# Finding uniqueness of PDE via. energy method

1. Mar 7, 2009

### Weilin Meng

1. The problem statement, all variables and given/known data
consider a solution such that:
$$-\triangle u + b\triangledown u + cu = f$$
in domain Ω

and

$$\delta u/\delta n=g$$
in domain δΩ

where b is a constant vector and c is a constant scalar. Show that if c is large enough compared to |b|, there is uniqueness

2. Relevant equations
Energy Method to find uniqueness
Divergence theorem

3. The attempt at a solution

Assume two solutions u1 and u2. and v = u1 - u2
by linearity: $$-div(\triangle v) + b\triangledown v + cv = 0$$ in Ω
and $$\delta v/\delta n=\triangledown v\cdot n=0$$ in δΩ

Multiply the PDE by v and integrate over Ω:

$$\int_{\Omega }^{} -div(\triangle v)v + b\triangledown vv + cv^{2} = 0=\int_{\partial \Omega }^{}\triangledown v\cdot nvd\sigma +\int_{\Omega }^{}|\triangledown v|^{2}+b\triangledown vv+cv^{2}$$

(used divergence theorem to change the first integral on the right hand side.)

since $$\delta v/\delta n=\triangledown v\cdot n=0$$

The integral becomes:

$$\int_{\Omega }^{}|\triangledown v|^{2}+b\triangledown vv+cv^{2} = 0$$

if c is large enough, then it is the same as saying that b is very small. So we can consider b = 0 and therefore: $$b\triangledown vv = 0$$

so we get :

$$c\int_{\Omega}^{}v^{2} = -\int_{\Omega }^{}|\triangledown v|^{2}$$

which is only possible if v=0

therefore u1-u2 = 0
and u1=u2, and we have uniqueness

I'm not too sure about the last few steps. Do you think it's valid?

Last edited: Mar 7, 2009
2. Mar 7, 2009

### maze

Looks all good except for the missing steps in this part:

To rigorously go from the first equation to the second, you need that the integral over $\Omega$ of $v b \cdot \nabla v$ is not unboundedly larger than the integral of the function v2. (do you see why?) More specifically, we want

$$| \int_\Omega v b \cdot \nabla v | \le C \int_\Omega v^2$$

With some integration by parts you can get the left hand side of this to be equal to a constant times the integral of v2 over the boundary. So you need some way to relate the integral of v2 over the boundary to the integral of v2 over the inside - I would suggest using a trace theorem.

3. Mar 7, 2009

### Weilin Meng

I don't see why. So it's not ok to just subtract cv^2 to the other side?

4. Mar 7, 2009

### maze

Consider if you had solutions u1, u2, u3, ... corresponding to some subtracted versions v1, v2, v3, ...

What would happen if $|b \cdot \nabla v1| = 1|v1|$, $|b \cdot \nabla v2| = 2|v2|$, $|b \cdot \nabla v3| = 3|v3|$, ...

Remember, you would like to have some condition that says "the solution is unique if |b| < const*c". That const could be very large and you don't even have to know what it is, but it can't be infinite.

Last edited: Mar 7, 2009
5. Mar 7, 2009

### Weilin Meng

Hmm. I thought that with the question saying that "If c is large enough compared to b", that was saying that c was so much larger, that we can just ignore the b part of the equation, which is why I set b = 0.

But now I think I see that I am considering only one case, and I need a more general situation. That's where the inequality comes in?

6. Mar 7, 2009

### maze

Yeah, so c can be much much larger than b, but how much larger can't depend on which particular solution v you are considering.

Your idea is basically good though, and needs only a little modification. Instead of
$$c\int_{\Omega}^{}v^{2} = -\int_{\Omega }^{}|\triangledown v|^{2}$$

$$(c-\alpha)\int_{\Omega}^{}v^{2} \le -\int_{\Omega }^{}|\triangledown v|^{2}$$

where the constant α depends on b and Ω, but not v. Then you choose c so that (c-α) is positive.

Last edited: Mar 7, 2009
7. Mar 7, 2009

### Weilin Meng

ok, i'm a little confused of how to go from: $$\int_{\Omega }^{}|\triangledown v|^{2}+b\triangledown vv+cv^{2} = 0$$

to
$$(c-\alpha)\int_{\Omega}^{}v^{2} \le -\int_{\Omega }^{}|\triangledown v|^{2}$$

since we're not exactly eliminating b anymore...I'm trying to work backworks and convert the inequality to the original integral, but I can't get a to be independent of v

8. Mar 7, 2009

### maze

Ok, supposing that
$$| \int_\Omega v b \cdot \nabla v | \le \alpha \int_\Omega v^2$$

could you see how to get there?

9. Mar 7, 2009

### Weilin Meng

$$\int_{\Omega }^{}|\triangledown v|^{2}+b\triangledown vv+cv^{2} = 0$$

I tried to do integration by parts on $$b\triangledown vv$$
In hopes of trying to eliminate $$|\triangledown v|^{2}$$

But no luck....you mentioned the trace theorem. Is that necessary? I am not familiar with it.

10. Mar 7, 2009

### maze

All of those little tricks - integration by parts, trace theorem, etc - are needed to show that
$$| \int_\Omega v b \cdot \nabla v | \le \alpha \int_\Omega v^2$$

We will get to that.

