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Practice Problem #3

jradford

Here's number 3 from the practice problems. Could someone walk me through it?

Recall that the solution to Laplace's equation Δu = 0 on the unit square with boundary conditions u(x,0)=f(x),u(x,1)=g(x), and u(0,y)=u(1,y)=0 is u(x,y)=n=1(cnenπy+dnenπy)sin(nπx), where cn+dn are the Fourier sine coefficients of f and cnenπ+dnenπ are the Fourier sine coefficients of g. Use this to solve Laplace's equation on the unit square subject to the boundary conditions u(x,0)=1 and u(1,y)=u(x,1)=u(0,y)=0. Note: It is sufficient to specify cn and dn as the unique solutions to a pair of equations.

wbartoli

Here is my understanding of the problem. Can someone else confirm this? It's a lot less complicated if you don't try to solve for cn and dn explicitly. Given that cn and dn are the Fourier sine coefficients of f, we can set up an equation to solve for the coefficients as we have done before.
Instead of an=20f(x)sin(nπx) dx, the equation becomes for$=1$
cn+dn=210f(x)sin(nπx) dx
In this case, we have two coefficients that we are solving for, instead of one. Since f(x)=1,
cn+dn=210sin(nπx) dx
and
cn+dn=2nπcos(nπx)|10=2nπ(cos(nπ)1).
Thus,
cn+dn=2nπ((1)n1)







Using a similar process, set cnenπ+dnenπ equal to the Fourier computation for g, so
cnenπ+dnenπ=210g(x)sin(nπx) dx.
However, g(x)=0, so
cnenπ+dnenπ=0.


Without solving for cn and dn directly, the unique solutions cn and dn can be found in the following system of equations:
cn+dn=2nπ((1)n1)
cnenπ+dnenπ=0

jradford

Thanks so much, Massey! That makes perfect sense.

wbartoli

For everyone following along, this problem looks like an abbreviated version of another problem here on Discourse (http://www.marksmath2.org/t/relatively-easy-laplace-equation-on-a-square/96/), where @shill2 has solved for cn and dn.