# How to continue boundary condition functions in order to use the d'Alembert Formula for PDEs

I am trying to use the d'Alembert formula for the wave equation with the following boundary conditions:

$$u_{tt} = u_{xx}, \hspace{0.5cm} x<3, t>0,\\ u_x(3,t)=0, \hspace{0.5cm} t >0 \\ u(x,0) = \begin{cases} (x-2)^2 x^2 , \hspace{0.5cm} x \in [0,2] \\ 0 , \hspace{0.5cm} x <0 , x \in (2,3) \end{cases} \\ u_t (x,0) = \begin{cases} x \sin(\pi x), \hspace{0.5cm} x \in [-2,0] \\ 0 , \hspace{0.5cm} x<-2 , x\in (0,3) \end{cases}.$$

The d'Alembert formula, for $$u_{tt}=a^2 u_{xx}$$, with boundary conditions $$\varphi_0,\varphi_1$$ is the following: $$u(x,t) = \frac{\varphi_0 (x-at)+\varphi_0 (x+at)}{2} +\frac{1}{2a} \int\limits_{x-at}^{x+at} \varphi_1 (\lambda)d\lambda.$$

More precisely, I know that this formula assumes that the boundary conditions $$\varphi_0, \varphi_1$$ are defined over all reals. (The spatial variable $$x \in (-\infty ,+\infty).)$$

So in order to use the formula, I have to continue the functions $$u(x,0), u_t(x,0)$$ in some way, so that the solution that I obtain with the formula is a solution to this particular problem. How do I do that?

• There's not a unique way to do it., assuming the spatial domain is really meant to be the whole line instead of say $(-\infty,3]$.
– Ian
Feb 8, 2022 at 22:40
• Im looking for a way. Not every possible way.(or a special one) Feb 8, 2022 at 23:30
• I mean you can technically choose anything. But it looks like just extending by zero would be $C^2$? That would be the naive answer.
– Ian
Feb 8, 2022 at 23:42

I have come up with something... Lets think that we have found a solution $$u(x,t)$$ via the d'Alembert's formula $$$$\label{eq:dalamber} u(x,t) = \frac{u(x-t,0)+u(x+t,0)}{2}+\frac{1}{2}\int\limits_{x-t}^{x+t} u_t (\xi,0) d\xi$$$$ Ofcourse this formula assumes that the functions $$u(x-t,0)=\varphi_0(x-t)$$ and $$u(x+t,0)=\varphi_1(x+t)$$ are defined over all the reals. But our given functions $$\varphi_0, \varphi_1 : (-\infty, 3) \to \mathbb R$$ are not. And so we will have to find suitable continuations of these functions $$\hat{\varphi_0}(x),\hat{\varphi_1}(x): \mathbb R \to \mathbb R,$$ i.e. functions for which $$\hat{\varphi}_0(x)\Big\vert_{(-\infty,3)} = \varphi_0(x), \hspace{1cm} \hat{\varphi}_1(x)\Big\vert_{(-\infty,3)} = \varphi_1(x).$$ Lets see what the boundary condition $$u_x(3,t) =0, t>0$$ has to say about the continuations $$\hat{\varphi}_0(x),\hat{\varphi}_1(x).$$ Differentiating the d'Alembert's formula, and using Leibniz rule for differentiating under the integral sign we get: \begin{align*} u_x(x,t) &= \frac{1}{2}\frac{\partial}{\partial x} (\hat{\varphi}_0(x-t)+\hat{\varphi}_0(x+t))+\frac{1}{2} \left[\int\limits_{x-t}^{x+t} \frac{\partial \hat{\varphi}_1(\xi)}{\partial x} d\xi +\hat{\varphi}_1(x+t)-\hat{\varphi}_{1}(x-t) \right] \\ &=\frac{\hat{\varphi}'_0(x-t)+\hat{\varphi}'_0(x+t)}{2}+\frac{\hat{\varphi}_1(x+t)-\hat{\varphi}_1(x-t)}{2}. \end{align*} Therefore, since $$\xi$$ is not dependent of $$x$$ or $$t,$$ $$0 = u_x(3,t) = \frac{\hat{\varphi}'_0(3-t)+\hat{\varphi}'_0(3+t)}{2}+\frac{\hat{\varphi}_1(3+t)-\hat{\varphi}_1(3-t)}{2}.$$ So in order to $$u_x(3,t)=0,t>0$$ we need to have
$$\begin{array}{|l} \frac{\hat{\varphi}'_0(3-t)+\hat{\varphi}'_0(3+t)}{2} =0 \\ \frac{\hat{\varphi}_1(3+t)-\hat{\varphi}_1(3-t)}{2} =0 \end{array}, t >0.$$ More precisely \begin{align*} \varphi_1 (x)&=\hat{\varphi}_1 (6-x) \hspace{0.5cm}, x \in (-\infty, 3), \\ \varphi_0(x) &=-\hat{\varphi}'_0 (6-x) \Leftrightarrow \int \varphi'_0(x) dx =\int \hat{\varphi}'_0 (6-x) d(6-x), x \in (-\infty,3), \\ \varphi_0(x) &= \hat{\varphi}'_0(6-x) +C, \hspace{0.5cm} x \in (-\infty,3). \end{align*} So suitable continuations are $$\hat{\varphi_0}(x) = \begin{cases} \varphi_0 (x), \hspace{0.5cm} \text{за} \hspace{0.5cm} x \in (-\infty, 3), \\ \varphi_0(6-x), \hspace{0.5cm} \text{за} \hspace{0.5cm} x \in (3,+\infty] \end{cases}, \hspace{0.5cm} \hat{\varphi}_1(x) = \begin{cases} \varphi_1(x), \hspace{0.5cm} x \in (-\infty,3), \\ \varphi_1(6-x), \hspace{0.5cm} x \in (3,+\infty) \end{cases}.$$ $$\hat{u}(x,t) = \frac{\tilde{\varphi}_0(x-t)+\tilde{\varphi}_0(x+t)}{2}+ \frac{1}{2}\int\limits_{x-t}^{x+t} \tilde{\varphi}_1(\xi) d\xi, \hspace{0.5cm} \hat{u}(x,t)\Big\vert_{x<3,t>0} = u(x,t).$$