Using finite difference to solve DE

  • Thread starter jdawg
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  • #1
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Homework Statement


d2T/dx2 = 5*(dT/dx) - 0.1*x = 0
T(0) = 50
T(10) = 400
(Δx) = 2


I've figured out how to do these problems when Δx = 1, but when it equals any other number it goes wrong.
I know you start by plugging in the algebraic approximations for the differential elements, I think maybe my problem is the nodes?


Homework Equations


d2T/dx2 = (Ti+1 * Ti + Ti-1)/((Δx))
dT/dx = (Ti+1 - Ti-1)/(2*Δx)

The Attempt at a Solution


node1 = 2; node2 = 4; node3 = 6; node4 = 8;

For node1:
(T2 - 2*T1 + T0)/(22) + 5*((T2 - T0)/(2*2)) - 0.1*(2) = 0

End up with ======> -0.5*T1 + 1.5*T2 = 50.2

For node2:
(T3 - 2*T2 + T1)/(22) + 5*((T3 - T1)/(2*2)) - 0.1*(4) = 0

End up with =======> -T1 - 0.25*T2 + 1.5*T3 = 0.4

For node3:
(T4 - 2*T3 + T2)/(22) + 5*((T4 - T2)/(2*2)) - 0.1*(6) = 0

End up with ========> -T2 - 0.5*T3 + 1.5*T4 = 0.6

For node4:
(T5 - 2*T4 + T3)/(22) + 5*((T5 - T3)/(2*2)) - 0.1*(8) = 0

End up with ========> -T3 - 0.5*T4 = -724.2

I put all the coefficients into a matrix and its tridiagonal, which is good I think. But when I try to plot it in matlab it gives me a crazy looking zigzag, which I'm pretty sure isn't correct.

If someone could point out what I'm doing wrong I would really appreciate it! :)
 
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Answers and Replies

  • #2
Simon Bridge
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For arbitrary ##\Delta x## you have: ##x_n = x_0 + n\Delta x## ... this is the lynchpin.
Tri-diagonal matrix is correct: that's what you get for a second derivative.

If you only had the second derivative, then the matrix would have (1,-2,1) on the diagonal, multiplied by a constant.
You should be able to write out the matrices for each term and just add them up.
 
  • #3
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I still don't really understand.
 
  • #4
Simon Bridge
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I still don't really understand.
It is best to understand the techniques you are applying instead of just going through the motions by rote.

Note: post #1, under "relevant equations", check that first expression.

If ##D^2_iy## indicates the second derivative of ##y(x)## evaluated at point ##x=x_i : i\in\mathbb{Z}## then:
$$D^2_iy \approx \frac{y_{i+1}-2y_{i}+y_{i-1}}{h^2}: y_i=y(x_i)$$
... here I used ##h=\Delta x## to avoid typing lots of deltas.

However, this expression may be inconsistent with the second expression in the same section.
If ##D_iy\approx\frac{1}{2h}\big(y_{i+1}-y_{i-1}\big)## then: ##D_i^2y \approx \frac{1}{4h}\big(D_{i+1}y-D_{i-1}y\big)##
... so it kinda helps to know where you are coming from.

This would make the matrix for a second order DE potentially quin-diagonal. But I don't think it's a fatal problem since these are only approximations anyway.

Now, using this notation...
I can define a vector ##\vec x = \big(x_0,x_1,x_2,\cdots x_n,\cdots x_{N-2},x_{N-1}\big)^t : x_n=x_0+nh## ... for the N discrete points you want to do your evaluation over. (The "##^t##" indicates a transpose so this is defined to be a column vector.)

I can derive, from that, other vectors like: ##\vec y = \big(y_0,y_1, \cdots y_n, cdots y_{N-1}\big)^t:y_n=y(x_n)## and, similarly ##\vec y' = \big(y'(x_0)\cdots\big)^t## etc.

You follow me so far?

When I look at the relationships, I can see that ##\vec y' = D\vec y## where ##D## is a matrix that looks like this:
$$D=\frac{1}{2h}\begin{bmatrix}
0 & 1 & 0 &\cdots\\
-1 & 0 & 1 & 0 & \cdots\\
0 & -1 & 0 & 1 & 0 & \cdots\\
\cdots & 0 & -1 & 0 & 1 & 0 & \cdots\\
& & \ddots & \ddots & \ddots & \ddots & \ddots & &\\
\end{bmatrix}$$ ... ie, this is a tri-diagonal matrix where the diagonal elements are (-1,0,1) times a constant (1/2h).

This follows from your definition in post #1 (section 2, 2nd equation). You should check, if you have not already.
OTOH: If we follow the definition of the derivative given in math text books you get a tri-diagonal matrix with elements (0,-1,1)/h.

... however you do it, it also follows that ##\vec y'' = D\vec y' = D(D\vec y) = D^2\vec y##

Thus any DE can, with a bit of care, be rewritten as a matrix equation.

Yours is (from post #1):
d2T/dx2 = 5*(dT/dx) - 0.1*x = 0
Which is ##D^2\vec T = 5D\vec T - 0.1\vec x ##
Where ##h=2, T(0)=T(x_0)=T_0=50, T(10)=T(x_{175})=T_{175} = 400##

Rearrange, you get ##A\vec T = \vec x## where ##A=50D-10D^2##.
The solution, therefore, amounts to finding the inverse of the matrix A.
 
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