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Use a signed power test for a Laplacian eigenfunction. For $\gamma \ge 2$, test the Dirichlet Laplacian eigenfunction equation with $|u{|}^{\gamma -2}u$. Its derivative is $(\gamma -1)|u{|}^{\gamma -2}Du$; at $\gamma =2$ the test function is just $u$. Integration by parts gives the exact energy estimateSet $v=|u{|}^{\gamma /2}$. It belongs to the zero-boundary Sobolev space and obeys the Sobolev chain rule:For $\gamma >2$, the power map is $C^1$ with bounded derivative on the bounded range of $u$. For $\gamma =2$, the absolute value map is Lipschitz, and $D|u|=\mathrm{sgn}(u)Du$ almost everywhere; on the zero set, use the fact that a gradient of a Sobolev function vanishes on a level set. These facts justify the identity even when $u$ changes sign.

Apply the given Sobolev inequality with the Sobolev conjugate exponent $2\kappa =2n/(n-2)$:Testing with a signed power raises the integrability exponent from $\gamma$ to $\gamma \kappa$. Absorbing the factor $1/4$ into $C(n)$ gives precisely the requested inequality.

Put $a(y)=(1-y^2{)}^2$. Multiply $u_{xx}={\partial }_y(a{u}_y)$ by $2u_x$ and integrate over $-1<y<1$. An integration by parts givesbecause $a(1)=a(-1)=0$. Therefore the energy estimate is in fact the conservation lawIf $u(0,y)=u_x(0,y)=0$, then differentiating the first identity in $y$ also gives $u_y(0,y)=0$, so the conserved nonnegative energy is zero. Hence $u_x=0$ and $(1-y^2)u_y=0$ throughout the open strip. There $|y|<1$, so both derivatives vanish; connectedness and the initial value now give

For a smooth function $g$ define its spherical meanThe Kirchhoff formuladefines a smooth solution of the three-dimensional wave equation for all positive and negative $t$ and has the prescribed data at $t=0$.

For uniqueness, apply the local energy estimate to the difference of two solutions on a backward light cone. Its energy at the cone tip is bounded by the zero initial energy on the cone base, so the difference and its derivatives vanish. Covering spacetime by such cones proves uniqueness among $C^2$ solutions.

Compact support is unnecessary for existence or uniqueness: the sphere in the Kirchhoff formula is compact for each $(t,x)$, so arbitrary smooth data suffice, and the cone-energy proof is local.

SetThen the forced equation is the abstract Cauchy problemFor $Z_0\in \mathcal{H}$ and $F\in C([0,T];\mathcal{H})$, a mild solution of an abstract Cauchy problem is a function $Z\in C([0,T];\mathcal{H})$ satisfying the variation-of-constants formula

Suppose $F(s)\in D(A)$ and both $F$ and $AF$ are continuous. If also $Z_0\in D(A)$, then the closedness of $A$ permits differentiation under the Bochner integral:Thus $Z\in C^1([0,\infty );\mathcal{H})$ and $Z^{\prime }=AZ+F$. The assumption $Z_0\in D(A)$ is necessary here: a unitary group has no smoothing, so the conditions on $F$ alone cannot make $U(t)Z_0$ differentiable for arbitrary $Z_0\in \mathcal{H}$.

For the resulting strong solution, skew symmetry of $A$ gives the energy estimateIntegration yields

PutThe assumed energy estimate for the linear wave equation and the first nonlinear estimate give, for $w\in X_{b,\tau }$,Choose $b\ge 2C_{0}D$ and then choose $\tau >0$ so small that $C_0\beta {\tau }^{1/2}{b}^2\le b/2$. Then $A$ maps the closed ball $X_{b,\tau }$ into itself.

For $w,\tilde{w}\in X_{b,\tau }$, the difference $Aw-A\tilde{w}$ solves the linear equation with zero Cauchy data and forcing $F(w)-F(\tilde{w})$. The second nonlinear estimate therefore givesShrinking $\tau$ once more makes the coefficient strictly smaller than one. Since $X_{\tau }$ is a Banach space and its closed ball is complete, $A$ is a contraction on $X_{b,\tau }$.

The first equation determines $u_m$ linearly from ${\theta }_m$:Substitution into the second equation gives an ordinary differential equation on the finite-dimensional space ${\tilde{H}}_m$. Its right-hand side is polynomial, and therefore locally Lipschitz continuous. The Picard-Lindelöf theorem supplies a unique local solution. The energy estimate in part (iii) bounds ${\theta }_m$ on every finite time interval, so the finite-dimensional continuation criterion rules out finite-time escape. The solution is consequently unique on every interval $[0,T]$.

For a smooth Dirichlet Laplacian eigenfunction, testing $-\Delta u=\lambda u$ with $|u{|}^{\gamma -2}u$ for $\gamma \ge 2$ gives the displayed energy estimate. The Sobolev chain rule for $v=|u{|}^{\gamma /2}$ then yields $\int |Dv{|}^2=\lambda {\gamma }^2/[4(\gamma -1)]\int |u{|}^{\gamma }$. The absolute value and the sign in the test function permit nodal changes. At $\gamma =2$, use the Lipschitz absolute-value chain rule and the fact that a gradient of a Sobolev function vanishes on a level set.

For a periodic or boundary-tangent divergence-free vector field $u$, integration by parts givesConsequently $\int (u\cdot \nabla f)f=0$, the cancellation underlying many fluid energy estimates.

For a smooth semilinear wave equation whose nonlinearity is a smooth function of the solution and its first derivatives, local Sobolev space theory gives a continuation criterion: a finite maximal existence time requires divergence of a sufficiently high solution norm. A bounded base norm $H^{s+1}\times H^s$, with $s>n/2$, allows restarting local existence. Higher regularity persists by differentiated energy estimates. Precise norms can be weakened for particular nonlinearities.

Positive $\epsilon$ adds diffusion to a one-dimensional scalar conservation law. For a sufficiently decaying smooth solution, the divergence structure gives $\frac{1}{2}\frac{d}{dt}\Vert u{\Vert }_2^2+\epsilon \Vert u_x{\Vert }_2^2=0$, so the $L^2$ energy estimate is uniform in $\epsilon$. If $|F^{\prime }|\le M$, testing against $-u_{xx}$ also yields $\Vert u_x(t){\Vert }_2^2\le e^{M^{2}t/\epsilon }\Vert u_x(0){\Vert }_2^2$. Thin travelling waves explain why a uniform gradient bound generally fails.

If $\square u=F$ with d'Alembert operator $\square =-{\partial }_t^2+\Delta$, then $\Vert \partial u(t){\Vert }_2\le \Vert \partial u(0){\Vert }_2+{\int }_0^t\Vert F(s){\Vert }_{2}ds$, where $\partial u=(u_t,\nabla u)$. Test the wave equation against $u_t$, use integration by parts, and apply the Cauchy-Schwarz inequality. The same energy estimate applies to a finite system of wave equations.