Uniqueness of solution to the wave function

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fluidistic
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Homework Statement


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



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.
 

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  • #2
gabbagabbahey
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I don't understand the part "the difference tends to 0 at infinity". What difference?

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

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.

First prove that the energy is conserved by calculating [itex]\partial_t E[/itex]. Then take a look at the quantity [itex]u_3\equiv u_2-u_1[/itex] :wink:
 
  • #3
fluidistic
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They mean that [tex]\lim_{x\to\infty}\left[u_2(t,x)-u_1(t,x)\right]=0[/tex]. In other words, the two solutions tend towards the same value as [itex]x\to\infty[/itex]



First prove that the energy is conserved by calculating [itex]\partial_t E[/itex]. Then take a look at the quantity [itex]u_3\equiv u_2-u_1[/itex] :wink:

Thanks so much for the clarification and the nice tip! It feels good to be unstuck for a moment!
I've reached [tex]\partial _t E(t)=\partial _t \int _{\mathbb{R}^3} \triangle u dV[/tex]. 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 [tex]\partial _t \int _{\mathbb{R}^3} (\partial _t u)^2 dV[/tex]. 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
gabbagabbahey
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Thanks so much for the clarification and the nice tip! It feels good to be unstuck for a moment!
I've reached [tex]\partial _t E(t)=\partial _t \int _{\mathbb{R}^3} \triangle u dV[/tex]. 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 [tex]\partial _t \int _{\mathbb{R}^3} (\partial _t u)^2 dV[/tex]. 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!

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

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.

[tex]\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[/tex]

Use the product rule to carry out the differentiation...what do you get?
 
  • #5
fluidistic
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Neither of those expressions look correct...how are you getting them?:confused:
My work: [tex]\partial _t E(t)=\frac{1}{2} \partial _t \int _{\mathbb{R}^3} (\partial _t u)^2+\nabla u \cdot \nabla u dV[/tex].
But according to the wave equation, [tex]\frac{\partial ^2 u}{\partial t^2}=\triangle u[/tex]. So replacing the first or second term yields my 2 results. What's wrong?
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.

[tex]\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[/tex]

Use the product rule to carry out the differentiation...what do you get?
Hmm, I get [tex]\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[/tex].
I used the relation [tex]\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 )[/tex].
 
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  • #6
gabbagabbahey
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My work: [tex]\partial _t E(t)=\frac{1}{2} \partial _t \int _{\mathbb{R}^3} (\partial _t u)^2+\nabla u \cdot \nabla u dV[/tex].
But according to the wave equation, [tex]\frac{\partial ^2 u}{\partial t^2}=\triangle u[/tex]. So replacing the first or second term yields my 2 results. What's wrong?

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

Hmm, I get [tex]\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[/tex].
I used the relation [tex]\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 )[/tex].

That's better. You can also use the product rule to show that [itex]\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
fluidistic
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Well, [itex](\partial_t u)^2\neq\partial_t^2 u[/itex] and [itex]\nabla u \cdot \nabla u\neq \triangle u[/itex], so I don't see how you are replacing either term.
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. :smile:


That's better. You can also use the product rule to show that [itex]\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.
In other words, a rule that involves [tex]\vec a \cdot \vec \nabla \cdot \phi[/tex] 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.
 
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  • #8
gabbagabbahey
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In other words, a rule that involves [tex]\vec a \cdot \vec \nabla \cdot \phi[/tex] for example?


Is this a typo? You are looking for a rule that involves [itex]\textbf{a}\cdot\mathbf{\nabla}\varphi[/itex]. The expression [itex]\mathbf{\nabla}\cdot\varphi[/itex] represents the divergence of [itex]\varphi[/itex] (not the gradient), and the divergence of a scalar is undefined/meaningless.


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.

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

[tex]\mathbf{\nabla}\cdot(\varphi\textbf{a})=\textbf{a}\cdot\mathbf{\nabla}\varphi+\varphi(\mathbf{\nabla}\cdot \textbf{a})[/tex]

Use that with [itex]\textbf{a}=\mathbf{\nabla}u[/itex] and [itex]\varphi=\partial_t u[/itex].
 
