Solving partial differential equation in 2d with 3 boundary conditionsConsider the following pde:−Δu=0in (0, π)×(0, 4) with boundary conditions:u(0,y)=u(π,y) for y∈(0,4)andu(x,0)=cos⁡xsin⁡x for x∈(0,π)andu(x,4)=cos⁡xsin⁡x+4 for x∈(0,π)

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Solving partial differential equation in 2d with 3 boundary conditionsConsider the following pde:−Δu=0in (0, π)×(0, 4) with boundary conditions:u(0,y)=u(π,y) for y∈(0,4)andu(x,0)=cos⁡xsin⁡x for x∈(0,π)andu(x,4)=cos⁡xsin⁡x+4 for x∈(0,π)

Jamie Maldonado · Accepted Answer

Step 1The periodic boundary condition tells you that the solution is&amp;nbsp;π&amp;nbsp;-periodic in x-direction, or at least can be interpreted as such if suitably extended outside the rectangle. This means that in that direction you get a Fourier series expansionu(x,y)=a0(y)2+∑ak(y)cos⁡(2kx)+bk(y)sin⁡(2kx).This gives the same result as the separation approach, but starting from a different direction.Comparing the conditions on the upper and lower boundaries gives&amp;nbsp;b1(0)=b1(4)=12,&amp;nbsp;&amp;nbsp;a0(4)=8&amp;nbsp;all other coefficients are zero at&amp;nbsp;y=0&amp;nbsp;and&amp;nbsp;y=4. Further,−Δu(x,y)=−a0&amp;nbsp;″(y)2+∑k[4k2ak(y)−ak&amp;nbsp;″(y)]cos⁡(2kx)+[4k2bk(y)−bk&amp;nbsp;″(y)]sin⁡(2kx).Now comparing coefficients and solving the simple DE givesa0(y)=2y,b1(y)=cosh(2(y−2))2cosh(4),all other coefficient functions being zero.

Solving partial differential equation in 2d with 3

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