Finite difference method problem with solving an equation

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Aldo
Aldo 2016-11-8
评论： Torsten 2016-11-10
Trying to use Finite difference method, to write the equation in AT = b matrices. But I don't know how to write FDM on that type of equation, please see image.

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采纳的回答

Torsten
Torsten 2016-11-8
(k(x_(i+1/2))*(T(x_(i+1))-T(x_(i)))/(x_(i+1)-x_(i))-k(x_(i-1/2))*(T(x_(i))-T(x_(i-1)))/(x_(i)-x_(i-1)))/(x_(i+1/2)-x_(i-1/2)) = Q(x_i)
with
x_(i+1/2)=(x_(i+1)+x_(i))/2
x_(i-1/2)=(x_(i)+x_(i-1))/2
Best wishes
Torsten.

更多回答（5 个）

Torsten
Torsten 2016-11-8
[d/dx(k*dT/dx)](x) ~ (k(x+h/2)*(dT/dx)(x+h/2) - k(x-h/2)*(dT/dx)(x-h/2))/h
Approximations for (dT/dx)(x+h/2) and dT/dx(x-h/2) are
(dT/dx)(x+h/2) ~ (T(x+h)-T(x))/h
(dT/dx)(x-h/2) ~ (T(x)-T(x-h))/h
Now insert in the first equation.
The finite-difference approximation in my first response was more general because it took into account non-equidistant grids (i.e. h is not fixed over the complete interval). But note that I missed the minus-sign in front of the approximaton for d/dx(k*dT/dx).
Best wishes
Torsten.
Torsten
Torsten 2016-11-9
-(k(x_(i+1/2))*y_(i+1)-(k(x_(i+1/2))+k(x_(i-1/2)))*y_(i)+k(x_(i-1/2))*y_(i-1))/h^2 = Q(x_(i))
with
x_(i+1/2)=(x_(i)+x_(i+1))/2
x_(i-1/2)=(x_(i-1)+x_(i))/2
for i=2,...,(n-1)
y(1) and y(n) are given by your boundary conditions.
So to incorporate the boundary conditions, you can add the equations
y(1) = 300
y(n) = 410
to the system from above.
The complete system to be solved then looks like
y(1) = 300
-(k(x_(i+1/2))*y_(i+1)-(k(x_(i+1/2))+k(x_(i-1/2)))*y_(i)+k(x_(i-1/2))*y_(i-1))/h^2 = Q(x_(i)) (i=2,...,n-1)
y(n) = 410
Best wishes
Torsten.
Aldo
Aldo 2016-11-8
Aren't you suppose to use this? Where did you get x_(i+1/2)=(x_(i+1)+x_(i))/2 x_(i-1/2)=(x_(i)+x_(i-1))/2 from?
Best regards Aldo :D
Aldo
Aldo 2016-11-8
Can't figure out a way to get the equation to look like this. And shouldn't (dT/dx)(x+h/2) ~ (T(x+h)-T(x))/h, be divided by 2h?
Best regards
Aldo
Aldo 2016-11-9
clear all; close all, clf;
T0=300; Tslut=410;
N=100;
h=410/N; n=N-1; x=h*(0:n)'; X=[0;x;410];
k=2+x/7; Q=260*exp(-(x-410/2).^2);
A= sparse(n);
for i=1:n-1, A(i,i)=(2*k/h^2); end
for i=1:n-1, A(i,i+1)= (k/h^2); A(i+1,i)= (k/h^2);end
b= Q
(1)= b(1)+(k/h^2)*T0
b(n)= b(n)+(k/h^2)*Tslut
T= A/b
I get this error "Subscripted assignment dimension mismatch. Error in uppgift5 (line 9) for i=1:n-1, A(i,i)=(2*k/h^2); end "
can you see how I can resolve it?
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Aldo
Aldo 2016-11-10
编辑：Aldo 2016-11-10
Yes you are right it doesn't take much time to calculate. However, how do you get T specifically at x=1.5? I am interested in finding the value of T at point 1.5
Best Regards Aldo :D
Torsten
Torsten 2016-11-10
Take N=340 - then T at 1.5 is T(151) :-)
For arbitrary N,
T15 = interp1(x,T,1.5);
Best wishes
Torsten.

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