve numerically this kind of PDE system anyway.
differential-equations numerical-integration error boundary-conditions
I tried to NDSolve the PDE system
$$partial_t w =xcdot wquadquadpartial_z x=w$$
for
$$(t,z)in[0,1]times[0,pi]$$
with boundary conditions
$$x(t,0)=w(t,0)=w(t,pi)=0$$
and initial conditions
$$w(0,z)=sin zquadquad x(0,z)=1-cos z$$
Here's my code:
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == Sin[z], x[0, z] == 1 - Cos[z],
w[t, 0] == 0, w[t, π] == 0, x[t, 0] == 0}, {w , x}, {t, 0,
1}, {z, 0, π}]
Mathematica displays the following warning:
"NDSolve::mconly: For the method NDSolve`IDA, only machine real code
is available. Unable to continue with complex values or beyond
floating-point exceptions."
I would appreciate any help on how to overcome this error or solve numerically this kind of PDE system anyway.
differential-equations numerical-integration error boundary-conditions
differential-equations numerical-integration error boundary-conditions
edited Nov 27 '18 at 15:45
asked Nov 27 '18 at 13:29
user61386
add a comment |
add a comment |
1 Answer
1
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oldest
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This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
1}, {z, 0, [Pi]},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MinPoints" -> 80, "MaxPoints" -> 100,
"DifferenceOrder" -> "Pseudospectral"}}];
{ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}
answered Nov 27 '18 at 14:44
Alex Trounev
6,1201419
Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
– user61386
Nov 27 '18 at 15:05
@user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
– Alex Trounev
Nov 27 '18 at 15:22
My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
– user61386
Nov 27 '18 at 20:59
Sorry, where does this equation come from?
– Alex Trounev
Nov 27 '18 at 21:12
1
Initial and boundary conditions must be consistent.
– Alex Trounev
Nov 28 '18 at 10:28
|
show 5 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
1}, {z, 0, [Pi]},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MinPoints" -> 80, "MaxPoints" -> 100,
"DifferenceOrder" -> "Pseudospectral"}}];
{ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}
answered Nov 27 '18 at 14:44
Alex Trounev
6,1201419
Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
– user61386
Nov 27 '18 at 15:05
@user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
– Alex Trounev
Nov 27 '18 at 15:22
My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
– user61386
Nov 27 '18 at 20:59
Sorry, where does this equation come from?
– Alex Trounev
Nov 27 '18 at 21:12
1
Initial and boundary conditions must be consistent.
– Alex Trounev
Nov 28 '18 at 10:28
|
show 5 more comments
This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
1}, {z, 0, [Pi]},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MinPoints" -> 80, "MaxPoints" -> 100,
"DifferenceOrder" -> "Pseudospectral"}}];
{ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}
answered Nov 27 '18 at 14:44
Alex Trounev
6,1201419
Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
– user61386
Nov 27 '18 at 15:05
@user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
– Alex Trounev
Nov 27 '18 at 15:22
My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
– user61386
Nov 27 '18 at 20:59
Sorry, where does this equation come from?
– Alex Trounev
Nov 27 '18 at 21:12
1
Initial and boundary conditions must be consistent.
– Alex Trounev
Nov 28 '18 at 10:28
|
show 5 more comments
This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
1}, {z, 0, [Pi]},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MinPoints" -> 80, "MaxPoints" -> 100,
"DifferenceOrder" -> "Pseudospectral"}}];
{ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}
answered Nov 27 '18 at 14:44
Alex Trounev
6,1201419
This is a quasilinear hyperbolic system of equations. Not all initial data is valid, w=0 should be excluded from the initial data. An example of solving the problem
s = NDSolve[{D[w[t, z], t] == w[t, z]*x[t, z],
D[x[t, z], z] == w[t, z], w[0, z] == 1, x[0, z] == 1 - Cos[z],
w[t, 0] == 1, w[t, [Pi]] == 1, x[t, 0] == 0}, {w, x}, {t, 0,
1}, {z, 0, [Pi]},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MinPoints" -> 80, "MaxPoints" -> 100,
"DifferenceOrder" -> "Pseudospectral"}}];
{ContourPlot[Evaluate[w[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "w",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic],
ContourPlot[Evaluate[x[t, z] /. s], {t, 0, 1}, {z, 0, [Pi]},
Contours -> 20, ColorFunction -> Hue, PlotLabel -> "x",
FrameLabel -> {"t", "z"}, PlotLegends -> Automatic]}
answered Nov 27 '18 at 14:44
Alex Trounev
6,1201419
answered Nov 27 '18 at 14:44
Alex Trounev
6,1201419
answered Nov 27 '18 at 14:44
Alex Trounev
6,1201419
answered Nov 27 '18 at 14:44
Alex Trounev
6,1201419
6,1201419
Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
– user61386
Nov 27 '18 at 15:05
@user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
– Alex Trounev
Nov 27 '18 at 15:22
My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
– user61386
Nov 27 '18 at 20:59
Sorry, where does this equation come from?
– Alex Trounev
Nov 27 '18 at 21:12
1
Initial and boundary conditions must be consistent.
– Alex Trounev
Nov 28 '18 at 10:28
|
show 5 more comments
Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
– user61386
Nov 27 '18 at 15:05
@user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
– Alex Trounev
Nov 27 '18 at 15:22
My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
– user61386
Nov 27 '18 at 20:59
Sorry, where does this equation come from?
– Alex Trounev
Nov 27 '18 at 21:12
1
Initial and boundary conditions must be consistent.
– Alex Trounev
Nov 28 '18 at 10:28
Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
– user61386
Nov 27 '18 at 15:05
Thanks for great help! Also: how can I see that boundary condition $w(t,0)=0$ is invalid?
– user61386
Nov 27 '18 at 15:05
@user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
– Alex Trounev
Nov 27 '18 at 15:22
@user61386 It is necessary to bring the system to a second order equation using $x=w_t/w$ and then $x_z=w=(w_t/w)_z$
– Alex Trounev
Nov 27 '18 at 15:22
My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
– user61386
Nov 27 '18 at 20:59
My only problem is that with the initial conditions used above, i.e. $w(0,z)=1$ and $x(0,z)=1-cos z$, the PDE system equation $partial_t x =w$ seems to fail for $t=0$.
– user61386
Nov 27 '18 at 20:59
Sorry, where does this equation come from?
– Alex Trounev
Nov 27 '18 at 21:12
Sorry, where does this equation come from?
– Alex Trounev
Nov 27 '18 at 21:12
1
1
Initial and boundary conditions must be consistent.
– Alex Trounev
Nov 28 '18 at 10:28
Initial and boundary conditions must be consistent.
– Alex Trounev
Nov 28 '18 at 10:28
|
show 5 more comments
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