But first, assuming that the above equation is true, can you get from
A: $$\int_{\Omega }^{}|\triangledown v|^{2}+v b \cdot \triangledown v+cv^{2} = 0$$

to

B: $$(c-\alpha)\int_{\Omega}^{}v^{2} \le -\int_{\Omega }^{}|\triangledown v|^{2}$$

Hint: if you are doing a lot of tricks, you're trying too hard! It's straightforward!!

11. Mar 7, 2009

### Weilin Meng

assuming that the above equation is true...I take equation A:
$$\int_{\Omega }^{}|\triangledown v|^{2}+v b \cdot \triangledown v+cv^{2} = 0$$

I then move things around and take the absolute value:
$$|\int_{\Omega }^{}b \cdot \triangledown vv| = \int_{\Omega }^{}|\triangledown v|^{2}+cv^{2}$$

I plug it into the above equation and I get equation B!

So from equation B, if we set (c-a) to be positive, the inequality can only hold with v = 0?

Ok if we have that...how do we get the above equation?

12. Mar 7, 2009

### maze

Great!

So now we want to show that

$$| \int_\Omega v b \cdot \nabla v | \le \alpha \int_\Omega v^2$$

Before we get started algebraically, give a few thoughts as to what this inequality means intuitively. For any solution v, the gradient of v can't be "too big" compared to the original v (when integrated). Just take a few seconds to think about this to build your intuition.

Ok, now lets get started. Look at the intequality. On the left side, there is a derivative (the gradient) of v, but on the right side there is only v's and no derivatives. To go from the left to the right, we have to get rid of the derivatives. How to do this? Integration by parts, of course!

Your mission, if you choose to accept it, is to take
$$\int_\Omega v b \cdot \nabla v$$

and convert it into something else that doesn't have any derivatives in it. (hint: express $b \cdot \nabla v = \sum_i b_i \frac{\partial}{\partial x_i} v_i$ and integrate each term by parts). What happens?

Once you get rid of the derivatives we can go to the next step.

Last edited: Mar 7, 2009
13. Mar 8, 2009

### Weilin Meng

I'm not sure if I did this right, but this is what I got so far and it looks like a mess and i did this assuming we're in R^3:
$$\int v(b_1\partial v_1/\partial x_1+b_2\partial v_2/\partial x_2+b_3\partial v_3/\partial x_3)=(vv_1b_1-\int v_1\partial v/\partial x)+(vv_2b_2-\int v_2\partial v/\partial x)+(vv_3b_3-\int v_3\partial v/\partial x)$$

I want to say that that equals to:
$$bv^{2}-b\int v\triangledown v$$

Last edited: Mar 8, 2009
14. Mar 8, 2009

### maze

Ahh so that is integration by parts for Ω 1D. In many dimensions, it goes: (n normal vector with components ni)

$$\int_\Omega v \frac{\partial}{\partial x_i} u = \int_{\partial \Omega} u v n_i - \int_\Omega u \frac{\partial}{\partial x_i} v$$

Give it another try with this one.

Last edited: Mar 8, 2009
15. Mar 8, 2009

### Weilin Meng

ok, so would the integration be this?

$$\int_{\partial \Omega }^{}v^{2}n_i-\int_{\Omega }^{}v\frac{\partial v}{\partial x_i}$$

16. Mar 8, 2009

### maze

yeah sorry i was editing in the meantime 1 sec. I need to check my work to be careful about things here. The end result is that you get v^2 integrated over the boundary.

17. Mar 8, 2009

### Weilin Meng

Ok well, im going to come back to this tomorrow...it's already 3 am lol...I have another similar question with non-linear equations...but that won't take nearly as long.

good night!

18. Mar 8, 2009

### maze

Yeah ok you should get

$$\int_\Omega v \nabla v \cdot b = \int_{\partial \Omega} v^2 b \cdot n - \int_\Omega v \nabla v \cdot b$$

But then look: the last term on the right is minus the term on the left, so you can bring it over to get

$$\int_\Omega v \nabla v \cdot b = \frac{1}{2}\int_{\partial \Omega} v^2 b \cdot n$$

so

$$| \int_\Omega v \nabla v \cdot b | \le \frac{1}{2}|b|\int_{\partial \Omega} v^2$$

You should actually compute this in coordinates, since this sort of thing occurs all the time and it is useful to be comfortable with the manipulations. Also compute it if for no other reason than to check and make sure what I've written here is right... you can see I've made mistakes in the calculation a couple times already ... (-:

Tomorrow give some thought about how this relates to the inequality we are trying to prove.

Last edited: Mar 8, 2009
19. Mar 8, 2009

### Weilin Meng

Question, where did the n disappear to?

Ok so right now we have the left side of the inequality...somehow we need to convert the right to domain omega, and get constant a which depends on b. Does b/2 = a?

20. Mar 8, 2009

### maze

The n got bounded out, remember |b.n| <= |b| since n has length 1

$$|\frac{1}{2}\int_{\partial \Omega} v^2 b \cdot n | \le \frac{1}{2}\int_{\partial \Omega} |v^2 b \cdot n| = \frac{1}{2}\int_{\partial \Omega} v^2 |b \cdot n|$$

b/2 = a is a good guess, but not quite. We will pick up another constant from going to the boundary to the whole domain.

Last edited: Mar 8, 2009