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  • #9
fluidistic
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Is this a typo? You are looking for a rule that involves [itex]\textbf{a}\cdot\mathbf{\nabla}\varphi[/itex]. The expression [itex]\mathbf{\nabla}\cdot\varphi[/itex] represents the divergence of [itex]\varphi[/itex] (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

[tex]\mathbf{\nabla}\cdot(\varphi\textbf{a})=\textbf{a}\cdot\mathbf{\nabla}\varphi+\varphi(\mathbf{\nabla}\cdot \textbf{a})[/tex]

Use that with [itex]\textbf{a}=\mathbf{\nabla}u[/itex] and [itex]\varphi=\partial_t u[/itex].
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 [tex]2 (\nabla(\partial_t u) \cdot \nabla u) =2 (\nabla u \nabla \partial _t u + \partial _t u \triangle u)[/tex].
I will try to work this out.
 
  • #10
gabbagabbahey
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I get that [tex]2 (\nabla(\partial_t u) \cdot \nabla u) =2 (\nabla u \nabla \partial _t u + \partial _t u \triangle u)[/tex].

That doesn't look quite right...try it again and be careful.
 
  • #11
fluidistic
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That doesn't look quite right...try it again and be careful.
I've retried the whole exercise.
Is the following true?: [tex]2 (\nabla(\partial_t u) \cdot \nabla u)=2 \left [ \vec \nabla \cdot (\underbrace {\partial _t u}_{scalar} \underbrace {\nabla u}_{vector}) \right ][/tex].
If so, I reach that this is equal to [tex]2 \left [ \partial _t u \vec \nabla \cdot \nabla u + \nabla (\partial _t u) \cdot \nabla u \right ][/tex]. 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.
 
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  • #12
gabbagabbahey
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I've retried the whole exercise.
Is the following true?: [tex]2 (\nabla(\partial_t u) \cdot \nabla u)=2 \left [ \vec \nabla \cdot (\underbrace {\partial _t u}_{scalar} \underbrace {\nabla u}_{vector}) \right ][/tex].

No, you are missing a term.

Oh wait, div (grad (any vector))=0, right?

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 [itex]\textbf{a}=\mathbf{\nabla}u[/itex] and [itex]\varphi=\partial_t u[/itex] into the vector identity posted earlier gives you

[tex]\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[/tex]

So, [itex]\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u[/itex]___?
 
  • #13
fluidistic
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No, you are missing a term.
Are you sure? To me it looks like the first term of your formula (except for the factor 2 of course)
[tex]\mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right)=\mathbf{\nabla}u\cdot\ma thbf{\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[/tex]
,

No, the curl of a gradient is zero. The divergence of a gradient is defined as the Laplacian.
Oh boy... how can I error so much? It was so obvious...
Let's go through this part together. Straight substitution of [itex]\textbf{a}=\mathbf{\nabla}u[/itex] and [itex]\varphi=\partial_t u[/itex] into the vector identity posted earlier gives you

[tex]\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[/tex]

So, [itex]\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u[/itex]___?

Actually I had started with the same expression and I reached all the terms you have but in different orders. My work was [tex]\vec \nabla \cdot (\partial _t u \nabla u)=\partial _t u \vec \nabla \cdot \nabla u + \nabla (\partial _t u) \cdot \nabla u[/tex]. I don't know if my work is equivalent to yours. Namely, if my work equals [tex]\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u+(\partial_t u)\triangle u[/tex]. I'd appreciate a feedback about the orders of the terms.
Continuing your work, the next step is [tex]\nabla (\partial _t u) \cdot \nabla u + (\partial _t u) \frac{\partial ^2 u}{\partial t^2}[/tex]. I guess I made an error, I just made a step.
 
  • #14
gabbagabbahey
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Are you sure? To me it looks like the first term of your formula (except for the factor 2 of course) ,

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.


My work was [tex]\vec \nabla \cdot (\partial _t u \nabla u)=\partial _t u \vec \nabla \cdot \nabla u + \nabla (\partial _t u) \cdot \nabla u[/tex]. I don't know if my work is equivalent to yours. Namely, if my work equals [tex]\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u+(\partial_t u)\triangle u[/tex]. I'd appreciate a feedback about the orders of the terms.

