Find a particular solution v(x,y) to the following Laplace equation and use it to transform the following equation into one that is homogeneous - but has non-zero boundary conditions. Then solve the resulting equation and use it to obtain the solution for u.
Uxx+Uyy=4 0
U(x,0)=0
U(x,H)=0
Uy(0,y)=0
Uy(L,y)=0
Just all around stumped. Having issues figuring out how to get rid of the 4? Or get a particular solution to equal that.
Uxx+Uyy=4 0
U(x,H)=0
Uy(0,y)=0
Uy(L,y)=0
Just all around stumped. Having issues figuring out how to get rid of the 4? Or get a particular solution to equal that.
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Let V(x,y) = U(x,y) - (x^2 + y^2). Then Vx = Ux - 2x, Vxx = Uxx - 2, and by symmetry Vyy = Uyy - 2. Therefore Vxx + Vyy = (Uxx - 2) + (Vyy - 2) = Uxx + Uyy - 4 = 0. Now
U(x,0) = 0 implies V(x,0) = -x^2
U(x,H) = 0 implies V(x,H) = -(x^2 + H^2)
Uy(0,y) = 0 implies Vy(0,y) = -2y
Uy(L,y) = 0 implies Uy(L,y) = -2y.
So we have the equivalent problem
Vxx + Vyy = 0, 0
V(x,0) = -x^2
V(x,H) = -(x^2 + H^2)
Vy(0,y) = -2y
Vy(L,y) = -2y.
Try separation of variables with this problem first. Once V(x,y) has been found, your solution will be U(x,y) = V(x,y) + (x^2 + y^2).
U(x,0) = 0 implies V(x,0) = -x^2
U(x,H) = 0 implies V(x,H) = -(x^2 + H^2)
Uy(0,y) = 0 implies Vy(0,y) = -2y
Uy(L,y) = 0 implies Uy(L,y) = -2y.
So we have the equivalent problem
Vxx + Vyy = 0, 0
V(x,H) = -(x^2 + H^2)
Vy(0,y) = -2y
Vy(L,y) = -2y.
Try separation of variables with this problem first. Once V(x,y) has been found, your solution will be U(x,y) = V(x,y) + (x^2 + y^2).