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Using the method of separation of variables, how can I separate each X,Y,Z if the differential equation has a function of R(x,y,z)?

Example:

$ R_{xx} + R_{yy} + R_{zz} = 0 $

I understand how to apply the method if R is only a funtion of X and Y, but when it comes to three variables, I am completely lost.

user109334
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  • What do you mean by separation of variables? What is stopping you from trying a solution in the form $R(x,y,z) = X(x) \cdot Y(y) \cdot Z(z)$? Also, shouldn't it be $R_{xx}$? – Mark Fantini Nov 16 '13 at 22:03
  • Yes, it should be $R_{xx}$. But then when I apply that assumption, I end up with $X''YZ + XY''Z + XYZ'' = 0$. How can I separate these three variables if I only have two sides of the equation? – user109334 Nov 16 '13 at 22:04

1 Answers1

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The argument parallels the two variable case. Setting

$R(x, y, z) = X(x)Y(y)Z(z), \tag{1}$

we have

$X_{xx}(x)Y(y)Z(z) + X(x)Y_{yy}(y)Z(z) + X(x)Y(y)Z_{zz}(z) = 0, \tag{2}$

and dividing through by $X(x)Y(y)Z(z)$ we obtain

$X_{xx} / X + Y_{yy} / Y + Z_{zz} / Z = 0, \tag{3}$

which we write as

$X_{xx} / X = -Y_{yy} / Y - Z_{zz} / Z. \tag{4}$

Now we note that, since the two sides depend upon different independent variables, there must be a constant, call it $-k_x^2$, to which they are each equal, thus:

$X_{xx} / X = -k_x^2, \tag{5}$

or

$X_{xx} + k_x^2X = 0, \tag{5A}$

and

$Y_{yy} / Y + Z_{zz} / Z = k_x^2. \tag{6}$

Having separated out the $x$ dependence, we write (6) as

$Y_{yy} / Y = k_x^2 - Z_{zz} / Z, \tag{7}$

and once again observe that the two sides depend on different independent variables, so again each must equal some constant value, call it $-k_y^2$ this time:

$Y_{yy} / Y = -k_y^2 = k_x^2 - Z_{zz} / Z, \tag{8}$

which leads to

$Y_{yy} + k_y^2Y = 0 \tag{9}$

and

$Z_{zz} + k_z^2Z = 0, \tag{10}$

where we have set

$k_z^2 = -(k_x^2 + k_y^2). \tag{11}$

It should be noted that

$k_x^2 + k_y^2 + k_z^2 = 0, \tag{12}$

so that at least one of the three numbers $k_x, k_y, k_z$ must be complex. In the typical case occurring in practical applications, the $k_x, k_y, k_z$ are either real or pure imaginary, leading to solutions of (5A), (9), (10) which are respectively periodic or exponential, again analogous to the two-dimensional case.

Finally, it is worth noting that the techniques outlined above easily extend to the $n$-dimensional case of the equation

$\sum_1^n R_{x_jx_j} = 0; \tag{13}$

if we set

$R = \prod_1^nX_j(x_j), \tag{13A}$

we obtain $n$ equations of the form

$d^2X_j / dx_j^2 + k_j^2X_j = 0 \tag{14}$

with

$\sum_1^nk_j^2 = 0; \tag{15}$

the details are easy to execute and left to the reader. As is well-known, the solutions $X_j(x_j$) are of the form

$X_j(x_j) = a_+e^{ik_jx_j} + a_-e^{-ik_jx_j} \tag{16}$

for suitably chosen $a_\pm$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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