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| - | <h1>Bistability Analysis</h1>
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| - | <p>The literal meaning of bifurcation is,m "division into two parts, branches, or aspects" (Answers). In mathematics, this 'division' is in reference to when a non-linear equation suddenly switches from having one solution to having multiple solutions when a parameter of the system is varied. The point at which bifurcation is referred to as the 'bifurcation point.' If a system is bistable it can be stable, or at rest in two distinct states.</p>
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| - | <br>
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| - | <center>
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| - | <p>bistability1.png</p>
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| - | <p><b>figure 1.</b> This is a graphical representation of bistability. Ball 2 can be equally 'happy' in either the 'one' or the 'three' positions. ~Once it calls into one of these states, ball 2 will not bin inclined to leave this state, hence making it a 'stable state.'</p>
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| - | <br>
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| - | <h3>Why are they important?</h3>
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| - | <p>Analysing the stability of a system is ket to the success of a project. In biological terms, bistability represetns the potential for our system to switch between the expression fo gree and cyan fluarescent proteins. Our goal is to analyse the parameter space in order to determione which paramenters andre needed to allow for bistability in the system.
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| - | <br>
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| - | a bifurcation diagram has branches which define the possbile states a system can rest in. By analysing the bifurcation plots, we can determing how far apart the two 'branches,' or stable states, are. the further apart the branches, the harder it will be to attacin a switch between the two stable states, however the switch also becomes more stable with increased distance between the branchs. Therefore, we want to manipulate the parameters in the system to guarantee stable bistability.
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| - | <h3>How do you analyse the stability of a system?</h3>
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| - | <p>In order to analyse the stability of the system, we utilised XPPAuto, C programming and MATLAB. We will attempt to detail the maths behind what the programme does computationally.
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| - | <br>
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| - | The process involved with general stability is as follws. We have taken a simple example system that is easy to analyse in order to demonstrate the general process:
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| - | <br>
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| - | Take the following example system of two differential equations:
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| - | <br>
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| - | bistability2.png
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| - | <b>1. Set the differential equations equal to zero:</b>
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| - | <br>
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| - | bistability3.png
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| - | <br>
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| - | <b>2. Find the fixed points of the system of differential equations:</b>
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| - | There is a fixed point at x=y=0. By observation, we can see that there are fixed points at x=0, y=2, and x=3, y=0. Finally, by solving this system of equations, we see that there is also a fixed point at x=1 and y=1.
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| - | <br>
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| - | <b>3. Evaluate the Jacobian matrix of the system at these fixed points:</b>
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| - | <br>
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| - | If we label the top fuction <i>f1</i> and the bottom <i>f2</i>, the Jacobian matrix of the system will be of the firm:
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| - | <br>
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| - | bistability4.png
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| - | <br>
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| - | When we evaluate the Jacobian matrix for our example system of equations, we get:
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| - | <br>
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| - | bistability5.png
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| - | <br>
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| - | <b>4. Using the Jacobian matrix, find the corresponding eigenvalues:</b>
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| - | <u>Definition</u>: an eigenvalue is a value λ such that Ax = λx, where x is said to be an eigenvector of A. In order to solve for λ, solve the following equation:
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| - | <br>
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| - | bistability6.png
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| - | <br>
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| - | Where det is the deterinant of the matrix and I is the identity matrix. THis is known as the characteristic equation of a matrix.
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| - | <br>
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| - | In the case of our example system, the eigenvalues are:
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| - | <br>
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| - | <center>bistability7.png</center>
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| - | <br>
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| - | <b>5. Take the signs of these eigenvalues in order to analyse the stability:</b>
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| - | <br>
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| - | The eigenvalues of a system give important information about the stability of the system. The stability of each of the fixed points can be analysed based on the following criteria:
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| - | <center>
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| - | <table>
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| - | <tr>
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| - | <td>
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| - | Positive real and complex parts
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| - | </td>
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| - | <td>
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| - | = unstable
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| - | </td>
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| - | </tr>
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| - | <tr>
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| - | <td>
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| - | Negative real and complex parts
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| - | </td>
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| - | <td>
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| - | = stable
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| - | </td>
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| - | </tr>
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| - | <tr>
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| - | <td>
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| - | Positive real part
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| - | </td>
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| - | <td>
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| - | = unstable spiral (spiraling out)
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| - | </td>
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| - | </tr>
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| - | <tr>
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| - | <td>
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| - | Negative real part
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| - | </td>
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| - | <td>
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| - | stable spiral (spiralling in)
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| - | </td>
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| - | </tr>
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| - | </table>
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| - | </center>
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| - | <br>
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| - | So, for our ixample system, it can be deduced that the stability at the fixed poitns is as follows:
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| - | <br>
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| - | <center>
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| - | <table>
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| - | <tr>
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| - | <td>
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| - | One positive and one negative eigenvalue at (0,0) means that
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| - | </td>
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| - | <td>
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| - | <b>(0,0)</b> is a <b>saddle node</b>
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| - | </td>
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| - | <td>
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| - | </tr>
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| - | <tr>
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| - | <td>
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| - | Two negative eigenvalues at (0,2) means that
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| - | </td>
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| - | </tr>
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| - | <tr>
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| - | <td>
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| - | <b>(0,2)</b> is a <b>stable node</b>
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| - | </td>
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| - | </tr>
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| - | <tr>
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| - | <td>
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| - | Two negative eigenvalues at (3,0) means that
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| - | </td>
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| - | <td>
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| - | <b>(3,0)</b> is a <b>stable node</b>
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| - | </tr>
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| - | <tr>
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| - | <td>
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| - | One positive and one negative eigenvalue at (1,1) means that
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| - | </td>
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| - | <td>
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| - | <b>(1,1)</b> is a <b>saddle node</b>
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| - | </td>
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| - | </tr>
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| - | </table>
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| - | </center>
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| - | <br>
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| - | <h3>Simulations involved</h3>
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| - | <br>
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| - | <center>
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| - | bistability8.png
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| - | <p><b>Figure 2.</b> This is a bifurcation diagram that we generated using Maple. It ranges over the parameter space of P1 (GFP) and C (a dimensionless group of parameters) and plots the fixed points o the system at those values.
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| - | </center
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| - | Where there are three fixed points for a given value of C1, the system will exhibit bistability. we also know that the further apart the two outer 'prongs' the more stable the two definitive fixed states will be, however thsi makes it more difficult to switch between states.
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| - | </p>
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