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I am reading about the one dimensional heat equation and boundary conditions. This is what the author writes about the left hand end of the rod:

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Can someone explain to me why the slope must be positive at $x=0$? If the rod is hotter shouldn't it loose temperature, i.e. negative slope?

Alex M.
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DoubleOseven
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1 Answers1

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The slope in the $t$ direction is negative because the temp is dropping over time. But in the $x$ direction, the change in heat is positive, because things are cooler to the left.

B. Goddard
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  • Then the question is: why is $u_x (L, t) \le 0$ (understood as directional derivative at $x=L$ from the left), given that $x=L$ is the maximum? – Alex M. Oct 09 '16 at 12:16
  • This is at the right end of the rod, and there, things are cooler to the right. When $t=0$, the graph of the temperature distribution is some sort of discontinuous function where there's a step up at $x=0$ and a step down at $x=L$. A short time later, these steps are smoothed over, but there's still a sharp slope up at the left and a sharp slope down at the right. – B. Goddard Oct 09 '16 at 12:23
  • So you are saying that the derivative that I am talking about is the one from the right (where there is no rod), not from the left? If so, then the text is a bit idiotic (a statement supported by the abundance of colours and gloss in that book). Unfortunately, I believe that formulations like this are to be expected from authors whose books reach the 10th edition... – Alex M. Oct 09 '16 at 12:34
  • The derivative at $x=0$ is at the left. The derivative at $x=L$ is at the right. There is no rod to the left of $x=0$ or to the right of $x=L$, but there is "surrounding medium", so the heat has a place to go. (Yeah, non-narcissistic textbook authors are rare.) – B. Goddard Oct 09 '16 at 12:42
  • I am confused. Is it right or wrong? I still dont get it – DoubleOseven Oct 09 '16 at 13:06
  • I think it's right. Perhaps if you draw a picture of the initial temperature, including the surrounding medium, and then imagine the temp one second later, you'll see the derivatives. – B. Goddard Oct 09 '16 at 13:25