Certainly! Let's go through the step-by-step derivation and explanation of each identity from the list. I'll focus on clarity and show the reasoning using vector calculus rules. --- ### 1. **\(\text{div}(\Phi \mathbf{u}) = \Phi\, \text{div}\,\mathbf{u} + \mathbf{u} \cdot \nabla \Phi\)** **Explanation:** Apply the **product rule** for divergence (Leibniz rule): \[ \text{div}(\Phi \mathbf{u}) = \nabla \cdot (\Phi \mathbf{u}) \] Expand using the product rule for divergence: \[ = (\nabla \Phi) \cdot \mathbf{u} + \Phi (\nabla \cdot \mathbf{u}) \] Or, rearranged: \[ = \Phi\, \text{div}\,\mathbf{u} + \mathbf{u} \cdot \nabla \Phi \] **This is the identity (1.288).** --- ### 2. **\(\text{div}(\Phi \mathbf{A}) = \Phi\, \text{div}\,\mathbf{A} + \mathbf{A} \cdot \nabla \Phi\)** **Explanation:** \(\mathbf{A}\) is a tensor, but the rule is similar: \[ \text{div}(\Phi \mathbf{A}) = \nabla \cdot (\Phi \mathbf{A}) = (\nabla \Phi) \cdot \mathbf{A} + \Phi (\nabla \cdot \mathbf{A}) \] Which is: \[ = \Phi\, \text{div}\,\mathbf{A} + \mathbf{A} \cdot \nabla \Phi \] **This is identity (1.289).** --- ### 3. **\(\text{div}(\mathbf{A}^T \mathbf{u}) = \text{div}\,\mathbf{A} \cdot \mathbf{u} + \mathbf{A} : \text{grad}\, \mathbf{u}\)** **Explanation:** Let \(\mathbf{A}\) be a tensor and \(\mathbf{u}\) a vector. Recall: \(\text{div}(\mathbf{A}^T \mathbf{u}) = \nabla \cdot (\mathbf{A}^T \mathbf{u})\). By product rule for tensors and vectors: \[ \text{div}(\mathbf{A}^T \mathbf{u}) = (\nabla \cdot \mathbf{A}) \cdot \mathbf{u} + \mathbf{A} : \nabla \mathbf{u} \] **This is identity (1.290).** --- ### 4. **\(\text{div}(\mathbf{A}\mathbf{B}) = \text{grad} \mathbf{A} : \mathbf{B} + \mathbf{A} \cdot \text{div} \mathbf{B}\)** **Explanation:** Product rule for divergence of tensor product: \[ \text{div}(\mathbf{A}\mathbf{B}) = (\nabla \mathbf{A}) : \mathbf{B} + \mathbf{A} \cdot (\nabla \cdot \mathbf{B}) \] **This is identity (1.291).** --- ### 5. **\(\text{div}(\mathbf{u} \times \mathbf{v}) = \mathbf{v} \cdot \text{curl} \mathbf{u} - \mathbf{u} \cdot \text{curl} \mathbf{v}\)** **Explanation:** Use vector calculus identity: \[ \nabla \cdot (\mathbf{u} \times \mathbf{v}) = \mathbf{v} \cdot (\nabla \times \mathbf{u}) - \mathbf{u} \cdot (\nabla \times \mathbf{v}) \] **This is identity (1.292).** --- ### 6. **\(\text{div}(\mathbf{u} \otimes \mathbf{v}) = (\nabla \mathbf{u}) \mathbf{v} + \mathbf{u} (\nabla \cdot \mathbf{v})\)** **Explanation:** The divergence of an outer product: \[ \nabla \cdot (\mathbf{u} \otimes \mathbf{v}) = (\nabla \mathbf{u}) \mathbf{v} + \mathbf{u} (\nabla \cdot \mathbf{v}) \] **This is identity (1.292).** --- ### 7. **\(\nabla (\Psi \Phi) = (\nabla \Phi) \Psi + \Phi \nabla \Psi\)** **Explanation:** This is the standard product rule for gradients: \[ \nabla (\Psi \Phi) = (\nabla \Psi) \Phi + \Psi (\nabla \Phi) \] So, as written: \[ = (\nabla \Phi) \Psi + \Phi \nabla \Psi \] **This is identity (1.293).** --- ### 8. **\(\nabla (\Phi \mathbf{u}) = \mathbf{u} \otimes \nabla \Phi + \Phi \nabla \mathbf{u}\)** **Explanation:** Product rule for gradients of vector fields: \[ \nabla (\Phi \mathbf{u}) = \mathbf{u} \otimes \nabla \Phi + \Phi \nabla \mathbf{u} \] **This is identity (1.293).