dsolve code is not running

syms x f(x) g(x) h(x) M2 M3 a1 k K  L A Rd  Br Pr Nb Nt G Le H
eqn = [ diff(f,2) - L*diff(f)*diff(g) + A*a1*exp(L*g) == 0,...
        (K/k+Rd)*diff(g,2) + (K/k)*(M2+M3)*diff(h)*diff(g) + exp(-L*g)*Br*(diff(f))^2/a1 + Pr*(Nb*diff(h)*diff(g) + Nt*(diff(g))^2) + H*g == 0,...
        diff(h,2) +(Nt/Nb)*diff(g,2) + G*Pr*Le*(1-h) == 0 ];
cond = [ f(0) == 0,g(0) == 0,h(0) == 0,  f(1) == 0,g(1) == 0,h(1) == 0 ];
F1 = dsolve(eqn,cond);     f1 = F1.f; g1 = F1.g; h1 = F1.h;     disp(collect([f1 g1 h1],x))

Answer 1

[Walter Roberson]

syms x f(x) g(x) h(x) M2 M3 a1 k K  L A Rd  Br Pr Nb Nt G Le H
eqn = [ diff(f,2) - L*diff(f)*diff(g) + A*a1*exp(L*g) == 0,...
        (K/k+Rd)*diff(g,2) + (K/k)*(M2+M3)*diff(h)*diff(g) + exp(-L*g)*Br*(diff(f))^2/a1 + Pr*(Nb*diff(h)*diff(g) + Nt*(diff(g))^2) + H*g == 0,...
        diff(h,2) + (Nt/Nb)*diff(g,2) + G*Pr*Le*(1-h) == 0 ];
cond = [ f(0) == 0,g(0) == 0,h(0) == 0,  f(1) == 0,g(1) == 0,h(1) == 0 ];
reshape(eqn(x),[],1)

ans = 

F1 = dsolve(eqn, cond);     f1 = F1.f; g1 = F1.g; h1 = F1.h;     disp(collect([f1 g1 h1],x))

Warning: Unable to find symbolic solution.

Dot indexing is not supported for variables of this type.

Error in sym/subsref (line 902)
R_tilde = builtin('subsref',L_tilde,Idx);

[Walter Roberson]

diff(h,2) +(Nt/Nb)*diff(g,2) + G*Pr*Le*(1-h) == 0
          ^
          

That is not addition. That is "unary plus". Just like

[1 -2]

is not subtraction, you can use + instead of - in expressions the same way you use negative sign; for example,

[+1 -2 +2 -1]

would be perfectly understandable in context.

Spacing is important. Inside [] and {}, unless you are inside (), if there is whitespace between the previous term and this term, and there is + or - that is not followed by whitespace before the next term, then the previous term ends. To signify addition or subtraction, you need to either have whitespace after the character or else you need to have no whitespace before the character

[+1-2 + 2 -1]
 U S  A   U

U -> Unary operator

S -> subtraction

A -> addition

and two terms would result, (1-2+2) and (-1)

[Walter Roberson]

Next thing to note is that you cannot provide two boundary conditions for the same function differential order. f(0)==0, fine by itself. f(1)==1, fine by itself. f(0)==0 and f(1)==1 together, will not work. The boundary conditions are used only to try to find a value for the constant of integration, and the code cannot reason about them further than that.

But in your case, it does not matter: even if you remove the boundary conditions, the system is too hard for MATLAB to solve.

[MINATI PATRA]

Thanks

Walter

for your valuable suggestion. I think because of hghly nonlinearity, dsolve is unable to solve. Any way, I have solved with bvp code.

Answer 2

[KALYAN ACHARJYA]

The issues are here

f1=F1.f; 
g1=F1.g;
h1=F1.h;

Here, lets say for first line of the code

f1=F1.f;

Whre F1 is the sym variable with size 1x4, whereas f is the sym function

>> whos F1
  Name      Size            Bytes  Class    Attributes
  F1        1x4                 8  sym                
>> whos f
  Name      Size            Bytes  Class     Attributes
  f         1x1                 8  symfun  

What exactly are you trying to do in these lines?

Variable Access??

[MINATI PATRA]

I want to find f g h in symbolic form

Answer 3

[randerss simil]

 M2 M3 a1 k K  L A Rd  Br Pr Nb Nt G Le H

Assign values for above variables instead of symbolic variables,

As you are using only three equations, with 3 unknowns in x, f, g, h

Then you can solve it,

[randerss simil]

dsolve expects, as many equations as the number of symoblic variables that constitute equations.

[randerss simil]

Try with ode45 function