FE 1D method and hat functions

[Carla1985]

Hi all, I'm doing a project on the finite elements method and am struggling to understand a part of it.

I have defined the hat functions as:

\[
\phi_i(x) =
\begin{cases}
\frac{x-x_{i-1}}{h} & \text{if } x_{i-1}\leq x<x_i \\
\frac{x_{i+1}-x}{h} & \text{if } x_i\leq x<x_{i+1}\\
0 & \text{otherwise}
\end{cases}
\]

I know this means that for each $\phi_i(x_i)$ is 1 when i=j and 0 the rest of the time. The part I'm stuck on is the next step:

\[
u_h(x) = \sum_{j=1}^{n-1}c_j\phi_j(x)
\]

How do I sum these functions in their general form?

Regards
Carla

 

[I like Serena]

Hi all, I'm doing a project on the finite elements method and am struggling to understand a part of it.

I have defined the hat functions as:

\[
\phi_i(x) =
\begin{cases}
\frac{x-x_{i-1}}{h} & \text{if } x_{i-1}\leq x<x_i \\
\frac{x_{i+1}-x}{h} & \text{if } x_i\leq x<x_i{i+1}\\
0 & otherwise
\end{cases}
\]

I know this means that for each $\phi_i(x_i)$ is 1 when i=j and 0 the rest of the time. The part I'm stuck on is the next step:

\[
u_h(x) = \sum_{j=1}^{n-1}c_j\phi_j(x)
\]

How do I sum these functions in their general form?

Regards
Carla

Hey Carla! ;)

I think it would help if you visualize it.

Suppose we pick $h=1$ and $c_1=2, c_2=3,c_3=1$.
Can you draw each $\phi_i$?
And then draw the combined $u_h$?

What does $u_h$ look like?
Can you come up with a general form for its formula?

 

[Carla1985]

Hey Carla! ;)

I think it would help if you visualize it.

Suppose we pick $h=1$ and $c_1=2, c_2=3,c_3=1$.
Can you draw each $\phi_i$?
And then draw the combined $u_h$?

What does $u_h$ look like?
Can you come up with a general form for its formula?

I know the c's scale each of the hats so at i=1 there would be a point of height 2, at i=2 a point at height 3 etc, combining these would give a piecewise polynomial of straight lines between each point, so a line from (1,2) to (2,3) etc. I'm not sure how I get a general form from that though

 

[I like Serena]

I know the c's scale each of the hats so at i=1 there would be a point of height 2, at i=2 a point at height 3 etc, combining these would give a piecewise polynomial of straight lines between each point, so a line from (1,2) to (2,3) etc. I'm not sure how I get a general form from that though

Indeed, if we pick $x$ between $x_i$ and $x_{i+1}$, we can find $u_h(x)$ as a linear interpolation between $c_i$ and $c_{i+1}$.
A linear interpolation is the same as a weighted average:
$$u_h(x) = \frac{(x-x_i)c_{i+1} + (x-x_{i+1})c_{i}}{x_{i+1}-x_{i}}\qquad\text{if }x_i \le x < x_{i+1}$$
Or:
$$u_h(x) = \frac{x-x_i}{h}c_{i+1} + \frac{x-x_{i+1}}{h}c_{i}\qquad\quad\text{if }x_i \le x < x_{i+1}$$

 

[Carla1985]

Indeed, if we pick $x$ between $x_i$ and $x_{i+1}$, we can find $u_h(x)$ as a linear interpolation between $c_i$ and $c_{i+1}$.
A linear interpolation is the same as a weighted average:
$$u_h(x) = \frac{(x-x_i)c_{i+1} + (x-x_{i+1})c_{i}}{x_{i+1}-x_{i}}\qquad\text{if }x_i \le x < x_{i+1}$$
Or:
$$u_h(x) = \frac{x-x_i}{h}c_{i+1} + \frac{x-x_{i+1}}{h}c_{i}\qquad\quad\text{if }x_i \le x < x_{i+1}$$

Ok, I think I understand that, though I'm still struggling to get my head around working with a function that has more than one equation. For example I have differentiations and multiplications eg $\phi^1_1\phi^1_2$ and $f\phi_1$ etc. How can I do those sort of calculations without one equation?

 

[I like Serena]

Ok, I think I understand that, though I'm still struggling to get my head around working with a function that has more than one equation. For example I have differentiations and multiplications eg $\phi^1_1\phi^1_2$ and $f\phi_1$ etc. How can I do those sort of calculations without one equation?