** --- ### 9. **\(\nabla (\mathbf{u} \cdot \mathbf{v}) = (\nabla \mathbf{u})^T \mathbf{v} + (\nabla \mathbf{v})^T \mathbf{u}\)** **Explanation:** Gradient of a dot product: \[ \nabla (\mathbf{u} \cdot \mathbf{v}) = (\nabla \mathbf{u})^T \mathbf{v} + (\nabla \mathbf{v})^T \mathbf{u} \] **This is identity (1.293).** --- ### 10. **\(\text{curl}(\Phi \mathbf{u}) = (\nabla \Phi) \times \mathbf{u} + \Phi \nabla \times \mathbf{u}\)** **Explanation:** Product rule for curl: \[ \nabla \times (\Phi \mathbf{u}) = (\nabla \Phi) \times \mathbf{u} + \Phi (\nabla \times \mathbf{u}) \] **This is identity (1.294).** --- ### 11. **\(\text{curl}(\mathbf{u} \times \mathbf{v}) = \mathbf{u}\, \text{div}\,\mathbf{v} - \mathbf{v}\, \text{div}\,\mathbf{u} + (\mathbf{v} \cdot \nabla) \mathbf{u} - (\mathbf{u} \cdot \nabla) \mathbf{v}\)** **Explanation:** Curl of a cross product: \[ \nabla \times (\mathbf{u} \times \mathbf{v}) = \mathbf{u} (\nabla \cdot \mathbf{v}) - \mathbf{v} (\nabla \cdot \mathbf{u}) + (\mathbf{v} \cdot \nabla) \mathbf{u} - (\mathbf{u} \cdot \nabla) \mathbf{v} \] **This is identity (1.294).** --- ## **Summary Table** | Identity | Step/Explanation | Formula | |---|---|---| | 1 | Product rule, scalar & vector | \(\text{div}(\Phi \mathbf{u}) = \Phi\, \text{div}\,\mathbf{u} + \mathbf{u} \cdot \nabla \Phi\) | | 2 | Product rule, scalar & tensor | \(\text{div}(\Phi \mathbf{A}) = \Phi\, \text{div}\,\mathbf{A} + \mathbf{A} \cdot \nabla \Phi\) | | 3 | Product rule, tensor transpose & vector | \(\text{div}(\mathbf{A}^T \mathbf{u}) = \text{div}\,\mathbf{A} \cdot \mathbf{u} + \mathbf{A} : \text{grad}\, \mathbf{u}\) | | 4 | Product rule, tensors | \(\text{div}(\mathbf{A}\mathbf{B}) = \text{grad} \mathbf{A} : \mathbf{B} + \mathbf{A} \cdot \text{div} \mathbf{B}\) | | 5 | Divergence of cross product | \(\text{div}(\mathbf{u} \times \mathbf{v}) = \mathbf{v} \cdot \text{curl} \mathbf{u} - \mathbf{u} \cdot \text{curl} \mathbf{v}\) | | 6 | Divergence of tensor (outer) product | \(\text{div}(\mathbf{u} \otimes \mathbf{v}) = (\text{grad} \mathbf{u}) \mathbf{v} + \mathbf{u} \text{div} \mathbf{v}\) | | 7 | Gradient of scalar product | \(\nabla (\Psi \Phi) = (\nabla \Phi) \Psi + \Phi \nabla \Psi\) | | 8 | Gradient of scalar and vector | \(\nabla (\Phi \mathbf{u}) = \mathbf{u} \otimes \nabla \Phi + \Phi \nabla \mathbf{u}\) | | 9 | Gradient of dot product | \(\nabla (\mathbf{u} \cdot \mathbf{v}) = (\nabla \mathbf{u})^T \mathbf{v} + (\nabla \mathbf{v})^T \mathbf{u}\) | | 10 | Curl of scalar times vector | \(\text{curl}(\Phi \mathbf{u}) = (\nabla \Phi) \times \mathbf{u} + \Phi \text{curl} \mathbf{u}\) | | 11 | Curl of cross product | \(\text{curl}(\mathbf{u} \times \mathbf{v}) = \mathbf{u}\, \text{div}\,\mathbf{v} - \mathbf{v}\, \text{div}\,\mathbf{u} + (\mathbf{v} \cdot \nabla) \mathbf{u} - (\mathbf{u} \cdot \nabla) \mathbf{v}\) | --- If you'd like a full step-by-step breakdown for a specific one, let me know which one!