How to evaluate a point in a solved differential equation in matlab?

Hello!

I have a problem with  solving a DE with my matlab code, or more specifically I don't know how to continue solve a problem.

The question is to solve this equation and evaluate T at x=1.6.

The dif.equation is as follows:

% -d(k*dT/dx)/dx=Q(x)
% where k=2+x/7 ; Q=260*e^-(x-L/2)^2  ; where 0<=x<=L

I have solved it with finite difference methd, and it seems to work. How do I evaluate T at x=1.6? from here?

Here is what I got so far in the code:

%Given values: 
x0=0; 
xn=3.6; 
y0=320; 
yn=450; 
L=3.6;


n=40; 
h=(xn-x0)/n;
x=x0+h*[1:n-1];

Q = @(x) 260.*exp(-(x-L/2).^2);
alfa= @(x) 1/(14*h)-2/(h^2)-x./(7*h^2);
beta= @(x) 4/h^2+((2/(7*h^2)).*x);
gamma= @(x) -1/(14*h)-x./(7*h^2)-(2/h^2);

A1=diag(alfa(x(2:n-1)),-1);
A2=diag(beta(x));
A3=diag(gamma(x(1:(n-2))),1);
A=A1+A2+A3;

b=Q(x);
b(1)= b(1)-alfa(x0)*y0;
b(n-1)=b(n-1)-gamma(xn)*yn;

y=A\b';  

xx=[x0 x xn]
yy=[y0 y' yn]
plot(xx,yy)
xlabel('Length'); ylabel('Temperature')

Answer 1

[Torsten]

y16 = interp1(xx,yy,1.6)

[Maria Karlsson Osipova]

Thank you! However, I forgot to mention that I need to do error analysis on the given value.

[Torsten]

Ok. In order to prevent interpolation errors, include x=1.6 in your discretization (i.e. 1.6 should be part of

x=x0+h*[1:n-1];

).

E.g. if n=36*i, y(16*i) gives T at x=1.6 where "i" can be chosen as 1,2,3,4,...

Best wishes

Torsten.

[Bruno Luong]

"I need to do error analysis on the given value."

What kind of error analysis?

[Torsten]

Most probably how abs(y_num-y_exact) at x=1.6 behaves as h->0.

[Bruno Luong]

I'm not aware if such local error estimate exists, do you?

There is a lot of formula to estimate residual error in some global Banach's/Hilbert norm

|y-y_h| <= C*h^alpha |y_h|.

Not sure how such local norm |y-y_h|(x=1.6) would need as input to get an error bound.

That's why my question.