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

[tex]\mathbf{\nabla}\cdot(\mathbf{\nabla}u)\equiv \triangle u[/tex]

Continuing your work, the next step is [tex]\nabla (\partial _t u) \cdot \nabla u + (\partial _t u) \frac{\partial ^2 u}{\partial t^2}[/tex].

That wasn't quite what I meant. I meant isolate the [itex]\mathbf{\nabla} (\partial _t u) \cdot \mathbf{\nabla} u[/itex] term on one side of the equation, so you can substitute it into your equatiion for [itex]\partial_t E[/itex]
 
  • #15
fluidistic
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I'm all confused!
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.
I trust you although I'm not sure where exactly, but it doesn't matter.




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

[tex]\mathbf{\nabla}\cdot(\mathbf{\nabla}u)\equiv \triangle u[/tex]
Ok good, yeah I meant what I am supposed to have.


That wasn't quite what I meant. I meant isolate the [itex]\mathbf{\nabla} (\partial _t u) \cdot \mathbf{\nabla} u[/itex] term on one side of the equation, so you can substitute it into your equatiion for [itex]\partial_t E[/itex]
I don't see where I can substitute for [tex]\mathbf{\nabla} (\partial _t u) \cdot \mathbf{\nabla} u[/tex].
I have that [itex]\mathbf{\nabla} (\partial _t u) \cdot \mathbf{\nabla} u = \nabla \cdot \left [ (\partial _t u ) \nabla u \right ] - (\partial _t u )\triangle u[/itex].

While [tex]\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[/tex]...
 
  • #16
gabbagabbahey
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While [tex]\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[/tex]...

Take a look at post #6 (and #5)...that's not what you have.
 
  • #17
fluidistic
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Take a look at post #6 (and #5)...that's not what you have.
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 [tex]\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[/tex].

Because we've showed that [tex] \partial _t (\nabla u \cdot \nabla u)=2 \left [\mathbf{\nabla }(\partial_t u) \cdot \mathbf{ \nabla }u+(\partial_t u)\triangle u \right ][/tex].
 
  • #18
gabbagabbahey
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Because we've showed that [tex] \partial _t (\nabla u \cdot \nabla u)=2 \left [\mathbf{\nabla }(\partial_t u) \cdot \mathbf{ \nabla }u+(\partial_t u)\triangle u \right ][/tex].

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

In post #5, you showed

[tex]\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[/tex]

since [itex]\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 )[/itex]


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

That means,

[tex]\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[/tex]

Now you want to use the vector product rule to replace [itex]\nabla(\partial_t u) \cdot \nabla u[/itex]
 
  • #19
fluidistic
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But we didn't show that...you really need to keep better track of your calculations
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
fluidistic
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Ok back.

In post 6 you wrote [tex]\partial_t( (\nabla u \cdot \nabla u))=2 (\partial_t\nabla u \cdot \nabla u)=2 (\nabla(\partial_t u) \cdot \nabla u)[/tex]. Using the vector identity we saw in wikipedia, you found [tex]\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[/tex].

It seems I'm confused about if [tex]\nabla(\partial_t u) \cdot \nabla u=\mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right)[/tex]. 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 [tex]\int _{\mathbb{R}} \left [(\partial_t^2 u)(\partial_t u)+\nabla(\partial_t u) \cdot \nabla u \right] dV[/tex].
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.
 
  • #21
gabbagabbahey
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Using the vector identity we saw in wikipedia, you found [tex]\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[/tex].

Doesn't this tell you that [itex]\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u=\mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right)-(\partial_t u)\triangle u[/itex]?

It seems I'm confused about if [tex]\nabla(\partial_t u) \cdot \nabla u=\mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right)[/tex]. It seems that no according to your last post. Can you confirm that?

You can confirm that yourself now can't you? :wink:
 
  • #22
fluidistic
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Doesn't this tell you that [itex]\mathbf{\nabla}(\partial_t u)\cdot\mathbf{\nabla}u=\mathbf{\nabla}\cdot\left((\partial_t u)\mathbf{\nabla}u\right)-(\partial_t u)\triangle u[/itex]?



