Numerical Solutions

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

• Section 1.2 – Solutions and Initial Value Problems

• Section 1.4 – The Approximation Method of Euler

Expected Educational Results

• Objective 4–1: I can estimate the solution to an IVP using Euler’s Method.

• Objective 4–2: I can use the Existence and Uniqueness Theorem to determine if a first-order IVP has a unique solution.

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$$y$-value when given an initial point $y(x_0)=y_0$$y(x_0)=y_0$ 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 $n$$n$th-order DE is finding a solution to the DE on some interval $(a,b)$$(a,b)$ that satisfies $n$$n$ initial conditions.

IVP Example

An example of a first-order IVP is ${\displaystyle y^{\prime }=x(y+1)}$$\displaystyle y^{\,\prime}=x(y+1)$, $y(x_0)=y_0$$y(x_0)=y_0$. From Calculus I, we know the first derivative describes the slopes of curves. So ${\displaystyle y^{\prime }=x(y+1)}$$\displaystyle y^{\,\prime}=x(y+1)$ describes the slope of all curves $y(x)$$y(x)$ everywhere on the $xy$$xy$-plane where certain conditions are met. The initial condition $y(x_0)=y_0$$y(x_0)=y_0$ is the point $(x_0,y_0)$$(x_0,y_0)$. When solving an IVP, we want to find the curve $y(x)$$y(x)$ that passes through the point $(x_0,y_0)$$(x_0,y_0)$.

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 $(x_0,y_0)$$(x_0,y_0)$.

2. There is a unique solution to the IVP, i.e., there is only one solution curve that passes through the point $(x_0,y_0)$$(x_0,y_0)$.

3. The solution to the IVP is not unique, i.e., there is more than one solution curve that passes through the point $(x_0,y_0)$$(x_0,y_0)$.

Existence and Uniqueness Theorem

Given an IVP of the form ${\displaystyle \frac{dy}{dx}=f(x,y)}$$\displaystyle\frac{dy}{dx}=f(x,y)$, $y(x_0)=y_0$$y(x_0)=y_0$. If $f$$f$ and ${\displaystyle \frac{\partial f}{\partial y}}$$\displaystyle\frac{\partial f}{\partial y}$ are continuous on some rectangle ${\displaystyle R=\{(x,y)|a<x<b,c<y<d\}}$$\displaystyle R=\left\{(x,y)|a<x<b,c<y<d\right\}$ that contains the point $(x_0,y_0)$$(x_0,y_0)$, then the IVP has a unique solution on some interval about $x$$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)$$f(x,y)$ is continuous on some open rectangle containing the point $(x_0,y_0)$$(x_0,y_0)$.

   • Determine if ${\displaystyle \frac{\partial f}{\partial y}}$$\displaystyle\frac{\partial f}{\partial y}$ is continuous on the same open rectangle containing the point $(x_0,y_0)$$(x_0,y_0)$.

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

   • Since $f(x,y)$$f(x,y)$ and ${\displaystyle \frac{\partial f}{\partial y}}$$\displaystyle\frac{\partial f}{\partial y}$ are both continuous on some open rectangle containing the point $(x_0,y_0)$$(x_0,y_0)$, then the IVP has a unique solution on some interval about $x$$x$.

   • Since $f(x,y)$$f(x,y)$ is not continuous on some open rectangle containing the point $(x_0,y_0)$$(x_0,y_0)$, then the IVP does not have a unique solution on some interval about $x$$x$.

   • Since ${\displaystyle \frac{\partial f}{\partial y}}$$\displaystyle\frac{\partial f}{\partial y}$ is not continuous on some open rectangle containing the point $(x_0,y_0)$$(x_0,y_0)$, then the IVP does not have a unique solution on some interval about $x$$x$.

Investigation 01

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

1. ${\displaystyle x{y}^{\prime }=y+3x^2}$$\displaystyle xy^{\,\prime}=y+3x^2$, $y(0)=0$$y(0)=0$

2. ${\displaystyle y^{\prime }=\frac{x}{y+1}}$$\displaystyle y^{\,\prime}=\frac{x}{y+1}$, $y(0)=0$$y(0)=0$

3. ${\displaystyle x{y}^{\prime }-y=x^2\cos (x)}$$\displaystyle xy^{\,\prime}-y=x^2\cos{(x)}$, $y(0)=-3$$y(0)=-3$

4. ${\displaystyle y^{\prime }=1+y^2}$$\displaystyle y^{\,\prime}=1+y^2$, ${\displaystyle y\left(\frac{\pi }{2}\right)=0}$$\displaystyle y\left(\frac{\pi}{2}\right)=0$

5. ${\displaystyle y^{\prime }=x-y+1}$$\displaystyle y^{\,\prime}=x-y+1$, $y(1)=2$$y(1)=2$

6. ${\displaystyle y^{\prime }=x+\sqrt{y}}$$\displaystyle y^{\,\prime}=x+\sqrt{y}$, $y(1)=0$$y(1)=0$

7. ${\displaystyle y^{\prime }=\frac{y}{2x}}$$\displaystyle y^{\,\prime}=\frac{y}{2x}$, $y(0)=1$$y(0)=1$

8. ${\displaystyle x^{\prime }=\frac{x}{t}+3t}$$\displaystyle x^{\,\prime}=\frac{x}{t}+3t$, $x(1)=0$$x(1)=0$

9. ${\displaystyle t{y}^{\prime }-y=t^2\cos (t)}$$\displaystyle ty^{\,\prime}-y=t^2\cos{(t)}$, $y(0)=-3$$y(0)=-3$

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