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I have this initial value problem:

$u_t + xu_x = -u^2$, with $u(x,0)=1$.

So from this we have

$ \dfrac{dt}{1} = \dfrac{dx}{u} = \dfrac{du}{u^2}$

so $\dfrac{du}{dt} = -u^2 \implies u=\dfrac{1}{t+w}$

and $\dfrac{dx}{dt} = u \implies x=tu+z \implies z=x-tu$

where $w$,$z$ are constants. My problem is that I'm not too sure how to illustrate $u(x,t)$ as , since this is implicit.

tellap
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1 Answers1

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Your computations are incorrect. Thew equation should be $$ \dfrac{dt}{1} = \dfrac{dx}{x} = \dfrac{du}{u^2}. $$ Observe that the equation for $x$ is different. The solution is $$ x=C\,e^t,\quad u=\frac{1}{t+w}. $$ From here and the initial condition you find $$ u(x,t)=\frac{1}{1+t}. $$

Julián Aguirre
  • 76,354
  • Thank you for that, I knew I was going wrong somewhere with it being implicit. – tellap Mar 25 '15 at 21:36
  • So I end up having $u(x,t) = \dfrac{1}{t+w(z)}$. When I plug in the initial value, I end up with $w(x)=1$. Do I just substitute this back into $u(x,t)$ or into $z=x-\mathrm{log}(t+w)$ seeing as we would have $w(z)$ in $u(x,t)$? – tellap Mar 25 '15 at 21:46
  • I have edited my answer, because it was wrong. – Julián Aguirre Mar 26 '15 at 11:48