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I tried to solve the following problem:

Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is a smooth function and that $u$ is related to $t$ and $x$ by the eqution:

$$u=f(x-ut)$$

Which of the following differential equations does $u$ satisfy?

(a) $u_t-uu_x=0$

(b) $u_x-u^2u_t=0$

(c) $u_x+tu_t=0$

(d) $u_t+u_x(u_t)^2=0$

(e) $u_t+uu_x=0$

My attempt is to write the equation as $u-f(x-ut)=0$ and differentiate it using Chain Rule, see if I can obtain some information from that:

differentiate with respect to $x$ : $$u_x-(f_x \cdot x_x+f_u\cdot u_x+f_t\cdot t_x)=0$$ $$u_x-(f_x+f_u\cdot u_x)=0$$ and since $u_x=f_x$ and $f_u=u_u=1$ we have : $$-u_x=0$$ and Similarly if I differentiate with respect to $t$ I get $-u_t=0$

I think there must be somewhere I got wrong, but I couldn't figure it out.

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  • $\begingroup$ From your first part follows that $u_x=f_x+f_u u_x$, but immediately after that you say $u_x=f_x$. Do you see the error you make? Also, you made an error, it should be $u_x-(f_x-f_u u_x=0)$. $\endgroup$ – User123456789 Jun 16 '17 at 4:22
  • $\begingroup$ Oh I see the problem, but I am still struggling : if the functions $u$ and $f$ are equal, then why not $u_x=f_x$? $\endgroup$ – Alan Lao Jun 16 '17 at 5:50
  • $\begingroup$ You are struggling with distinction between partial and total derivatives $\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}$. $\endgroup$ – User123456789 Jun 16 '17 at 6:00
  • $\begingroup$ Oh, you help me refreshing my memory from Calculus course! thanks very much @DaanvdWoude ! $\endgroup$ – Alan Lao Jun 16 '17 at 6:11
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$f$ is a function of a single variable $\xi=x-ut$.

$$ u_x=f'(x-ut)(1-u_x t) $$

$$ u_x(1+tf'(\xi))=f'(\xi) $$

$$ u_x=\frac{f'(\xi)}{1+tf'(\xi)} $$

$$ u_t=f'(\xi)(-u_t t-u) $$

$$ u_t(1+f'(\xi)t)=-uf'(\xi) $$

$$ u_t=-u\frac{f'(\xi)}{1+tf'(\xi)} $$

recalling what the expression we found for $u_x$ and substituting into the equation above.

$$ u_t=-uu_x $$

$$ u_t+uu_x=0 $$

all the PDE's given can be solved via the method of characteristics. For instance

$$ u_x-u^2u_t $$

gives rise to the differential system

$$ \frac{\mathbb{d}x}{1}=\frac{\mathbb{d}t}{-u^2},\;\;\;\mathbb{d}u=0 $$

$$ t+u^2x=C_1,\;\;\;u=C_2,\;\;\;C_2=\varphi(C_1) $$

$$ u=\varphi(t+u^2x) $$

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