Author Topic: Concept: Where does the (-) come from  (Read 784 times)

Rachel Mandel

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Concept: Where does the (-) come from
« on: January 22, 2019, 04:29:32 PM »
Still confused about where the negative sign comes in the Constant Coefficients and Variable Coefficients section. For example, how I derive the example in the variable section would be...

Ut + tUx     = 0
dx + tut      = 0
dx              = -tut
x                = -1/2t2
x + 1/2t2  = 0

Victor Ivrii

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Re: Concept: Where does the (-) come from
« Reply #1 on: January 22, 2019, 04:49:49 PM »
Do you mean $u_t + tu_x=0$? Then $\frac{dt}{1} = \frac{dx}{t}=\frac{du}{0}$. The first equation implies $$dx=tdt \implies x=\frac{1}{2}t^2+C\implies x-\frac{1}{2}t^2=C.$$

Rachel Mandel

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Re: Concept: Where does the (-) come from
« Reply #2 on: January 22, 2019, 10:05:47 PM »
Yes, I do mean that equation. When I try it myself however, (lines 2 to 3 of my initial post) the math comes out that dx = -tdt.

Wanying Zhang

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Re: Concept: Where does the (-) come from
« Reply #3 on: January 22, 2019, 10:53:03 PM »
I guess that since $u$ is of $x$ and $x$ is of $t$, they are not independent and you need to use $\partial$ in your steps. Also, there should be chain rule in this case, so you cannot simply take integrals as in your second equation.

Victor Ivrii

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Re: Concept: Where does the (-) come from
« Reply #4 on: January 23, 2019, 07:10:24 AM »
Yes, I do mean that equation. When I try it myself however, (lines 2 to 3 of my initial post) the math comes out that dx = -tdt.
You need to read Section 2.1 of textbook to understand that equation $au_t+bu_x=f$ requires equation of integral curves $\frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}$. What is on your post are incomprehensibly  written senseless manipulations.