Boundary Value Problems/Introduction to BVPs

Introduce Boundary value problems for a single independent variable.

• 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.

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

${\displaystyle y^{(n)}+p_{n-1}(x)y^{(n-1)}+...+p_{1}(x)y'(x)+p_{0}y(x)=f(x)}$

and ${\displaystyle n}$ additional equations specifying conditions on the solution and the derivatives at a point ${\displaystyle x_{0}\in I}$

${\displaystyle y^{(n-1)}(x_{0})=b_{n-1}}$, ..., ${\displaystyle y'(x_{0})=b_{1}}$, ${\displaystyle y(x_{0})=b_{0}}$

Example:

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

Solution:

${\displaystyle \int y'dx=\int xdx}$

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

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

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

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

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

Solution:

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

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

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

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

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

${\displaystyle {\begin{array}{c}r_{1}=4\\r_{2}=1\end{array}}}$

See the Wikipedia link for more on Initial Value Problems

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

${\displaystyle \scriptstyle {\frac {d^{2}x}{dt^{2}}}+p(t){\frac {dx}{dt}}+q(t)x(t)=f(t)}$ with ${\displaystyle \scriptstyle a_{0}x(a)+a_{1}x'(a)=g}$ and ${\displaystyle \scriptstyle b_{0}x(b)+b_{1}x'(b)=h}$ on the interval ${\displaystyle \scriptstyle I_{ab}=\{x|a\leq t\leq b\}}$

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

See the wikipedia topic

Boundary Value Problems