Can it be possible to use a for loop to precise the code

M = 0.2; k = 1; L = 0.1; Pr = 2; G = 2; S1 = 0.1; Ec = 0.1; fw = 1; R = 0.3; a5 = 2; a6 = 3; a7 = 4;
a8 = 10; a9 = 1.72;  a10 = 8; a11 = 2; a12 = 3;
xb = 5;    %%%  xb = Inf;
syms x f0(x) g0(x) h0(x) f1(x) g1(x) h1(x) f2(x) g2(x) h2(x) f3(x) g3(x) h3(x) f4(x) g4(x)  h4(x)
eqn0 = [ diff(f0,3) == 0, diff(g0,2) == 0, diff(h0,2) == 0 ];
cond0 = [ f0(0) == fw, subs(diff(f0),0) == 1, g0(0) == 0, h0(0) == 1, subs(diff(f0),xb) == 0,  g0(xb) == 0, h0(xb) == 0 ];
F0 = dsolve(eqn0,cond0);     f0 = F0.f0; g0 = F0.g0; h0 = F0.h0;    
%%% The following lines need to be in squeezed so that it will save time (using for loop) 
eqn1 = [ diff(f1,3) + a5*( f0*diff(f0,2) - diff(f0)^2 + k*diff(g0) ) - a6*M^2*diff(f0) + a7*L*h0 == 0, G*diff(g1,2) - 2*g0 - diff(f0,2) == 0, (a9+R)*diff(h1,2) + Pr*( a10*f0*diff(h0) + S1*h0 + Ec*( a11*diff(f0,2)^2 + a12*M^2*diff(f0)^2 ) ) == 0  ];
cond1 = [ f1(0) == 0, subs(diff(f1),0) == 0, g1(0) == 0, h1(0) == 0, subs(diff(f1),xb) == 0,  g1(xb) == 0,  h1(xb) == 0 ];
F1 = dsolve(eqn1,cond1);     f1 = F1.f1; g1 = F1.g1; h1 = F1.h1;     
eqn2 = [ diff(f2,3) + a5*( f1*diff(f1,2) - diff(f1)^2 + k*diff(g1) ) - a6*M^2*diff(f1) + a7*L*h1 == 0,G*diff(g2,2) - 2*g1 - diff(f1,2) == 0, (a9+R)*diff(h2,2) + Pr*( a10*f1*diff(h1) + S1*h1 + Ec*( a11*diff(f1,2)^2 + a12*M^2*diff(f1)^2 ) ) == 0 ];
cond2 = [ f2(0) == 0, subs(diff(f2),0) == 0, g2(0) == 0, h2(0) == 0, subs(diff(f2),xb) == 0,  g2(xb) == 0,  h2(xb) == 0 ];
F2 = dsolve(eqn2,cond2);     f2 = F2.f2; g2 = F2.g2; h2 = F2.h2;   
eqn3 = [ diff(f3,3) + a5*( f2*diff(f2,2) - diff(f2)^2 + k*diff(g2) ) - a6*M^2*diff(f2) + a7*L*h2 == 0,G*diff(g3,2) - 2*g2 - diff(f2,2) == 0, (a9+R)*diff(h3,2) + Pr*( a10*f2*diff(h2) + S1*h2 + Ec*( a11*diff(f2,2)^2 + a12*M^2*diff(f2)^2 ) ) == 0 ];
cond3 = [ f3(0) == 0, subs(diff(f3),0) == 0, g3(0) == 0, h3(0) == 0, subs(diff(f3),xb) == 0,  g3(xb) == 0,  h3(xb) == 0 ];
F3 = dsolve(eqn3,cond3);     f3 = F3.f3; g3 = F3.g3; h3 = F3.h3;      
eqn4 = [ diff(f4,3) + a5*( f3*diff(f3,2) - diff(f3)^2 + k*diff(g3) ) - a6*M^2*diff(f3) + a7*L*h3 == 0, G*diff(g4,2) - 2*g3 - diff(f3,2) == 0, (a9+R)*diff(h4,2) + Pr*( a10*f3*diff(h3) + S1*h3 + Ec*( a11*diff(f3,2)^2 + a12*M^2*diff(f3)^2 ) ) == 0  ];
cond4 = [ f4(0) == 0, subs(diff(f4),0) == 0, g4(0) == 0, h4(0) == 0, subs(diff(f4),xb) == 0,  g4(xb) == 0,  h4(xb) == 0 ];
F4 = dsolve(eqn4,cond4); f4 = F4.f4; g4 = F4.g4; h4 = F4.h4; 
f = f0 + f1 + f2 + f3 + f4; g = g0 + g1 + g2 + g3 + g4; h = h0 + h1 + h2 + h3 +h4;    
vpa(f)

ans = 

Answer 1

[Walter Roberson]

