# One-Dimensional Mathematical Model for Kayak Propulsion
## 3 Analytical Solutions for Periodic Propulsion
### Square Sine Stroke  
One way to consider the periodic propulsion is modeling them with a trigonometric function. [PERSON] et al. [8] did this with a _sine_ function but they only analyzed one stroke, which allowed them to avoid the negative value of this function in the second half of its period, which would represent the second stroke. This disadvantage forces us to use its absolute value and complicates the mathematical development considerably. Additionally, its time average is not favorable to our interests. Thus, in our case, we resort to a square sine function, \(f=f_{0}sin^{2}(wt)\), where \(f_{0}\) is the force peak applied to the paddle, and \(w=2\pi\cdot v\) the angular frequency which accounts for the stroke frequency \(v\). Our equation has the following form:  
\[m\dot{v}+dv^{2}=f_{0}sin^{2}(wt)=\frac{f_{0}}{2}(1-cos(2 wt)). \tag{17}\]  
This identity divides the force in two terms, on which there is a constant force of magnitude equal to half the peak force \(\frac{f_{0}}{2}\) and another periodical force with twice the frequency of the original force. Let \(f^{\prime}=f_{0}/2\). We group the terms to obtain the following:  
\[m\dot{v}=f^{\prime}\left(1-\frac{d}{f^{\prime}}v^{2}\right)-f^{\prime}\cdot cos (2 wt). \tag{18}\]  
We make the following change in variable:  
\[u=\sqrt{\frac{d}{f^{\prime}}}\cdot v\qquad\frac{du}{dt}=\sqrt{\frac{d}{f^{ \prime}}}\frac{dv}{dt} \tag{19}\]  
and that leaves us with the following:  
\[m\sqrt{\frac{f^{\prime}}{d}}\dot{u}=f^{\prime}(1-u^{2})-f^{\prime}\cdot cos(2 wt). \tag{20}\]  
Let us now assume that the solution separates into the sum of two functions as follows:  
\[u=g_{1}+g_{2}\qquad\quad u^{2}=g_{1}^{2}+g_{2}^{2}+2g_{1}g_{2}\qquad\quad\dot{ u}=\dot{g_{1}}+\dot{g_{2}}. \tag{21}\]  
Then, we have the following:  
\[m\sqrt{\frac{f^{\prime}}{d}}(g_{1}+g_{2})=f^{\prime}(1-g_{1}^{2}-g_{2}^{2}-2g _{1}g_{2})-f^{\prime}\cdot cos(2 wt). \tag{22}\]  
Confronting the first term of the left member with the first two terms of the right member, we observe the following:  
\[m\sqrt{\frac{f^{\prime}}{d}}\dot{g_{1}}=f^{\prime}(1-g_{1}^{2}). \tag{23}\]  
That is an analogous equation to the solved one in the case of a constant force. Now, we have \(f^{\prime}=f_{0}/2\) and the solution for this part is the following:\[g_{1}=tanh\Bigg{(}\sqrt{\frac{d}{f^{\prime}}}\left(\frac{f^{\prime}}{m}t+K^{\prime} \right)\Bigg{)}. \tag{24}\]  
Eliminating these three terms from the equation, we are left with the following:  
\[m\sqrt{\frac{f^{\prime}}{d}}\delta_{2}=-f^{\prime}(g_{2}^{2}+2g_{1}g_{2})-f^{ \prime}\cdot cos(2 wt). \tag{25}\]  
Here, we assume that the cross term \(2g_{1}\cdot g_{2}\) can be neglected. The resulting equation can be addressed by introducing the Fourier series from this point onward:  
\[f(t)=a_{0}/2+\sum_{n}a_{n}sin(w_{n}\cdot t)+\sum_{n}b_{n}cos(w_{n}\cdot t). \tag{26}\]  
Restricted to the sine Fourier series, in this particular case we have the following:  
\[m\dot{v}=f^{\prime}\left(1-\frac{d}{f^{\prime}}v^{2}\right)-f^{\prime}\cdot cos (2 wt). \tag{18}\]0  
\[m\dot{v}=f^{\prime}\left(1-\frac{d}{f^{\prime}}v^{2}\right)-f^{\prime}\cdot cos (2 wt). \tag{18}\]1  
Since the proposed force has frequency \(2w\), it is logical to assume that the solution has a strongly dominant term at the same frequency. Under this assumption, we can neglect the cross terms in \(g_{2}^{2}\) and solve it in a much simpler way. We will come back to this in a later section.