A Uniqueness Theorem

Equations (5.3.5) subject to given initial and boundary conditions have a unique solution.

Assume that there exist two solutions $u^{(1)}_i$ and $u^{(2)}_i$ of eqn. (5.3.5) subject to the same initial and boundary conditions. Then

$$w_i =\ u^{(1)}_i - u^{(2)}_i\$$

$$\rho_0\frac{\partial^2w_i}{\partial t^2} = (\lambda + \mu )w_{k,ki} + \mu w_{i,jj},$$ (5.5.1)

and, on the boundary, either

$$T^\ast_{ij}n_j = \lambda w_{k,k}n_i + \mu (w_{i,j} + w_{j,i})n_j = 0$$ (5.5.2)

or

$$w_i = 0,$$ (5.5.3)

and

$$w_i(X_j,0) = 0,\ \dot w_i (X_{j,}0) = 0.$$ (5.5.4)

Taking the inner product of (5.5.1) with $\dot w_i(X_j,t)$, integrating the resulting equation over the region $R$ occupied by the body in the reference configuration and by using the divergence theorem, we arrive at

$$\int T^\ast_{ij}n_j\dot w_idA - \int T^\ast_{ij}\dot w_{j,i} dV = \frac{d}{dt}\int\rho_0\frac{\dot w_i\dot w_i}{2}dV.$$ (5.5.5)

The first integral vanishes because of (5.5.2) and (5.5.3). The second integral can be rearranged to read

$$\frac{d}{dt}\int\left[\frac{\lambda}{2}w_{k,k}w_{i,i} + \mu e^\ast_{ij}e^\ast_{ij}\right]dV,$$

in which $e^\ast_{ij} = (w_{i,j} + w_{j,i})/2$. Thus eqn. (5.5.5) becomes

$$\frac{d}{dt}\int\left[\frac{\lambda}{2}e^\ast_{kk}e^\ast_{ii} + \mu e^\ast_{ij}e^\ast_{ij} + \frac{\rho_0}{2}\dot w_i\dot w_i\right]dV = 0.$$ (5.5.6)

Integrating this and making use of the initial conditions (5.5.4) for $w_i$, we obtain

$$\int\left[\frac{\lambda}{2}e^\ast_{kk}e^\ast_{ii} + \mu e^\ast_{ij}e^\ast_{ij} + \frac{\rho_0}{2}\dot w_i\dot w_i\right] dV = 0.$$ (5.5.7)

Noting that the integrand can be written as

$$\left(\frac{\lambda}{2} + \frac{\mu}{3}\right)e^\ast_{kk}e^\ast_{ii}+\mu e^{\ast d}_{ij} e^{\ast d}_{ij} + \frac{\rho_0}{2}\dot w_i\dot w_i$$ (5.5.8)

in which $e^{\ast d}_{ij} = e^\ast_{ij} - e^\ast_{kk} \displaystyle\frac{\delta_{ij}}{3}$, we see that every term in (5.5.8) is positive provided that

$$3\lambda + 2\mu > 0,\ \mu > 0.$$ (5.5.9)

We will henceforth assume that $\lambda$ and $\mu$ satisfy (5.5.9). Thus for eqn. (5.5.7) to hold,

$$e^{\ast d}_{ij} = 0,\ e^\ast_{ii} = 0,\ \dot w_i = 0.$$ (5.5.10)

Since $w_i = 0$ initially, therefore, $w_i (X_j,t) = 0$ which implies that $u^{(1)}_i(X_j,t) = u^{(2)}_i (X_j,t)$.

Thus if somehow one can find a solution of eqn. (5.3.5) that satisfies the prescribed initial and boundary conditions, then that is the only solution of eqn. (5.3.5). There are very few dynamic problems that have been solved.

In a static problem or, more appropriately, in a quasi-static problem, the left-hand side of eqn. (5.3.5) becomes zero and one needs only the prescribed boundary conditions. Thus the difference solution $w_i = u^{(1)}_i - u^{(2)}_i$ will satisfy

$$0 = (\lambda + \mu )w_{k,ki} + \mu w_{i,jj},$$ (5.5.11)

and either (5.5.2) or (5.5.3) on the boundary. Taking the inner product of (5.5.11) with $w_i$, integrating the resulting equation over $R$, and using the divergence theorem we arrive at

$$\int T^\ast_{ij} n_jw_idA - \int T^\ast_{ij} w_{j,i}dV = 0.$$ (5.5.12)

The first integral vanishes because of (5.5.2) and (5.5.3) and the second integral can be written as

$$\int\left[\left(\frac{\lambda}{2} + \frac{\mu}{3}\right) e^\ast_{kk} e^\ast_{ii} + \mu e^{\ast d}_{ij} e^{\ast d}_{ij}\right] dV = 0.$$ (5.5.13)

For this equation to hold,

$$e^{\ast d}_{ij} = 0,\ e^\ast_{ii} = 0,$$

and hence

$$e^\ast_{ij} = 0.$$ (5.5.14)

A solution of eqn. (5.5.14) given in Section 3.14 is

$$w_i = a_i + b_{ij} X_j$$ (5.5.15)

in which $a_i$ and $b_{ij} = -b_{ji}$ are constants. If displacements are prescribed at three noncolinear points on the boundary, then $w_i = 0$ and the solution of the given boundary value problem is unique. However, if surface tractions are prescribed on all of the boundary, then different solutions of the same boundary value problem can differ at most by a rigid body motion. Even though the two displacement fields differ by a rigid body motion, the stress fields and the strain fields obtained from such displacement fields are identical.