**Proof plan.** The proof proceeds in four parts. First, we show that the regularity and [boundary](/page/Boundary) conditions on $g$ and $h$ force the Fourier sine coefficients $\hat{g}_n$ and $\hat{h}_n$ to decay as $O(n^{-4})$ and $O(n^{-3})$ respectively, via repeated [integration by parts](/theorems/210). This decay is fast enough that the [series](/page/Series) defining $u$, together with its second-order derivatives in $t$ and $x$, converge uniformly by the [Weierstrass M-Test](/theorems/272). [Interchange of Limits and Derivatives](/theorems/260) then justifies passing derivatives through the sum, so that each term satisfying the PDE implies the sum does as well. The initial conditions are recovered from the definition of the Fourier sine coefficients. Uniqueness is proved by an energy argument using the [Gauss-Green Theorem](/theorems/28).
**Step 1: Fourier coefficient decay.**
[claim:Decay Of Sine Fourier Coefficients Under Regularity]
Let $L > 0$ and $\mu_n := n\pi/L$. If $g \in C^4([0,L])$ with $g(0) = g(L) = g''(0) = g''(L) = 0$, then
\begin{align*}
\hat{g}_n := \frac{2}{L}\int_0^L g(x)\sin(\mu_n x) \, d\mathcal{L}^1(x)
\end{align*}
satisfies $|\hat{g}_n| \leq C/n^4$ for some constant $C > 0$ independent of $n$. Similarly, if $h \in C^3([0,L])$ with $h(0) = h(L) = 0$, then $|\hat{h}_n| \leq C'/n^3$.
[/claim]
[proof]
We compute $\hat{g}_n$ by four successive integrations by parts. At each step the integrated term $[\cdots]_0^L$ is evaluated using the boundary conditions.
**First integration by parts.** Integrate $g(x)\sin(\mu_n x)$ with $u = g$, $dv = \sin(\mu_n x) \, dx$:
\begin{align*}
\int_0^L g(x)\sin(\mu_n x) \, d\mathcal{L}^1(x) &= \left[-\frac{g(x)\cos(\mu_n x)}{\mu_n}\right]_0^L + \frac{1}{\mu_n}\int_0^L g'(x)\cos(\mu_n x) \, d\mathcal{L}^1(x).
\end{align*}
The boundary term is $-g(L)\cos(n\pi)/\mu_n + g(0)/\mu_n = 0$ since $g(0) = g(L) = 0$.
**Second integration by parts.** Integrate $g'(x)\cos(\mu_n x)$ with $u = g'$, $dv = \cos(\mu_n x) \, dx$:
\begin{align*}
\frac{1}{\mu_n}\int_0^L g'(x)\cos(\mu_n x) \, d\mathcal{L}^1(x) &= \frac{1}{\mu_n}\left[\frac{g'(x)\sin(\mu_n x)}{\mu_n}\right]_0^L - \frac{1}{\mu_n^2}\int_0^L g''(x)\sin(\mu_n x) \, d\mathcal{L}^1(x).
\end{align*}
The boundary term is $g'(L)\sin(n\pi)/\mu_n^2 - g'(0)\sin(0)/\mu_n^2 = 0$ since $\sin(n\pi) = \sin(0) = 0$.
**Third integration by parts.** Integrate $g''(x)\sin(\mu_n x)$:
\begin{align*}
-\frac{1}{\mu_n^2}\int_0^L g''(x)\sin(\mu_n x) \, d\mathcal{L}^1(x) &= -\frac{1}{\mu_n^2}\left[-\frac{g''(x)\cos(\mu_n x)}{\mu_n}\right]_0^L + \frac{1}{\mu_n^3}\int_0^L g'''(x)\cos(\mu_n x) \, d\mathcal{L}^1(x) \\
&= \frac{1}{\mu_n^3}\int_0^L g'''(x)\cos(\mu_n x) \, d\mathcal{L}^1(x),
\end{align*}
where the boundary term vanishes because $g''(0) = g''(L) = 0$.
**Fourth integration by parts.** Integrate $g'''(x)\cos(\mu_n x)$:
\begin{align*}
\frac{1}{\mu_n^3}\int_0^L g'''(x)\cos(\mu_n x) \, d\mathcal{L}^1(x) &= \frac{1}{\mu_n^3}\left[\frac{g'''(x)\sin(\mu_n x)}{\mu_n}\right]_0^L - \frac{1}{\mu_n^4}\int_0^L g^{(4)}(x)\sin(\mu_n x) \, d\mathcal{L}^1(x).
