Let \(g\) be \(2\pi\)-periodic and continuous on \([-\pi,\pi]\), with Fourier series \[ \frac{a_0}{2}+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx). \] Let \(f\) be \(2\pi\)-periodic and satisfy \[ f''(x)+k f(x)=g(x), \] where \(k\ne n^2\) for \(n=1,2,3,\ldots\). Find the Fourier series of \(f\) and prove that it converges everywhere.
Solution
Write the Fourier coefficients of \(f\) as \[ \frac{A_0}{2}+\sum_{n=1}^\infty(A_n\cos nx+B_n\sin nx). \]
1. The Fourier coefficients satisfy\[ kA_0=a_0, \qquad (k-n^2)A_n=a_n, \qquad (k-n^2)B_n=b_n \]for every \(n\geq1\).
Since \(f''=g-kf\) is continuous and \(f\) is periodic, \(f\) is \(C^2\) and \(f'\) is also \(2\pi\)-periodic. Integrating the differential equation over \([-\pi,\pi]\) gives \[ \int_{-\pi}^{\pi}f''(x)\,dx=0, \] so comparison of the constant Fourier coefficients gives \(kA_0=a_0\).
For \(n\geq1\), two integrations by parts, using periodicity of \(f\) and \(f'\), give \[ \frac1\pi\int_{-\pi}^{\pi}f''(x)\cos(nx)\,dx=-n^2A_n \] and \[ \frac1\pi\int_{-\pi}^{\pi}f''(x)\sin(nx)\,dx=-n^2B_n. \] Taking the cosine and sine Fourier coefficients of \(f''+kf=g\) therefore yields \[ (k-n^2)A_n=a_n, \qquad (k-n^2)B_n=b_n. \]
2. If \(k\neq0\), the Fourier series of \(f\) is\[ \boxed{ \frac{a_0}{2k} +\sum_{n=1}^\infty \frac{a_n\cos(nx)+b_n\sin(nx)}{k-n^2} }. \]If \(k=0\), necessarily \(a_0=0\), and the Fourier series is\[ \boxed{ c-\sum_{n=1}^\infty \frac{a_n\cos(nx)+b_n\sin(nx)}{n^2} }, \qquad c\coloneqq\frac1{2\pi}\int_{-\pi}^{\pi}f(x)\,dx. \]
The hypothesis \(k\neq n^2\) for \(n\geq1\) makes every nonconstant denominator in step 1 nonzero. Thus \[ A_n=\frac{a_n}{k-n^2}, \qquad B_n=\frac{b_n}{k-n^2}. \] If \(k\neq0\), step 1 also gives \(A_0=a_0/k\), which yields the first displayed series.
If \(k=0\), the constant relation in step 1 gives \(a_0=0\), while \(A_0\) is not determined by the differential equation. Since the constant term in the Fourier series is the mean of \(f\), one has \(A_0/2=c\), giving the second displayed series.
3. In either case, the displayed Fourier series converges absolutely and uniformly on \(\RR\).
Let \(M=\max_{[-\pi,\pi]}\abs g\). The Fourier coefficients of \(g\) satisfy \[ \abs{a_n}\leq2M, \qquad \abs{b_n}\leq2M. \] For all sufficiently large \(n\), \[ \abs{k-n^2}\geq\frac{n^2}{2}. \] Hence, when \(k\neq0\), \[ \frac{\abs{a_n}+\abs{b_n}}{\abs{k-n^2}} \leq\frac{8M}{n^2} \] for all sufficiently large \(n\). When \(k=0\), the corresponding bound is \[ \frac{\abs{a_n}+\abs{b_n}}{n^2}\leq\frac{4M}{n^2}. \] Since \(\sum n^{-2}\) converges and \(\abs{\sin(nx)},\abs{\cos(nx)}\leq1\), the Weierstrass \(M\)-test gives absolute and uniform convergence.
4. The Fourier series converges everywhere to \(f\).
3. Uniform convergence permits termwise integration against \(1\), \(\cos(nx)\), and \(\sin(nx)\), so \(F\) has the Fourier coefficients displayed in step
2. These are exactly the Fourier coefficients \(A_n,B_n\) of \(f\) from step
1. Thus the continuous periodic function \(h=f-F\) has every Fourier coefficient equal to zero.
By Fejér’s theorem, the Cesàro means of the Fourier series of a continuous periodic function converge uniformly to that function. Every Cesàro mean of the Fourier series of \(h\) is identically zero, so \(h=0\). Hence \(F=f\) everywhere.
5. Q.E.D.
Step 2 gives the Fourier series, and steps <1>3–<1>4 prove its everywhere convergence to \(f\).