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!