The periodic ramp is an odd function, so $a_0=a_n=0$. Integration by parts givesThus the Fourier series of a periodic sawtooth isThe periodic function is smooth at $x=\pi /2$, so the Dirichlet-Jordan convergence theorem permits evaluation there. Even-index terms vanish and odd-index sines alternate in sign, giving

Write $h_n=a_n\cos (n\pi x/L)+b_n\sin (n\pi x/L)$. For the partial sum $S_N=a_0/2+{\sum }_{n=1}^N{h}_n$, ordinary multiplication givesOn $[-L,L]$, the integrals of every nonconstant sine or cosine vanish. Fourier orthogonality further gives $\int \cos (m\pi x/L)\cos (n\pi x/L)dx=L{\delta }_{mn}$ and the same identity for two sines, while every sine-cosine integral is zero. ThusFor square-integrable $f$, its Fourier partial sums converge to $f$ in L2 space, so their squared norms converge to the squared norm of $f$. Passing to the limit proves Parseval's identity:Working with finite partial sums before taking the norm limit avoids an unjustified interchange of an infinite double series and integration.

Every harmonic in this Fourier series has period $2L$. Since the first-harmonic coefficient is nonzero, the fundamental period cannot be smaller: translating by a period $T$ must multiply that nonzero complex coefficient by $e^{i\pi T/L}=1$. Thus, taking $L>0$,

Put $\xi =x-vt$ and $U(\xi )=A{\mathrm{sech}}^2\xi$. For $A\ne 0$, differentiation gives $U^{\prime \prime }=4U-6U^2/A$ and therefore $U^{(3)}=4U^{\prime }-(12/A)U{U}^{\prime }$. The Korteweg-De Vries equation reduces toSince $U$ is nonconstant, both coefficients must vanish. Thus the nonzero solitary wave isThere is also the trivial solution $A=0$, for which $v$ has no effect. The negative amplitude follows from the minus sign of the nonlinear term; it agrees with the sech-squared solitary wave of the KdV equation.

The repeated characteristic root is $-1$, matching the exponential forcing. Put $y=e^{-x}w$; then $y^{\prime \prime }+2y^{\prime }+y=e^{-x}{w}^{\prime \prime }$, so $w^{\prime \prime }=8$. The zero initial conditions give $w(0)=w^{\prime }(0)=0$, henceThe extra quadratic factor is the resonance associated with the double root.

The complementary characteristic roots are $-1,-2$. Substituting $A\sin x+B\cos x$ into the differential operator gives $(A-3B)\sin x+(3A+B)\cos x$, so a particular solution is $\sin x-3\cos x$. The initial conditions then require $C+D=3$ and $C+2D=1$, giving

The characteristic equation $(r-3)(r-4)=0$ gives complementary terms $C{e}^{3x}+D{e}^{4x}$. A linear particular solution is $12x+7$, since its substitution gives $144x$. The initial conditions give $C+D=-7$ and $3C+4D=-12$, hence