You can confirm that yourself now can't you? :wink:

I get it. First question: yes.
Second question: of course not.

I reach [tex]\partial _t E(t)=\int _{\mathbb{R}^3} (\partial ^2 _t u) (\partial _t u)+ \nabla \cdot ((\partial _t u)\nabla u)-(\partial _t u) \triangle u dV=\int _{\mathbb{R}^3} \nabla \cdot ((\partial _t u ) \nabla u ) dV[/tex]. I'm not confident since I don't see how this could equate 0. Did I mixed up things once again?!
 
  • #23
gabbagabbahey
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That's better!:smile:

Now, isn't there some theorem that tells you how to integrate the divergence of a vector field over any volume?:wink:
 
  • #24
fluidistic
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That's better!:smile:

Now, isn't there some theorem that tells you how to integrate the divergence of a vector field over any volume?:wink:

Gauss theorem or the divergence theorem.
So [tex]\int _{\mathbb{R}^3} \nabla \cdot ((\partial _t u) \nabla u) dV=\int _{S^2} (\partial _t u) \nabla u d \vec S[/tex] where [tex]S^2[/tex] means the surface of a sphere.
Is this right? I still don't see how it can equal 0.
For it to be 0, I believe [tex]\nabla u[/tex] should be orthogonal to [tex]\vec S=\hat n S[/tex] which points in the same direction than any radius of the sphere S. So [tex]\nabla u[/tex] should be tangent to any sphere whose center is the same as [tex]S^2[/tex].
I know I'm not explaining well myself, but it's the best I can.
However, I don't see it to be true. I've no idea what is the direction of [tex]\nabla u[/tex].
 
  • #25
gabbagabbahey
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Why do you think you have to integrate over the surface of a sphere? That's not really what the divergence theorem says, is it?
 
  • #26
fluidistic
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Why do you think you have to integrate over the surface of a sphere? That's not really what the divergence theorem says, is it?

Because I thought that [tex]\mathbb{R}^3[/tex] could be thought as an infinite radius sphere, S would have been its surface. I think I never dealt with an infinite region like [tex]\mathbb{R}^3[/tex]. So I don't know how to apply it in the case of an infinite volume.
 
  • #27
gabbagabbahey
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The divergence theorem tells you to integrate over the surface that bounds the volume. When the volume is all space, you must choose a surface that bounds all space.

One such choice is a spherical surface of radius [itex]r[/itex], in the limit that [itex]r\to\infty[/itex].

What do you know about the behavior of a wavefunction at infinity?
 
  • #28
fluidistic
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The divergence theorem tells you to integrate over the surface that bounds the volume. When the volume is all space, you must choose a surface that bounds all space.

One such choice is a spherical surface of radius [itex]r[/itex], in the limit that [itex]r\to\infty[/itex].

What do you know about the behavior of a wavefunction at infinity?

Ok I get the picture, I wasn't so so wrong.
I'm almost sure the wave function tends to 0 at infinity. Is it complicated to show this? What is the physical meaning of such an assertion? Does that mean that the energy flux density (Poynting vector) decreases with the distance of emission of the wave?
 
  • #29
gabbagabbahey
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The easiest assumption to make is that the wavefunction is normalizable. That means that [itex]\int_{\mathbb{R}^3}u^2 dV[/itex] must be finite, and hence [itex]u[/itex] must fall off at least as fast as [itex]1/r[/itex] at large distances from the source of the wave. (otherwise that integral will diverge)

That tells you that [itex]\mathbf{\nabla}u[/itex] must fall off at least as fast [itex]1/r^2[/itex] and so

[tex]\lim_{r\to\infty} r^2u\mathbf{\nabla}u=0[/itex]

which is pretty much the exact limit that occurs in your surface integral.
 
