# Difference between an IVP and BVP

• May 15th 2011, 01:31 AM
bugatti79
Difference between an IVP and BVP
FOlks I have googled with this entry 'difference between ivp and bvp' but did not find any comprehensive information.
The question is a part of past exam paper.

Explain the difference between an IVP and BVP in the context of PDE's giving explicit examples of each.

Thanks
• May 15th 2011, 04:15 AM
kjchauhan
Difference between an Initial Value Problem(IVP) and a Boundary Value Problem(BVP):
- IVP: all the values needed to solve the particular problem are specified at a single point.
- BVP: all the values needed to solve the particular problem are specified at different points.
• May 15th 2011, 07:10 AM
Sudharaka
Quote:

Originally Posted by bugatti79
FOlks I have googled with this entry 'difference between ivp and bvp' but did not find any comprehensive information.
The question is a part of past exam paper.

Explain the difference between an IVP and BVP in the context of PDE's giving explicit examples of each.

Thanks

Dear bugatti79,

Well I think this will clarify your doubts, Boundary value problem - Wikipedia, the free encyclopedia.
• May 15th 2011, 08:26 AM
HallsofIvy
The terms "initial value" and "boundary value" come from physics application where, for example, determining the motion of an object under a given force where we are typically given the position and speed at t= 0 (of course, it not necessary that the point at which we are given all additional conditions be 0) while problems where we are given values at two or or more points are problems where we are looking for the values inside a boundary.

There is an important mathematical difference between these kinds of problems: if we have a second order differential equation with given initial values whether or not a unique solution exists depends only on the equation, not the initial conditions, while whether or not there exists a solution to the same equation, with given boundary conditions, depends upon the boundary conditions as well.

example: the intial value problem y''+ y= 0, y(0)= A, y'(0)= B has the unique solution y(t)= A cos(t)+ B sin(t). But the boundary value problem y''+ y= 0, y(0)= 0, y(pi)= 1 has NO solution while the bvp y''+ y= 0, y(0)= 0, y(pi)= 0 has an infinite number of solutions.

Think of it this way: you are firing a rifle down a range. An initial value problem would be where you fix the rifle at a given point and raise the barrel at a given angle. When you pull the trigger the bullet will follow a specific path. A boundary value problem would be where you fix the rifle at a given point and want to hit a target at a given distance down the range, by adjusting the barrel's angle. You might NOT be able to do that at all- the target might not be within the rifles range. Or there might not be a unique solution-if the target is within range, there can be two different angles that will give the same distance.
• May 15th 2011, 09:19 AM
bugatti79
Quote:

Originally Posted by HallsofIvy
The terms "initial value" and "boundary value" come from physics application where, for example, determining the motion of an object under a given force where we are typically given the position and speed at t= 0 (of course, it not necessary that the point at which we are given all additional conditions be 0) while problems where we are given values at two or or more points are problems where we are looking for the values inside a boundary.

There is an important mathematical difference between these kinds of problems: if we have a second order differential equation with given initial values whether or not a unique solution exists depends only on the equation, not the initial conditions, while whether or not there exists a solution to the same equation, with given boundary conditions, depends upon the boundary conditions as well.

example: the intial value problem y''+ y= 0, y(0)= A, y'(0)= B has the unique solution y(t)= A cos(t)+ B sin(t). But the boundary value problem y''+ y= 0, y(0)= 0, y(pi)= 1 has NO solution while the bvp y''+ y= 0, y(0)= 0, y(pi)= 0 has an infinite number of solutions.

Think of it this way: you are firing a rifle down a range. An initial value problem would be where you fix the rifle at a given point and raise the barrel at a given angle. When you pull the trigger the bullet will follow a specific path. A boundary value problem would be where you fix the rifle at a given point and want to hit a target at a given distance down the range, by adjusting the barrel's angle. You might NOT be able to do that at all- the target might not be within the rifles range. Or there might not be a unique solution-if the target is within range, there can be two different angles that will give the same distance.

Excellent! Thats pretty clear! Thanks to all.
• May 16th 2011, 01:16 AM
Ackbeet
Quote:

Originally Posted by HallsofIvy
The terms "initial value" and "boundary value" come from physics application where, for example, determining the motion of an object under a given force where we are typically given the position and speed at t= 0 (of course, it not necessary that the point at which we are given all additional conditions be 0) while problems where we are given values at two or or more points are problems where we are looking for the values inside a boundary.

There is an important mathematical difference between these kinds of problems: if we have a second order differential equation with given initial values whether or not a unique solution exists depends only on the equation, not the initial conditions, while whether or not there exists a solution to the same equation, with given boundary conditions, depends upon the boundary conditions as well.

example: the intial value problem y''+ y= 0, y(0)= A, y'(0)= B has the unique solution y(t)= A cos(t)+ B sin(t). But the boundary value problem y''+ y= 0, y(0)= 0, y(pi)= 1 has NO solution while the bvp y''+ y= 0, y(0)= 0, y(pi)= 0 has an infinite number of solutions.

Think of it this way: you are firing a rifle down a range. An initial value problem would be where you fix the rifle at a given point and raise the barrel at a given angle. When you pull the trigger the bullet will follow a specific path. A boundary value problem would be where you fix the rifle at a given point and want to hit a target at a given distance down the range, by adjusting the barrel's angle. You might NOT be able to do that at all- the target might not be within the rifles range. Or there might not be a unique solution-if the target is within range, there can be two different angles that will give the same distance.

Second bugatti79's comments. Very clear. Nice analogy. (Clapping)