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The 3+1D wave equation for spherically symmetric waves is $$\frac{\partial^2 u}{\partial t^2} = c^2 \left( \frac{\partial^2 u}{\partial r^2} + \frac{2}{r} \frac{\partial u}{\partial r} \right) $$ where $u=u(t,r)$ and the initial conditions are $$u(0,r)=g(r)$$ $$u_t(0,r)=h(r) . $$ If we let $v=ru$, we get $$\frac{\partial^2 v}{\partial t^2} = c^2 \frac{\partial^2 v}{\partial r^2} $$ which is the one spatial dimensional wave equation, where the initial conditions are $$v(0,r)=rg(r)$$ $$v_t(0,r)=rh(r) . $$ References:
"The Mathematical Theory of Huygens' Principle", B.B. Baker and E. T. Copson, 1987, p9
"The Mathematical Theory of Wave Motion", G. R. Baldock and T.Bridgeman, 1981, p31
https://en.wikipedia.org/wiki/Wave_equation#Spherical_waves

PART B) If $\phi(t,x)$ is a solution to the 1+1D wave equation and if the initial conditions $\phi(0,x)$ and $\phi_t(0,x)$ are given, then D'Alembert's Formula for the 1+1D wave equation gives
$$\phi(t,x)= \frac 12[ \phi(0,x-ct)+ \phi(0,x+ct) ]+ \frac1{2c} \int_{x-ct}^{x+ct} \phi_t(0,y)dy . $$ Letting $g(x)=\phi(0,x)$ and $h(x)= \phi_t(0,x)$ (with $c=1$, so $ct=t$) this is commonly written $$\phi(t,x)= \frac 12[ g(x-t)+ g(x+t) ]+ \frac12 \int_{x-t}^{x+t} h(y)dy . $$ References:
http://mathworld.wolfram.com/dAlembertsSolution.html
http://en.wikipedia.org/wiki/D%27Alembert%27s_formula
http://www.jirka.org/diffyqs/htmlver/diffyqsse32.html
http://people.uncw.edu/hermanr/pde1/dAlembert/dAlembert.htm

THE QUESTIONS:

1) Then can we use D'Alembert's Formula for the 1+1D wave equation with initial conditions to get the solution to the spherically symmetric 3+1D wave equation with initial conditions?

2) If so, what is the physical (or intuitive) meaning of the integral term in D'Alemberts solution in the three spatial dimensional case?

UPDATE: The following references may indicate the answer to question 1) is yes.

math.ualberta.ca 337week0405.pdf after equation 180: "Thus the equation for $U$ can be solved using the formula we have discussed" {here $U$ is the 3D solution in $rU$ and the formula is D'Alemberts}.

Stanford Univ waveequation3.pdf page 4 Lemma 3 and page 5.

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    $\begingroup$ Related question by OP: physics.stackexchange.com/q/267899/2451 $\endgroup$ – Qmechanic Jul 18 '16 at 21:08
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    $\begingroup$ In order for a d'Alembert-like solution to work, you need (1) an initial delta function in position to propagate outward in a shell proportional to $\delta(r-vt)$ and (2) an initial delta function in velocity to make a 'shockwave' of displcement also propagating outward at $v$, as I wrote in my answer to the linked question. $\endgroup$ – knzhou Jul 18 '16 at 22:45
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    $\begingroup$ I think that means it should also work for the n-dimensional wave equation, up to various powers of $r$ appearing. But there may be a subtlety in even dimensions, following this question. I'm not sure exactly what breaks down. $\endgroup$ – knzhou Jul 18 '16 at 22:46
  • $\begingroup$ Related question: math.stackexchange.com/q/1570407 $\endgroup$ – user45664 Jul 30 '16 at 16:37

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