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

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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 
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Chris DeGroot
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Have you looked into the method of characteristics?


May 17, 2012, 00:25 

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May 17, 2012, 00:32 

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Chris DeGroot
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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 

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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 

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Chris DeGroot
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Yes that looks correct.


May 17, 2012, 01:02 

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May 17, 2012, 02:34 

#8  
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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 
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Filippo Maria Denaro
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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 

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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 

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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 

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Filippo Maria Denaro
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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 

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Quote:
u_t + uu_x + uu_y = 0 True? 

May 17, 2012, 13:20 

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May 17, 2012, 13:28 

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Chris DeGroot
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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  
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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! 

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