\end{align*}
The boundary term vanishes since $\sin(n\pi) = \sin(0) = 0$. The remaining [integral](/page/Integral) is bounded by $\|g^{(4)}\|_{L^1([0,L])}$. Therefore:
\begin{align*}
|\hat{g}_n| = \frac{2}{L}\left|\int_0^L g(x)\sin(\mu_n x) \, d\mathcal{L}^1(x)\right| \leq \frac{2}{L\mu_n^4}\|g^{(4)}\|_{L^1} = \frac{2L^3}{n^4\pi^4}\|g^{(4)}\|_{L^1}.
\end{align*}
For $h \in C^3([0,L])$ with $h(0) = h(L) = 0$, we perform three integrations by parts on $\hat{h}_n$.
**First integration by parts.**
\begin{align*}
\int_0^L h(x)\sin(\mu_n x) \, d\mathcal{L}^1(x) = \left[-\frac{h(x)\cos(\mu_n x)}{\mu_n}\right]_0^L + \frac{1}{\mu_n}\int_0^L h'(x)\cos(\mu_n x) \, d\mathcal{L}^1(x).
\end{align*}
The boundary term is $-h(L)\cos(n\pi)/\mu_n + h(0)/\mu_n = 0$ since $h(0) = h(L) = 0$.
**Second integration by parts.**
\begin{align*}
\frac{1}{\mu_n}\int_0^L h'(x)\cos(\mu_n x) \, d\mathcal{L}^1(x) = \frac{1}{\mu_n}\left[\frac{h'(x)\sin(\mu_n x)}{\mu_n}\right]_0^L - \frac{1}{\mu_n^2}\int_0^L h''(x)\sin(\mu_n x) \, d\mathcal{L}^1(x).
\end{align*}
The boundary term vanishes since $\sin(n\pi) = \sin(0) = 0$.
**Third integration by parts.**
\begin{align*}
-\frac{1}{\mu_n^2}\int_0^L h''(x)\sin(\mu_n x) \, d\mathcal{L}^1(x) = -\frac{1}{\mu_n^2}\left[-\frac{h''(x)\cos(\mu_n x)}{\mu_n}\right]_0^L + \frac{1}{\mu_n^3}\int_0^L h'''(x)\cos(\mu_n x) \, d\mathcal{L}^1(x).
\end{align*}
The boundary term is $\frac{1}{\mu_n^3}[h''(L)\cos(n\pi) - h''(0)]$, which is bounded by $\frac{2\|h''\|_\infty}{\mu_n^3}$. The remaining integral is bounded by $\|h'''\|_{L^1}/\mu_n^3$. Therefore:
\begin{align*}
|\hat{h}_n| = \frac{2}{L}\left|\int_0^L h(x)\sin(\mu_n x) \, d\mathcal{L}^1(x)\right| \leq \frac{C'}{n^3}
\end{align*}
for a constant $C' > 0$ depending on $L$, $\|h''\|_\infty$, and $\|h'''\|_{L^1}$.
[/proof]
**Step 2: [Uniform convergence](/page/Uniform%20Convergence) of the series and its derivatives.**
Write $u_n(t,x) := \left[\hat{g}_n \cos(\mu_n c t) + \frac{\hat{h}_n}{\mu_n c}\sin(\mu_n c t)\right]\sin(\mu_n x)$. For any $(t,x) \in [0,T] \times [0,L]$:
\begin{align*}
|u_n(t,x)| \leq |\hat{g}_n| + \frac{|\hat{h}_n|}{\mu_n c} \leq \frac{C}{n^4} + \frac{C'}{n^3 \cdot n\pi/(cL)} = O(n^{-4}).
\end{align*}
Since $\sum_{n=1}^\infty n^{-4} < \infty$, the [Weierstrass M-Test](/theorems/272) gives uniform and absolute convergence of $\sum u_n$ on $[0,T] \times [0,L]$.
