
[Sponsors] 
May 16, 2012, 23:39 
Exact solution for 2D inviscid burgers equation.

#1 
New Member
Join Date: May 2012
Posts: 5
Rep Power: 6 
I write a code for numerical method for 2D inviscid burgers equation:
u_t + (1/2u^2)_x + (1/2u^2)_y = 0, initial function: u(0, x) = sin(pi*x) but I don't know how to solve the exact solution for it. And I need it to complete the test for my code. Anybody who can tell me how to obtain the exact solution for it? Thanks very much! 

May 17, 2012, 00:19 

#2 
Senior Member
Chris DeGroot
Join Date: Nov 2011
Location: Canada
Posts: 388
Rep Power: 8 
Have you looked into the method of characteristics?


May 17, 2012, 00:25 

#3 
New Member
Join Date: May 2012
Posts: 5
Rep Power: 6 

May 17, 2012, 00:32 

#4 
Senior Member
Chris DeGroot
Join Date: Nov 2011
Location: Canada
Posts: 388
Rep Power: 8 
I don't know off hand the solution to the problem, I was just making the suggestion to look into method of characteristics since it is used for hyperbolic PDE. See if you can find a copy of "An Introduction to Nonlinear Partial Differential Equations" by Logan. If I remember correctly it covers the Burger's equation.


May 17, 2012, 00:47 

#5  
New Member
Join Date: May 2012
Posts: 5
Rep Power: 6 
Quote:
PS: I see someone says that for 1D case: Ut + (0.5u^2)x = 0, and the initial function is u(0, x) = sin(pi*x); and the exact solution will be: u = sin(pi*(x  ut)) Am I right? 

May 17, 2012, 00:57 

#6 
Senior Member
Chris DeGroot
Join Date: Nov 2011
Location: Canada
Posts: 388
Rep Power: 8 
Yes that looks correct.


May 17, 2012, 01:02 

#7 
New Member
Join Date: May 2012
Posts: 5
Rep Power: 6 

May 17, 2012, 02:34 

#8  
Senior Member
cfdnewbie
Join Date: Mar 2010
Posts: 557
Rep Power: 12 
I
Quote:
I am speculating here, but to me it looks like you took the solution to a linear scalar transport with speed a and just replaced a by u in the solution? Could that be? 

May 17, 2012, 03:41 

#9 
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 2,938
Rep Power: 34 
The Burgers equation
du/dt + u*du/dx =0 implies that the velocity is constant along particolar curves, called "characteristic lines". To see that, write the total differential of u(x,t): du= dt* du/dt + dx *du/dx = dt* (du/dt + dx/dt *du/dx) Therefore you see that u=constant along the line in the x,t plane that satisfies dx/dx=u. Being u constant, you have straight lines of different slopes. In the 2D case, you have some new issue, for example you have two equations for u and v or in some other cases the 2D Burgers equations is factorized 

May 17, 2012, 11:26 

#10 
Senior Member
Chris DeGroot
Join Date: Nov 2011
Location: Canada
Posts: 388
Rep Power: 8 
In the 1D case you have: u_t + uu_x = 0
This implies: du/dt = u_t + dx/dt*u_x = 0 Then choose: dx/dt = u so that the two equations above are equivalent. This leaves you with the following ODE IVP: dx/dt = u du/dt = 0 x(0) = xi u(xi,0)=sin(pi*xi) Solving and applying IC: x = u*t + k(xi) = u*t + xi u = l(xi) = sin(pi*xi) So your solution in implicit form is: x = u*t + xi u = sin(pi*xi) Perhaps it doesn't make sense to combine them together since u would be then defined as a function of u. I started running through the 2D calculation and realized your initial condition is not a function of y. Thus it is not a wellposed problem. Do you mean u(0, x, y) = sin(pi*x), i.e. u does not change with y? If so, you end up with an equation for your second characteristic as: y = u*t + eta, but your initial condition is not a function of eta, so your solution actually doesn't change from the 1D case as far as I can tell. Actually this makes sense since u_y would start out as zero and will never change from this value. You will just have advection in the xdirection. 

May 17, 2012, 11:42 

#11 
Senior Member
Chris DeGroot
Join Date: Nov 2011
Location: Canada
Posts: 388
Rep Power: 8 
Actually since u = sin(pi*xi) you can simplify a but further to:
x = sin(pi*xi)*t + xi u = sin(pi*xi) However, plotting the characteristics given by the first equation, they are definitely going to cross each other leading to shocks. Does everyone agree with this statement? 

May 17, 2012, 12:34 

#12 
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 2,938
Rep Power: 34 
However, I don't understand your 2D case... it should imply a set of two equations:
Ut + UUx +VUy=0 Vt + UVx +VVy=0 each one says U and V are constant along the curves dx/dt=U and dy/dt=V, thus what do you mean for 2D Burgers equation? 

May 17, 2012, 12:56 

#13  
Senior Member
Chris DeGroot
Join Date: Nov 2011
Location: Canada
Posts: 388
Rep Power: 8 
Quote:
u_t + uu_x + uu_y = 0 True? 

May 17, 2012, 13:20 

#14 
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 2,938
Rep Power: 34 

May 17, 2012, 13:28 

#15 
Senior Member
Chris DeGroot
Join Date: Nov 2011
Location: Canada
Posts: 388
Rep Power: 8 
Okay maybe you are right, but you must agree that is what the person who posted stated at the beginning of this thread. I think you are right that physically it does not make sense as an advection equation. For scalar advection you could have:
c_t + uc_x + vc_y = 0 where [u v] is a known velocity field advecting c. Or if it is a momentumtype equation (as I think it is intended to be) it would have to be u_t + uu_x + vu_y = 0 v_t + uv_x + vv_y = 0 as you said before. I think the poster needs to clarify what exactly is being solved. 

May 17, 2012, 13:55 

#16  
Senior Member
cfdnewbie
Join Date: Mar 2010
Posts: 557
Rep Power: 12 
Quote:
That's true. however, if the initial conditions are just u=sin(f(x)), as stated (and assuming periodic bcs), then the solution will just be the developing shock solution found from the Cole Hopf transform... Another idea if a solution to the 2d burgers is too hard to find: Use the method of manufactured solution to validate the code! 

Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Question about boundary condition of 2D Burgers equation  effort8  Main CFD Forum  0  January 29, 2012 21:55 
inviscid burgur's equation  morteza08  Main CFD Forum  2  August 13, 2010 08:45 
exact solution of acoustic equations  pran  Main CFD Forum  0  February 14, 2002 03:46 
Exact 2D NS solution for benchmarking?  Tony  Main CFD Forum  4  July 31, 2001 14:50 
CFL Condition  Matt Umbel  Main CFD Forum  14  January 12, 2001 15:34 