A weak solution satisfies an integrated identity in which derivatives are transferred to test functions. It can therefore exist with fewer classical derivatives.
For a partial differential equation whose operator is nonincreasing in the Hessian variable, a continuous viscosity subsolution satisfies whenever a smooth test function touches it from above. A viscosity supersolution satisfies the reverse inequality for tests touching from below. A viscosity solution satisfies both conditions. For , these conditions are respectively and . They make sense even when has no classical second derivatives.
Let a continuous function satisfy both viscosity inequalities for the Laplace operator. On any relatively compact ball, take its harmonic replacement with matching boundary values. If has a positive maximum, the function still lies below somewhere for sufficiently small positive , while matching it on the boundary. At an interior maximum of their difference it is a valid upper test, but its Laplacian is , a contradiction. The opposite perturbation excludes a negative minimum. Thus throughout the ball and is a smooth harmonic function. A locally defined test is extended near its touching point using a smooth cutoff function when global tests are required.
A weak-strong uniqueness principle says that a weak solution with the same data as a sufficiently regular classical solution agrees with that solution throughout its interval of regular existence, within the specified weak class. The conclusion is class-dependent: all bounded weak solutions of a prescribed linear transport field satisfy it, while nonlinear conservation laws require an admissibility condition such as an entropy solution condition.
For with homogeneous Dirichlet boundary conditions, smooth uniformly elliptic principal coefficients, and initial data , a space-time weak solution has zero lateral trace, trace , and satisfies
for every space-time test function with zero lateral trace and . The velocity initial condition is encoded by the last term; an arbitrary space-time function need not have an trace of its time derivative. The equation itself gives , so has a continuous representative.
A weak formulation multiplies a differential equation by a test function, integrates, and uses integration by parts to move derivatives away from the unknown function. Natural boundary conditions appear in the resulting boundary terms.
The Galerkin method seeks an approximate solution in a finite-dimensional subspace and requires the equation's residual to be orthogonal to that subspace. Uniform energy estimates and compactness can then produce a weak solution as the subspaces become dense.
The Petrov-Galerkin method chooses an approximate solution in a finite-dimensional trial vector space and tests a weak formulation against a possibly different test space . For a bilinear form and a linear functional , it imposes for every . Taking recovers the usual Galerkin method. Different test functions can alter stability and accuracy; the Galerkin orthogonality is then against the test space and need not be an orthogonal projection in an energy norm.
A Fourier truncation retains a finite set of boundary-compatible Fourier modes and projects the PDE onto their span by the Galerkin method. Nonlinear products can generate unretained harmonics, so their coefficients are discarded by projection rather than declared zero pointwise. Agreement with a perturbation expansion through cubic order requires retention of every quadratically forced mode that feeds back on the critical mode at cubic order.
// Destination: dynamical-systems.bigb
If embeds compactly into and embeds continuously into , then boundedness in together with a suitable time-derivative bound in makes a family relatively compact in . A standard case is
If and for a dense continuous embedding and some , then has a representative in . Indeed, its pairing with each element of a dense subset of is absolutely continuous, and the uniform bound extends continuity to every element of .
If is weakly continuous and an energy equality makes continuous, then is strongly continuous. This follows from the Radon-Riesz theorem applied whenever .

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