Boundary Value Problems/Introduction to BVPs

Objective

Introduce Boundary value problems for a single independent variable.

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.

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, $y$,

$y^{(n)}+p_{n-1}(x)y^{(n-1)}+\mathrm{...}+p_1(x)y^{\prime }(x)+p_{0}y(x)=f(x)$

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

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

Example:

The differential equation is $y^{\prime }=x$ (First order differential equation.) and the initial condition at $x=0$ is given as $y(0)=1$ .

Solution:

$\int y^{\prime }dx=\int xdx$

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

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

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

$y^{\prime }=xy$ and the initial condition at $x=0$ is given as $y(0)=3$ .

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 $y^{″}+5y^{\prime }+4y=0$ (Second order differential equation.) and the two initial conditions at $x=0$ given as $y(0)=1,y^{\prime }(0)=-2$ .

Solution:

Assume the solution has the form $y=e^{rx}$

$y^{\prime }=r{e}^{rx},y^{″}=r^2{e}^{rx}$

$y^{″}+5y^{\prime }+4y=r^2{e}^{rx}+5r{e}^{rx}+4e^{rx}$

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

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

$\begin{array}{c}{r}_1=4\\ r_2=1\end{array}$

See the Wikipedia link for more on Initial Value Problems

Two point BVPs for an ODE

Begin with second order DEs, $x^{″}=f(t,x,x^{\prime })$, with conditions on the solution at $t=a$ and $t=b$.

$\frac{d^{2}x}{d{t}^2}+p(t)\frac{dx}{dt}+q(t)x(t)=f(t)$ with $a_{0}x(a)+a_1{x}^{\prime }(a)=g$ and $b_{0}x(b)+b_1{x}^{\prime }(b)=h$ on the interval $I_{ab}=\{x|a\le t\le b\}$

Example

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

See the wikipedia topic

Boundary Value Problems