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About mattiao

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  1. Dear Joigus, finally I have been working again on this topic after several weeks. While I confirm that with a system of 2 non linear equations and one parameter K, adding the condition of \[ \boldsymbol{\nabla}J = 0 \] works to find the maximum K for which the system has solution, when I try to apply it to a system of 3 equations, this method doesn't work. I do an example to work on it: \[ \begin{cases} 4x^3 - y + 4xy^2k = 0 \\ 4x^2y + 4y^3 - x - k = 0 \\ 2z + p = 0 \end{cases} \] First of all I have trying to iterate k and I found that the system has solution until
  2. Yes, now I would like to solve a system with 4 variables and 1 parameter, so in total 4 equations + the equation to close it maximizing the parameter. Unfortunately no... but if it helps we can consider it linear as first step and once solved, to move on non-linear.
  3. Dear Joigus, I have been thinking and working at this topic and I have done some progresses! I have extended your intuition to systems with more equations. So if we have a system of 3 unknows and 1 parameter of which we want to find the higher value that makes the system solvable, we can use the parameter as an unknows and add a 4th equation. To write the 4th equation we can consider the gradient of the other 3 equations. We can do \( \vec{v_{1}} = \boldsymbol{\nabla}f_{1}\wedge\boldsymbol{\nabla}f_{2} \), so \( \vec{v_{1}} \) is the vector tangent to the curve created by the i
  4. Joigus, I think using \( \boldsymbol{\nabla}f_{1}\wedge\boldsymbol{\nabla}f_{2}=0\) as 3rd equation of the system, is the correct solution! I have tried with several systems of 2 equations and it always working! In the next days I will try also with systems with more than 2 equations... I will keep you update, thanks.
  5. Yes joigus, I think we start to be on the right way. So, as you already understood, my system becomes: \[ \begin{cases} −x^2 + 10x − 1/2y^2 + 6y − K = 0 \\ −2x^2 + 20x + 3/2y^2 − 18y = 0 \\ (−2x+10)−(−4x+20)=0 \end{cases} \] And like you said, from the 3rd equation I get \(x=5\). Substituting this in the 2nd equation, we get \(y=4.367\) or \(y=7.633\). Finally substituting x and y in the 1st equation, we get \(k=41.667=\frac{125}{3} \) . So for this example imposing the 3rd equation as the difference of the derivates respect x, works. You can try to do the same
  6. Hi Joigus, sorry if I haven't been clear enough, I'll try to explain better. It's true that the equation in the example can be solved exactly, but as I explained, that was just an example, my system is done be 4 equations and it can't be solve analitically, it has to be solve with numeri methods (Newton-Rapson in my case). I try to change example, anyway keeping just 2 equations to start: \[ \begin{cases} 30sin(x/4)+100sin(x/10)-x − 1/2y^2 + 6y −k= 0 \\ 60sin(x/4)+200sin(x/10)-2x + 3/2y^2 − 18y = 0 \end{cases} \] I am trying to use k as a new variable and to add
  7. I am considering K as an independent variable, so our system becomes: \[ F(x, y, k) = 0 \] The third equation is the Jacobian of the first equation respect the first variable (x) minus the Jacobian of the second equation rw respect again the first variable (x). So for the example I put above, it becomes: \[ (-2x+10) - (-4x+20) = 0 \] The reason why add this 3rd equation is that solving this system with Newton-Rapson I got the result I was looking for. Otherwise I don't see how to get it using numerical methods... P.S.: sorry but I am not able to write formu
  8. ok, very good, but now let's complicate things... This is the analytical solution of this system. In my case I have much more complex systems and I can't analytically solve, so I use a Newton-Raphson method. I found I can get the same result (K=125/3), adding a third equation to the system and using k as a variable: df1/dx - df2/dx = 0 (the derivate respect x of the first equation minus the derivate respect x of the second equation) This makes sense, because it imposes that the intersections of 2 surfaces represented by the first 2 equations of the system with the pla
  9. Thanks Sensei, the result is correct (K=125/3), but I have't clear how you found it. Can you explain, please?
  10. Dear all, I have a system of N non-linear equations in N variables and 1 parameter. The system has solutions until the parameter is less than a certain value. I am interested in finding the maximum value of the parameter for which the system has solutions. Could you help me to find a way to do that? I have tried to use the parameter as an extra variable of the system and to add one more equation, but I struggle to find the new equation... I try to give you an example in case I have not been clear enough: −x^2 + 10x − 1/2y^2 + 6y − K = 0 −2x^2 + 20x + 3/2y^2 − 18y = 0 How do
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