I am having a problem solving second order differential equation in matlab. The equation is: d2y/dx2=n-p; where n and p both are functions of x. Boundary conditions y(0)=0 and y(100)=0.8. When I try to solve using bvp4c it shows matrix dim must agree

I am having a problem solving second order differential equation in matlab. The equation is:

   d2y/dx2=n-p; where n and p both are functions of x.

Boundary conditions y(0)=0 and y(100)=0.8. When I try to solve using bvp4c it shows matrix dimensions must agree. I have understood the problem but dont know how to solve. My function

function yprime=bvpfunction(x,y)
yprime=[y(2);n-p]

%n and p are both functions of x.

Answer 1

[Walter Roberson]

You do not have an 'end' matching your 'function' statement, so we know that your bvpfunction code cannot possibly be a nested function that might have access to a variable shared with a defining outer function.

We also know that you have no "global" statements that might import n or p.

The fact that you received an error about matrix dimensions instead of an error about undefined function or variable tells us that n and p must both somehow be defined anyhow.

From this we can deduce that your n and p are functions. And we can deduce that you are calling the functions without any input. In this situation, your code is equivalent to

yprime=[y(2);n()-p()]

and you are then running into problems because when you call n and p without input, the sizes of their outputs are not compatible with each other, or else the size of their output is not compatible with y(2)

Probably you want

yprime=[y(2);n(x)-p(x)]

[Md Mesbahus Saleheen]

Actually I have n(x) and p(x) and I want to send these values in this function.How can I send these values and then solve. the equation is, d2y/dx2=n(x)-p(x) I have n(x)and p(x) for 100 values of x. That means n(x) and p(x) are both 1x100 matrix. So how can I solve y for y(0)= and y(100)=0.8 boundary conditions

[Walter Roberson]

Are n and p formula, or are they vectors with known x values and corresponding output values? Are they functions or are they lookup tables? Or are they functions that for each x return a 1 x 100 vector?

[Md Mesbahus Saleheen]

n and p are vectors with known x values

x=linspace(0,100,100)
sol1=bvp4c(@fun,@bc,bvpinit);
sol2=bvp4c(@fun1,@bc1,bvpinit);
n=sol1.y;
p=sol2.y;

[Walter Roberson]

function yprime = bvpfunction(t, y, x, n, p)
  napprox = interp1(x, n, t, 'linear', 'extrap');
  papprox = interp1(x, p, t, 'linear', 'extrap');
  yprime=[y(2); napprox - papprox];

and then in the place where you previously passed @bvpfunction to the ode routines, now pass

@(t, y) bvpfunction(t, y, x, n, p)

where x, n, p are the values you show.

Note: extrapolation outside of the table might not be very accurate, but if you cannot constrain the t range then you might need it. If you can constrain the t range then the 'extrap' should not do any harm; you could remove the 'extrap' if you are sure the t will always be within the x range.

[Torsten]

>n and p are vectors with known x values
>x=linspace(0,100,100)
>sol1=bvp4c(@fun,@bc,bvpinit);
>sol2=bvp4c(@fun1,@bc1,bvpinit);
>n=sol1.y;
>p=sol2.y;

So you solve another two bvp's to get the values for n and p ?

In this case, you should solve all three bvps (for n, p and y) together to get higher efficiency and accuracy.

Best wishes

Torsten.