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I don't really see the difference, because in both case we need to determine y and the values of the constants. The only difference is that we give the value of y and y' in the former and the value of either 2 y or 2 y' in the latter.

I solve both problems the same way. I don't really understand the theory, I guess.

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9 Answers 9

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For a simple example (second order ODE), an initial value problem would say $y(a)=p$, $y'(a)=q$.

A boundary value problem would specify $y(a)=p$, $y(b)=q$.

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    $\begingroup$ I like this definition. Want to double check whether this strictly always true? Does a IVP always only have information at a single point? And is there ever boundary value problem that includes information at a derivative? $\endgroup$
    – Ben
    Nov 3, 2022 at 9:33
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An initial value problem is how to aim my gun. A boundary value problem is how to aim my gun so that the bullet hits the target.

Qualitatively the methods of solution are sometimes different, because Taylor series approximate a function at a single point, i.e. at 0.

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Initial Value Problems:

In initial value problems, we are given the value of function $y(x)$ and its derivative $y'(x)$ at the same point ( initial point ) sy at $x = 0$ i.e $y(0)= xi1$ and $y'(0)= x_2$.

Boundary Value Problems:

In boundary value problem, we are given the value of function $y(x)$ at two different points, i.e $y(a)= x_1$ and $y(b)= x_2$.

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Initial Value Problems:

Initial value problem does not require to specify the value at boundaries, instead it needs the value during initial condition. This usually apply for dynamic system that is changing over time as in Physics. An example, to solve a particle position under differential equation, we need the initial position and also initial velocity. Without these initial values, we cannot determine the final position from the equation.

Boundary Value Problems: In contrast, boundary value problems not necessarily used for dynamic system. Instead, it is very useful for a system that has space boundary. An example would be shape from shading problem in computer vision. To determine surface gradient from the PDE, one should impose boundary values on the region of interest.

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Initial value problem will be given initial conditions. But the boundary value problem contains boundary conditions like y(x1) and y(x2).

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Initial valye problems are those,which are related to the initial conditions of a question and no limit is used. While boundry vaue problems are those in which limit is used i.e an interval is given !

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In initial value problem values are given according to initial stages such as when there is initial stage means at zero time the Velocity and Acceleration have zero values similarly in initial value problems the points given according to zero value of that function and its derivative. Where in boundary value problem the end points are non zero element means at t>0 the conditions given will called the boundary value problems. For Example.

G(0)=0 , G'(0)=0 I,C's (Initial conditions)

G(2)=4 , G'(1)=3 B,C's (Boundary conditions)

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In initial boundary value problem, there are many values for independent variables to find the dependent variables variables and on other hand there are one value for independent variable (x) to find the dependent variables (y).

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In initial value problem we always want to determine the value of f(x)and f'(x) at initial point it may be 0 or something else but initial like f(1)=3 and f'(1)=2 then we can determine the constant.But in boundry value problem the condition will in form of a interval i.e.f(0)=3,f(2)=5

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    $\begingroup$ Welcome to MSE. What's the point of providing an answer to a question posted years ago and which has an accepted answer? $\endgroup$ Nov 25, 2017 at 14:36
  • $\begingroup$ @José Carlos Santos, Maybe to add extra details $\endgroup$
    – ESCM
    Mar 18, 2021 at 17:46

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