Question

For u(x,y) defined on the square domain of 0 ≤ x ≤ 1 and 0 ≤ y ≤1 ,

solve the PDE ² /x² + ² /y² = 0 , with the boundary
conditions, (i) u(0, y) = 0 (ii) u(1, y)

Answer 1

To solve the given PDE, you can use the method of separation of variables. Assume the solution is of the form u(x, y) = X(x)Y(y). Substituting this into the PDE gives X''(x)/X(x) = -Y''(y)/Y(y). Solving the resulting ordinary differential equations and applying the boundary conditions yields the solution u(x, y) = ∑(Aₙ sin(nπx)e^(nπy) + Bₙ cos(nπx)e^(nπy)), where Aₙ and Bₙ are constants.

Step-by-step explanation:

To solve the given PDE 2⁄x² + 2⁄y² = 0, we can use the method of separation of variables. Assume the solution is of the form u(x, y) = X(x)Y(y). Substituting this into the PDE gives X''(x)/X(x) = -Y''(y)/Y(y). Since the left side depends only on x and the right side depends only on y, the only way for this equation to hold true is if both sides are equal to a constant.

Letting the constant be -λ², we have X''(x)/X(x) = λ² and Y''(y)/Y(y) = λ². Solving these ordinary differential equations separately gives X(x) = sin(λx) or cos(λx) and Y(y) = Ae^(λy) + Be^(-λy) (where A and B are constants).

Applying the boundary conditions u(0, y) = 0 and u(1, y) = 0, we find that X(0) = 0 and X(1) = 0. This leads to the condition sin(λ) = 0, which implies λ = nπ (where n is an integer). Therefore, the solution to the PDE is u(x, y) = ∑(Aₙ sin(nπx)e^(nπy) + Bₙ cos(nπx)e^(nπy)), where Aₙ and Bₙ are constants.