NDsolve with ODE-PDE
$begingroup$
Kindly I hope to know what is the wrong here. How to use NDsolve for a coupled ODE-PDE differential equations
NDSolve[{F''[t]+F[t]==0,
F[0]==0,F'[3*Pi]==1,D[u[t,x],t]==D[u[t,x],x,x],u[0,x]==0,u[t,0]==5*F[t],u[t,5]==0},
{F,u},{t,0,3*Pi},{x,0,5}];
I have received the following messages
Function::fpct: Too many parameters in {t,x} to be filled from
Function[{t,x},0][t].
NDSolve::ndode: The equations {(F^[Prime])[3
[Pi]]==1,F[t]+(F^[Prime][Prime])[t]==0} are not differential
equations or initial conditions in the dependent variables {u}.
differential-equations
edited Jan 15 at 11:55
b.gatessucks
18k23369
asked Jan 15 at 11:54
EszaEsza
161
$endgroup$
add a comment |
$begingroup$
Kindly I hope to know what is the wrong here. How to use NDsolve for a coupled ODE-PDE differential equations
NDSolve[{F''[t]+F[t]==0,
F[0]==0,F'[3*Pi]==1,D[u[t,x],t]==D[u[t,x],x,x],u[0,x]==0,u[t,0]==5*F[t],u[t,5]==0},
{F,u},{t,0,3*Pi},{x,0,5}];
I have received the following messages
Function::fpct: Too many parameters in {t,x} to be filled from
Function[{t,x},0][t].
NDSolve::ndode: The equations {(F^[Prime])[3
[Pi]]==1,F[t]+(F^[Prime][Prime])[t]==0} are not differential
equations or initial conditions in the dependent variables {u}.
differential-equations
edited Jan 15 at 11:55
b.gatessucks
18k23369
asked Jan 15 at 11:54
EszaEsza
161
$endgroup$
1
$begingroup$
Functions must be defined in the whole area{t,0,3*Pi},{x,0,5}. It is necessary to separate the ODE from the PDE. Or calculate the functionF[t,x], but then the problem loses its meaning
$endgroup$
– Alex Trounev
Jan 15 at 12:23
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that
$endgroup$
– Esza
Jan 15 at 17:38
add a comment |
$begingroup$
Kindly I hope to know what is the wrong here. How to use NDsolve for a coupled ODE-PDE differential equations
NDSolve[{F''[t]+F[t]==0,
F[0]==0,F'[3*Pi]==1,D[u[t,x],t]==D[u[t,x],x,x],u[0,x]==0,u[t,0]==5*F[t],u[t,5]==0},
{F,u},{t,0,3*Pi},{x,0,5}];
I have received the following messages
Function::fpct: Too many parameters in {t,x} to be filled from
Function[{t,x},0][t].
NDSolve::ndode: The equations {(F^[Prime])[3
[Pi]]==1,F[t]+(F^[Prime][Prime])[t]==0} are not differential
equations or initial conditions in the dependent variables {u}.
differential-equations
edited Jan 15 at 11:55
b.gatessucks
18k23369
asked Jan 15 at 11:54
EszaEsza
161
$endgroup$
Kindly I hope to know what is the wrong here. How to use NDsolve for a coupled ODE-PDE differential equations
NDSolve[{F''[t]+F[t]==0,
F[0]==0,F'[3*Pi]==1,D[u[t,x],t]==D[u[t,x],x,x],u[0,x]==0,u[t,0]==5*F[t],u[t,5]==0},
{F,u},{t,0,3*Pi},{x,0,5}];
I have received the following messages
Function::fpct: Too many parameters in {t,x} to be filled from
Function[{t,x},0][t].
NDSolve::ndode: The equations {(F^[Prime])[3
[Pi]]==1,F[t]+(F^[Prime][Prime])[t]==0} are not differential
equations or initial conditions in the dependent variables {u}.
