Uniqueness of solution to the wave function

[fluidistic]

Could you help me a little more?
Now I have to look at u_3=u_2-u_1. I know that u_3(0,x)=0 and that \partial _t u_3 (0,x)=0.
I have the fact that \int _{{S^2 (r)}_{r\to \infty}} (\partial _t u) \nabla u d \vec S =0 \Rightarrow E(t,x)=K \in \mathbb{R}.

 

[gabbagabbahey]

Well, what is E(t=0) for u_3?

 

[fluidistic]

Well, what is E(t=0) for u_3?

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 u_1 and u_2? I don't know how to justify it properly.

 

[gabbagabbahey]

E is indeed zero (although you may want to show me your proof of this, so that I can check your reasoning), so

\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

What kind of numbers (e.g. real, complex, imaginary, positive, negative etc.) are the quantities (\partial _t u_3)^2 and \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?

 

[fluidistic]

E is indeed zero (although you may want to show me your proof of this, so that I can check your reasoning)

\partial _t u_3 = \partial _t u_2 - \partial _t u_1. In t=0, all terms are worth 0.
Hence the first term of the integrand of \int _{\mathbb{R}^3} \frac{1}{2} \left [(\partial _t u_3)^2 +\mathbf{\nabla} u_3 \cdot \mathbf{\nabla} u_3 \right] dV is worth 0.
Now we've showed that the second term of the integrand is worth 2 (\nabla(\partial_t u_3) \cdot \nabla u_3). But we just showed that \partial _t u_3=0 when t=0, hence all this term is also worth 0. And so E(t=0,x)=0, \forall t and \forall x.

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 positive if real since we're dot producting a vector with itself. We have (a,b,c) \cdot (a,b,c)=a^2+b^2+c^2\geq 0.
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 u_3=0... Is that done?!

 

[gabbagabbahey]

Now we've showed that the second term of the integrand is worth 2 (\nabla(\partial_t u_3) \cdot \nabla u_3).

We did?!... Certainly, if u_3(0,x)=0, you can also say \mathbf{\nabla}u_3(0,x)=0...that's all you need in order to show the second term vanishes.

In order to say that the integral will be zero for all t, you also need to show that u_3 satisfies the wave equation, so that you can apply the proof of E being constant.

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.

I'd say it is is safe to assume that u_3 is real-valued (and hence so are its derivatives). However, (\partial_t u_3)^2 doesn't necessarily have to be positive, it could also be zero.

Thus the only way for the E to be 0 is that both terms are actually 0.

Right.

Oh... thus u_3=0

You can't directly conclude that from the fact that (\partial_t u_3)^2 and\mathbf{\nabla}u_3\cdot\mathbf{\nabla}u_3 are both zero. However, the fact that (\partial_t u_3)^2=0 does allow you to conclude that u_3 is constant in time. And you also know that u_3(t=0,x)=0 so yes, u_3=0 and hence u_1=u_2 and thus thereis only one unique solution to the given wave equation.

 

[fluidistic]

In order to say that the integral will be zero for all LaTeX Code: t , you also need to show that LaTeX Code: u_3 satisfies the wave equation, so that you can apply the proof of LaTeX Code: E being constant.

Ok, what about this argument: "u_1 and u_2 are solutions to the WE, hence any linear combination of them also is a solution to the WE; thus u_3 also satisfies it."?
Thanks for the rest.

 

[gabbagabbahey]

"u_1 and u_2 are solutions to the WE, hence any linear combination of them also is a solution to the WE; thus u_3 also satisfies it."

Exactly.

 

[fluidistic]

Exactly.

Ok. A big thank you for everything for all the help you provided me and all the time taken to try to teach me a lot of things. I'm going to copy the full solution on a sheet of paper once again and try to digest it completely.

I have a similar problem although harder where the energy is not given and I have to deduce it; so you might see me again posting in the same forum!