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?


1 Answer 1


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.

  • $\begingroup$ Thank you for the clarification. I have added more info which would better explain my doubt. $\endgroup$
    – user115625
    Apr 27, 2020 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. $\endgroup$ Apr 27, 2020 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, 2020 at 13:41
  • $\begingroup$ @user115625 Based on what you have shown, the prime indicates a time derivative at $t=0$ $\endgroup$ Apr 28, 2020 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, 2020 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.