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Suppose $-u^{\prime \prime }+u+q(u)=f$ and $-v^{\prime \prime }+v+q(v)=g$ on a bounded interval, the solutions have the same Dirichlet boundary conditions, and $q$ is a nondecreasing function. Multiplying the difference equation by $w=u-v$ and using integration by parts givesThe nonlinear integral is nonnegative. Young's inequality for products gives the displayed energy estimate. It proves uniqueness and continuous dependence on forcing. The same argument holds for nodal values when integration by parts is replaced by discrete summation by parts.

The conjugate-power constrained product maximum proves Young's inequality for products by rescaling. For nonnegative $x,y$ and conjugate exponents $p,q>1$, let $A=x^p/p+y^q/q$. If $A=0$, both variables vanish. Otherwise the displayed normalized variables satisfy $X^p/p+Y^q/q=1$, hence $XY\le 1$. Since $1/p+1/q=1$, this means $xy/A\le 1$, proving $xy\le x^p/p+y^q/q$. Equality in the constrained maximum requires $X=Y=1$, equivalent to $x^p=y^q$; the zero case gives the same equality criterion. This approach simultaneously gives the sharp constant and every equality case.

Write $w=u-v$ and $r=f-g$. Subtraction of the two boundary value problems gives $-w^{\prime \prime }+w+u^3-v^3=r$, with $w(0)=w(1)=0$. Multiply by $w$ and apply integration by parts:Consequently the required energy identity isThe nonlinear term is nonnegative pointwise, sinceThis is the monotonicity energy estimate for a semilinear Dirichlet problem for the monotone function $s\mapsto s^3$. Apply Young's inequality for products, $rw\le (r^2+w^2)/2$, inside the last integral and discard the nonnegative nonlinear term. Moving $\frac{1}{2}{\int }_0^1{w}^{2}dx$ to the left givesIn particular, equal forcing gives $w=0$, proving uniqueness and a quantitative continuous dependence on forcing estimate for this boundary value problem.

Let $f_i=f(ih)$ and $g_i=g(ih)$, and define the scaled forward difference operator, backward difference operator and second-order central difference byThe two finite difference methods are the nonlinear systemswith the prescribed Dirichlet boundary conditions $U_0=V_0=A$ and $U_N=V_N=B$. These equations determine the interior nodal values.

For clarity distinguish the interior and edge inner products and their norms:The bracketed norm includes the last edge $i=N$. Let $W=U-V$, so $W_0=W_N=0$, and put $q_i=D_x^{-}{W}_i$ for $1\le i\le N$. Direct telescoping gives discrete summation by parts:Subtract the two nonlinear systems, multiply each equation by $h{W}_i$ and sum. The resulting discrete energy identity isThe nonlinear summands are nonnegative by the factorization in (a). Using Young's inequality for products at every node givesThis also proves uniqueness of each discrete solution. Existence can be seen by minimizing, with the endpoints fixed, the strictly convex energy functionalIt is a coercive function of its finitely many interior values, so it attains a minimum. Its first derivatives are $h(-D_x^{+}{D}_x^{-}{W}_i+W_i+W_i^3-f_i)$; thus its minimizer solves the scheme. The same argument applies to $g$.

Use $u_i=u(x_i)$ for the sampled exact solution, and let $e_i=U_i-u_i$. The Dirichlet boundary conditions imply $e_0=e_N=0$. The definition of the truncation error $\phi$ gives the exact discrete equationSubtracting it from the numerical equation therefore givesThe sign on the right is positive because ${\phi }_i$ was defined as the difference approximation minus the exact second derivative. Multiply by $h{e}_i$, sum over the interior nodes and apply discrete summation by parts:Again $(U_i^3-u_i^3)e_i=e_i^2(U_i^2+U_i{u}_i+u_i^2)\ge 0$. By Young's inequality for products, $(\phi ,e{)}_h\le \frac{1}{2}\Vert \phi {\Vert }_h^2+\frac{1}{2}\Vert e{\Vert }_h^2$. HenceNo fourth derivative is needed for this step: the assumed twice continuous differentiability suffices to define $\phi$ and obtain the discrete energy estimate.

Use a normalization proof of Young's inequality. For arbitrary $x,y\ge 0$, putIf $A=0$, both coordinates are zero and the desired inequality is equality. If $A>0$, defineThen $X^p/p+Y^q/q=1$. Part (c) gives $XY\le 1$, and the conjugate exponent identity gives $XY=xy/A$. Therefore Young's inequality for products isThe unique equality case in part (c) says $X=Y=1$. Translated back, and including the zero case, this gives the sharp conditionConversely, if both powers equal $r\ge 0$, then $xy=r^{1/p+1/q}=r$ and the right side is $r(1/p+1/q)=r$, directly verifying equality.