differential-equations
differential-equations
edited Jan 15 at 11:55
b.gatessucks
18k23369
asked Jan 15 at 11:54
EszaEsza
161
edited Jan 15 at 11:55
b.gatessucks
18k23369
asked Jan 15 at 11:54
EszaEsza
161
edited Jan 15 at 11:55
b.gatessucks
18k23369
edited Jan 15 at 11:55
b.gatessucks
18k23369
edited Jan 15 at 11:55
b.gatessucks
18k23369
18k23369
asked Jan 15 at 11:54
EszaEsza
161
asked Jan 15 at 11:54
EszaEsza
161
asked Jan 15 at 11:54
EszaEsza
161
161
1
$begingroup$
Functions must be defined in the whole area{t,0,3*Pi},{x,0,5}. It is necessary to separate the ODE from the PDE. Or calculate the functionF[t,x], but then the problem loses its meaning
$endgroup$
– Alex Trounev
Jan 15 at 12:23
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that
$endgroup$
– Esza
Jan 15 at 17:38
add a comment |
1
$begingroup$
Functions must be defined in the whole area{t,0,3*Pi},{x,0,5}. It is necessary to separate the ODE from the PDE. Or calculate the functionF[t,x], but then the problem loses its meaning
$endgroup$
– Alex Trounev
Jan 15 at 12:23
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that
$endgroup$
– Esza
Jan 15 at 17:38
1
1
$begingroup$
Functions must be defined in the whole area
{t,0,3*Pi},{x,0,5}. It is necessary to separate the ODE from the PDE. Or calculate the function F[t,x], but then the problem loses its meaning$endgroup$
– Alex Trounev
Jan 15 at 12:23
$begingroup$
Functions must be defined in the whole area
{t,0,3*Pi},{x,0,5}. It is necessary to separate the ODE from the PDE. Or calculate the function F[t,x], but then the problem loses its meaning$endgroup$
– Alex Trounev
Jan 15 at 12:23
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that
$endgroup$
– Esza
Jan 15 at 17:38
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that
$endgroup$
– Esza
Jan 15 at 17:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can solve for F and use that solution to solve for u:
solF = NDSolve[{F''[t] + F[t] == 0, F[0] == 0, F'[3*Pi] == 1}, {F}, {t, 0, 3*Pi}][[1]];
ff = F /. solF;
Plot[ff[t], {t, 0, 3*Pi}, AxesLabel -> {"t", "F[t]"}]
solu = NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*ff[t], u[t, 5] == 0}, {u}, {t, 0, 3*Pi}, {x, 0, 5}][[1, 1]];
Plot3D[Evaluate[u[t, x] /. solu], {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> {"t", "x", "u[t,x]"}]
answered Jan 15 at 12:17
kglrkglr
181k10200413
$endgroup$
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that?
$endgroup$
– Esza
Jan 15 at 17:38
add a comment |
$begingroup$
To solve the equations jointly, we need to define the problem so that there is a Cauchy problem. I will propose the option that the solution of the system of equations coincides with that obtained by another method by @kglr
sol = NDSolve[{D[F[t, x], t, t] + F[t, x] == D[F[t, x], x, x],
F[0, x] == 0, Derivative[1, 0][F][0, x] == -1,
Derivative[0, 1][F][t, 0] == 0, Derivative[0, 1][F][t, 5] == 0,
D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*F[t, 0], u[t, 5] == 0}, {F, u}, {t, 0, 3*Pi}, {x, 0,
5}];
{Plot3D[F[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "F"],
Plot3D[u[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "u"]}
answered Jan 15 at 19:01
Alex TrounevAlex Trounev
6,8481420
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can solve for F and use that solution to solve for u:
solF = NDSolve[{F''[t] + F[t] == 0, F[0] == 0, F'[3*Pi] == 1}, {F}, {t, 0, 3*Pi}][[1]];
ff = F /. solF;
Plot[ff[t], {t, 0, 3*Pi}, AxesLabel -> {"t", "F[t]"}]
solu = NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*ff[t], u[t, 5] == 0}, {u}, {t, 0, 3*Pi}, {x, 0, 5}][[1, 1]];
Plot3D[Evaluate[u[t, x] /. solu], {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> {"t", "x", "u[t,x]"}]
answered Jan 15 at 12:17
kglrkglr
181k10200413
$endgroup$
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that?
$endgroup$
– Esza
Jan 15 at 17:38
add a comment |
$begingroup$
You can solve for F and use that solution to solve for u:
solF = NDSolve[{F''[t] + F[t] == 0, F[0] == 0, F'[3*Pi] == 1}, {F}, {t, 0, 3*Pi}][[1]];
ff = F /. solF;
Plot[ff[t], {t, 0, 3*Pi}, AxesLabel -> {"t", "F[t]"}]
solu = NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*ff[t], u[t, 5] == 0}, {u}, {t, 0, 3*Pi}, {x, 0, 5}][[1, 1]];
Plot3D[Evaluate[u[t, x] /. solu], {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> {"t", "x", "u[t,x]"}]
answered Jan 15 at 12:17
kglrkglr
181k10200413
$endgroup$
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that?
$endgroup$
– Esza
Jan 15 at 17:38
add a comment |
$begingroup$
You can solve for F and use that solution to solve for u:
solF = NDSolve[{F''[t] + F[t] == 0, F[0] == 0, F'[3*Pi] == 1}, {F}, {t, 0, 3*Pi}][[1]];
ff = F /. solF;
Plot[ff[t], {t, 0, 3*Pi}, AxesLabel -> {"t", "F[t]"}]
solu = NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*ff[t], u[t, 5] == 0}, {u}, {t, 0, 3*Pi}, {x, 0, 5}][[1, 1]];
Plot3D[Evaluate[u[t, x] /. solu], {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> {"t", "x", "u[t,x]"}]
answered Jan 15 at 12:17
kglrkglr
181k10200413
$endgroup$
You can solve for F and use that solution to solve for u:
solF = NDSolve[{F''[t] + F[t] == 0, F[0] == 0, F'[3*Pi] == 1}, {F}, {t, 0, 3*Pi}][[1]];
ff = F /. solF;
Plot[ff[t], {t, 0, 3*Pi}, AxesLabel -> {"t", "F[t]"}]
solu = NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*ff[t], u[t, 5] == 0}, {u}, {t, 0, 3*Pi}, {x, 0, 5}][[1, 1]];
Plot3D[Evaluate[u[t, x] /. solu], {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> {"t", "x", "u[t,x]"}]
answered Jan 15 at 12:17
kglrkglr
181k10200413
answered Jan 15 at 12:17
kglrkglr
181k10200413
answered Jan 15 at 12:17
kglrkglr
181k10200413
answered Jan 15 at 12:17
kglrkglr
181k10200413
181k10200413
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that?
