I've been trying to find the solution to the 3-D wave equation for vector fields, that is:

$$\nabla^2 \vec{u}(\vec{x}, t) = \frac{1}{c^2} \frac{\partial^2 \vec{u}(\vec{x}, t)}{\partial t^2}$$

Given initial conditions $$\vec{u}(\vec{x},0)=\vec{\Phi}(\vec{x})$$ $$\left. \frac{\partial\vec{u}(\vec{x},t)}{\partial t}\right|_{t=0}=\vec{\Psi}(\vec{x}) $$

Whatever sources I found so far take me to Kirchoff's formula, which is the solution for the 3D wave equation for scalar fields. I've been trying to obtain the solution for vector fields from that, considering that each Cartesian component of a vector field is a scalar.

I've gotten so far [$B(\vec{x},ct)$ being the sphere with center at $\vec{x}$ and with a radius $ct$]:

[Corrected: $ct \vec{\Psi}(\vec{y})$ to $t \vec{\Psi}(\vec{y})$]

$$\vec{u}(\vec{x}, t) = \frac {1} {4 \pi c^2t^2} \oint_{\partial B(\vec{x}, ct)} [\vec{\Phi}(\vec{y}) + t \vec{\Psi}(\vec{y}) + \mathbb{X}]\space dS(\vec{y})$$ Where using Cartesian coordinates $i, j, k$, $$\mathbb{X} = (\nabla\Phi_i(\vec{y})\cdot(\vec{y}-\vec{x}))\hat{i} + (\nabla\Phi_j(\vec{y})\cdot(\vec{y}-\vec{x}))\hat{j} + (\nabla\Phi_k(\vec{y})\cdot(\vec{y}-\vec{x}))\hat{k}$$

My vector math is quite shaky -- so I'd like to ask:

  1. Is my derivation correct so far?
  2. Can $\mathbb{X}$ or $\oint \mathbb{X}$ be simplified further, and made independent of the coordinate system?

Any corrections or pointers to some reference material with the right answer will be appreciated.


Thanks to @WillieWong here is the expression I was trying to get to: $$\vec{u}(\vec{x}, t) = \frac {1} {4 \pi c^2t^2} \oint_{\partial B(\vec{x}, ct)} \left( \vec{\Phi}(\vec{y}) + t \vec{\Psi}(\vec{y}) + \left[ \left(\vec{y} - \vec{x}\right)\cdot\nabla\right]\vec{\Phi}(\vec{y}) \right)\space dS(\vec{y})$$

  • $\begingroup$ No, it is not right. What's up with the factor of $\sqrt{3}$? And why the vector $\vec{q}$? If you examine the last term inside the integrand, you've now made the gradient of a scalar function, and taken its dot product with a vector. This makes it a scalar, not a vector! $\endgroup$ Commented Feb 4, 2013 at 9:05
  • $\begingroup$ Oops! @Willie Wong you are right of course! What I am actually looking for is $$(\nabla \Phi_i(\vec{y})\cdot (\vec{y} - \vec{x}))\hat{i} + \nabla (\Phi_j(\vec{y})\cdot (\vec{y} - \vec{x}))\hat{j} + (\nabla \Phi_k(\vec{y})\cdot (\vec{y} - \vec{x})\hat{k} $$ $\endgroup$
    – Avijit
    Commented Feb 4, 2013 at 9:23

1 Answer 1


Applying the scalar Kirchoff's formula, you have that for $\vec{u} = (u_1, \ldots, u_d)$ that for each component $i$

$$ u_i(\vec{x},t) = \frac{1}{4\pi c^2 t^2} \oint_{\partial B(\vec{x},ct)} \left[ ct \Psi_i(\vec{y}) + \Phi_i(\vec{y}) + (\vec{y} - \vec{x}) \cdot \nabla \Phi_i(y) \right] \mathrm{d}S(\vec{y}) $$

So that putting everything in its place you have that

$$ \vec{u}(\vec{x},t) = \frac{1}{4\pi c^2 t^2} \oint_{\partial B(\vec{x},ct)} \left\{ ct \vec{\Psi}(\vec{y}) + \vec{\Phi}(\vec{y}) + \left[(\vec{y} - \vec{x}) \cdot \nabla\right] \vec{\Phi}(y) \right\} \mathrm{d}S(\vec{y}) $$


$$ \left[(\vec{y} - \vec{x}) \cdot \nabla\right] \vec{\Phi}(y) = \sum_{j = 1}^3(y_j - x_j)\partial_j \vec{\Phi}(y) $$

with the right hand side evaluated component-wise.


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