Applying the method of lines to a partial differential equation and using Runge-kutta method
0
By method of lines I converted the PDE
u_t=u_{xx},
with the initial and boundary conditions
u(0,t)=u(0.1,t)
u(0,x)=sin(2*pi*x)
to a system of ODEs, as follows
function ut=pde1(t,u)
%
% Problem parameters
% global ncall
xl=0.0;
xr=1.0;
d=10;
a=10;
%
% PDE
n=length(u);
h=((xr-xl)/(n-1));
for i=1:n
if(i==1) ut(i)=0.0;
elseif(i==n) ut(i)=0;
else ut(i)=d*(u(i+1)-2.0*u(i)+u(i-1))/h^2;
end
end
ut=ut';
%
Now I am solving the ODEs ut by Runge-kutta method as follows:
%Initial condition
n=500;
for i=1:n
u0(i)=sin((2*pi)*(i-1)/(n-1));
end
t0=0.0;
tf=0.1;
tout=linspace(t0,tf,n);
y=zeros(n,length(tout));
y(:,1)=u0';
h=1/n;
for i = 1 : length(tout)
t=tout(i);
k1 = pde_1(t,y(:,i));
k2 = pde_1(t+h/2, y(:,i)+h*k1/2);
k3 = pde_1(t+h/2, y(:,i)+h*k2/2);
k4 = pde_1(t+h, y(:,i)+h*k3);
y(:,i+1) = y(:,i) + h*(k1 + 2*k2 + 2*k3 + k4)/6;
end
y
plot(tout,y(:,1))
but the outputs are NAN! any help?
numerical-methods heat-equation runge-kutta-methods
asked Jan 3 at 21:17
Ahmad M.o Kassef
1
New contributor
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
0
By method of lines I converted the PDE
u_t=u_{xx},
with the initial and boundary conditions
u(0,t)=u(0.1,t)
u(0,x)=sin(2*pi*x)
to a system of ODEs, as follows
function ut=pde1(t,u)
%
% Problem parameters
% global ncall
xl=0.0;
xr=1.0;
d=10;
a=10;
%
% PDE
n=length(u);
h=((xr-xl)/(n-1));
for i=1:n
if(i==1) ut(i)=0.0;
elseif(i==n) ut(i)=0;
else ut(i)=d*(u(i+1)-2.0*u(i)+u(i-1))/h^2;
end
end
ut=ut';
%
Now I am solving the ODEs ut by Runge-kutta method as follows:
%Initial condition
n=500;
for i=1:n
u0(i)=sin((2*pi)*(i-1)/(n-1));
end
t0=0.0;
tf=0.1;
tout=linspace(t0,tf,n);
y=zeros(n,length(tout));
y(:,1)=u0';
h=1/n;
for i = 1 : length(tout)
t=tout(i);
k1 = pde_1(t,y(:,i));
k2 = pde_1(t+h/2, y(:,i)+h*k1/2);
k3 = pde_1(t+h/2, y(:,i)+h*k2/2);
k4 = pde_1(t+h, y(:,i)+h*k3);
y(:,i+1) = y(:,i) + h*(k1 + 2*k2 + 2*k3 + k4)/6;
end
y
plot(tout,y(:,1))
but the outputs are NAN! any help?
numerical-methods heat-equation runge-kutta-methods
asked Jan 3 at 21:17
Ahmad M.o Kassef
1
New contributor
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to the Mathematics StackExchange! What language are you programming in? Also, you might get a better answer from StackOverflow; while the Runge-kutta is a numerical algorithm, your focus appears to be on the implementation rather than the mathematical properties, its correctness, or proving theorems with the method.
– Larry B.
Jan 3 at 21:43
Your code deviates in some relevant points from your equations. Is it reallyu(0,t)=u(0.1,t)or did you want periodic boundariesu(0,t)=u(1,t)or zero boundariesu(0,t)=u(1,t)=0? The last one you enforced with your implementation. Do you get a useful result if you decrease the factordor the time steph?
– LutzL
2 days ago
add a comment |
0
0
0
By method of lines I converted the PDE
u_t=u_{xx},
with the initial and boundary conditions
u(0,t)=u(0.1,t)
u(0,x)=sin(2*pi*x)
to a system of ODEs, as follows
function ut=pde1(t,u)
%
% Problem parameters
% global ncall
xl=0.0;
xr=1.0;
d=10;
a=10;
%
% PDE
n=length(u);
h=((xr-xl)/(n-1));
for i=1:n
if(i==1) ut(i)=0.0;
elseif(i==n) ut(i)=0;
else ut(i)=d*(u(i+1)-2.0*u(i)+u(i-1))/h^2;
end
end
ut=ut';
%
Now I am solving the ODEs ut by Runge-kutta method as follows:
%Initial condition
n=500;
for i=1:n
u0(i)=sin((2*pi)*(i-1)/(n-1));
end
t0=0.0;
tf=0.1;
tout=linspace(t0,tf,n);
y=zeros(n,length(tout));
y(:,1)=u0';
h=1/n;
for i = 1 : length(tout)
t=tout(i);
k1 = pde_1(t,y(:,i));
k2 = pde_1(t+h/2, y(:,i)+h*k1/2);
k3 = pde_1(t+h/2, y(:,i)+h*k2/2);
k4 = pde_1(t+h, y(:,i)+h*k3);
y(:,i+1) = y(:,i) + h*(k1 + 2*k2 + 2*k3 + k4)/6;
end
y
plot(tout,y(:,1))
but the outputs are NAN! any help?
