- Boundary value problem
In

mathematics , in the field ofdifferential equation s, a**boundary value problem**is adifferential equation together with a set of additional restraints, called the**boundary conditions**. A solution to a boundary value problem is a solution to the differential equation which also satisfies the boundary conditions.Boundary value problems arise in several branches of physics as any physical differential equation will have them. Problems involving the

wave equation , such as the determination ofnormal mode s, are often stated as boundary value problems. A large class of important boundary value problems are theSturm-Liouville problem s. The analysis of these problems involves theeigenfunction s of adifferential operator .To be useful in applications, a boundary value problem should be well posed. This means that given the input to the problem there exists a unique solution, which depends continuously on the input. Much theoretical work in the field of

partial differential equation s is devoted to proving that boundary value problems arising from scientific and engineering applications are in fact well-posed.Among the earliest boundary value problems to be studied is the

Dirichlet problem , of finding theharmonic function s (solutions toLaplace's equation ); the solution was given by theDirichlet principle .**Initial value problem**A more mathematical way to picture the difference between an initial value problem and a boundary value problem is that an initial value problem has all of the conditions specified at the same value of the independent variable in the equation (and that value is at the lower boundary of the domain, thus the term "initial" value). On the other hand, a boundary value problem has conditions specified at the extremes of the independent variable. For example, if the independent variable is time over the domain [0,1] , an initial value problem would specify a value of $y(t)$ and $y\text{'}(t)$ at time $t=0$, while a boundary value problem would specify values for $y(t)$ at both $t=0$ and $t=1$.

If the problem is dependent on both space and time, then instead of specifying the value of the problem at a given point for all time the data could be given at a given time for all space. For example, the temperature 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. Whereas in the middle of a still pond if somebody taps the water with a known force that would create a ripple and give us an initial condition.**Types of boundary value problems**If the boundary gives a value to the

normal derivative of the problem then it is aNeumann boundary condition . For example, if there is a heater at one end of an iron rod, then energy would be added at a constant rate but the actual temperature would not be known.If the boundary gives a value to the problem then it is a

Dirichlet boundary condition . For example, if one end of an iron rod is held at absolute zero, then the value of the problem would be known at that point in space.If the boundary has the form of a curve or surface that gives a value to the normal derivative and the problem itself then it is a

Cauchy boundary condition .Aside from the boundary condition, boundary value problems are also classified according to the type of differential operator involved. For an

elliptic operator , one discusseselliptic boundary value problem s. For anhyperbolic operator , one discusseshyperbolic boundary value problems . These categories are further subdivided into linear and various nonlinear types.**Example**Consider the ordinary differential equation :$y"(x)+y(x)=0\; ,$to be solved for the unknown function $y(x)$. Impose the boundary conditions

:$y(0)=0,\; y(pi/2)=2.$

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

:$y(x)\; =\; A\; sin(x)\; +\; B\; cos(x).,$

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

**ee also****Related mathematics:**

*initial value problem

*differential equation s

*Green's function s

*Stochastic processes and boundary value problems

*Examples of boundary value problems

*Sturm-Liouville theory

*Dirichlet boundary condition

*Neumann boundary condition

*Sommerfeld radiation condition

*Cauchy boundary condition **Physical applications:**

*wave s

*normal modes

*electrostatics

*Laplace equation

*potential theory **Numerical algorithms:**

*shooting method **References*** A. D. Polyanin and V. F. Zaitsev, "Handbook of Exact Solutions for Ordinary Differential Equations (2nd edition)", Chapman & Hall/CRC Press, Boca Raton, 2003. ISBN 1-58488-297-2.

* A. D. Polyanin, "Handbook of Linear Partial Differential Equations for Engineers and Scientists", Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9.**External links*** [

*http://eqworld.ipmnet.ru/en/solutions/lpde.htm Linear Partial Differential Equations: Exact Solutions and Boundary Value Problems*] at EqWorld: The World of Mathematical Equations.

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