# Uniqueness of solution to the wave function

1. Mar 29, 2010

### fluidistic

1. The problem statement, all variables and given/known data
Demonstrate that if $$u_1$$ and $$u_2$$ are solutions of the wave equation $$\frac{\partial ^2 u}{\partial t^2} - \triangle u=0$$ such that $$u_1 (0,x)=u_2(0,x)$$, $$\partial _t u_1 (0,x)=\partial _t u_2(0,x)$$ and such that the difference "tends to 0 at infinity" sufficiently quickly, then $$u_1=u_2$$.
Hint: First prove that the following energy is conserved: $$E(t)=\int _{\mathbb{R}^3} \frac{1}{2} \left [ (\partial _t u)^2 +\nabla u \cdot \nabla u \right ] dV$$.

2. The attempt at a solution
Nothing concrete.
I don't understand the part "the difference tends to 0 at infinity". What difference?
Anyway, even assuming that the energy is conserved, I've absolutely no idea about what to do. I'm stuck on the this first exercise since a week. I don't ask for an answer, but rather any push/help.
I don't even know how to start proving that this energy is conserved.

2. Mar 29, 2010

### gabbagabbahey

They mean that $$\lim_{x\to\infty}\left[u_2(t,x)-u_1(t,x)\right]=0$$. In other words, the two solutions tend towards the same value as $x\to\infty$

First prove that the energy is conserved by calculating $\partial_t E$. Then take a look at the quantity $u_3\equiv u_2-u_1$

3. Mar 29, 2010

### fluidistic

Thanks so much for the clarification and the nice tip! It feels good to be unstuck for a moment!
I've reached $$\partial _t E(t)=\partial _t \int _{\mathbb{R}^3} \triangle u dV$$. I'd like to show it's worth 0, but I'm not sure how to. My intuition tells me to use the divergence theorem, but I don't think I can apply it.
Anyway, it's also worth $$\partial _t \int _{\mathbb{R}^3} (\partial _t u)^2 dV$$. So if either expression is worth 0 then I've showed that the energy is conserved.
If you feel like pushing me once more, don't hesitate!

4. Mar 29, 2010

### gabbagabbahey

Neither of those expressions look correct...how are you getting them?

Instead, just realize that the integral is over spacial coordinates, while the derivative is over time. Since the two are independent, it doesn't matter which order you carry out the integration and differentiation.

$$\partial_t E(t)=\partial_t \int _{\mathbb{R}^3} \frac{1}{2} \left [ (\partial _t u)^2 +\nabla u \cdot \nabla u \right ] dV =\int _{\mathbb{R}^3} \partial_t\left(\frac{1}{2} \left [ (\partial _t u)^2 +\nabla u \cdot \nabla u \right ] \right)dV$$

Use the product rule to carry out the differentiation...what do you get?

5. Mar 29, 2010

### fluidistic

My work: $$\partial _t E(t)=\frac{1}{2} \partial _t \int _{\mathbb{R}^3} (\partial _t u)^2+\nabla u \cdot \nabla u dV$$.
But according to the wave equation, $$\frac{\partial ^2 u}{\partial t^2}=\triangle u$$. So replacing the first or second term yields my 2 results. What's wrong?
Hmm, I get $$\int _{\mathbb{R}^3} \left ( \frac{\partial ^2 u}{\partial t^2} \right ) \left ( \frac{\partial u}{\partial t} \right )+\partial _t (\nabla u \cdot \nabla u)dV$$.
I used the relation $$\partial _t \left [ \left ( \frac{\partial u}{\partial t} \right ) \left ( \frac{\partial u}{\partial t} \right ) \right ]=2 \left ( \frac{\partial ^2 u}{\partial t^2} \right ) \left ( \frac{\partial u}{\partial t} \right )$$.

Last edited: Mar 29, 2010
6. Mar 29, 2010

### gabbagabbahey

Well, $(\partial_t u)^2\neq\partial_t^2 u$ and $\nabla u \cdot \nabla u\neq \triangle u$, so I don't see how you are replacing either term.