%%% The following lines need to be in squeezed so that it will save time (using for loop)

Using a for loop is only faster than "unrolling" to individual statements if

• the amount of time spent parsing the unrolled code exceeds the amount of overhead of indexing and looping; or
• the Execution Engine is able to find code patterns that it can optimize native code for

With the small number of lines, and the small number of iterations implied (5 at most), and taking into account that you cannot put symbolic functions into normal vectors (you need to put them into cell arrays instead), I suspect that you cannot save enough execution time by looping to make a notable difference. And the Execution Engine cannot do much to optimize symbolic operations beyond what you could already do by using, for example,

S0 = sym(0);

%...
cond2 = [ f2(S0) == S0, subs(diff(f2),S0) == S0, g2(S0) == S0, h2(S0) == S0, subs(diff(f2),xb) == S0,  g2(xb) == S0,  h2(xb) == S0 ];

to reduce the time needed to detect the numeric 0 and convert it to symbolic 0 repeatedly.

Symbolic calculations are done inside a separate process, inside the MuPAD-based Symbolic Engine, that does not use the Execution Engine (and cannot take advantage of multiple cores or of GPUs). It is those operations that are slow; looping or not in this code is timing noise compared to the time spent in the Symbolic Engine.

[Sushanth]

%%  Can I write something like:
S0 = sym(0); xb = sym(5);
for k = 1:3
    eqn = [ diff(f,3) + a5*( f0*diff(f0,2) - diff(f0)^2 + k*diff(g0) ) - a6*M^2*diff(f0) + a7*L*h0 == 0, G*diff(g1,2) - 2*g0 - diff(f0,2) == 0, (a9+R)*diff(h,2) + Pr*( a10*f0*diff(h0) + S1*h0 + Ec*( a11*diff(f0,2)^2 + a12*M^2*diff(f0)^2 ) ) == 0  ];
    cond = [ f1(S0) == S0, subs(diff(f1),S0) == S0, g1(S0) == S0, h1(S0) == S0, subs(diff(f1),xb) == S0,  g1(xb) == S0,  h1(xb) == S0 ];
    F = dsolve(eqn,cond);     f(k) = F.f; 
end
f = f0 + f1 + f2 + f3 

[Walter Roberson]

%this might need some tweaks
S0 = sym(0); xb = sym(5);
f = {f0, f1, f2, f3, f4};
g = {g0, g1, g2, g3, g4};
h = {h0, h1, h2, h3, h4};

N = length(f);
F = cell(1,N);
for k = 1:N-1
    df0 = diff(f{k}); d2f0 = diff(df0);
    df1 = diff(f{k+1}); d3f1 = diff(df1,2);
    dg0 = diff(g{k});
    dh0 = diff(h{k});
    d2g1 = diff(g{k+1},2);
    d2h1 = diff(h{k+1},2);
    eqn = [ d3f1 + a5*( f{k}*d2f0 - df0^2 + k*dg0 ) - a6*M^2*df0 + a7*L*h0 == S0, 
            G*d2g1 - 2*g{k} - d2f0 == S0,
            (a9+R)*d2h1 + Pr*( a10*f{k}*dh0 + S1*h{k} + Ec*( a11*d2f0^2 + a12*M^2*df0^2 ) ) == 0  ];
    cond = [ f{k+1}(S0) == S0,
             df1(S0) == S0,
             g{k+1}(S0) == S0,
             h{k+1}(S0) == S0,
             df1(xb) == S0,
             g{k+1}(xb) == S0,
             h{k+1}(xb) == S0 ];
    sol = dsolve(eqn,cond);
    F(k) = sol.("f" + k); 
end
f = sum(F) 

This might be better coding practice that reduces the chance of error in constructing the various equations -- but that does not mean that it will be any faster.

