Finite element method (proof question)

[sara_87]

Homework Statement

this is part of a theorem for the error estimate for the model problem for finite element method.
i have to prove the following inequality:
$$\left|u'(x)-\tilde{u}'_h(x)\right|\leq max_{0\leq(y)\leq1}\left|u''(y)\right|$$

Homework Equations

$$\tilde{u_h}$$ is the interpolant of u, where u is the solution of D (BVP that depends on x).
h is the mesh size.
(it will help a lot if you know about finite element method to understand this question)

The Attempt at a Solution

I don't know how to start but we were told that the mean value theorem is helpful:

u'($$\varphi$$)=$$\frac{u(x_{i+1})-u(x_i)}{x_{i+1}-x_{i}}$$
where x_i to x_(i+1) is one element.
but i don't know how this would help.
I think maybe i don't understand the inequality properly

 

[Mark44]

Homework Statement

this is part of a theorem for the error estimate for the model problem for finite element method.
i have to prove the following inequality:
$$\left|u'(x)-\tilde{u}'_h(x)\right|\leq max_{0\leq(y)\leq1}\left|u''(y)\right|$$

Homework Equations

$$\tilde{u_h}$$ is the interpolant of u, where u is the solution of D (BVP that depends on x).
h is the mesh size.
(it will help a lot if you know about finite element method to understand this question)

The Attempt at a Solution

I don't know how to start but we were told that the mean value theorem is helpful:

u'($$\varphi$$)=$$\frac{u(x_{i+1})-u(x_i)}{x_{i+1}-x_{i}}$$
where x_i to x_(i+1) is one element.
but i don't know how this would help.
I think maybe i don't understand the inequality properly

I took one class in Finite Element Analysis, many years ago (maybe before you were born!), so am not an expert.

Your hint will be useful here, I believe.
$$u' (\varphi)~=~\frac{u(x_{i+1})-u(x_i)}{x_{i+1}-x_{i}}$$

This is equivalent to
$$u(x_{i+1})-u(x_i)~=~u'(\varphi)(x_{i+1}-x_{i})$$

What you want is $$u'(x_{i+1})-u'(x_i)$$
You can figure out something to replace that difference, can't you?

 

[sara_87]

thanks,
do i replace it with:
$$u''(\varphi)(x_{i+1}-x_{i})$$
but don't we need to do something about the $$\tilde{u_{h}}'$$

 

[Mark44]

Can you tell me a bit more about $$\tilde{u_{h}}'$$
?
In your first post you described $$\tilde{u_{h}}$$, not $$\tilde{u_{h}}'$$.

Also, the phi that appears in the MVT is some number between x_i and x_{i + 1}, so you'll need to take that into account in the expression for u''.

 

[sara_87]

$$\tilde{u}'_h$$ is the derivative with respect to x of $$\tilde{u}_h$$

how shall i take into account the phi
(sorry if I am asking stupid questions, i just find this quite hard)

 

[Mark44]

No problem about asking questions - I don't find them stupid at all. As far as how to take into account the phi, right now, I don't know.

Maybe you can help me remember what it is you're working with in FEA. You have some function u(x) and a mesh at which you know the function values u(x_i). There isn't a way to calculate values of u(x) for values of x not in the mesh. If there were, you wouldn't need the machinery of FEA. Do you also know the derivative values at those points, u'(x_i)? I suspect that you don't. The formula for u(x) is unknown, so at points not in the mesh, you use some interpolating polynomial to get these points. Since you have a formula for the interpolating polynomial (I'm assuming it's a polynomial, but maybe it doesn't have to be one), you can get a formula for $$\tilde{u}'(x)$$. (I have omitted the h subscript for convenience.)

Is all of this correct?

The inequality you want to prove says that the distance between u'(x) and the derivative of the interpolating polynomial is bounded by the max. value of the |u''(y)|.

You haven't said anything about y or why the max is over values of y between 0 and 1. How does y figure into things?

 

[sara_87]

yes what you said is right.
y is just the value at the y-axis
the basis function for the construction of V_h (where V_h is the vector space)
is:
$$y=\phi_j(x_i)=1,$$ if i=j
and 0 otherwise

that's why y is between 0 and 1.

 

[Mark44]

So wouldn't y be either 0 or 1, not values between the two?

Any way, I'll look tonight to see if I can find the text I used, and see if that gives me any insights.

 

[sara_87]

it's like, a line on the x-axis then makes a mountain (triangle) shape where the peak (x_i) is at one, then goes back down to zero (at x_(i+1)). so between x_i and x_(i+1), there could be values of y between 0 and 1.

thank you.

 

[Mark44]

I'm still looking into this...

Another question that comes to mind is the significance of u''(y) on the right side of the inequality you need to prove. From what you've said, y is a value between 0 and 1. I don't understand why we would want to evaluate u'' on this interval.

The values for y are still not clear to me, either. In post 7 you said

$$y=\phi_j(x_i)=1,$$
and 0 otherwise

and in post 9 you said

there could be values of y between 0 and 1.

The first quote implies that y will be either 0 or 1, but nothing in between. The second quote contradicts this.

I looked for the text I used in the class I took, but didn't find anything. If there was a text, I guess I got rid of it long ago, but in fact, the class might have been a seminar and there wasn't a text. None of my other texts go into any detail about FEA, so I don't have any book resources.

The direction I plan to go is to sketch a graph of u(x) on the interval [x_{i -1}, x_{i + 1}]. That graph will be a slightly squiggly curve. I will also sketch a graph of the interpolating function $$\tilde{u}$$, connecting the three known points (x_i-1, u(x_i-1), (x_i, u(x_i), and (x_i+1, u(x_i+1) with straight lines.

I'll then look at u'(x) and $$\tilde{u}'$$ at those points, and see how I can use them to approximate u''(x).

So that's my plan. I will also notify the other homework helpers and mentors to see if they have any ideas.