📂Partial Differential Equations

One-Dimensional D'Alembert's Formula

Review¹

Suppose the Cauchy problem for the wave equation is given as follows.

$$ \begin{align*} u_{tt}-u_{xx}&= 0 && \text{in } \mathbb{R}^2=\mathbb{R}_{x} \times \mathbb{R}_{t} \\ u=g,\quad u_{t}=&\h && \text{on } \mathbb{R}\times\left\{t=0\right\} \end{align*} $$

Then $g \in C^2(\mathbb{R}), h\in C^1(\mathbb{R})$. Let’s define $u(x,t)$ as follows.

$$ \begin{equation} u(x,t)=\dfrac{1}{2} \left[ g(x+t)+g(x-t) \right] + \dfrac{1}{2} \int_{x-t}^{x+t}h(y)dy \quad \forall\ (x,t)\in \mathbb{R}^2 \end{equation} $$

Then, $u\in C^2(\mathbb{R}^2)$ is the solution to the given Cauchy problem.

Explanation

$(1)$ is called the d’Alembert’s formula. The term 1-dimensional refers to the dimension of space.

Proof

Let’s represent the given differential equation as follows.

$$ (\partial_{t}+\partial_{x})(\partial_{t}-\partial_{x})u=u_{tt}-u_{xx}=0 \quad \text{in } \mathbb{R}^{2} $$

And let’s define $v$ as follows.

$$ v=(\partial_{t}- \partial_{x})u=u_{t}-u_{x} \quad \in C^1( \mathbb{R}^2) $$

Then, the given differential equation becomes the following homogeneous transport equation.

$$ v_{t}+v_{x}=0 \quad \text{in } \mathbb{R}^2 $$

Then, if we say $a(\xi):=v(\xi,0), \xi \in \mathbb{R}$, $v(x,t)=a(x-t)$ holds. By the definition of $v$, the following is satisfied.

$$ u_{t}-u_{x}=a(x-t) \quad \forall\ (x,t)\in \mathbb{R}^2 $$

This is a non-homogeneous transport equation. Since $u(x,0)=g(x)$, the solution to the non-homogeneous transport equation is given as follows.

$$ \begin{align*} u(x,t) =&\ g(x+t)+\int_{0}^t a(x+(s-t)(-1)-s)ds \\ =&\ g(x+t) + \int_{0}^t a(x+t-2s)ds \\ =&\ g(x+t) +\dfrac{1}{2}\int_{x-t}^{x+t}a(y)dy \end{align*} $$

The third equality holds by substituting $-2s+x+t=y$. Since $g \in C^2$ and $v \in C^1$, differentiating $u$ with respect to $t$ yields the following.

$$ u_{t}(x,t)=g^{\prime}(x+t)+\dfrac{1}{2}\big( a(x+t)+a(x-t)\big) $$

By the assumption, $u_{t}(x,0)=h(x)$ holds.

$$ h(x)=g^{\prime}(x)+a(x) \ \implies \ a(x)=h(x)-g^{\prime}(x) $$

Therefore, $u(x,t)$ is as follows.

$$ \begin{align*} u(x,t) =&\ g(x+t)+\dfrac{1}{2}\int_{x-t}^{x+t} \left[ h(y)-g^{\prime}(y) \right] dy \\ =&\ \dfrac{1}{2}\left[ g(x+t)+g(x-t) \right] + \dfrac{1}{2} \int_{x-t}^{x+t}h(y)dy \quad \forall \ (x,t)\in \mathbb{R}^2 \end{align*} $$

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