Finding physical boundary conditions
Physical boundary conditions (BCs) are needed to choose the physically appropriate solution or solutions from the range of mathematically correct ones. In the radioactive decay example earlier on, there was an infinite number of mathematically correct solutions (all with different coefficients $a_0$). In a particular physical context we would sensibly choose the one where the coefficient is the number of atoms of radioactive material present at the beginning.
Here is another example: What is the distance, $x$, an object falls under the influence of gravity during time, $t$? Careful - $x$ is the dependent variable here; it's $x(t)$!
| Physical notation | Generic notation | |||
|---|---|---|---|---|
| falling distance: | $x$ | - | dep. var.: | $y$ |
| time: | $t$ | - | indep. var.: | $x$ |
| acceleration due to gravity: | $g$ | - | coefficient: | $c$ |
| ODE to solve: | $\frac{{\rm d}^2x}{{\rm d}t^2}=g$ | - | $\frac{{\rm d}^2y}{{\rm d}x^2}=c$ | |
| Integrate: | $\frac{{\rm d}x}{{\rm d}t}=\int{g{\rm d}t}=gt+v_0$ | - | $\frac{{\rm d}y}{{\rm d}x}=\int{c{\rm d}x}=cx+a$ | |
| and integrate again: | $x(t)=\int{(gt+v_0){\rm d}t}$ | - | $y(x)=\int{(cx+a){\rm d}x}$ | |
| $=\frac{1}{2}gt^2+v_0t+x_0$ | - | $=\frac{1}{2}cx^2+ax+b$ | ||
Each of the two integration steps creates an integration constant ($v_0$ and $x_0$ or $a$ and $b$ in this example). The constant $v_0$ is added to $gt$ (units ms-2·s); it therefore has to be a velocity. The constant $x_0$ is added to $gt^2$ (units ms-2·s2); so it is a length. Since we are solving for the position as a function of time, $x(t)$, it makes sense to identify these constants as the velocity and position at the start, i.e. $x_0=x(t=0)$ and $v_0=v(t=0)$. Boundary conditions that refer to a point in time are sometimes called intitial conditions.
Each boundary condition is in fact an integration constant.
-> Each integration step causes the need for one boundary condition.
-> To solve an ODE of n-th order, we need to supply n boundary conditions.
For PDEs, it is sufficient to have as many BCs as the sum of the orders with respect to all independent variables. However, sometimes fewer are needed. This occurs when integrations with respect to different variables can be carried out in the same step so that the resulting BCs can be combined into one.
Next, we'll turn to PDEs and their use in physics.