how to write matlab bvp4c code for moving boundary surface problem?

hello, I am new in the bvp4c method of Matlab. I want to find multiple graphs for different value of 'r' for the following problem. Equation is f''' +ff''=0 Boundary condition is f(0)=0, f'(0)=1-r, f'(infnty)=r

Answer 1

[Stephan]

Hi,

i recommend to start simple and work step by step:

1. Have a read in the documentation of bvp4c
2. Try to implement your model with a single fixed value for r and get a valid solution
3. Once you have this, use a for loop to vary r like needed for your purposes and save all the results in a Matrix in different columns or rows
4. Plot the results by using the saved values

Problems by doing this? --> Come back and ask a specific question including your code so far.

Best regards

Stephan

[Anil Gautam]

function cortell
global r
r=[0]
solinit = bvpinit(linspace(0,10,1000),[0 1- r r]);
options = bvpset('stats','on');

sol = bvp4c(@equation,@bvpbc,solinit,options);
eta = sol.x;
u = sol.y;
clf reset
 plot(eta,u(3,:));
        hold on;
        axis([0 4 0 1.3]);
        xlabel('\eta')
        ylabel('f''(\eta)')

shg

% --------------------------------------------------------------------------
function dy=equation(~,y)
%dy=zeros(4,1); 
dy=[y(2);
    y(3);
    -y(1)*y(3);
    %---------------------------------------------------------------------
     ];

function res = bvpbc(y0,yinf)
global r
res = [y0(1)
       y0(2)-1+ r
       yinf(2)- r
      ];

    if true
      % code
    end

this is code . how I can I get the different graph for different value of 'r' . In other word, for r=0.0, 0.25,0.75 how i can plot graph for f on y axis and eta on x axis.. please suggest.

[Stephan]

See the answer from Torsten...

Answer 2

[Torsten]

function main

r=[0 0.02 0.04]
for i=1:numel(r)
  r_actual = r(i);  
  solinit = bvpinit(linspace(0,10,10000),[0 1-r_actual 0]);
  options = bvpset('stats','on');

    sol{i} = bvp4c(@equation,@(y0,yinf)bvpbc(y0,yinf,r_actual),solinit,options);
  end

for i=1:numel(r)
  eta = sol{i}.x;
  u = sol{i}.y;
  plot(eta,u(3,:));
  hold on;
end  

end

% --------------------------------------------------------------------------
function dy=equation(~,y)
%dy=zeros(4,1); 
dy=[y(2);
    y(3);
    -y(1)*y(3);
    %---------------------------------------------------------------------
     ];
end

function res = bvpbc(y0,yinf,r_actual)
res = [y0(1)
       y0(2)-1+ r_actual
       yinf(2)- r_actual
      ];
end

[Anil Gautam]

Thanks, it works properly. thanks a lot. I want to know something more. how we can relate this initial guess solinit = bvpinit(linspace(0,10,10000),[0 1-r_actual 0]);

with following boundary condition res = [y0(1) y0(2)-1+ r_actual yinf(2)- r_actual ] why it is written 0 at the place of r_actual in this line [0 1-r_actual 0]

[Torsten]

Try this code - it uses the solution of the last step as initial guess for the next:

function main
global ix

r=[0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0];
r_actual = r(1);  
ir=1;
  solinit = bvpinit(linspace(0,10,1000),@(x)guess(x,ir,[0 1-r_actual r_actual]));
  options = bvpset('stats','on');
  ix = 0;
    sol{1} = bvp4c(@equation,@(y0,yinf)bvpbc(y0,yinf,r_actual),solinit,options);
for ir=2:numel(r)
  r_actual = r(ir); 
  solinit = bvpinit(sol{ir-1}.x,@(x)guess(x,ir,sol{ir-1}.y));
  options = bvpset('stats','on');
  ix = 0;
    sol{ir} = bvp4c(@equation,@(y0,yinf)bvpbc(y0,yinf,r_actual),solinit,options);
end

for ir=1:numel(r)
  eta = sol{ir}.x;
  u = sol{ir}.y;
  plot(eta,u(3,:));
  hold on;
end  

end

% --------------------------------------------------------------------------
function dy=equation(~,y)
%dy=zeros(4,1); 
dy=[y(2);
    y(3);
    -y(1)*y(3);
    %---------------------------------------------------------------------
     ];
end

function res = bvpbc(y0,yinf,r_actual)
res = [y0(1)
       y0(2)-1+ r_actual
       yinf(2)- r_actual
      ];
end  

function yini=guess(x,ir,sol)
global ix
ix = ix + 1;
if ir==1
  yini(1)=sol(1);
  yini(2)=sol(2);
  yini(3)=sol(3);
else
  yini(1)=sol(1,ix);
  yini(2)=sol(2,ix);
  yini(3)=sol(3,ix);
end
end

[Anil Gautam]

thanku very much. thanks a lot