 Solution

ID: past-exam-of-the-finite-element-methods-for-partial-differential-equations-course-of-the-university-of-oxford/2023/3/a/solution

ε (v) = \frac{1}{2} (\nabla v + (\nabla v)^{T})

and the divergence are linear in

v

. Consequently the given

a

a (v, v) = 2 μ ∥ ε (v) ∥_{0}^{2} + λ ∥ div v ∥_{0}^{2} \geq 0.

∥ \cdot ∥_{0}

here uses the Frobenius inner product pointwise and then the L2 norm. All expressions are finite for

v \in H^{1} (Ω; R^{3})

Expand the quadratic term at

θ v + (1 - θ) w

. The linearity of

ℓ

cancels its contribution to the difference, giving

θ J (v) + (1 - θ) J (w) - J (θ v + (1 - θ) w) = \frac{θ ( 1 - θ )}{2} (a (v, v) - 2 a (v, w) + a (w, w)) = \frac{θ ( 1 - θ )}{2} a (v - w, v - w) = θ (1 - θ) (μ ∥ ε (v - w) ∥_{0}^{2} + \frac{λ}{2} ∥ div (v - w) ∥_{0}^{2}) .

For

0 \leq θ \leq 1

, this is nonnegative. Therefore

J (θ v + (1 - θ) w) \leq θ J (v) + (1 - θ) J (w),

J

is a convex functional on the whole Sobolev space. No boundary condition is needed for this conclusion. The argument uses nonnegativity of the quadratic term; uniqueness on the zero-boundary Sobolev space will instead use the Korn inequality in part (b).

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