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Initial displacement specifies the configuration of a wave equation at the starting time. For a second-order evolution equation it is paired with the initial velocity. Specifying only the boundary forcing does not determine these initial data.

On the unit square, these data determine only the zero solution of the heat equation $u_y=u_{xx}$. The squared $L^2$ norm decreases because its derivative is $-2\int u_x^{2}dx$, and its initial value is zero. For Laplace's equation, $\sin (\pi x/2)\sinh (\pi y/2)$ is a nonzero solution with the same three data; adding zero Dirichlet data on the top edge restores uniqueness by Green's first identity. For the wave equation $u_{yy}=u_{xx}$, $\sin (\pi x/2)\sin (\pi y/2)$ is a counterexample; the missing datum is initial velocity. Adding $u_y(x,0)=0$ restores uniqueness by conservation of $\frac{1}{2}\int (u_x^2+u_y^2)dx$. Thus the type of equation determines what information is needed on an unprescribed edge.

This is the wave equation, again taking $y$ as time. Zero is not unique: for $k=\pi /2$,satisfies $u_{xx}=u_{yy}$ and all the stated boundary conditions. The initial displacement is zero but the initial velocity is not prescribed.

A sufficient extra condition is $u_y(x,0)=0$. For the resulting homogeneous data use wave energyThe equation and integration by parts give $E^{\prime }(y)=[u_x{u}_y{]}_0^1=0$. At $x=1$ the factor $u_x$ vanishes, and at $x=0$ the factor $u_y$ vanishes because $u(0,y)=0$. The initial displacement implies $u_x(x,0)=0$, so the extra initial velocity condition gives $E(0)=0$. Consequently $u_x=u_y=0$ and $u=0$. With the stated boundary regularity, integrate the local energy conservation law over inset rectangles and let their edges approach the boundary; the energy and flux terms converge by continuity of the first derivatives.

Let $\Omega$ be the solid exterior to the cavity, and choose $\mathbf{n}$ outward from the solid, hence into the cavity on its boundary $S$. Subtraction of the incident solution from the total solution gives the scattered displacement field equationsThe traction-free cavity condition for the total field gives ${\sigma }_{ij}^s{n}_j=-{\sigma }_{ij}^0{n}_j$ on $S$. The scattered field has zero initial displacement and initial velocity before the pulse reaches the cavity and satisfies an outgoing radiation condition, excluding an additional incoming elastic wave. Equivalently one seeks the causal response to this prescribed boundary traction. Although the incident plane pulse extends across an infinite plane, the causal scattered field from the bounded cavity has finite propagation distance at each finite time.

To prove energy uniqueness for traction-driven elasticity, take the difference $\mathbf{w}$ of two solutions with the same initial conditions, body force, boundary traction and outgoing condition. It satisfiesFor a large enclosing ball $B_L$, multiply by ${\dot{w}}_i$ and integrate over $\Omega \cap B_L$. Integration by parts givesSymmetry of stress replaces the last derivative by ${\dot{e}}_{ij}(w)$, and major symmetry of the elastic stiffness tensor makes that integral the derivative of the strain energy. ThereforeThe cavity contribution is zero. On any fixed finite time interval choose $L$ beyond the causal support of the scattered difference; the outer contribution is then zero as well. This gives conservation of its nonnegative elastic energy. Positive mass density and positive strain energy imply that zero initial energy forces $\dot{\mathbf{w}}=0$ everywhere; the zero initial displacement field then gives $\mathbf{w}=0$. The same argument works for finite-energy outgoing solutions by taking the limit of enclosing surfaces and excluding incoming energy. Hence the causal scattered field ${\mathbf{u}}^s$ is unique. Specifying the incident field and cavity traction alone without the initial and outgoing conditions would not establish uniqueness.

Let $w$ be the difference of two sufficiently regular solutions of the wave equation with the prescribed data. By linearity, $w$ has zero initial displacement, zero initial velocity, and homogeneous Dirichlet boundary conditions. Define its wave energy byFor $\alpha =0$, differentiation under the integral sign and integration by parts yieldThe boundary term vanishes because $w=0$ at both fixed endpoints implies $w_t=0$ there. The zero initial data give $E(0)=0$, so $w_t=w_x=0$ for all times. Thus $w$ is spatially constant, and its zero boundary values give $\boxed{w=0}$. This proves uniqueness for the undamped wave equation.

For $\alpha =3$, let $F_n$ denote the time factor normalized by $F_n(0)=0$, $F_n^{\prime }(0)=1$. By superposition the Fourier sine series solution isAt $t=0$, its initial velocity is ${\sum }_{n\ge 1}{b}_n\sin (nx)={\sin }^{2}x$. Multiply by $\sin (mx)$, integrate on $[0,\pi ]$, and use the permitted interchange and Fourier orthogonality. This yields the requested integral expressionUnder $x\mapsto \pi -x$, ${\sin }^{2}x$ is unchanged and $\sin (n(\pi -x))=(-1{)}^{n+1}\sin (nx)$. Hence every even-index Fourier coefficient is zero, including $b_2$.

For odd $n$, use ${\sin }^{2}x=(1-\cos 2x)/2$ and the product-to-sum formula:ThusIn particular $b_1=8/(3\pi )$ and $b_3=-8/(15\pi )$. An explicit version of the solution of a differential equation isOnly the first, overdamped normal mode, the third, critically damped normal mode, and the odd underdamped normal modes of index at least five occur. No odd-index Fourier coefficient vanishes.

At $x=L/2$, the integration interval first moves through positive arguments, then crosses zero, and finally contains all of $[-L,L]$. The result isThe final plateau is the integral of the entire initial velocity profile divided by $2c$.

For $c>0$ and $L>0$, D'Alembert's formula for the wave equation on the full real line isFor the specified initial data, define the cumulative initial-velocity integralThen $\boxed{y(x,t)=[A(x+ct)-A(x-ct)]/(2c)}$. The initial velocity jumps at $\pm L$, so the resulting solution is interpreted piecewise classically, or as a weak wave-equation solution across the corresponding characteristic curves.