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How will I get the solution in the form of integration $$ \phi (0,t)=\frac{R^{3}}{2}\frac{A}{\sqrt{\pi }}\int_{0}^{\infty }k^2e^{-R^{2}k^{2}/4}\cos (\sqrt{k^2+2} t)\ dk. $$ from the equation, when $d=3$ $$\phi_{tt} - \frac{1}{r^{d-1}}\frac{\partial}{\partial r} \left(r^{d-1} \frac{\partial \phi}{\partial r}\right) + 2m^2 \phi = 0.$$

Is it related to the profile?

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  • $\begingroup$ No offence, but it is my impression that you might be reading something more complex than you can handle right now. You really need to step up your knowledge on special functions, separation of variables, eigenfunctions and eigenvalues, Fourier and Laplace transforms, etc. if you want to fully understand the maths that are being used in the articles you are trying to read. $\endgroup$ – Pragabhava Jul 8 '13 at 4:17
  • $\begingroup$ Actually, I apologize!!! I understand how do you do maths but due to lack of practice,sometimes I feel difficulties to guess further step. And thanks for your suggestion.I'm also reviewing these topics and studying what necessary but it's true when I come here I find better solution than text book because here is an option to discuss. $\endgroup$ – Complex Guy Jul 8 '13 at 7:24
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First of all, taking the variables $$ t = \frac{t_0}{m}, \qquad r = \frac{r_0}{m}, $$ the PDE becomes

$$ \phi_{tt} - \frac{1}{r^{d-1}}\frac{\partial}{\partial r} \left(r^{d-1} \frac{\partial \phi}{\partial r}\right) + 2 \phi = 0 $$ where we have lost the sub-indexes in the notation.

Now, using separation of variables $\phi(t,r) = A(t) \Phi(r)$, we end up with $$ A''(t) \Phi(r) - \frac{A(t)}{r^{d-1}}\frac{d}{d r} \left(r^{d-1} \frac{d \Phi(r)}{d r}\right) + 2 A(t)\Phi(r) = 0. $$ Dividing by $A(t)\Phi(r)$ and rearranging, $$ \frac{A''(t)}{A(t)} + 2 = \frac{1}{r^{d-1}\Phi(r)}\frac{d}{d r} \left(r^{d-1} \frac{d \Phi(r)}{d r}\right). $$ The left hand size is a function of $t$, while the right hand is a function of $r$. The only way for the equality to hold is that both functions are equal to a constant (which I'll call -$k^2$), hence \begin{align} A''(t) + (2 + k^2)A(t) &= 0,\\ \\ \frac{1}{r^{d-1}}\frac{d}{d r} \left(r^{d-1} \frac{d \Phi(r)}{d r}\right) + k^2 \Phi(r) &=0. \end{align} The first equation is easily solved, yielding $$ A(t) = A_0 \cos \omega t + B_0 \sin \omega t $$ where $\omega = \sqrt{2 + k^2}$ is called the dispersion relation. For the radial equation, taking $$ \Phi(r) = r^\beta \chi(r), $$ we have $$ \chi''(r) + \frac{d - 1 + 2\beta}{r}\chi'(r) + \left(k^2 + \frac{\beta(\beta + d - 2)}{r^2}\right)\chi(r) = 0. $$ Let $\beta = 1 - \frac{d}{2}$, $$ \chi''(r) + \frac{1}{r} \chi'(r) + \left(k^2 - \frac{(1-\frac{d}{2})^2}{r^2}\right)\chi(r) = 0. $$ Taking $r = \frac{\rho}{k}$ and $\chi(r) = \hat{\chi}(\rho)$, $$ \hat{\chi}''(\rho) + \frac{1}{\rho} \hat{\chi}'(\rho) + \left(1 - \frac{(1-\frac{d}{2})^2}{\rho^2}\right)\hat{\chi}(\rho) = 0, $$ which is the Bessel differential equation of order $1-\frac{d}{2}$, and has as solution $$ \hat{\chi}(\rho) = c_1 J_{1-\frac{d}{2}}(\rho) + c_2 Y_{-1+\frac{d}{2}}(\rho), $$ where $J_\alpha$ and $Y_\alpha$ are the Bessel function of first and second kind, of order $\alpha$.

