1
$\begingroup$

I'm trying to find the $B_{nm}$ that satisfy: $$\sum_{m=0}^\infty \sum_{n=1}^\infty B_{nm}\sin(nx)\cos(my) = 1$$ for $0

Multiplying both sides by $\sin(n'x)\cos(m'y)$ and integrating, the LHS becomes:

$$ \\B_{nm}\int_0^\pi \sin(nx)\sin(n'x)dx\int_0^\pi \cos(my)\cos(m'y)dy =$$ $\frac{\pi^2}{4}B_{n'm'}\space$ for $\space m' \ne 0$

$\frac{\pi^2}{2}B_{n'0}\space$ for $\space m' = 0$

The RHS is: $$\int_0^\pi\sin(n'x)dx\int_0^\pi \cos(m'y)dy = 0$$ Therefore, $B_{nm} =0$

I can't think of another way to solve this problem. If I multiply through by $\sin(n'x)\sin(m'y)$ and integrate instead, I get:

$$ \\B_{nm}\int_0^\pi \sin(nx)\sin(n'x)dx\int_0^\pi \cos(my)\sin(m'y)dy =$$ $\sum_{n=1}^\infty\frac{\pi n((-1)^{n+m}-1)}{2(m^2-n^2)}B_{n'm'}\space$

which seems impossible to isolate $B_{nm}$

  • 0
    Are the $|B_{nm}|$ summable? – 2017-02-25

2 Answers 2

1

Partial answer:

If you restrict the $B_{mn}$ to be square summable the it is not possible to do what you want.

To see why, let $\phi(x) = \sum_{n_1 =0}^\infty \sum_{n_2 =1}^\infty B_{n_1 n_2} \sin (n_1 x_1) \cos (n_2 x_2)$ and suppose that the condition holds. (The issue is that $n_2$ starts at $1$.)

Then we have $\phi(x) = \operatorname{sgn} x_1$ ae. on $[-\pi,\pi]^2$ hence for $x_1 \neq 0$ and $m \ge 1$ we have $\int_{-\pi}^\pi \cos (mt) \phi((x_1,t)) dt = \int_{-\pi}^\pi \cos (mt) \operatorname{sgn} x_1 dt =0$ from which we get $\sum_{n_1 =0}^\infty B_{n_1 m} \sin (n_1 x_1) = 0$ for ae. $x_1$. It follows from this that $B_{n_1 m} = 0$ for all $n_1 \ge 0, m \ge 1$, which is a contradiction.

More directly, you could expand $\phi$ in terms of the basis formed from the orthogonal functions $x \mapsto b_1(n_1 x_1) b_2 (n_2 x_2)$ where $b_1,b_2 \in \{\cos,\sin \}$ and note that the coefficients corresponding to $n_2=0$ are not all zero.

0

Hint: Have another look at the $n'=1$ term on the right-hand side. And then look at the other odd-$n'$ terms.

  • 0
    I'm not sure that I follow your hint, but I believe I see a way to do it. If $m' = 0$, then $B_{n'0} = \frac{4}{n \pi}$ for odd $n$, and $B_{n'0} = 0$ for even $n$. So the answer becomes: $u(x,y)= \sum_{n=1}^\infty \frac{4}{\pi} \frac{1}{2n-1} \sin((2n - 1)x)$ – 2017-02-25
  • 0
    Yes: when $n$ is odd, there are fewer arches of the sine below the axis than above on $[0,\pi]$, so the integral cannot be zero. On the other hand, the $y$ part of the series can be written down immediately because Fourier series are unique, and $m=0$ gives the $1$ by itself. – 2017-02-25
  • 0
    But if $m=0$, the solution no longer has a y-dependence. I was given boundary conditions for a PDE that demand the solution have a y-dependence. Is there another way around that, or am I missing something? – 2017-02-25
  • 0
    Ah, you didn't specify that. I'll have another look. – 2017-02-25
  • 0
    @infinitylord What boundary conditions do you have? – 2017-02-25
  • 0
    Actually, I just realized they are satisfied! The boundary conditions are $u(0,y) = u(\pi,y) = 0$ and $u_y(x, 0) = u_y(x, \pi) = 0$. The $y $ boundary conditions are satisfied automatically since $u_y(x,y) = 0$ – 2017-02-25
  • 0
    Good, then you can try starting a solution that only depends on $x$, which makes the calculations a bit simpler. – 2017-02-25