That's better. You can also use the product rule to show that $\partial_t( (\nabla u \cdot \nabla u))=2 (\partial_t\nabla u \cdot \nabla u)=2 (\nabla(\partial_t u) \cdot \nabla u)[/tex] There is a vector product rule that involves the dot product of a vector with the gradient of a scalar; you can use that to simplify things even further. 7. Mar 29, 2010 ### fluidistic I stand corrected, I see my errors. Thanks for point this out. I know I look like a beginner, but I think that's what I am. In other words, a rule that involves $$\vec a \cdot \vec \nabla \cdot \phi$$ for example? I'm currently looking at http://en.wikipedia.org/wiki/List_of_vector_identities and http://en.wikipedia.org/wiki/Vector_calculus_identities, but I still can't find anything I could recognize. Last edited by a moderator: Apr 24, 2017 8. Mar 29, 2010 ### gabbagabbahey Is this a typo? You are looking for a rule that involves [itex]\textbf{a}\cdot\mathbf{\nabla}\varphi$. The expression $\mathbf{\nabla}\cdot\varphi$ represents the divergence of $\varphi$ (not the gradient), and the divergence of a scalar is undefined/meaningless.

Near the end of the 2nd link, under the section entitled "Product of a scalar and a vector", is the rule

$$\mathbf{\nabla}\cdot(\varphi\textbf{a})=\textbf{a}\cdot\mathbf{\nabla}\varphi+\varphi(\mathbf{\nabla}\cdot \textbf{a})$$

Use that with $\textbf{a}=\mathbf{\nabla}u$ and $\varphi=\partial_t u$.

Last edited by a moderator: Apr 24, 2017
9. Mar 29, 2010

### fluidistic

I'm not sure why I wrote this, although I understand what makes sense and what not. It wasn't a typo but an error of mine of not taking care of what I write.
Thanks for your time for helping me and having check out the wikipedia's links. Ok now I see it.
I get that $$2 (\nabla(\partial_t u) \cdot \nabla u) =2 (\nabla u \nabla \partial _t u + \partial _t u \triangle u)$$.
I will try to work this out.

10. Mar 29, 2010

### gabbagabbahey

That doesn't look quite right...try it again and be careful.

11. Mar 30, 2010

### fluidistic

I've retried the whole exercise.
Is the following true?: $$2 (\nabla(\partial_t u) \cdot \nabla u)=2 \left [ \vec \nabla \cdot (\underbrace {\partial _t u}_{scalar} \underbrace {\nabla u}_{vector}) \right ]$$.
If so, I reach that this is equal to $$2 \left [ \partial _t u \vec \nabla \cdot \nabla u + \nabla (\partial _t u) \cdot \nabla u \right ]$$. This can even simplify further, right?

Edit: Oh wait, div (grad (any vector))=0, right? This indeed simplifies if all is right.
Oh no... this simplifies to the original equation. Now I don't see why I used the vectorial idendity.

Last edited: Mar 30, 2010
12. Mar 30, 2010

### gabbagabbahey

No, you are missing a term.

No, the curl of a gradient is zero. The divergence of a gradient is defined as the Laplacian.

Let's go through this part together. Straight substitution of $\textbf{a}=\mathbf{\nabla}u$ and $\varphi=\partial_t u$ into the vector identity posted earlier gives you

$$\mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right)=\mathbf{\nabla}u\cdot\mathbf{\nabla}(\partial_t u)+(\partial_t u)(\mathbf{\nabla}\cdot\mathbf{\nabla}u)=\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u+(\partial_t u)\triangle u$$

So, $\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u$___?

13. Mar 30, 2010

### fluidistic

Are you sure? To me it looks like the first term of your formula (except for the factor 2 of course)
,

Oh boy... how can I error so much? It was so obvious...
Actually I had started with the same expression and I reached all the terms you have but in different orders. My work was $$\vec \nabla \cdot (\partial _t u \nabla u)=\partial _t u \vec \nabla \cdot \nabla u + \nabla (\partial _t u) \cdot \nabla u$$. I don't know if my work is equivalent to yours. Namely, if my work equals $$\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u+(\partial_t u)\triangle u$$. I'd appreciate a feedback about the orders of the terms.
Continuing your work, the next step is $$\nabla (\partial _t u) \cdot \nabla u + (\partial _t u) \frac{\partial ^2 u}{\partial t^2}$$. I guess I made an error, I just made a step.

14. Mar 30, 2010

### gabbagabbahey

You had an equation with only two terms (one on each side). I had an equation with 3 terms (two on the RHS), the extra term is non-zero, so yes, I'm sure.

Apart from some missing brackets (I interpret $\partial _t u \nabla u$ as $\frac{\partial}{\partial t}(u \nabla u)$, when you are supposed to have $\frac{\partial u}{\partial t}(\nabla u)$ instead), these two expressions are equivalent.

$$\mathbf{\nabla}\cdot(\mathbf{\nabla}u)\equiv \triangle u$$

That wasn't quite what I meant. I meant isolate the $\mathbf{\nabla} (\partial _t u) \cdot \mathbf{\nabla} u$ term on one side of the equation, so you can substitute it into your equatiion for $\partial_t E$

15. Mar 30, 2010

### fluidistic

I'm all confused!
I trust you although I'm not sure where exactly, but it doesn't matter.

