Introduction to BVPs

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[edit] Objective

Introduce Boundary value problems for a single independent variable.

[edit] Approach

What is a Boundary Value problem?

Solution of a Boundary Value Problem is directly related to solution of an Initial Value Problem. So let's review the material on IVPs first and then make the connection to BVPs.

Details of solving a two point BVP.

[edit] Initial Value Problems

For a single independent variable $x$ in an interval $I : a < x < b$ , an initial value problem consists of an ordinary differential equation including one or more derivatives of the dependent variable, $y$ ,

$y (n) + p n - 1 (x) y (n - 1) + ... + p 1 (x) y'(x) + p 0 y (x) = f (x)$

and $n$ additional equations specifying conditions on the solution and the derivatives at a point $x_0 \in I$

$\scriptstyle y^{(n-1)}(x_0) = b_{n-1}$ , ..., $\scriptstyle y'(x_0) = b_1$ , $\scriptstyle y(x_0) = b_0$

Example:

The differential equation is $\scriptstyle y' = x$ (First order differential equation.) and the initial condition at $\scriptstyle x=0$ is given as $\scriptstyle y(0)=1$ .

Solution:

$\scriptstyle \int y' dx = \int x dx$

$\scriptstyle y = \frac{x^2}{2} + C$ .

When , $\scriptstyle x=0$ $\scriptstyle 1 = C$ and $\scriptstyle y = \frac{x^2}{2} + 1$

Get out a piece of paper and try to solve the following IVP in a manner similar to the preceding example:

$\scriptstyle y' = xy$ and the initial condition at $\scriptstyle x=0$ is given as $\scriptstyle y(0)=3$ .

Once you have an answer (or are stuck) check your solution here. Click here for the solution: IVP-student-1

A second order ODE example:

The differential equation is $y'' + 5 y' + 4 y = 0$ (Second order differential equation.) and the two initial conditions at $x = 0$ given as $y (0) = 1, y'(0) = - 2$ .

Solution:

Assume the solution has the form $y = e r x$

$\scriptstyle y' = re^{rx}, y'' = r^2 e^{rx}$

$\scriptstyle y'' + 5y' + 4y = r^2 e^{rx} + 5 re^{rx} + 4e^{rx}$

$\scriptstyle 0 = r^2 e^{rx} + 5 re^{rx} + 4e^{rx}$

$\scriptstyle 0 = r^2 + 5 r + 4$ The characteristic polynomial. Solve for "r".

$\scriptstyle \begin{array}{c} r_1 = 4 \\ r_2 = 1 \end{array}$

See the Wikipedia link for more on Initial Value Problems

[edit] Two point BVPs for an ODE

Begin with second order DEs, $x'' = f (t, x, x')$ , with conditions on the solution at $t = a$ and $t = b$ .

$\scriptstyle \frac {d^2 x}{dt^2} + p(t) \frac {d x}{dt} + q(t) x(t) = f(t)$ with $\scriptstyle a_0x(a) +a_1x'(a)= g$ and $\scriptstyle b_0x(b) +b_1x'(b)= h$ on the interval $\scriptstyle I_{ab} = \{x | a \leq t \leq b \}$

[edit] Example

$\scriptstyle \frac {d^2 x}{dt^2} + 4 \frac {d x}{dt} + 2 x(t) = f(t)$ with $\scriptstyle x(0) = 0$ and $\scriptstyle x(1)=0$ on the interval $\scriptstyle I_{ab} = \{x | 0 \leq t \leq 1 \}$