For term-by-term [differentiation](/page/Derivative), we need uniform convergence of the derived series. Differentiating $u_n$ twice in $t$:
\begin{align*}
\partial_{tt} u_n(t,x) = -\mu_n^2 c^2 \left[\hat{g}_n \cos(\mu_n c t) + \frac{\hat{h}_n}{\mu_n c}\sin(\mu_n c t)\right]\sin(\mu_n x),
\end{align*}
so $|\partial_{tt} u_n(t,x)| \leq \mu_n^2 c^2 (|\hat{g}_n| + |\hat{h}_n|/(\mu_n c)) \leq C'' n^2 \cdot n^{-4} + C'' n^2 \cdot n^{-4} = O(n^{-2})$. Since $\sum n^{-2} < \infty$, the [Weierstrass M-Test](/theorems/272) gives uniform convergence of $\sum \partial_{tt} u_n$.
Differentiating $u_n$ twice in $x$:
\begin{align*}
\partial_{xx} u_n(t,x) = -\mu_n^2 \left[\hat{g}_n \cos(\mu_n c t) + \frac{\hat{h}_n}{\mu_n c}\sin(\mu_n c t)\right]\sin(\mu_n x),
\end{align*}
which satisfies the same bound $|\partial_{xx} u_n| = O(n^{-2})$, so $\sum \partial_{xx} u_n$ also converges uniformly by the [Weierstrass M-Test](/theorems/272).
By [Interchange of Limits and Derivatives](/theorems/260) (applied to the partial sums $S_N(t, x) := \sum_{n=1}^N u_n(t, x)$, which are $C^2$ in $t$ with $S_N'' = \sum_{n=1}^N \partial_{tt} u_n$ converging uniformly and $S_N(0, x)$ converging), the sum $t \mapsto \sum_n u_n(t, x)$ is $C^2$ in $t$ and $\partial_{tt} u(t, x) = \sum_n \partial_{tt} u_n(t, x)$. Since this holds for every $x \in [0, L]$, $\partial_{tt} u = \sum \partial_{tt} u_n$ on $[0, T] \times [0, L]$.
The argument for $x$-derivatives is analogous: fix any $t \in [0, T]$ and consider $x \mapsto u_n(t, x)$. The partial sums $S_N(t, \cdot) = \sum_{n=1}^N u_n(t, \cdot)$ are $C^2$ in $x$, $S_N(t, 0) = 0$ converges, and $S_N''(\cdot) = \sum_{n=1}^N \partial_{xx} u_n(t, \cdot)$ converges uniformly in $x$ on $[0, L]$ (since $|\partial_{xx} u_n| = O(n^{-2})$). By [Interchange of Limits and Derivatives](/theorems/260), $\partial_{xx} u = \sum \partial_{xx} u_n$.
**Step 3: The PDE, boundary conditions, and initial conditions.**
Each term $u_n$ satisfies the PDE. To verify this, compute directly:
\begin{align*}
\partial_{tt} u_n - c^2 \partial_{xx} u_n &= -\mu_n^2 c^2 \left[\hat{g}_n \cos(\mu_n ct) + \frac{\hat{h}_n}{\mu_n c}\sin(\mu_n ct)\right]\sin(\mu_n x) \\
&\quad + c^2 \mu_n^2 \left[\hat{g}_n \cos(\mu_n ct) + \frac{\hat{h}_n}{\mu_n c}\sin(\mu_n ct)\right]\sin(\mu_n x) = 0.
\end{align*}
Since we may differentiate term by term (Step 2), $\partial_{tt} u - c^2 \partial_{xx} u = \sum_{n=1}^\infty (\partial_{tt} u_n - c^2 \partial_{xx} u_n) = 0$.
For the boundary conditions: $\sin(\mu_n \cdot 0) = 0$ and $\sin(\mu_n L) = \sin(n\pi) = 0$ for every $n$, so $u_n(t, 0) = u_n(t, L) = 0$. Since the series converges uniformly, $u(t, 0) = \sum u_n(t, 0) = 0$ and $u(t, L) = 0$.
For the initial displacement: at $t = 0$, $\cos(\mu_n c \cdot 0) = 1$ and $\sin(\mu_n c \cdot 0) = 0$, so $u(0, x) = \sum_{n=1}^\infty \hat{g}_n \sin(\mu_n x)$. This is the Fourier sine expansion of $g$ on $[0, L]$ with coefficients $\hat{g}_n$. Since $g \in C^4([0,L])$ with $g(0) = g(L) = 0$ and $|\hat{g}_n| = O(n^{-4})$, the series $\sum \hat{g}_n \sin(\mu_n x)$ converges uniformly (by the [Weierstrass M-Test](/theorems/272)) to a [continuous](/page/Continuity) [function](/page/Function) that agrees with $g$ in $L^2([0,L])$ (since $\{\sqrt{2/L}\sin(\mu_n x)\}_{n \geq 1}$ is a complete orthonormal system in $L^2([0,L])$ and the coefficients are exactly the Fourier sine coefficients of $g$). A continuous function determined by its $L^2$-class is unique, so $u(0,x) = g(x)$ pointwise.
