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I have the PDE

$$\frac{\partial{u}}{\partial{t}} + u \frac{\partial{u}}{\partial{x}} = −2u$$

Written explicitly, I thought the characteristic equations were $\frac{\partial{t}}{\partial{r}} = 1$, $\frac{\partial{x}}{\partial{r}} = u$, and $\frac{\partial{u}}{\partial{r}} = -2u$

But the textbook says that the characteristic equations are $\frac{\partial{x}}{\partial{r}} = 1$, $\frac{\partial{u}}{\partial{r}} = -2u$, which I believe is implicit form

Why is there this difference? Are these two forms equivalent? Is one correct and the other incorrect? If they are equivalent, then why does the textbook use the latter characteristic equations instead of the former?

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ What do you mean by implicit form? What is defined implicitly? An application of the chain rule (where $x, t$ are dependent on a parameter $r$) shows that \begin{align} \frac{du(x(r),t(r))}{dr} &= \frac{\partial u}{\partial x} \frac{dx}{dr} + \frac{\partial u}{\partial t} \frac{dt}{dr} \\ &= u \frac{\partial u}{\partial x} + \frac{\partial u}{\partial t} \\ &= -2u \\ \implies \frac{dt}{dr} &= 1, \quad \frac{dx}{dr} = u, \quad \frac{du}{dr} = -2u \end{align} as you gave. So what you wrote is correct. $\endgroup$
    – Matthew Cassell
    Aug 19, 2018 at 3:44

1 Answer 1

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$$\frac{\partial{u}}{\partial{t}} + u \frac{\partial{u}}{\partial{x}} = −2u$$

The three characteristic differential equations are :

$$\frac{dt}{dr} = 1 \quad;\quad \frac{dx}{dr} = u \quad;\quad\frac{du}{dr} = -2u$$ which can be written on condensed form : $$\frac{dt}{1} = \frac{dx}{u} = \frac{du}{-2u} = dr$$

This shows that the three equations are not independent. Any one of them is related to the two others. Equivalently they are related to any linear combination of the three equations.

SOLVING :

A first characteristic equation comes from $\frac{dx}{u} = \frac{du}{-2u}$ : $$u+2x=c_1$$ A second characteristic equation comes from $\frac{dt}{1} = \frac{du}{-2u}$ : $$ue^{2t}=c_2$$ The general solution of the PDE expressed on the form of implicite equation is : $$\Phi(u+2x\:,\:ue^{2t})=0$$ $\Phi$ is an arbitrary function of two variables, to be determined according to some boundary and/or initial conditions.

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  • $\begingroup$ Thanks for the answer. What I'm wondering is, if they're equivalent, then why do we work from the condensed form $\frac{dt}{1} = \frac{dx}{u} = \frac{du}{-2u} = dr$, instead of the form $\frac{dt}{dr} = 1 \quad;\quad \frac{dx}{dr} = u \quad;\quad\frac{du}{dr} = -2u$? [...] $\endgroup$
    – The Pointer
    Aug 19, 2018 at 8:28
  • $\begingroup$ [...] For instance, If we worked from $\frac{dt}{dr} = 1 \quad;\quad \frac{dx}{dr} = u \quad;\quad\frac{du}{dr} = -2u$, then we would have (using separation of variables) $t = r + c_1$ from $\dfrac{dt}{dr} = 1$, $u = c_2e^{-2r}$ from $\dfrac{du}{dr} = -2u$, and then $\dfrac{dx}{dr} = u = c_2e^{-2r}$ which gives $x = c_2e^{-2r}$ using separation of variables again (if I did my integration properly). [...] $\endgroup$
    – The Pointer
    Aug 19, 2018 at 8:34
  • $\begingroup$ [...] These are supposedly equivalent, but we can see that we've seemingly gotten different values depending on whether we used the condensed form $\frac{dt}{1} = \frac{dx}{u} = \frac{du}{-2u} = dr$ or $\frac{dt}{dr} = 1 \quad;\quad \frac{dx}{dr} = u \quad;\quad\frac{du}{dr} = -2u$? $\endgroup$
    – The Pointer
    Aug 19, 2018 at 8:39
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    $\begingroup$ You got $u=c_2e^{-2r}$ and $t=r+c_1$ thus $u=c_2e^{-2t+2c_1}=C_2e^{-2t}$ with $C_2=c_2e^{2c_1}$. That is exactly the characteristic equation found in my answer : $$ue^{2t}=C_2$$ Also, you got $x=c_2e^{-2r}$ which is not correct. Integrating $\frac{dx}{dr}=c_2e^{-2r}$ gives $x=-\frac12 c_2e^{-2r}+c_3=-\frac12 u+c_3$. And with $2c_3=C_1$ , this is exactly the characteristic equation found in my answer : $$u+2x=C_1$$ Of course, the two methods give the same result. $\endgroup$
    – JJacquelin
    Aug 19, 2018 at 9:05
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    $\begingroup$ The results are the same, but expressed on two different forms : One gives the result on parametric form (parameter $r$). The other gives the result on the form of implicit equation; They are equivalent since eliminating the parameter leads to the implicit equation. $\endgroup$
    – JJacquelin
    Aug 19, 2018 at 9:14

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