# Can value of the variable be substituted in partial derivatives before taking the derivative?

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?

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.
• @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. Apr 27, 2020 at 18:24
• @user115625 Based on what you have shown, the prime indicates a time derivative at $t=0$ Apr 28, 2020 at 13:45
• 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? Apr 28, 2020 at 14:13