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







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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 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


















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







share|cite|improve this question







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 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
















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







share|cite|improve this question







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






share|cite|improve this question







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.











share|cite|improve this question







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.









share|cite|improve this question




share|cite|improve this question






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




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 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




















  • 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


















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












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