# Solve the partial differential equation $u_t + uu_x=0$ [duplicate]

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Solve the following partial differential equation $u_t + uu_x=0$ with $u=u(x,t)$ and $u(x,0)=x$.

I am having trouble in applying the SIDE CONDITION.

• The Characteristics are $dx/dt$=$u$, here u is constant along the characteristics.
• Along the characteristics $dx/dt=g(x_o)$
• Solution is $x=g(X_o)t + d$, where d is an arbitary constant
• But note at $t=0$, $x=x_0$...Hence $d=x_0$
• The Characteristics is $x=g(x_0)t + x_0$, these are straight lines with variable slope

I am not sure on what to do after this point.

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## marked as duplicate by 5pm, JohnD, Sasha, Asaf Karagila, MicahFeb 18 '13 at 5:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See math.stackexchange.com/questions/305727/… and the link posted by Willie Wong in the comments. –  user53153 Feb 17 '13 at 21:16
I have checked this but i still can't seem to see how to use the side condition –  Yenny Chen Feb 17 '13 at 21:48

## 1 Answer

This PDE is called the (inviscid) Burgers' equation. The characteristic equations are the system of ODEs $${dx\over dt}=u, \quad {du\over dt}=0,$$ which has solution $$u=C_1, \quad x=ut+C_2,$$ where $C_1,C_2$ are constants. But $C_1$ is a function of $C_2$, so $$u=C_1\implies u=C_1(C_2)=C_1(x-ut).$$ To determine, $C_1$ note that the side condition requires $$u(x,0)=C_1(x-u(x,0)\cdot 0)=x\implies C_1(x)=x.$$ Thus, $u=x-ut$ and we conclude $$u(x,t)={x\over 1+t}.$$

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