Let's analyze whether the given displacements can be a solution to a linear elasticity problem for isotropic materials with **no body force**. ## **Step 1: Displacement Components** Given: \[ u = c(x^2 + 10) \] \[ v = 2cyz \] \[ w = c(-xy + z^2) \] where \(c\) is a constant. --- ## **Step 2: Check Compatibility with Linear Elasticity Equilibrium** For isotropic materials, **with no body force**, the *Navier’s equations* (equilibrium equations) are: \[ \mu \nabla^2 \mathbf{u} + (\lambda + \mu) \nabla(\nabla \cdot \mathbf{u}) = 0 \] where \(\lambda\) and \(\mu\) are Lamé's constants. This means, for each displacement component: \[ \nabla^2 u_i + \frac{\lambda + \mu}{\mu} \frac{\partial}{\partial x_i} (\nabla \cdot \mathbf{u}) = 0 \] --- ## **Step 3: Compute \(\nabla \cdot \mathbf{u}\) (Divergence of Displacement)** \[ \nabla \cdot \mathbf{u} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} \] Compute each: - \(\frac{\partial u}{\partial x} = c \cdot 2x = 2cx\) - \(\frac{\partial v}{\partial y} = 2cz\) - \(\frac{\partial w}{\partial z} = c \cdot 2z = 2cz\) So, \[ \nabla \cdot \mathbf{u} = 2c x + 2c z + 2c z = 2c x + 4c z \] --- ## **Step 4: Calculate Laplacians** Calculate \(\nabla^2 u\), \(\nabla^2 v\), and \(\nabla^2 w\): ### For \(u\): \[ u = c(x^2 + 10) \] \[ \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \] - \(\frac{\partial^2 u}{\partial x^2} = c \cdot 2 = 2c\) - \(\frac{\partial^2 u}{\partial y^2} = 0\) - \(\frac{\partial^2 u}{\partial z^2} = 0\) So, \[ \nabla^2 u = 2c \] ### For \(v\): \[ v = 2c y z \] - \(\frac{\partial^2 v}{\partial x^2} = 0\) - \(\frac{\partial^2 v}{\partial y^2} = 0\) - \(\frac{\partial^2 v}{\partial z^2} = 0\) So, \[ \nabla^2 v = 0 \] ### For \(w\): \[ w = c(-x y + z^2) \] - \(\frac{\partial^2 w}{\partial x^2} = 0\) - \(\frac{\partial^2 w}{\partial y^2} = 0\) - \(\frac{\partial^2 w}{\partial z^2} = c \cdot 2 = 2c\) So, \[ \nabla^2 w = 2c \] --- ## **Step 5: Check Equilibrium Equations** Each component must satisfy: \[ \mu \nabla^2 u_i + (\lambda + \mu) \frac{\partial}{\partial x_i} (\nabla \cdot \mathbf{u}) = 0 \] ### For \(u\) (\(x\)-component): \[ \mu \nabla^2 u + (\lambda + \mu) \frac{\partial}{\partial x} (\nabla \cdot \mathbf{u}) = 0 \] \[ \mu (2c) + (\lambda + \mu) \frac{\partial}{\partial x}(2cx + 4cz) = 0 \] \[ \frac{\partial}{\partial x}(2c x + 4c z) = 2c \] So, \[ 2c\mu + 2c(\lambda + \mu) = 2c(\lambda + 2\mu) \] Set this to zero: \[ 2c(\lambda + 2\mu) = 0 \implies c = 0 \text{ (unless all material constants are zero, which is unphysical)} \] ### For \(v\) (\(y\)-component): \[ \mu (0) + (\lambda + \mu) \frac{\partial}{\partial y}(2cx + 4cz) = 0 \] \[ \frac{\partial}{\partial y}(2c x + 4c z) = 0 \] So, \[ 0 = 0 \] (Always satisfied.) ### For \(w\) (\(z\)-component): \[ \mu (2c) + (\lambda + \mu) \frac{\partial}{\partial z}(2c x + 4c z) = 0 \] \[ \frac{\partial}{\partial z}(2c x + 4c z) = 4c \] So, \[ 2c\mu + 4c(\lambda + \mu) = 2c\mu + 4c\lambda + 4c\mu = 4c\lambda + 6c\mu \] Set to zero: \[ 2c\mu + 4c(\lambda + \mu) = 0 \implies 2\mu + 4\lambda + 4\mu = 4\lambda + 6\mu \] So, \[ c(4\lambda + 6\mu) = 0 \implies c = 0 \text{ (again, unless all constants are zero)} \] --- ## **Step 6: Conclusion** The only way for these displacement fields to satisfy the equilibrium equations **for all \(x, y, z\)** (with nonzero Lamé constants) is if \(c = 0\), i.e., **zero displacement everywhere**. --- ## **Final Answer** ### **No, the given displacement fields cannot be a solution to a linear elasticity problem for isotropic materials with no body force, unless \(c = 0\) (trivial solution).** --- **Explanation:** The displacement must satisfy the equilibrium equations of linear elasticity. For the given fields, the equilibrium can only be satisfied for the trivial case \(c = 0\). For any nonzero \(c\), the equations are not satisfied.