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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))
采纳的回答
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)
F1 = dsolve(eqn, cond); f1 = F1.f; g1 = F1.g; h1 = F1.h; disp(collect([f1 g1 h1],x))
3 个评论
Walter Roberson
2021-2-12
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
2021-2-12
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
2021-2-12
更多回答（2 个）
KALYAN ACHARJYA
2021-2-12
编辑：KALYAN ACHARJYA
2021-2-12
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??
randerss simil
2021-2-12
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,
2 个评论
randerss simil
2021-2-12
dsolve expects, as many equations as the number of symoblic variables that constitute equations.
randerss simil
2021-2-12
Try with ode45 function
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