For the initial velocity: differentiating the series term by term (justified by Step 2, since the first time-derivative series has terms bounded by $O(n^{-3})$):
\begin{align*}
\partial_t u_n(0, x) = \left[-\hat{g}_n \mu_n c \sin(\mu_n c \cdot 0) + \hat{h}_n \cos(\mu_n c \cdot 0)\right]\sin(\mu_n x) = \hat{h}_n \sin(\mu_n x).
\end{align*}
Therefore $\partial_t u(0, x) = \sum_{n=1}^\infty \hat{h}_n \sin(\mu_n x)$. This is the Fourier sine expansion of $h$ on $[0, L]$ with coefficients $\hat{h}_n$. Since $h \in C^3([0,L])$ with $h(0) = h(L) = 0$ and $|\hat{h}_n| = O(n^{-3})$, the series converges uniformly by the [Weierstrass M-Test](/theorems/272) to a continuous function that agrees with $h$ in $L^2([0,L])$ (since $\hat{h}_n$ are the Fourier sine coefficients of $h$ with respect to the complete orthonormal system $\{\sqrt{2/L}\sin(\mu_n x)\}_{n \geq 1}$). A continuous function determined by its $L^2$-class is unique, so $\partial_t u(0, x) = h(x)$ pointwise.
**Step 4: Uniqueness via energy conservation.**
Suppose $u$ and $v$ both solve the Cauchy-Dirichlet problem. Define $w := u - v$. Then $w$ solves the homogeneous problem:
\begin{align*}
\begin{cases}
\partial_{tt} w - c^2 \partial_{xx} w = 0, & (t,x) \in (0,T) \times (0,L), \\
w(t,0) = w(t,L) = 0, & t \in (0,T), \\
w(0,x) = 0, \quad \partial_t w(0,x) = 0, & x \in [0,L].
\end{cases}
\end{align*}
Define the energy
\begin{align*}
E(t) := \frac{1}{2}\int_0^L \left[|\partial_t w(t,x)|^2 + c^2|\partial_x w(t,x)|^2\right] d\mathcal{L}^1(x).
\end{align*}
Differentiating under the integral sign (justified since $w \in C^2$):
\begin{align*}
\frac{dE}{dt} = \int_0^L \left[\partial_t w \cdot \partial_{tt} w + c^2 \partial_x w \cdot \partial_{xt} w\right] d\mathcal{L}^1(x).
\end{align*}
For the second term, integrate by parts in $x$ using the [Gauss-Green Theorem](/theorems/28) (in one dimension, this is integration by parts):
\begin{align*}
\int_0^L c^2 \partial_x w \cdot \partial_{xt} w \, d\mathcal{L}^1 = c^2 \bigl[\partial_x w \cdot \partial_t w\bigr]_0^L - \int_0^L c^2 \partial_{xx} w \cdot \partial_t w \, d\mathcal{L}^1.
\end{align*}
The boundary term vanishes: since $w(t, 0) = 0$ for all $t$, differentiating in $t$ gives $\partial_t w(t, 0) = 0$; similarly $\partial_t w(t, L) = 0$. Therefore:
\begin{align*}
\frac{dE}{dt} = \int_0^L \partial_t w \cdot \bigl[\partial_{tt} w - c^2 \partial_{xx} w\bigr] \, d\mathcal{L}^1 = 0,
\end{align*}
since $w$ solves the homogeneous [wave equation](/page/Wave%20Equation). So $E(t) = E(0)$ for all $t \in [0, T]$.
At $t = 0$, $w(0, x) = 0$ gives $\partial_x w(0, x) = 0$, and $\partial_t w(0, x) = 0$ by hypothesis. Therefore $E(0) = 0$, which implies $E(t) = 0$ for all $t$. Since the integrand in $E(t)$ is non-negative and continuous, both $\partial_t w = 0$ and $\partial_x w = 0$ on $[0,T] \times [0,L]$. Hence $w$ is constant, and $w(0, x) = 0$ gives $w \equiv 0$, so $u = v$.
