Seeking periodic solutions with Fourier

Question

So I am looking to find the solutions of : $$-u''+u=\cos(6x)$$ I did go through all the usual stages: Fourier transform, inversion theorem (since the resulting $\hat u \in L^1)$, then Jordan lemma + Residue theorem to calculate the inverse transform and as you would expect I end up with the solution in terms of a convolution.

The problem is that the answer I am provided is completely different, and the hint tells me to use the "method for researching periodic solutions through the Fourier series". I'm not sure what that entails, so I would really appreciate if someone could guide me through it.

Thank you.

Answer 1

The standard way is to try a solution in the form $$ u=A\sin(6x)+B\cos(6x). $$ Putting this into the equation you will get $$ 36A\sin(6x)+36B\cos(6x)+A\sin(6x)+B\cos(6x)=\cos(6x). $$ This will give you a system of two equations in two unknowns for $A$ and $B$.

For a Fourier series, one has to write $$ u(x)=\sum_{n=-\infty}^\infty a_ne^{i2\pi n x}. $$ By a direct substitution this gives $$ \sum_{n=-\infty}^\infty 2\pi n^2 a_ne^{i2\pi n x}+\sum_{n=-\infty}^\infty a_ne^{i2\pi n x}=\frac{1}{2}e^{-i6x}+\frac{1}{2}e^{i6x}. $$ We can invert this by multiplying with $e^{-i2\pi mx}$ and integrating between $0$ and $2\pi$ yielding $$ (2\pi m^2+1)a_m=\frac{1}{4\pi}\int_0^{2\pi}e^{-2i\pi mx}e^{6ix}+\frac{1}{4\pi}\int_0^{2\pi}e^{-2i\pi mx}e^{-6ix}. $$ This can be easily worked out to obtain $a_m$.

Answer 2

Let's try to find a particular solution of $$ -u''+u=\cos(6x) \tag1$$ Since $\cos(6x)$ has period $\pi/3$, let's see if there is a solution also with period $\pi/3$. This proposed solution would have Fourier series $$ u(x) = a_0 + \sum_{k=1}^\infty \big(a_k\cos(6kx)+b_k\sin(6kx)\big) $$ Now $$ u'(x) = \sum_{k=1}^\infty\big( -6ka_k\sin(6kx)+6kb_k\cos(6kx)\big) \\ u''(x) = \sum_{k=1}^\infty\big( -36ka_k\cos(6kx)-36kb_k\sin(6kx)\big) $$ Then we get $$ -u''(x)+u(x) = a_0+ \sum_{k=1}^\infty\big((a_k-36ka_k)\cos(6kx)+(-36kb_k+b_k)\sin(6kx)\big) $$ This is supposed to equal $\cos(6x)$, so we equate coefficients. $a_0 = 0$. For $k > 1$ we get $$ (a_k-36ka_k)=0,\quad (-36kb_k+b_k)=0 $$ so $a_k = 0, b_k = 0$. Finally, for $k=1$ we get $$ a_1-36a_1 = 1,\quad -36b_1+b_1=0 $$ so $a_1 = -1/35, b_1 = 0$. Our solution is $-\cos(6x)/35$.

Answer 3

First of all, thank you very much both of you for your precious input. I also found something myself that I'd like to add to the table, albeit I'm not entirely sure I'm using it correctly.

Clearly, $T=\pi/3$. Let's now consider $\xi_k=k{2\pi \over T}=6k$

We have that the characteristic polynomial of the DE is: $P(\lambda)=-\lambda^2+1$

Now we observe that $P(i\xi_k)=-(i\xi_k)^2+1=36k^2+1\not=0 $ for $k\in Z$

So the equation has a solution of period $\pi/3$ given by: $$\hat u_k={c_k \over 36k^2+1}$$with $c_k$ being the Fourier coefficients of $cos(6x)$ given by the general formula $$c_k={1\over T} \int_0^Tf(x)e^{-i{2\pi \over T}kx}dx=$$

$${3 \over \pi}\int_0^{\pi \over3}cos(6x)e^{-i6kx}dx={3 \over \pi}\int_0^{\pi \over3}{e^{i6x}+e^{-i6x} \over 2}e^{-i6kx}dx=$$

$$={3\over 2\pi}\int_0^{\pi \over 3}e^{i6x(1-k)}+e^{-i6x(1+k)}={3\over 2\pi}\biggl(\bigg[{e^{i6x(1-k)}\over i6(1-k)} \bigg]_0^{\pi\over 3}-\bigg[{e^{-i6x(1+k)}\over -i6(1+k)} \bigg]_0^{\pi\over 3}\bigg)...$$

and so on - I'm not gonna write all the math since it would take ages. In essence, that seems to be the Fourier transform of $cos(6x)$ using as variable the initial $\xi_k=k{2\pi \over T}=6k$. Eventually we reach: $$c_{k}={1\over 2} \rightarrow k=\pm1$$ $$c_{k}=0 \rightarrow k\not=\pm1$$ And we finally get to the solution given by $$u(x)={1 \over 2(36k^2+1)}(e^{6x}+e^{-6x})={1\over 37}cos(6x)$$

Again, I'm not entirely certain of the legality of this method, but at least the result seems to be correct so that's something...