Hi all !
I'm new to PDE's and I would like to get some help with this equation:
-x_2*u_x1 + x_3*u_x2 + x_1*u_x3 = 0;
I have included three pictures on how I'm trying to solve it but I can't get the solution.
I'm trying to find the functions of X(s) and then trying to find y(s).
I was watching lectures from Giovanni Bellettini and tried to follow his steps.
Can anybody help me?
The numerical solution for this is pretty straightforward, but in your case (analytical solution) it's not :)
you can use this book:
http://www.springer.com/br/book/9781461448082
Hi Mohammad Zandssalimy
Thank you for your answer. I will try the book you recomended.
Dear Ante, this linear PDE should be analitycally solved using the method of characteristics, see Ian Sneddon, Elements of Partial Differential Equations, Chapters 1 and 2. Gianluca
- The famous method for solving linear, first order, and homogeneus PDEs is the method of characteristics, where you transform your PDE in a set of ODEs. The number of ODEs depending on the case.
Ante,
by solving system of ordinary diff. equations:
dx1/dt= - x3 ; dx2/dt=x3 ; dx3/dt= x1
you get x1(t)=c1*exp(-t)+c2*exp(t/2)*cos(omega*t)+c3*exp(t/2)*sin(omega*t) etc. for x2 and x3,
where c1,c2 and c3 are integration constants and omega=sqrt(3)/2.
By denoting as C(t)=exp(t/2)*cos(omega*t), S(t)=exp(t/2)*sin(omega*t) and E(t)=exp(t/2),
you get the algebraic equation:
C2(t)+S2(t)=1/E2(t)
Eliminating C(t), S(t) and E(t), i.e. expressing them by x1(t), x2(t) and x3(t) and using algebraic equation mentioned above you should obtain three different constants of motion expressed as algebraic functions of x1, x2 and x3. Then u(x1,x2,x3) is an arbitrary function od these three constants of motion.
Dear Ante,
Please find attached my own attempt at this problem, though you should be aware of any other means of solution. Let me know if this is helpful, and do please inform me of any mistakes in my working or misunderstandings regarding the theory on my part.
Do also read the other texts that have been recommended to you, as they may offer other alternatives currently unknown to myself.
Best regards,
Stephen
Dear Stephen,
I have looked at your solution, and this is very helpfull.
I would like to take this oportunity to ask you something.
In the video (link beloow) , this guy says that c_2 is a function of c_1.
Can you explain why?
https://www.youtube.com/watch?v=LpHqrlrU5pM&index=12&list=PLGCj8f6sgswntUil8yzohR_qazOfYZCg_
Dear Ante,
I shall look at your video to be certain, but as you will see from my own three ODEs (utilising the Method of Characteristics), when you equate one to the other - such as the first to the second, the second to the third, and the first to the third (involving at least two different variables, x and y, x and z, or y and z) - you thus obtain characteristic equations and eventual solutions (for u = u(k)), each of which contains more than one variable, such as u=u[(1/y)Exp(xy)]. Because of this, one can see that such a function is governed by variables that are dependent on at least one other variable - so that one is a function of the other. For example, in u=u[(1/y)Exp(xy)], one should be able to appreciate that this solution is dependent on both x and y, and hence, so is the characteristic equation, (1/y)Exp(xy), that forms it. Also, the final solution that you obtain, such as u = u[(1/y)Exp(xy)], can often be expressed as c_1 = u[c_2] and c_2 = u[c_1]. So you may find that one solution is xExp[xz] = u[(1/y)Exp(xy)] for the above specific question that you asked. Therefore, this will undoubtedly be the reason why c_2 is a function of c_1.
I hope that this helps, but do feel free to ask me anything else.
Best regards,
Stephen
Sir, I think I understand now
I will definitelly ask you some questions when I'm going to be covering second order PDE.
Thank you again
No probs, Ante. Glad to have been able to help you. I look forward to your future questions on second order PDEs. Cheers for now. Stephen
Dear Ante,
I was not overly happy with the solutions that I sent to you on Tuesday, so I have read up a little more on the subject of Vector Analysis and Fluid Dynamics, particularly regarding Potential Functions. I now have an appropriate solution to your problem, which fits exactly with the solution returned by my mathematical software, so I feel more confident now that this version is definitely correct.
I trust that this will be a little more explanatory and helpful to you, though the previous workings may also still have some virtue.
Best regards,
Stephen
- Badges
- Science topic
- Similar topics
- Books