Uniqueness of solution to the wave function

In summary: so in a similar way to how you did yours, you can just replace the first and second terms in the integral with the corresponding derivatives and you'll get the same result.
  • #36
fluidistic said:
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].

We did?!... Certainly, if [itex]u_3(0,x)=0[/itex], you can also say [itex]\mathbf{\nabla}u_3(0,x)=0[/itex]...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 [itex]t[/itex], you also need to show that [itex]u_3[/itex] satisfies the wave equation, so that you can apply the proof of [itex]E[/itex] 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 [itex]u_3[/itex] is real-valued (and hence so are its derivatives). However, [itex](\partial_t u_3)^2[/itex] 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 [tex]u_3=0[/tex]

You can't directly conclude that from the fact that [itex](\partial_t u_3)^2[/itex] and[itex]\mathbf{\nabla}u_3\cdot\mathbf{\nabla}u_3[/itex] are both zero. However, the fact that [itex](\partial_t u_3)^2=0[/itex] does allow you to conclude that [itex]u_3[/itex] is constant in time. And you also know that [itex]u_3(t=0,x)=0[/itex] so yes, [itex]u_3=0[/itex] and hence [itex]u_1=u_2[/itex] and thus thereis only one unique solution to the given wave equation.
 
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  • #37
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: "[tex]u_1[/tex] and [tex]u_2[/tex] are solutions to the WE, hence any linear combination of them also is a solution to the WE; thus [tex]u_3[/tex] also satisfies it."?
Thanks for the rest.
 
  • #38
fluidistic said:
"[tex]u_1[/tex] and [tex]u_2[/tex] are solutions to the WE, hence any linear combination of them also is a solution to the WE; thus [tex]u_3[/tex] also satisfies it."

Exactly.:smile:
 
  • #39
gabbagabbahey said:
Exactly.:smile:

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!
 

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