6
$\begingroup$

I have a question, concerning one special solution of the following system. First I will sum up my results, to give a context for my question. My question will be seperated by a line, if you want to skip the introduction.

$\begin{equation} \begin{aligned} \mathrm{II:} \quad\\ \mathrm{III:} \quad\\ \end{aligned} \begin{aligned} \dot{x}_2 &= a x_0 - b x_2 + cx_3 + g x_0 x_3\\ \dot{x}_3 &= -d y_2 x_3 + b x_2 - cx_3 \\ \end{aligned} \end{equation}$

with the initial conditions

$x_0 = x_1(0) \quad \rightarrow \quad x_2(0) = x_3(0) = x_4(0) = 0$

First I solved the homogenous part of the system: \begin{equation} \left(\begin{matrix}\dot{x}_2\\\dot{x}_3\end{matrix}\right) = \underbrace{\left[\begin{matrix}- b & g x_{0} + c\\b & -\left( d y_{2} +c \right)\end{matrix}\right]}_{\substack{\hat{A}}}\left(\begin{matrix}x_2\\x_3\end{matrix}\right) + \underbrace{\left(\begin{matrix}a x_0\\0\end{matrix}\right)}_{\substack{\textbf{b}}} \end{equation}

\begin{equation} \begin{aligned} \\ \lambda_1 &= -\frac{1}{2}\left[ b+ c+d y_{2} + \sqrt{\left(b + c + d y_2 \right)^2 - 4 b \left(d y_2 - g x_0 \right)} \right]\\ \\ \lambda_2 &= -\frac{1}{2}\left[ b+ c+d y_{2} + \sqrt{\left(b + c + d y_2 \right)^2 - 4 b \left(d y_2 - g x_0 \right)} \right] \\ \\ \end{aligned} \begin{aligned} v_1 &= \left(\begin{matrix}- \frac{2\left(c + g x_{0}\right)}{- b + c + d y_{2} + \sqrt{\left(b + c + d y_2 \right)^2 - 4 b \left(d y_2 - g x_0 \right)}}\\1\end{matrix}\right) \\ v_2 &= \left(\begin{matrix}- \frac{2\left(c + g x_{0}\right)}{- b + c + d y_{2} - \sqrt{\left(b + c + d y_2 \right)^2 - 4 b \left(d y_2 - g x_0 \right)}}\\1\end{matrix}\right)\\ \end{aligned} \label{eq::modell6_eigenwerte_und_vektoren} \end{equation}

which is given by \begin{equation} \textbf{x}_h = c_1 \textbf{v}_1 e^{\lambda_1t} + c_2 \textbf{v}_2 e^{\lambda_2t} \label{eq::modell5_homogene_loesung} \end{equation}

I found the particular solution using the following approach

\begin{equation} x_p = \left(\begin{matrix}A_1\\A_2\end{matrix}\right) \quad \dot{\textbf{x}}_p = \hat{A} \textbf{x}_p + \textbf{b} \end{equation}

the solution of this equation is \begin{equation} \textbf{x}_p = \frac{ax_0}{\left(d y_2 - g x_0 \right)}\left(\begin{matrix}\frac{d y_2 + c}{b}\\1\end{matrix}\right) \end{equation}

Adding homogenous and particular solution and using the initial conditions gave me the following general solution of the system.

