@Neb - I see that for second case ∇√φ = φ. Can this be useful ?
No.
∇²φ = ∂²φ/∂x² + ∂²φ/∂y² + ∂²φ/∂z²
∇√φ = ∂√φ/∂xi+ ∂√φ/∂yj + ∂√φ/∂zk
One is a scalar, the other is a vector. One involves first derivatives, the other involves second derivatives. There is no reason for any arbitrary scalar function φ that this would be the case.
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No.
∇²φ = ∂²φ/∂x² + ∂²φ/∂y² + ∂²φ/∂z²
∇√φ = ∂√φ/∂xi+ ∂√φ/∂yj + ∂√φ/∂zk
One is a scalar, the other is a vector. One involves first derivatives, the other involves second derivatives. There is no reason for any arbitrary scalar function φ that this would be the case.