$endgroup$
– Esza
Jan 15 at 17:38
add a comment |
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that?
$endgroup$
– Esza
Jan 15 at 17:38
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that?
$endgroup$
– Esza
Jan 15 at 17:38
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that?
$endgroup$
– Esza
Jan 15 at 17:38
add a comment |
$begingroup$
To solve the equations jointly, we need to define the problem so that there is a Cauchy problem. I will propose the option that the solution of the system of equations coincides with that obtained by another method by @kglr
sol = NDSolve[{D[F[t, x], t, t] + F[t, x] == D[F[t, x], x, x],
F[0, x] == 0, Derivative[1, 0][F][0, x] == -1,
Derivative[0, 1][F][t, 0] == 0, Derivative[0, 1][F][t, 5] == 0,
D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*F[t, 0], u[t, 5] == 0}, {F, u}, {t, 0, 3*Pi}, {x, 0,
5}];
{Plot3D[F[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "F"],
Plot3D[u[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "u"]}
answered Jan 15 at 19:01
Alex TrounevAlex Trounev
6,8481420
$endgroup$
add a comment |
$begingroup$
To solve the equations jointly, we need to define the problem so that there is a Cauchy problem. I will propose the option that the solution of the system of equations coincides with that obtained by another method by @kglr
sol = NDSolve[{D[F[t, x], t, t] + F[t, x] == D[F[t, x], x, x],
F[0, x] == 0, Derivative[1, 0][F][0, x] == -1,
Derivative[0, 1][F][t, 0] == 0, Derivative[0, 1][F][t, 5] == 0,
D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*F[t, 0], u[t, 5] == 0}, {F, u}, {t, 0, 3*Pi}, {x, 0,
5}];
{Plot3D[F[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "F"],
Plot3D[u[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "u"]}
answered Jan 15 at 19:01
Alex TrounevAlex Trounev
6,8481420
$endgroup$
add a comment |
$begingroup$
To solve the equations jointly, we need to define the problem so that there is a Cauchy problem. I will propose the option that the solution of the system of equations coincides with that obtained by another method by @kglr
sol = NDSolve[{D[F[t, x], t, t] + F[t, x] == D[F[t, x], x, x],
F[0, x] == 0, Derivative[1, 0][F][0, x] == -1,
Derivative[0, 1][F][t, 0] == 0, Derivative[0, 1][F][t, 5] == 0,
D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*F[t, 0], u[t, 5] == 0}, {F, u}, {t, 0, 3*Pi}, {x, 0,
5}];
{Plot3D[F[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "F"],
Plot3D[u[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "u"]}
answered Jan 15 at 19:01
Alex TrounevAlex Trounev
6,8481420
$endgroup$
To solve the equations jointly, we need to define the problem so that there is a Cauchy problem. I will propose the option that the solution of the system of equations coincides with that obtained by another method by @kglr
sol = NDSolve[{D[F[t, x], t, t] + F[t, x] == D[F[t, x], x, x],
F[0, x] == 0, Derivative[1, 0][F][0, x] == -1,
Derivative[0, 1][F][t, 0] == 0, Derivative[0, 1][F][t, 5] == 0,
D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
u[t, 0] == 5*F[t, 0], u[t, 5] == 0}, {F, u}, {t, 0, 3*Pi}, {x, 0,
5}];
{Plot3D[F[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "F"],
Plot3D[u[t, x] /. sol, {t, 0, 3*Pi}, {x, 0, 5},
AxesLabel -> Automatic, PlotLabel -> "u"]}
answered Jan 15 at 19:01
Alex TrounevAlex Trounev
6,8481420
answered Jan 15 at 19:01
Alex TrounevAlex Trounev
6,8481420
answered Jan 15 at 19:01
Alex TrounevAlex Trounev
6,8481420
answered Jan 15 at 19:01
Alex TrounevAlex Trounev
6,8481420
6,8481420
add a comment |
add a comment |
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$begingroup$
Functions must be defined in the whole area
{t,0,3*Pi},{x,0,5}. It is necessary to separate the ODE from the PDE. Or calculate the functionF[t,x], but then the problem loses its meaning$endgroup$
– Alex Trounev
Jan 15 at 12:23
$begingroup$
Many thanks for your reply. But I want to solve them simultaneously. How I can do that
$endgroup$
– Esza
Jan 15 at 17:38