This is the end of a PDE (heat equation in 2D) I am trying to solve with bounds from $0 < x < L$ and $0 < y < H$. It is a Newmann condition problem (i.e. all derivatives of $x$ and $y$ at the end points are $0$). Actually I am not sure if that is a Newmann condition so that would be my first question. Here is my end solution

$$u(x,y,t) = \sum _{ m=0 }^{ \infty }{ \sum _{ n=0 }^{ \infty }{ A_{ m,n }\cos \left( \frac { m\pi x }{ L } \right) } \cos \left( \frac { n\pi y }{ H } \right) e^{\lambda k t} } $$ with the initial condition $u(x,y,0) = F(x,y)$ for some known function $F(x,y)$ and $\lambda > 0 $ (I don't think the expotential term matters). My second question is: can I split up the coefficients and bring out the constant term, i.e. write the solution like this $$u(x,y,t) = A_0 + \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ A_{ m,n }\cos \left( \frac { m\pi x }{ L } \right) } \cos\left( \frac { n\pi y }{ H } \right) e^{\lambda k t} }. $$ You can then invoke the initial condition and are left with $$u(x,y,0) =F(x,y) = A_0 + \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ A_{ m,n }\cos \left( \frac { m\pi x }{ L } \right) } \cos\left( \frac { n\pi y }{ H } \right) }. $$ Then how would I go about finding $A_0$ and $A_{m,n}$?

I know how to do this for a sin series. Here I can show you my steps: \begin{align} F(x,y)&=\sum_{m=1}^\infty{\sin{(m\pi x)}}\sum_{n=1}^\infty{\sin{(n\pi y)}} A_{m,n}\\ &=\sum_{m=1}^\infty{\left( \sum_{n=1}^\infty{ A_{m,n} \sin{(n\pi y)}} \right)}\sin{(m\pi x)}\\\ &=\sum_{m=1}^\infty{B_m}\sin{(m\pi x)} \end{align} and so $B_m = \frac{2}{L}\int_0^L F(x,y) \sin{(m\pi x)} dx$ and so you plug this back in and then you get a second integral (a double integral infact in terms of $x$ and $y$) of the coefficient $A_{m,n}$.

Edit: The answer I got for $A_{m,n} $ is as follows. Is this correct? $$ A_{m,n} = \frac{4}{HL}\int_0^H \int_0^L F(x,y) \cos{\left(\frac{m\pi x}{L}\right)} \cos{\left(\frac{n \pi y}{H}\right)}\, dx dy$$

  • $\begingroup$ I believe you do have Neumann conditions. You definitely can split the term $m = 0$ and $n = 0$ from the sum, but then you still need to keep terms corresponding to $m = 0$ and $n \ne 0$, and also terms corresponding to $m \ne 0$ and $n = 0$. Your sum should be $$ u(x, y, t) = A_{0,0} + \sum_{m=1}^\infty A_{m,0}\cos\left(\frac{m\pi x}L\right) e^{\lambda_{m,0}t} + \sum_{n=1}^\infty A_{0,n}\cos\left(\frac{n\pi y}H\right)e^{\lambda_{0,n}t} + \sum_{m=1}^\infty\sum_{n=1}^\infty \cos\left(\frac{m\pi x}L\right)\cos\left(\frac{n\pi y}H\right)e^{-\lambda_{m,n}t}$$ I can't really recommend doing this. $\endgroup$ – Tunococ Aug 23 '12 at 21:32
  • $\begingroup$ BTW, Fourier coefficients for $\cos$ are similar to $\sin$. The only special term is the constant term. I'm not sure what you're confused about. $\endgroup$ – Tunococ Aug 23 '12 at 21:35

$$F(x,y) = A_0 + \sum_{m=1}^{\infty}{\sum_{n=1}^{\infty}{ A_{m,n}\cos \left( \frac { m\pi x }{ L } \right) } \cos\left( \frac { n\pi y }{ H } \right) }$$

Multiply both sides through by $\cos(m'\pi x/L)\cos(n'\pi x/L)$ for arbitrary but fixed positive integers $m'$, $n'$ and integrate over $0<x<L$ and $0<y<H$. The orthogonality if these cosine functions on these intervals will cause all the terms in the series to vanish except when $m=m'$ and $n=n'$, resulting in $$ \small \int_0^H\int_0^L F(x,y)\cos(m'\pi x/L)\cos(n'\pi y/H)\,dx\,dy=A_{n'm'}\int_0^H\int_0^L \cos^2(m'\pi x/L)\cos^2(n'\pi y/H)\,dx\,dy $$ which implies $$ A_{n'm'}=\frac{\int_0^H\int_0^L F(x,y)\cos(m'\pi x/L)\cos(n'\pi y/H)\,dx\,dy}{\int_0^H\int_0^L \cos^2(m'\pi x/L)\cos^2(n'\pi y/H)\,dx\,dy}. $$ Since $n'$ and $m'$ were arbitrary dummy indices, we can change them to $n$ and $m$. Also, the integral in the denominator equals ${LH\over 4}$, so we get $$ A_{nm}={4\over LH}\int_0^H\int_0^L F(x,y)\cos(m\pi x/L)\cos(n\pi y/H)\,dx\,dy. $$


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