Incorrect use of '=' operator. Assign a value to a variable using '=' and compare values for equality using '=='.

D=0.1;
U=1;
K= 1*10^-6;
ci=3;
co=2;
L=1;
y(1) =C;
y(2) = dc/dx;
y(2)'=dc^2/dx^2;
function dcdx = bvpfcn(c,x) % euation to solve
dcdx= y(2)(U*y(2)+K*y(1))/D];
end
function res =bcfcn(ya,yb) % boundary conditions
res = [ya(1)
    yb(1)-L];
end
function g = guess(x) %initial guess for y and y'
g = [0 1];
end
plot(sol.C, Sol.x,'-0')

[KSSV]

This line:

y(2)'=dc^2/dx^2;

Should be repalced with:

y(2) = dc^2/dx^2;

The code is buggy and will not run. Many variables are not defined. What you want to achieve?

[Naveen Krish]

Can you check whether my code is right?

At steady state, this can be described by: 𝐷*𝑑 2𝐶/ 𝑑𝑥 2 − 𝑈*𝑑𝐶/𝑑𝑥 − 𝐾𝐶 = 0

two boundary conditions 𝐶 = 𝐶𝑖𝑛 @ 𝑥 = 0, and 𝐶 = 𝐶𝑜𝑢𝑡 @ 𝑥 = 𝐿,

parameter values

𝐷 = 0.1 m2 /s, 𝑈 = 1 m/s, 𝐾 = 1𝐸 − 6 s -1 , 𝐶𝑖𝑛 = 3 mg/L, 𝐶𝑜𝑢𝑡 = 2 mg/L, 𝐿 = 1.0 m

I need to solve the ode as BVP4C function and plot the profile of C vs x

[Walter Roberson]

y(2) = dc^2/dx^2;

probably not, should perhaps be

y(3) = dc^2/dx^2;

since y(2) was already assigned to.

[Naveen Krish]

Function definitions in a script must appear at the end of the file.

Move all statements after the "guess" function definition to before the first local function definition.

[Naveen Krish]

this error showed in my plot(sol.C, Sol.x,'-0')

[KSSV]

Many variables are not defined.....you have to put plot before the functions.

[Naveen Krish]

can you be more specific

[Walter Roberson]

put your calls to the boundary value solvers here, returning sol

plot(sol.C, sol.x,'-0')

function dcdx = bvpfcn(c,x) % equation to solve
    D = 0.1;
    U = 1;
    K = 1*10^-6;
    ci = 3;
    co = 2;
    L = 1;
    C = appropriate_value
    y1 = @(c) C * ones(size(c));
    y2 = @(c) dc/dx;
    dcdx = y2(U*y2(c) + K*y1(c))/D];
end
function res =bcfcn(ya,yb) % boundary conditions
    res = [ya(1)
        yb(1)-L];
end
function g = guess(x) %initial guess for y and y'
    g = [0 1];
end

You will need to fill out something more concrete for y2 . Remember, these functions do not know what the derivative of your function is until you tell it.

But.. if you define y(1) as C then the derivative of that is 0, so you might be implying that y(2) should be 0 ??

[Naveen Krish]

y1 = C, y2 = dc/dx, y2' = d''c/dx2 this is how the derivative should be right?

[Walter Roberson]

@Naveen Krish

No. In MATLAB, ' is an operator means "complex transpose". It cannot be part of the name of a variable.

x = [1+2i 3-4i]

x = 
   1.0000 + 2.0000i   3.0000 - 4.0000i

x'

ans = 
   1.0000 - 2.0000i
   3.0000 + 4.0000i

When you have numeric data, you can do numeric estimation of the derivative using gradient()

If you have the Symbolic toolbox, then you can use diff() to request symbolic differentiation

syms x
f = tan(1-x^2)

f = 

f__prime = simplify(expand(diff(f, x)))

f__prime = 

f__primeprime = simplify(diff(f__prime,x))

f__primeprime =