Show PDE equation holds for u(x,t)=f(x-t)

macdermat · **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?

Last time I asked someone gave me below answer:

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.

outermeasure · **Posted:** Fri, 3 Aug 2012 06:08:48 UTC

macdermat wrote:

Hi everyone,

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

Last time I asked someone gave me below answer:

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 u=f(x-t)

with respect to x (keeping t constant) gives $\dfrac{\partial u}{\partial x}=f'(x-t)$ .
Differentiating u=f(x-t)

with respect to t (keeping x constant) gives $\dfrac{\partial u}{\partial t}=-f'(x-t)$ .

S.O.S. Mathematics CyberBoard

Show PDE equation holds for u(x,t)=f(x-t)

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