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MathidiaMathematics Academy
Partial differential equations II
Lesson 28 of 75
Standard practice

Weak solutions

Objective. Derive weak formulations by integration by parts.

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  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

Weak solutions extend the notion of solving a PDE to functions that need not be differentiable in the classical sense, by requiring an integral identity to hold against all smooth test functions. Starting from a PDE like −u'' = f, multiply by a test function φ vanishing at the boundary and integrate by parts to move a derivative off u onto φ, giving ∫u'φ' dx = ∫fφ dx, which makes sense even if u' has jumps. A function u satisfying this identity for every admissible φ is called a weak solution, and it coincides with the classical solution whenever the classical solution exists and is smooth enough. Example: a piecewise-linear u with a kink can be a weak solution of −u'' = f even though u'' fails to exist classically at the kink.

Worked example

Derive the weak formulation of −u''(x) = f(x) on (0,1) with u(0) = u(1) = 0, for test functions φ vanishing at both endpoints.

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  1. 1Multiply the equation by φ and integrate: −∫₀¹ u''φ dx = ∫₀¹ fφ dx.
  2. 2Integrate the left side by parts: −∫₀¹ u''φ dx = −[u'φ]₀¹ + ∫₀¹ u'φ' dx.
  3. 3Since φ(0) = φ(1) = 0, the boundary term −[u'φ]₀¹ vanishes identically.
  4. 4The weak formulation is therefore ∫₀¹ u'φ' dx = ∫₀¹ fφ dx for all such φ.

Answer: ∫₀¹ u′φ′ dx = ∫₀¹ fφ dx

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Common mistake. Students often integrate by parts and drop the boundary terms without checking that the test function actually vanishes there; if φ does not vanish at the boundary, the boundary contribution must stay in the weak formulation.

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