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I was going through the D'Alembert's solution for the wave equation using this pdf from University of British Columbia (UBC, Canada). Here's the link: https://www.math.ubc.ca/~ward/teaching/m316/lecture21.pdf

$$u(x, t) =F(x−ct) +G(x+ct)\tag{21.18}$$

$u(x,0) =F(x) +G(x) =u_0(x)\tag{21.19}$

$\frac{\partial u}{\partial t}(x,0) =−cF′(x) +cG′(x) =v_0(x)\tag{21.20}$

where, $u_0(x)$ is initial position and $v_0(x)$ is initial velocity.

Then they go on to derive $u(x,t)$ in terms of the initial conditions. How can they put the value of $t=0$ and then differentiate $u$ partially with respect to $t$? As far as I know, we need to differentiate first and then substitute the value of the variable to find the value of the derivative at a particular point.

I think if we are partially differentiating a function wrt $x$ then it doesn't matter if substitute the value of $t$ before or after the differentiation as it is anyways constant wrt that derivative.

Edit: 1. It seems that they are just being sloppy with their notation and I am misunderstanding it. But, if that is so then won't $F'$ still be a function of $(x-ct)$ and $G'$ a function of $(x+ct)$? $−cF(x) +cG(x) =\int_{o}^{x} v_0(ξ)dξ+A ...(21.21)$

In the next equation (21.21), they integrate the functions wrt to $x$. How can this be done as $F'$ and $G'$ are functions of different variables.

  1. I did the above step assuming $\alpha=x-ct$, and $\beta=x+ct$. this makes equation (21.20):

$∂u/∂t(x,t)= dF(\alpha)/d\alpha * ∂\alpha/∂t + dG(\beta)/d(\beta)*∂\beta/∂t$ $∂u/∂t(x,t)= -c*dF(\alpha)/d\alpha + c*dG(\beta)/d(\beta)$

Now putting $t=0$: $∂u/∂t(x,0)= -c*dF(\alpha)/d\alpha + c*dG(\beta)/d(\beta) = v_0(x)$

Now, I am stuck as I don't know which variable $\alpha$ or $\beta$ should I integrate it with?

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Their motion means evaluating the partial derivative at $t=0$. You can easily verify this by taking the partial derivative with respect to $t$, and then plugging in $t=0$. You can also easily verify that they are not doing what you propose, as their partial derivative is not equal to $0$.

As for your second point, yes, the order there doesn't matter. If you are finding the partial derivative with respect to $x$, then you can explicitly plug in a value for $t$ before or after the derivative. The result will be the same.

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  • $\begingroup$ Thank you for the clarification. I have added more info which would better explain my doubt. $\endgroup$ – user115625 Apr 27 '20 at 18:14
  • $\begingroup$ @user115625 $\partial F/\partial t$ evaluated at $t=0$ is still a function of $x$, so you can still integrate it with respect to $x$. Or maybe I am not fully understanding you. $\endgroup$ – BioPhysicist Apr 27 '20 at 18:24
  • $\begingroup$ Can we integrate wrt x for F' = G'? I'm assuming the ' shows the total derivative which should be wrt (x-ct) for F and (x+ct) for G. $\endgroup$ – user115625 Apr 28 '20 at 13:41
  • $\begingroup$ @user115625 Based on what you have shown, the prime indicates a time derivative at $t=0$ $\endgroup$ – BioPhysicist Apr 28 '20 at 13:45
  • $\begingroup$ Won't we have to use chain rule for that derivative (as I've shown in my edit)? In that case, it should be a total derivative wrt $\alpha$ and $\beta$. Also, all the places where I've encountered prime, it used to represent total derivative. Isn't that the generally accepted convention? $\endgroup$ – user115625 Apr 28 '20 at 14:13

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