numerical-methods heat-equation runge-kutta-methods
asked Jan 3 at 21:17
Ahmad M.o Kassef
1
New contributor
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
By method of lines I converted the PDE
u_t=u_{xx},
with the initial and boundary conditions
u(0,t)=u(0.1,t)
u(0,x)=sin(2*pi*x)
to a system of ODEs, as follows
function ut=pde1(t,u)
%
% Problem parameters
% global ncall
xl=0.0;
xr=1.0;
d=10;
a=10;
%
% PDE
n=length(u);
h=((xr-xl)/(n-1));
for i=1:n
if(i==1) ut(i)=0.0;
elseif(i==n) ut(i)=0;
else ut(i)=d*(u(i+1)-2.0*u(i)+u(i-1))/h^2;
end
end
ut=ut';
%
Now I am solving the ODEs ut by Runge-kutta method as follows:
%Initial condition
n=500;
for i=1:n
u0(i)=sin((2*pi)*(i-1)/(n-1));
end
t0=0.0;
tf=0.1;
tout=linspace(t0,tf,n);
y=zeros(n,length(tout));
y(:,1)=u0';
h=1/n;
for i = 1 : length(tout)
t=tout(i);
k1 = pde_1(t,y(:,i));
k2 = pde_1(t+h/2, y(:,i)+h*k1/2);
k3 = pde_1(t+h/2, y(:,i)+h*k2/2);
k4 = pde_1(t+h, y(:,i)+h*k3);
y(:,i+1) = y(:,i) + h*(k1 + 2*k2 + 2*k3 + k4)/6;
end
y
plot(tout,y(:,1))
but the outputs are NAN! any help?
numerical-methods heat-equation runge-kutta-methods
numerical-methods heat-equation runge-kutta-methods
asked Jan 3 at 21:17
Ahmad M.o Kassef
1
New contributor
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 3 at 21:17
Ahmad M.o Kassef
1
New contributor
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 3 at 21:17
Ahmad M.o Kassef
1
New contributor
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 3 at 21:17
Ahmad M.o Kassef
1
asked Jan 3 at 21:17
Ahmad M.o Kassef
1
1
New contributor
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ahmad M.o Kassef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to the Mathematics StackExchange! What language are you programming in? Also, you might get a better answer from StackOverflow; while the Runge-kutta is a numerical algorithm, your focus appears to be on the implementation rather than the mathematical properties, its correctness, or proving theorems with the method.
– Larry B.
Jan 3 at 21:43
Your code deviates in some relevant points from your equations. Is it reallyu(0,t)=u(0.1,t)or did you want periodic boundariesu(0,t)=u(1,t)or zero boundariesu(0,t)=u(1,t)=0? The last one you enforced with your implementation. Do you get a useful result if you decrease the factordor the time steph?
– LutzL
2 days ago
add a comment |
Welcome to the Mathematics StackExchange! What language are you programming in? Also, you might get a better answer from StackOverflow; while the Runge-kutta is a numerical algorithm, your focus appears to be on the implementation rather than the mathematical properties, its correctness, or proving theorems with the method.
– Larry B.
Jan 3 at 21:43
Your code deviates in some relevant points from your equations. Is it reallyu(0,t)=u(0.1,t)or did you want periodic boundariesu(0,t)=u(1,t)or zero boundariesu(0,t)=u(1,t)=0? The last one you enforced with your implementation. Do you get a useful result if you decrease the factordor the time steph?
– LutzL
2 days ago
Welcome to the Mathematics StackExchange! What language are you programming in? Also, you might get a better answer from StackOverflow; while the Runge-kutta is a numerical algorithm, your focus appears to be on the implementation rather than the mathematical properties, its correctness, or proving theorems with the method.
– Larry B.
Jan 3 at 21:43
Welcome to the Mathematics StackExchange! What language are you programming in? Also, you might get a better answer from StackOverflow; while the Runge-kutta is a numerical algorithm, your focus appears to be on the implementation rather than the mathematical properties, its correctness, or proving theorems with the method.
– Larry B.
Jan 3 at 21:43
Your code deviates in some relevant points from your equations. Is it really
u(0,t)=u(0.1,t) or did you want periodic boundaries u(0,t)=u(1,t) or zero boundaries u(0,t)=u(1,t)=0? The last one you enforced with your implementation. Do you get a useful result if you decrease the factor d or the time step h?– LutzL
2 days ago
Your code deviates in some relevant points from your equations. Is it really
u(0,t)=u(0.1,t) or did you want periodic boundaries u(0,t)=u(1,t) or zero boundaries u(0,t)=u(1,t)=0? The last one you enforced with your implementation. Do you get a useful result if you decrease the factor d or the time step h?– LutzL
2 days ago
add a comment |
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Welcome to the Mathematics StackExchange! What language are you programming in? Also, you might get a better answer from StackOverflow; while the Runge-kutta is a numerical algorithm, your focus appears to be on the implementation rather than the mathematical properties, its correctness, or proving theorems with the method.
– Larry B.
Jan 3 at 21:43
Your code deviates in some relevant points from your equations. Is it really
u(0,t)=u(0.1,t)or did you want periodic boundariesu(0,t)=u(1,t)or zero boundariesu(0,t)=u(1,t)=0? The last one you enforced with your implementation. Do you get a useful result if you decrease the factordor the time steph?– LutzL
2 days ago