# D'Alembert Solution to 1+1D wave equation - integration step

I am working through d'Alembert's solution to the 1+1D wave equation using the substitution of canonical coordinates. I have an initial condition of: $$u_{t}(x,0) = g(x)$$ with a general solution containing the following two arbitrary functions: $$u(x,t) = \phi(x-ct) + \psi(x+ct)$$ These two functions are composite functions. I know I am supposed to take the partial derivative with respect to $$t$$ and set it equal to $$g(x)$$ to get the following: $$-c\phi'(x) + c\psi'(x) = g(x)$$ The book provides the above prime notation and does not specify what $$\phi$$ & $$\psi$$ are being differentiated by. The issue is the next step in the solution states to carry out the integration from $$x_0$$ to $$x$$ to obtain: $$-c\phi(x) + c\psi(x) = \int^{x}_{x_0}g(\xi)d\xi +K$$ I believe my issue comes from correctly differentiating the composite function. At first I tried to just take the partial derivative with respect to $$t$$, setting $$t=0$$, and got the following: $$-c\frac{\partial{\phi(x)}}{\partial{t}} +c\frac{\partial{\psi(x)}}{\partial{t}} = g(x)$$ When I then integrate with respect to $$x$$ (the only way to make the integral of $$g(\xi)$$ make sense), I get the following: $$-c\int\frac{\partial{\phi(x)}}{\partial{t}}d(x) + ...$$ which doesn't make sense to me. So I then looked into if I was doing the partial derivative of the composite function incorrectly. I found some general texts online about derivatives of composite functions and got this: $$h = g(f(x_1,x_2)) = g(u_1(x_1,x_2),u_2(x_1,x_2))$$ And using the chain rule for composite functions: $$\frac{\partial h}{\partial x_1} = \frac{\partial h}{\partial u_1} * \frac{\partial u_1}{\partial x_1} + \frac{\partial h}{\partial u_2} * \frac{\partial u_2}{\partial x_1}$$ When I try to apply it to the problem, I get the following: $$u(x,t) = g(f(x,t)) = g(\phi(x,t),\psi(x,t))$$ with the following derivative with respect to $$t$$: $$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \phi} * \frac{\partial \phi}{\partial t} + \frac{\partial u}{\partial \psi} * \frac{\partial \psi}{\partial t}$$ which simplifies to (might be incorrect): $$\frac{\partial u}{\partial t} = -c * \frac{\partial u}{\partial \phi} + c * \frac{\partial u}{\partial \psi}$$ I don't know how to find $$\frac{\partial u}{\partial \phi}$$ or $$\frac{\partial u}{\partial \psi}$$ and those derivatives don't make sense when I try to integrate with respect to $$x$$: $$-c\int\frac{\partial{u}}{\partial{\phi}}d(x) + ...$$ Can anyone show me where I am making a mistake in my understanding, and how to correctly differentiate $$u(x,t) = \phi(x-ct) + \psi(x+ct)$$ to get it into the correct form to where I can integrate with respect to $$x$$? Thank you!

• > "The book provides the above prime notation and does not specify what $\phi$ & $\psi$ are being differentiated by." No, you are misunderstanding this very much. Both functions are 1-variable functions; it just so happens that the one variable that they are functions thereof, are themselves a linear function of two variables. Commented Jan 3 at 17:04
• The wiki does a better job of defining new variables so that $x$ isn't of questionable identity. Commented Jan 3 at 17:07
• Thank you everyone for your comments. I understand now! Commented Jan 3 at 20:04

It is possible to understand $$\phi (x-ct)=\phi (\xi (t))$$ and $$\psi(x+ct)=\psi (\xi (t))$$, where $$\xi(t)=x-ct$$; to get:$$u_t(x,0)={\partial\over\partial t}\phi (\xi(t))\bigg |_{t=0}+{\partial\over\partial t}\psi(\xi(t))\bigg |_{t=0}=g(\xi(t))\bigg |_{t=0}.$$Applying the chain rule one gets: $${\partial\over\partial t}\phi (\xi(t))+{\partial\over\partial t}\psi (\xi(t))={\partial\phi\over\partial \xi}{\partial\xi\over\partial t}+{\partial\psi\over\partial\xi}{\partial\xi\over\partial t}$$ Hence, $$u_t(x,0)=-c{\partial\phi\over\partial\xi}+c{\partial\psi\over\partial\xi}=g(\xi)$$ Integrating with respect to $$\xi$$ gives:$$-c\phi (\xi)+c\psi (\xi)=\int_{x_0}^xg(\xi)\;d\xi\;+K.$$ Where $$K$$ is a constant of integration from the right hand side that has been moved over. Now since we have set $$t=0$$ it is clear that $$\xi(0)=x$$ which when substituted into the above expression gives the result as per the textbook you have been consulting.