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An initial value problem is a differential equations problem in which you are given the the value of the function and sufficient of its derivatives at ONE VALUE OF X. Now, if you have a second order equation, you are given the value of the function and its first derivative at some value of x. (That value of x being the…
Jerrold C William
updated on 05 Jun 2019
An initial value problem is a differential equations problem in which you are given the the value of the function and sufficient of its derivatives at ONE VALUE OF X.
Now, if you have a second order equation, you are given the value of the function and its first derivative at some value of x. (That value of x being the "initial" time- often it is x= 0 but doesn't have to be.)
A boundary value problem is a differential equations problem in which you are given the value of the function at several different values of x.
Typically, if you have a second order equation, you are given the value of the function at the endpoints of an interval. (Those endpoints being the "boundary".)
There is a critical theoretical difference between the two: if f(x,y) is well behaved (continuous in both x and y) then the initial value problem dy/dx= f(x,y) with y(x0)= y0 has a unique solution.
There is no such theorem for boundary value problems. For example, the differential equation d2y/dx2= -y With initial conditions y(0)= A, y'(0)= B has the unique solution y(x)= A cos(x)+ B sin(x) no matter what A and B are.
The differential equation d2y/dx2= -y with the boundary value conditions y(0)= 0, y( π π)= 0 has an infinite number of solutions (y= A sin(x) for any value of A). The differential equation d2y/dx2= -y with the boundary value conditions y(0)= 0, y( π π)= 1 has NO solution.
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