Numerical Solutions

Author: John J Weber III, PhD Corresponding Textbook Sections:

Expected Educational Results

Initial Value Problems

Solutions to ODEs

There are three general methods for solving ODEs that we will discuss in this course:

  1. Graphical solutions [A sketch or graph representing ALL potential solutions to an ODE]

    • Phase lines: first-order autonomous ODEs (See CPT_03_Direction_Fields);

    • Direction fields: first-order ODEs (See CPT_03_Direction_Fields);

    • Phase planes: systems of first-order ODEs (To be discussed in CPT_17_Phase_Planes).

  2. Numerical Solutions [A numerical computation to estimate a y-value when given an initial point y(x0)=y0 on a single unknown solution curve of an ODE:

    • Euler's Method (To be discussed in a project)

    • Heun's Method (To be discussed in a project);

    • Fourth-order Runge Kutta [RK4] Method (To be discussed in a project).

  3. Analytical Solutions (the majority of the methods discussed in the course)

Initial Value Problems (IVPs)

Definition: Initial Value Problem (IVP)

An IVP for an nth-order DE is finding a solution to the DE on some interval (a,b) that satisfies n initial conditions.

IVP Example

An example of a first-order IVP is y=x(y+1), y(x0)=y0. From Calculus I, we know the first derivative describes the slopes of curves. So y=x(y+1) describes the slope of all curves y(x) everywhere on the xy-plane where certain conditions are met. The initial condition y(x0)=y0 is the point (x0,y0). When solving an IVP, we want to find the curve y(x) that passes through the point (x0,y0).

When solving IVPs, there are three possibilities:

  1. There is no solution to the IVP, i.e., there is no solution curve that passes through the point (x0,y0).

  2. There is a unique solution to the IVP, i.e., there is only one solution curve that passes through the point (x0,y0).

  3. The solution to the IVP is not unique, i.e., there is more than one solution curve that passes through the point (x0,y0).

Existence and Uniqueness Theorem

Given an IVP of the form dydx=f(x,y), y(x0)=y0. If f and fy are continuous on some rectangle R={(x,y)|a<x<b,c<y<d} that contains the point (x0,y0), then the IVP has a unique solution on some interval about x.

Using the Existence and Uniqueness Theorem

Procedure

  1. Determine if both of the conditions of the theorem are met (make sure that you provide a supporting explanation):

    • Determine if f(x,y) is continuous on some open rectangle containing the point (x0,y0).

    • Determine if fy is continuous on the same open rectangle containing the point (x0,y0).

  2. State your conclusion: Use an appropriate sentence similar to one of the following statements

    • Since f(x,y) and fy are both continuous on some open rectangle containing the point (x0,y0), then the IVP has a unique solution on some interval about x.

    • Since f(x,y) is not continuous on some open rectangle containing the point (x0,y0), then the IVP does not have a unique solution on some interval about x.

    • Since fy is not continuous on some open rectangle containing the point (x0,y0), then the IVP does not have a unique solution on some interval about x.

Investigation 01

For each of the following linear IVPs, determine, if there is a unique solution:

  1. xy=y+3x2, y(0)=0

  2. y=xy+1, y(0)=0

  3. xyy=x2cos(x), y(0)=3

  4. y=1+y2, y(π2)=0

  5. y=xy+1, y(1)=2

  6. y=x+y, y(1)=0

  7. y=y2x, y(0)=1

  8. x=xt+3t, x(1)=0

  9. tyy=t2cos(t), y(0)=3

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Last Modified: Wednesday, 2 September 2020 11:48 EDT