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Formulating Functional Equations

In technical mechanics, it is often necessary to create mathematical models of mechanical systems to analyze their movements and forces. An important skill in this regard is to describe the movements and forces in the form of functional equations. A functional equation is a mathematical equation that describes a function and thus represents the relationship between an independent variable and a dependent variable.

A common scenario occurs when a function required for problem solving is only given graphically (for example, a moment distribution, acceleration diagram, etc.). The solution to such a problem is only possible if a functional equation can be established from the given graph.

Setting up Linear Functions

Linear functions describe straight lines. Our task is to establish the equation of the line based on a given graph.

What does this equation generally look like? Linear functions have the following form:

$$ \begin{aligned} y = m \cdot x + b\qquad \end{aligned} $$

m: slope
b: y-intercept

\(x\) and \(y\) are the points on the desired line, which means that for each \(x\) we substitute into the equation, we obtain the corresponding \(y\) value on that line. Therefore, what we need to determine are the values of m (slope of the line) and b (y-intercept of the line).

In typical problems in Technical Mechanics, we encounter one of the following cases. They differ based on the given values:

A point on the function and the y-intercept are given:

Often, we encounter cases where a problem statement or a graphical representation of the linear function provides the y-intercept (point of intersection with the y-axis) and, for example, the zero point of the function.

Example of a problem statement: At time t=0, a torque of 1000 Nm is applied, which linearly decreases to 0 Nm within 10 seconds.

The following graph illustrates another example function with a given point and y-intercept:

The graph depicts a function with a given point P(3,0) and the y-intercept (-4,0).
Fig. 1: Example linear function with a given point and y-intercept

Procedure for setting up the desired linear function:

  1. Read the y-intercept \(b\) directly. In the above graph, the y-intercept of the function is located at \(b=-4\)

  2. Rearrange the equation of the linear function (above) for \(m\),

    $$ \begin{aligned} y = m \cdot x + b\qquad \Rightarrow \qquad m=\dfrac{y-b}{x} \end{aligned} $$

    Substitute the known point \((x|y) = (3|0)\) and \(b=-4\) into the equation and calculate \(m\):

    $$ m=\dfrac{y-b}{x} = \dfrac{0-(-4)}{3}=\dfrac{4}{3} $$

  3. Therefore, the equation of the desired linear function in the example graph in Fig. 1 is:

    $$ \underline{\underline{y=f(x)=\frac{4}{3}x-4}} $$

Case 2: Two points of the function are given:

Often, we also encounter cases where a problem statement or a graphical representation of the linear function provides two points that are not the y-intercept.

Example of a problem statement: At time t=2s, a torque of 1000 Nm is applied, which linearly decreases to 500 Nm within 10 seconds.

The following graph illustrates another example function with two given points:

The graph depicts a function with the given points P1(-7, 10) and P2(11, -7).
Fig. 2: Example linear function with two given points

Procedure for setting up the desired linear function:

  1. Read the points. It doesn't matter which point is denoted as \(P_1(x_1|y_1)\) and which point is denoted as \(P_2(x_2|y_2)\). Therefore, from the above example, we can read either:

    \(x_1 = -7\), \(y_1=10\), \(x_2=11\), \(y_2 = -7\),

    However, it would also be correct for this solution recipe to consider:

    \(x_1 =11\), \(y_1=-7\), \(x_2=-7\), \(y_2 = 10\).

  2. The formula for calculating the slope \(m\) is:

    $$ \begin{aligned} m = \dfrac{y_2-y_1}{x_2-x_1} \end{aligned} $$

    Substituted into the formula as shown in the above graph:

    $$ m=\dfrac{(-7)-10}{11-(-7)}=\dfrac{-17}{18}=-\dfrac{17}{18} $$

    If we interchange the points \(P_1\) and \(P_2\), we obtain the same result as described above:

    $$ m=\dfrac{10-(-7)}{(-7)-11}=\dfrac{17}{-18}=-\dfrac{17}{18} $$

  3. Rearrange the equation of the linear function (above) for \(b\),

    $$ \begin{aligned} y = m \cdot x + b\qquad \Rightarrow \qquad b=y-mx \end{aligned} $$

    substitute one of the known points \(P_1(x_1|y_1)\) or \(P_2(x_2|y_2)\), along with the calculated slope \(m\), into the equation to calculate \(b\):

    $$ b=y-mx = 10 - (-\dfrac{17}{18}) \cdot (-7) = \dfrac{180-119}{18} = \dfrac{61}{18} $$

  4. Therefore, the equation of the desired linear function for the example graph is:

    $$ \underline{\underline{y=f(x)=-\frac{17}{18}x + \frac{61}{18}}} $$

Setting up Quadratic Functions (Parabolas)

Quadratic functions describe parabolas. In Technical Mechanics, our task is to establish the equation of the parabola based on a given graph.

What does this equation generally look like? Quadratic functions have the following general form:

$$ \begin{aligned} y = ax^2+bx+c\qquad \end{aligned} $$

\(a\): opening parameter, \(a\neq0\)
\(a\),\(b\),\(c\): constant factors

\(a>0\): Upward-opening parabola
\(a<0\): Downward-opening parabola

\(x\) and \(y\) are the points on the desired parabola, which means that for each \(x\) we substitute into the equation, we obtain the corresponding \(y\) value on that parabola. Therefore, what we need to determine are the values of \(a\), \(b\) and \(c\).

To directly determine the function from the general form of the quadratic equation, we need two points \(P_1\) and \(P_2\) on this function, as well as the opening parameter \(a\). This allows us to set up 2 equations and determine \(b\) and \(c\). However, such a problem rarely occurs in Technical Mechanics.

In typical problems in Technical Mechanics, we are given the vertex \(S\) of the parabola and another point:

The graph depicts a parabolic function with the vertex at S(0, 36) and the point P(10, 0) also given.
Fig. 3: Example quadratic function with a given vertex and another given point

The vertex \(S\), depending on the opening parameter \(a\), is the minimum (for an upward-opening parabola) or maximum (for a downward-opening parabola) point of the parabola.

When given the vertex and an additional point of the parabola, it is convenient to use the vertex form of the quadratic function instead of the general form. In this form, the vertex can be directly substituted as follows:

$$ \begin{aligned} y = a(x-x_S)^2+y_S\qquad \end{aligned} $$

\(a\): opening parameter, \(a\neq0\)
\(x_S\), \(y_S\): coordinates of the vertex \(S\)

\(a>0\): Upward-opening parabola
\(a<0\): Downward-opening parabola

Recipe for setting up the desired quadratic function using the vertex form

  1. Substitute the vertex \(S(x_S|y_S)\) into the vertex form.

    Example with \(S(x_S|y_S)=(0|36)\) from the above graph (Fig. 3):

    $$ \begin{aligned} y=a(x-x_S)^2+y_S = a(x-0)^2+36=ax^2+36 \end{aligned} $$

  2. Determine the opening parameter \(a\). Rearrange the equation found under 1. for \(a\) and substitute the given point. Example with \(P(x|y)=(10|0)\) from the above graph (Fig. 3):

    $$ y = ax^2+36 \quad \Rightarrow \quad a=\dfrac{y-36}{x^2}=\dfrac{0-36}{10^2}= -\dfrac{36}{100}=-0,36 $$

  3. Therefore, the desired quadratic function is given by:

    $$ \underline{\underline{y=f(x)=-0,36x^2+36}} $$
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