\begin{equation} \begin{aligned} x_1 &= x_0\\ x_2 & = \frac{ax_0}{\left(d y_2 - g x_0 \right)} \left[-\frac{\left(g x_{0} +c\right) \left(b + c + d y_{2} - \sqrt{\dots}\right)}{\sqrt{\dots}\left(- b + c + d y_{2} + \sqrt{\dots} \right)} e^{-\frac{1}{2} \left[b + c + d y_{2}+ \sqrt{\dots}\right]t} + \frac{\left(g x_{0} +c\right) \left(b + c + d y_{2} + \sqrt{\dots}\right)}{\sqrt{\dots}\left(- b + c + d y_{2} - \sqrt{\dots} \right)} e^{-\frac{1}{2} \left[b + c + d y_{2}- \sqrt{\dots}\right]t} + \frac{d y_2 + c}{b}\right]\\ x_3 &= \frac{ax_0}{\left(d y_2 - g x_0 \right)} \left[\frac{\left(b + c + d y_{2} - \sqrt{\dots}\right)}{2 \sqrt{\dots}}e^{-\frac{1}{2} \left[b + c + d y_{2} + \sqrt{\dots}\right]t} - \frac{ \left(b + c + d y_{2} + \sqrt{\dots}\right)}{2 \sqrt{\dots}}e^{-\frac{1}{2} \left[b + c + d y_{2} - \sqrt{\dots}\right]t} + 1 \right]\\ x_4 &=\frac{d y_2 a x_0}{\left(d y_{2} - g x_{0}\right)}\left[-\frac{ \left(b + c + d y_{2} - \sqrt{\dots}\right)}{ \sqrt{\dots}\left[b + c + d y_{2} + \sqrt{\dots}\right]}e^{-\frac{1}{2} \left[b + c + d y_{2} + \sqrt{\dots}\right]t}+ \frac{ \left(b + c + d y_{2} + \sqrt{\dots}\right)}{ \sqrt{\dots}\left[b + c + d y_{2} - \sqrt{\dots}\right]}e^{-\frac{1}{2} \left[b + c + d y_{2} - \sqrt{\dots}\right]t} + t +\frac{ \left(b + c+ dy_2 \right)}{b\left(d y_{2} - g x_{0}\right)}\right] \end{aligned} \label{eq::modell6_loesung} \end{equation} with \begin{equation} \sqrt{\dots} = \sqrt{\left( b + c + d y_2 \right)^2 - 4b \left(dy_2 - g x_0 \right)} \end{equation}

---------------------------------------------

Now I want to analyze the special case $\left(d y_{2} - g x_{0}\right)$. This solution is very interesting because the exponent of the e function, switches from positive to negative at this transition. As I get a zero-division error in my standard solution, I want to solve the system of equations again, with the new set of parameters. \begin{equation} \left(\begin{matrix}\dot{x}_2\\\dot{x}_3\end{matrix}\right) = \underbrace{\left[\begin{matrix}- b & c + d y_{2}\\b & - c - d y_{2}\end{matrix}\right]}_{\substack{\hat{A}}}\left(\begin{matrix}x_2\\x_3\end{matrix}\right) + \underbrace{\left(\begin{matrix}a x_0\\0\end{matrix}\right)}_{\substack{\textbf{b}}} \end{equation}

homogenous solution: \begin{equation} \begin{aligned} \\ \lambda_1 &= 0\\ \\ \lambda_2 &= - \left(b + c + d y_{2}\right)\\ \\ \end{aligned} \begin{aligned} v_1 &= \left(\begin{matrix}\frac{1}{b} \left(c + d y_{2}\right)\\1\end{matrix}\right)\\ v_2 &= \left(\begin{matrix}-1\\1\end{matrix}\right)\\ \end{aligned} \label{eq::modell6_eigenwerte_und_vektoren_spezialfall_1} \end{equation} \begin{equation} \textbf{x}_h = c_1 \textbf{v}_1 e^{\lambda_1t} + c_2 \textbf{v}_2 e^{\lambda_2t} \end{equation}

And now I have a problem finding the particular solution. I use the same approach \begin{equation} x_p = \left(\begin{matrix}A_1\\A_2\end{matrix}\right) \quad \dot{\textbf{x}}_p = \hat{A} \textbf{x}_p + \textbf{b} \end{equation} which gives me the following set of equation \begin{equation} \begin{aligned} \mathrm{I:} \quad\\ \mathrm{II:} \quad\\ \end{aligned} \begin{aligned} 0 &= -b A_1 + (d y_2 + c) A_2 + ax_0\\ 0 &= b A_1 - (d y_2 + c) A_2 \\ \end{aligned} \end{equation} which gives the result $A_1 = \frac{d y_2 + c}{b} A_2$ if $ax_0 = 0$.

Now I have a problem to interprete this solution because I didn't expected the condition $ax_0 = 0$ for my solution. Is there only a solution of the system for $\left(d y_{2} - g x_{0}\right)$ if $ax_0 = 0$. Or is there a more general solution with $ax_0 \neq 0$? Did I make a mistake during my calculation?

$\endgroup$
  • $\begingroup$ well, this is a mouthful. i like it! $\endgroup$ – Alexander Day Apr 24 '17 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.