Editing Boundary value problem (section)

==Explanation==
Boundary value problems are similar to [[initial value problem]]s. A boundary value problem has conditions specified at the extremes ("boundaries") of the independent variable in the equation whereas an initial value problem has all of the conditions specified at the same value of the independent variable (and that value is at the lower boundary of the domain, thus the term "initial" value). A '''boundary value''' is a data value that corresponds to a minimum or maximum input, internal, or output value specified for a system or component.<ref>{{Cite book|title=ISO/IEC/IEEE International Standard - Systems and software engineering|publisher=ISO/IEC/IEEE 24765:2010(E)|pages=vol., no., pp.1-418}}</ref>

For example, if the independent variable is time over the domain [0,1], a boundary value problem would specify values for <math>y(t)</math> at both <math>t=0</math> and <math>t=1</math>, whereas an initial value problem would specify a value of <math>y(t)</math> and <math>y'(t)</math> at time <math>t=0</math>.

Finding the temperature at all points of an iron bar with one end kept at [[absolute zero]] and the other end at the freezing point of water would be a boundary value problem.

If the problem is dependent on both space and time, one could specify the value of the problem at a given point for all time or at a given time for all space.

Concretely, an example of a boundary value problem (in one spatial dimension) is 
:<math>y''(x)+y(x)=0 </math>
to be solved for the unknown function <math>y(x)</math> with the boundary conditions

:<math>y(0)=0, \ y(\pi/2)=2.</math>

Without the boundary conditions, the general solution to this equation is

:<math>y(x) = A \sin(x) + B \cos(x).</math>

From the boundary condition <math>y(0)=0</math> one obtains
:<math>0 = A \cdot 0 + B \cdot 1</math>
which implies that <math>B=0.</math> From the boundary condition <math>y(\pi/2)=2</math> one finds
:<math>2 = A \cdot 1 </math>
and so <math>A=2.</math> One sees that imposing boundary conditions allowed one to determine a unique solution, which in this case is 
:<math>y(x)=2\sin(x). </math>