The spatial solution for every $k$ is $$ \Phi_k(r) = r^{1-\frac{d}{2}} \left(c_1 J_{1-\frac{d}{2}}(k r) + c_2 Y_{-1+\frac{d}{2}}(kr)\right). $$ In the case $d = 3$, the solutions can be written as $$ \Phi_k(r) = \hat{c}_1 j_0(k r) + \hat{c}_2 y_0(k r) = \hat{c}_1(k) \frac{\sin k r}{r} + \hat{c}_2(k) \frac{\cos k r}{r}, $$ where $j_0$ and $y_0$ are the Spherical Bessel Functions.

To determine the constants $\hat{c}_1(k)$ and $\hat{c}_2(k)$, first we note that $r = 0$ is part of the domain of the field, and $\Phi_k(r)$ must be well behaved at the origin. This means that $\hat{c}_2(k) = 0$. On the other hand, there is a perfectly valid solution for every $k >0$, then $$ \phi(t,r) = \int_0^\infty \hat{c}_1(k) \frac{\sin k r}{r} [A_0(k) \cos \omega t + B_0(k) \sin \omega t] d k. $$ To determine $b(k)$, $A_0(k)$ and $B_0(k)$, we use the conditions \begin{align} \phi(0,r) &= a_0 e^{-r^2/R^2}, \\ \phi_r(0,r) &= 0, \\ \phi_t(r,0) &=0, \end{align} and then $$ a_0 e^{-r^2/R^2} = \int_0^\infty \hat{c}_1(k) \frac{\sin(k r)}{r} dk, $$ where $\hat{c}_1(k)$ can be obtained using the Fourier Transform, yielding $$ \hat{c}_1(k) = \frac{2 a_0}{\pi} \Im\left\{\int_0^\infty r e^{-r^2/R^2 + ikr}dr\right\}. $$ This can be easily integrated, and then the solution is $$ \phi(t,r) = \frac{a_0 R^3}{2 \sqrt{\pi}} \int_0^\infty k e^{-R^2 k ^2/4} \frac{\sin k r}{r} \cos \omega t \,dk. $$

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  • $\begingroup$ You wrote d=3 but it seems for d=2, is it? $\Phi_k(r) = \hat{c}_1 j_0(k r) + \hat{c}_2 y_0(k r) = \hat{c}_1(k) \frac{\sin k r}{r} + \hat{c}_2(k) \frac{\cos k r}{r},$ $\endgroup$ – Complex Guy Jul 8 '13 at 20:37
  • $\begingroup$ @ComplexGuy Nope, the definition of Spherical Bessel functions is \begin{align}j_n(x) &= \sqrt{\frac{\pi}{2 x}} J_{n + \frac{1}{2}}(x) \\ \\ y_n(x) &= \sqrt{\frac{\pi}{2 x}} Y_{n + \frac{1}{2}}(x)\end{align} $\endgroup$ – Pragabhava Jul 9 '13 at 2:19
  • $\begingroup$ Ohh thanks yes :-) $\endgroup$ – Complex Guy Jul 9 '13 at 5:37
  • $\begingroup$ Hi, Can you please check the question which is part of this question.math.stackexchange.com/questions/479876/… $\endgroup$ – Complex Guy Aug 31 '13 at 17:54
  • $\begingroup$ Which boundary conditions have applied to get $a_0 e^{-r^2/R^2} = \int_0^\infty \hat{c}_1(k) \frac{\sin(k r)}{r} dk,$ equation? $\endgroup$ – Complex Guy Sep 10 '13 at 19:11

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