  • #30
fluidistic
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The easiest assumption to make is that the wavefunction is normalizable. That means that [itex]\int_{\mathbb{R}^3}u^2 dV[/itex] must be finite, and hence [itex]u[/itex] must fall off at least as fast as [itex]1/r[/itex] at large distances from the source of the wave. (otherwise that integral will diverge)

That tells you that [itex]\mathbf{\nabla}u[/itex] must fall off at least as fast [itex]1/r^2[/itex] and so

[tex]\lim_{r\to\infty} r^2u\mathbf{\nabla}u=0[/itex]

which is pretty much the exact limit that occurs in your surface integral.
Ok thanks a lot. I'll try to do the second part now. Just a little question, is it really needed that u falls off at least as fast as 1/r². Wouldn't the integral still converges for a fall off at a rate of say [tex]r^{\frac{3}{2}}[/tex]. Any r^x where x>1.

Edit: I misread your last post. Disregard this remark.
 
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  • #31
fluidistic
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Could you help me a little more?
Now I have to look at [tex]u_3=u_2-u_1[/tex]. I know that [tex]u_3(0,x)=0[/tex] and that [tex]\partial _t u_3 (0,x)=0[/tex].
I have the fact that [tex]\int _{{S^2 (r)}_{r\to \infty}} (\partial _t u) \nabla u d \vec S =0 \Rightarrow E(t,x)=K \in \mathbb{R}[/tex].
 
  • #32
gabbagabbahey
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Well, what is [itex]E(t=0)[/itex] for [itex]u_3[/itex]?
 
  • #33
fluidistic
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Well, what is [itex]E(t=0)[/itex] for [itex]u_3[/itex]?

If I'm not wrong: 0. Thus 0 for all t and all x. Does no energy mean no wave? So there's no difference between [tex]u_1[/tex] and [tex]u_2[/tex]? I don't know how to justify it properly.
 
  • #34
gabbagabbahey
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[itex]E[/itex] is indeed zero (although you may want to show me your proof of this, so that I can check your reasoning), so

[tex]\int _{\mathbb{R}^3} \frac{1}{2} \left [(\partial _t u_3)^2 +\mathbf{\nabla} u_3 \cdot \mathbf{\nabla} u_3 \right] dV=0[/tex]

What kind of numbers (e.g. real, complex, imaginary, positive, negative etc.) are the quantities [itex](\partial _t u_3)^2[/itex] and [itex]\mathbf{\nabla} u_3 \cdot \mathbf{\nabla} u_3[/itex]? When you add up (integrate) a bunch of those kinds of numbers, what is the only way that you can get zero as a result?
 
  • #35
fluidistic
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[itex]E[/itex] is indeed zero (although you may want to show me your proof of this, so that I can check your reasoning)
[tex]\partial _t u_3 = \partial _t u_2 - \partial _t u_1[/tex]. In t=0, all terms are worth 0.
Hence the first term of the integrand of [tex]\int _{\mathbb{R}^3} \frac{1}{2} \left [(\partial _t u_3)^2 +\mathbf{\nabla} u_3 \cdot \mathbf{\nabla} u_3 \right] dV[/tex] is worth 0.
Now we've showed that the second term of the integrand is worth [tex]2 (\nabla(\partial_t u_3) \cdot \nabla u_3)[/tex]. But we just showed that [tex]\partial _t u_3=0[/tex] when t=0, hence all this term is also worth 0. And so [tex]E(t=0,x)=0[/tex], [tex]\forall t[/tex] and [tex]\forall x[/tex].

What kind of numbers (e.g. real, complex, imaginary, positive, negative etc.) are the quantities LaTeX Code: (\\partial _t u_3)^2 and LaTeX Code: \\mathbf{\\nabla} u_3 \\cdot \\mathbf{\\nabla} u_3 ? When you add up (integrate) a bunch of those kinds of numbers, what is the only way that you can get zero as a result?
Real and positive I'd say. Real, not really sure why (I just seen in Born's book on Optics that some general time-harmonic fields are complex and that the real parts represent the fields), but energy has to be real... although I wouldn't be so surprised if it's a complex number whose real part represent the energy.
The first term is positive if it's a real number. The second term is also postive if real since we're dot producting a vector with itself. We have [tex](a,b,c) \cdot (a,b,c)=a^2+b^2+c^2\geq 0[/tex].
Thus the only way for the E to be 0 is that both terms are actually 0. (unless there are some complex numbers! But I'll wait your voice as why it's not possible to be in such a situation).
Oh... thus [tex]u_3=0[/tex]... Is that done????!!!!!!!!
 

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