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I am trying to solve the pde given by:

$$\rho_t + (\rho v)_x = 0$$

Subject to conditions $v = f(x-vt)$ and $\rho(x, 0) = g(x)$.

I have used the product rule on the second term to observe that

$$\rho_t + v\rho_x = -\rho v_x$$

which is a quasi-linear pde. Then we can form the system of ode's:

$$\frac{dx}{dt} = v$$ and $$\frac{d\rho}{\rho} = -v_xdt$$

It is not immediately clear at this point how best to proceed. Any hints appreciated.

Cheers

Victoria
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  • It may be useful to add the solution suggested by the text:

    $$\rho = \frac{g(x-vt)}{1 + tf'(x-vt)}$$

    It's the deriving of the solution I'm having trouble with.

     – Victoria Apr 29 '15 at 23:04
  • This answer is suggestive of a method of characteristics approach. – Paul Apr 29 '15 at 23:14
  • I thought so too. But if you can no longer treat $v$ as a constant, then where can you begin with the odes? – Victoria Apr 29 '15 at 23:15

1 Answers1

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Since $v=f(x-v_0t)$, $v_x=f'(x-v_0t)$ and $v_t=-v_0f'(x-v_0t)$, or $f_t=-v_0f_x$. This last equation gives you a substitution which allows you to integrate the second equation.

Note, I am using $v_0$ for the coefficient of $t$ in the $f$ equation, since 2 different things both designated by $v$ is confusing.

Paul
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  • It could be that both are $v$: For instance if $v$ solves Burgers' equation with initial data $f$, then the formula as in the OP is correct. I guess we'll have to wait for @william to clarify. – Jose27 Apr 29 '15 at 22:56
  • Yes, v solves Burger's equation. The pde I have given is one of a system of pde's, the first being $v_t + vv_x = 0$ for $v(x, 0) = f(x)$. Does that help clarify? – Victoria Apr 29 '15 at 23:01
  • I am therefore slightly concerned about the derivatives and integrals in your answer. Apologies for the lack of clarity. – Victoria Apr 29 '15 at 23:03