
3.1: Introduction to Boundary and Initial Conditions


As you all know, solutions to ordinary differential equations are usually not unique (integration constants appear in many places). This is of course equally a problem for PDE’s. PDE’s are usually specified through a set of boundary or initial conditions. A boundary condition expresses the behavior of a function on the boundary (border) of its area of definition. An initial condition is like a boundary condition, but then for the time-direction. Not all boundary conditions allow for solutions, but usually the physics suggests what makes sense. Let me remind you of the situation for ordinary differential equations, one you should all be familiar with, a particle under the influence of a constant force,

\begin{align} {\frac{\partial^2 x}{\partial t^2}} &= a. & \text{Which leads to} \\ {\frac{\partial x}{\partial t}} &= at + v_0, & \text{and}\\ x &= {\frac{1}{2}} a {t}^{2} + {v}_{0} t + {x}_{0}. & l. 21\\ \nonumber \end{align}

This contains two integration constants. Standard practice would be to specify $$\frac{\partial x}{\partial t}(t=0) = v_0$$ and $$x(t=0)=x_0$$. These are linear initial conditions (linear since they only involve $$x$$ and its derivatives linearly), which have at most a first derivative in them. This one order difference between boundary condition and equation persists to PDE’s. It is kind of obviously that since the equation already involves that derivative, we can not specify the same derivative in a different equation.

The important difference between the arbitrariness of integration constants in PDE’s and ODE’s is that whereas solutions of ODE’s these are really constants, solutions of PDE’s contain arbitrary functions.

Let me give an example. Take

$u = y f(x)$ then $\frac{\partial u}{\partial y} = f(x).$

This can be used to eliminate $$f$$ from the first of the equations, giving $u = y \frac{\partial u}{\partial y}$ which has the general solution $$u=yf(x)$$.

One can construct more complicated examples. Consider $u(x,y) = f(x+y) + g(x-y)$ which gives on double differentiation $\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2}= 0.$

The problem is that without additional conditions the arbitrariness in the solutions makes it almost useless (if possible) to write down the general solution. We need additional conditions, that reduce this freedom. In most physical problems these are boundary conditions, that describes how the system behaves on its boundaries (for all times) and initial conditions, that specify the state of the system for an initial time $$t=0$$. In the ODE problem discussed before we have two initial conditions (velocity and position at time $$t=0$$).

3.1: Introduction to Boundary and Initial Conditions is shared under a CC BY-NC-SA 2.0 license and was authored, remixed, and/or curated by Niels Walet via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.