User:Peewack/Eqs

Partial differential equations
Given a partial differential equation of a function


 * $$ F(x_1,x_2,\dots,x_n) $$

of n variables, it is sometimes useful to guess solution of the form


 * $$ F = F_1(x_1) \cdot F_2(x_2) \cdots F_n(x_n) $$

or


 * $$ F = f_1(x_1) + f_2(x_2) + \cdots + f_n(x_n) $$

which turns the partial differential equation (PDE) into a set of ODEs. Usually, each independent variable creates a separation constant that cannot be determined only from the equation itself.

When such a technique works, it is called a separable partial differential equation.

Example (I)
Suppose F(x, y, z) and the following PDE:


 * $$ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z} = 0 \qquad\qquad (1)$$

We shall guess


 * $$ F(x,y,z) = X(x) + Y(y) + Z(z)\qquad\qquad (2) $$

thus making the equation (1) to


 * $$ \frac{dX}{dx} + \frac{dY}{dy} + \frac{dZ}{dz} = 0 $$

(since $$\frac{\partial F}{\partial x} = \frac{dX}{dx} $$).

Now, since X ' (x) is dependent only on x and Y ' (y) is dependent only on y (so on for Z ' (z)) and that the equation (1) is true for every x, y, z it is clear that each one of the term is constant. More precisely,


 * $$ \frac{dX}{dx} = c_1 \quad \frac{dY}{dy} = c_2 \quad \frac{dZ}{dz} = c_3\qquad\qquad (3) $$

where the constants c1, c2, c3 satisfy


 * $$ c_1 + c_2 + c_3 = 0\qquad\qquad (4) $$

Eq. (3) is actually a set of three ODEs. In this case they are trivial and can be solved by simple integration, giving:


 * $$ F(x,y,z) = c_1 x + c_2 y + c_3 z + c_4\qquad\qquad (5)$$

where the integration constant c4 is determined by initial conditions.

Example (II)
Consider the differential equation


 * $$\nabla^2 v + \lambda v = {\partial^2 v \over \partial x^2} + {\partial^2 v \over \partial y^2} + \lambda v = 0.$$

First we seek solutions of the form


 * $$ v = X(x)Y(y).\,$$

Most solutions are not of that form, but other solutions are sums of (generally infinitely many) solutions of that form.

Substituting,


 * $$ {\partial^2\over\partial x^2} [X(x)Y(y)]+{\partial^2\over\partial y^2}[X(x)Y(y)]+\lambda X(x)Y(y)= $$


 * $$ = X(x)Y(y)+X(x)Y(y)+\lambda X(x)Y(y)= 0\,$$

Divide throughout by X(x)


 * $$ = {X(x)Y(y) \over X(x)}+{X(x)Y(y)\over X(x)}+{\lambda X(x)Y(y)\over X(x)}$$


 * $$ ={X(x)Y(y) \over X(x)}+Y(y)+\lambda Y(y) = 0$$

and then by Y(y)


 * $$ ={X(x)\over X(x)}+{Y(y)+\lambda Y(y)\over Y(y)} = 0$$

Now X&prime;&prime;(x)/X(x) is a function of x only, and (Y&prime;&prime;(y)+&lambda;Y(y))/Y(y) is a function of y only, so for their sum to be equal to zero for all x and y, they must both be constant. Thus,


 * $$ {X(x)\over X(x)} = k = -{Y(y)+\lambda Y(y)\over Y(y)}$$

where k is the separation constant. This splits up into ordinary differential equations


 * $${X''(x)\over X(x)} = k$$


 * $$X''(x) - k X(x)=0\,$$

and


 * $${Y''(y)+\lambda Y(y)\over Y(y)} =-k$$


 * $$Y''(y)+(\lambda+k) Y(y) =0\,$$

which we can solve accordingly. If the equation as posed originally was a boundary value problem, one would use the given boundary values. See that article for an example which uses boundary values.