Ok good, yeah I meant what I am supposed to have.

I don't see where I can substitute for $$\mathbf{\nabla} (\partial _t u) \cdot \mathbf{\nabla} u$$.
I have that $\mathbf{\nabla} (\partial _t u) \cdot \mathbf{\nabla} u = \nabla \cdot \left [ (\partial _t u ) \nabla u \right ] - (\partial _t u )\triangle u$.

While $$\partial _t E(t)= \int _{\mathbb{R}} \frac{1}{2} \left [ 2 \left ( \frac{\partial ^2}{\partial t^2} \cdot \frac{\partial u}{\partial t} \right )+2 [ \mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right) ] \left ] dV$$...

16. Mar 30, 2010

### gabbagabbahey

Take a look at post #6 (and #5)...that's not what you have.

17. Mar 30, 2010

### fluidistic

You're right, thanks.
I forgot to type "u" in the first expression. What I wrote in my last post doesn't make sense. I meant $$\partial _t E(t)= \int _{\mathbb{R}} \frac{1}{2} \left [ 2 \left ( \frac{\partial u ^2}{\partial t^2} \cdot \frac{\partial u}{\partial t} \right )+2 [ \mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right) ] \left ] dV$$.

Because we've showed that $$\partial _t (\nabla u \cdot \nabla u)=2 \left [\mathbf{\nabla }(\partial_t u) \cdot \mathbf{ \nabla }u+(\partial_t u)\triangle u \right ]$$.

18. Mar 30, 2010

### gabbagabbahey

But we didn't show that...you really need to keep better track of your calculations

In post #5, you showed

$$\partial _t E(t)=\int _{\mathbb{R}^3} \partial_t\left(\frac{1}{2} \left [ (\partial _t u)^2 +\nabla u \cdot \nabla u \right ] \right)dV= \int _{\mathbb{R}} \frac{1}{2} \left [ 2 (\partial_t^2 u)(\partial_t u)+ \partial _t (\nabla u \cdot \nabla u)\right] dV$$

since $\partial _t \left [ \left ( \frac{\partial u}{\partial t} \right ) \left ( \frac{\partial u}{\partial t} \right ) \right ]=2 \left ( \frac{\partial ^2 u}{\partial t^2} \right ) \left ( \frac{\partial u}{\partial t} \right )$

In post #6, I showed that $\partial_t( \nabla u \cdot \nabla u)=2 \nabla(\partial_t u) \cdot \nabla u$, via the ordinary product rule.

That means,

$$\partial _t E(t)= \int _{\mathbb{R}} \frac{1}{2} \left [ 2 (\partial_t^2 u)(\partial_t u)+2 \nabla(\partial_t u) \cdot \nabla u \right] dV= \int _{\mathbb{R}} \left [(\partial_t^2 u)(\partial_t u)+\nabla(\partial_t u) \cdot \nabla u \right] dV$$

Now you want to use the vector product rule to replace $\nabla(\partial_t u) \cdot \nabla u$

19. Mar 30, 2010

### fluidistic

Yes, sorry. I need to sleep. I'll try to redo the exercise on my own and if I'm stuck I'll check here.
Thanks for all... I think I should be able to manage it alone tomorrow (not that I will sleep more than 5 hours though).
I'll try it tomorrow.

20. Mar 31, 2010

### fluidistic

Ok back.

In post 6 you wrote $$\partial_t( (\nabla u \cdot \nabla u))=2 (\partial_t\nabla u \cdot \nabla u)=2 (\nabla(\partial_t u) \cdot \nabla u)$$. Using the vector identity we saw in wikipedia, you found $$\mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right)=\mathbf{\nabla}u\cdot\mathbf{\nabla}(\partial_t u)+(\partial_t u)(\mathbf{\nabla}\cdot\mathbf{\nabla}u)=\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u+(\partial_t u)\triangle u$$.

It seems I'm confused about if $$\nabla(\partial_t u) \cdot \nabla u=\mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right)$$. It seems that no according to your last post. Can you confirm that?
If they are not equal then I can't use the vector product rule we've used in order to replace the second term of the integrand in $$\int _{\mathbb{R}} \left [(\partial_t^2 u)(\partial_t u)+\nabla(\partial_t u) \cdot \nabla u \right] dV$$.
Now my calculations on my draft are clearer. I have slept an hour during the day (a nap) so I feel much fresher than yesterday.