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Ultimately my goal is to find a candidate for the weak solution beyond the time when the classical solution does not exist and determine conditions on the motion of the shock that guarantees it is a weak solution of the following: $$ u_t+u^2u_x=0, \ x\in \mathbb{R}, \ t>0, \text{with} \ u(x,0)=F(x). $$

The first thing I want to do is find the classical solution using the method of characteristics, but I am having a little difficulty in that process.

Any help?

EditPiAf
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User112358
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  • i think it will be nice to fix an $F$ first. find the solution for various choices of $F$ – abel Apr 12 '15 at 23:45
  • Well I am trying to follow example 3 section 14.1 from Strauss's Partial Differential Equations as well as my class notes. There we find the classical solution to Burger's equation for any F. I would like to keep fairly general and not pick a specific F. – User112358 Apr 12 '15 at 23:51
  • i don't have a copy of strauss. what kind of examples did he work out? – abel Apr 13 '15 at 00:15
  • https://zr9558.files.wordpress.com/2013/11/partial-differential-equations.pdf – User112358 Apr 13 '15 at 00:43

1 Answers1

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you can use the method of characteristics just as was done in the example 3 from section 14.1. define the characteristic curve through $x = a, t = 0$ by the solution $x(t; a)$ of $$\frac{dx}{dt} = u^2(x, t), x = a \text{ at } t = 0.\tag 1$$ then it follows that $$\frac{d}{dt}u(x(t; a), t) = u_t(x, t) + u_x \frac{dx}{dt} = u_t + u^2u_x = 0.$$ therefore $u$ is constant on the characteristics. that is $$u(x(t;a), t) = u(x(0; a), 0) = u(a, 0)= F(a).$$

now, going back to $(1),$ we see that the right hand side $u^2(x,t) = F^2(a)$ does not depend on $t.$ therefore $$x(t; a) = tF^2(a)+a, u(a+tF^2(a), t) = F(a). \tag 2$$

set $x = a + tF^2(a)$ and call the inverse function $a = G(x, t).$ then we have $$u(x, t) = F(G(x, t)). $$

abel
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