[Sushanth]

%% After running the above code the following error is coming:

Unable to perform assignment because value of type 'sym' is not convertible to 'cell'.

Error in line 35

F(k) = sol.("f" + k);

Caused by:

Error using cell

Conversion to cell from sym is not possible.

[Walter Roberson]

 F{k} = sol.("f" + k);

ummm, I will have to come up with a correction for the sum(). I will post that later.

[Walter Roberson]

There might still need to be some tweaks to how the solutions are extracted and to exactly where in F is written to, and to forming the sum.

But in the meantime, the reason that dsolve() are failing complaining about extra indeterminates, is because you are trying to give two boundary conditions for the same function, g1(0) == 0 and g1(xb) == 0, and h1(0) == 0 and h1(xb) == 0. dsolve() can never handle more than one boundary condition for the same derivative of the same function.

M = 0.2; k = 1; L = 0.1; Pr = 2; G = 2; S1 = 0.1; Ec = 0.1; fw = 1; R = 0.3; a5 = 2; a6 = 3; a7 = 4;
a8 = 10; a9 = 1.72;  a10 = 8; a11 = 2; a12 = 3;
xb = 5;    %%%  xb = Inf;
syms x f0(x) g0(x) h0(x) f1(x) g1(x) h1(x) f2(x) g2(x) h2(x) f3(x) g3(x) h3(x) f4(x) g4(x)  h4(x)
eqn0 = [ diff(f0,3) == 0, diff(g0,2) == 0, diff(h0,2) == 0 ];
cond0 = [ f0(0) == fw, subs(diff(f0),0) == 1, g0(0) == 0, h0(0) == 1, subs(diff(f0),xb) == 0,  g0(xb) == 0, h0(xb) == 0 ];
F0 = dsolve(eqn0,cond0);     f0 = F0.f0; g0 = F0.g0; h0 = F0.h0;    


%this might need some tweaks
S0 = sym(0); xb = sym(5);
f = {f0, f1, f2, f3, f4};
g = {g0, g1, g2, g3, g4};
h = {h0, h1, h2, h3, h4};

N = length(f);
F = cell(1,N);
for k = 1:N-1
    df0 = diff(f{k}); d2f0 = diff(df0);
    df1 = diff(f{k+1}); d3f1 = diff(df1,2);
    dg0 = diff(g{k});
    dh0 = diff(h{k});
    d2g1 = diff(g{k+1},2);
    d2h1 = diff(h{k+1},2);
    eqn = [ d3f1 + a5*( f{k}*d2f0 - df0^2 + k*dg0 ) - a6*M^2*df0 + a7*L*h0 == S0, 
            G*d2g1 - 2*g{k} - d2f0 == S0,
            (a9+R)*d2h1 + Pr*( a10*f{k}*dh0 + S1*h{k} + Ec*( a11*d2f0^2 + a12*M^2*df0^2 ) ) == 0  ]
    cond = [ f{k+1}(S0) == S0,
             df1(S0) == S0,
             g{k+1}(S0) == S0,
             h{k+1}(S0) == S0,
             df1(xb) == S0,
             g{k+1}(xb) == S0,
             h{k+1}(xb) == S0 ]
    sol = dsolve(eqn,cond);
    if isempty(sol)
        F{k} = sym(nan);
    else
        F{k} = sol.("f" + k);
    end
end

eqn(x) = 

cond = 

eqn(x) = 

cond = 

Warning: Number of indeterminates greater than number of equations. Trying to parameterize solutions in terms indeterminates.

Warning: Unable to find symbolic solution.

eqn(x) = 

cond = 

Warning: Number of indeterminates greater than number of equations. Trying to parameterize solutions in terms indeterminates.

Warning: Unable to find symbolic solution.

eqn(x) = 

cond = 

Warning: Number of indeterminates greater than number of equations. Trying to parameterize solutions in terms indeterminates.

Warning: Unable to find symbolic solution.

celldisp(F)

 
F{1} =
 

 
 
F{2} =
 

NaN

 
 
F{3} =
 

NaN

 
 
F{4} =
 

NaN

 
 
F{5} =
 
     []
 

%f = sum(F)