I'm not sure I understand what it is that you are struggling with.

It's a piece wise function.
The function is the same on each piece, just with a different $i$.
So we look at one piece at a time.
On one piece it is just one equation.

Anyway, what do you mean with those differentiations and multiplications?

Regards,
I like Serena

 

[Carla1985]

I'm not sure I understand what it is that you are struggling with.

It's a piece wise function.
The function is the same on each piece, just with a different $i$.
So we look at one piece at a time.
On one piece it is just one equation.

Anyway, what do you mean with those differentiations and multiplications?

Regards,
I like Serena

Sorry, I didn't explain very well. A further step is to calculate $\int^1_0\phi '_i\phi '_j\ dx$ but $\phi '_i$ can be $\frac{1}{h}$, $-\frac{1}{h}$ or $0$, same for $\phi '_j$.

I'm calculating these to build a matrix to solve for c's.

 

[I like Serena]

Sorry, I didn't explain very well. A further step is to calculate $\int^1_0\phi '_i\phi '_j\ dx$ but $\phi '_i$ can be $\frac{1}{h}$, $-\frac{1}{h}$ or $0$, same for $\phi '_j$.

I'm calculating these to build a matrix to solve for c's.

To calculate those, you will need to distinguish the cases:

1. $i=j$
2. $i$ and $j$ differ by $1$
3. $i$ and $j$ differ by more than $1$

What would the integral look like in each of those cases?

ILS x

 

[Carla1985]

To calculate those, you will need to distinguish the cases:

1. $i=j$
2. $i$ and $j$ differ by $1$
3. $i$ and $j$ differ by more than $1$

What would the integral look like in each of those cases?

ILS x

I'm not sure if its the right way of thinking about it but I think (taking h=1 for now) if i=j they are the squared terms so the differentials of 1 and then -1 are both squared hence both 1, so the graph would just be 1 between 0 and 1, then the area would be y=1. For i and j differ by 1, half of each hat would over lap, except one half would always be positive, the other negative so the area would be -1/2. Lastly if they differ by more than 1 it would be 0 as there would be no overlap. I think if h is different everything would be divided by h

Is that close?

 

[I like Serena]

I'm not sure if its the right way of thinking about it but I think (taking h=1 for now) if i=j they are the squared terms so the differentials of 1 and then -1 are both squared hence both 1, so the graph would just be 1 between 0 and 1, then the area would be 1. For i and j differ by 1, half of each hat would over lap, except one half would always be positive, the other negative so the area would be -1/2. Lastly if they differ by more than 1 it would be 0 as there would be no overlap. I think if h is different everything would be divided by h

Is that close?

Yep. Close.

Let's pick an arbitrary $h$.

Then we have:
$$\int_0^1 \phi_i'(x)\phi_j'(x) dx = \begin{cases}
\frac 1 h \cdot \frac 1 h \cdot h + -\frac 1 h \cdot -\frac 1 h \cdot h &\text{if }i=j \\
\frac 1 h \cdot -\frac 1 h \cdot h&\text{if }|i-j|=1 \\
0 &\text{otherwise}
\end{cases}$$

$$\int_0^1 \phi_i'(x)\phi_j'(x) dx = \begin{cases}
\frac 2 h &\qquad\qquad\qquad&\text{if }i=j \\
-\frac 1 h&&\text{if }|i-j|=1 \\
0 &&\text{otherwise}
\end{cases}$$

 

[Carla1985]

Oh actually everything would be x2 as only each half of the hat is in one interval, so if h=1 half the hat area would be 1 but so would the other half so it would be 2 altogether

- - - Updated - - -

Yep. Close.

Let's pick an arbitrary $h$.

Then we have:
$$\int_0^1 \phi_i'(x)\phi_j'(x) dx = \begin{cases}
\frac 1 h \cdot \frac 1 h \cdot h + -\frac 1 h \cdot -\frac 1 h \cdot h &\text{if }i=j \\
\frac 1 h \cdot -\frac 1 h \cdot h&\text{if }|i-j|=1 \\
0 &\text{otherwise}
\end{cases}$$

$$\int_0^1 \phi_i'(x)\phi_j'(x) dx = \begin{cases}
\frac 2 h &\qquad\qquad\qquad&\text{if }i=j \\
-\frac 1 h&&\text{if }|i-j|=1 \\
0 &&\text{otherwise}
\end{cases}$$

Thank you! Its all just clicked into place, having a break helped to think of it differently :)