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 Post subject: Show PDE equation holds for u(x,t)=f(x-t)Posted: Fri, 3 Aug 2012 04:01:46 UTC
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Joined: Fri, 9 Jan 2009 13:44:46 UTC
Posts: 4
Hi everyone,

Could someone please tell me how can I use chain rule to show solution (1) in below paragraph holds?

But I don't understand why we have f '(x-t) instead of both du/dx and du/dt? Does f '(x-t) mean du/dt or du/dx (since chain rule says du/dx=du/dt * dt/dx and du/dt=du/dx * dx/dt)?

(I have used d because I did not have the correct character on the keyboard).

Thank you very much.

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 Post subject: Re: Show PDE equation holds for u(x,t)=f(x-t)Posted: Fri, 3 Aug 2012 06:08:48 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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macdermat wrote:
Hi everyone,

Could someone please tell me how can I use chain rule to show solution (1) in below paragraph holds?

But I don't understand why we have f '(x-t) instead of both du/dx and du/dt? Does f '(x-t) mean du/dt or du/dx (since chain rule says du/dx=du/dt * dt/dx and du/dt=du/dx * dx/dt)?

(I have used d because I did not have the correct character on the keyboard).

Thank you very much.

No. f' is the derivative of f (as a function of one variable), and now evaluate this function at x-t to give f'(x-t). Also, you need to keep track of what you are keeping constant in those partial derivatives when using the chain rule.

Differentiating with respect to x (keeping t constant) gives .
Differentiating with respect to t (keeping x constant) gives .

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