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bob012345

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- I'm a bit rusty and just want a set of eyes to check me on this solution of the Helmholtz equation for specific given boundary conditions in a defined region of space. Thanks.

So given the Helmholtz equation $$\nabla^2 u(x,y,z) + k^2u(x,y,z)=0$$ we do the separation of variables $$u=u_x(x)u_y(y)u_z(z)= u_xu_yu_z$$ and ##k^2 = k_x^2 + k_y^2 +k_z^2## giving three separate equations; $$\nabla^2_x u_x+ k_x^2 u_x=0$$ $$\nabla^2_y u_y+ k_y^2 u_y=0$$ $$\nabla^2_z u_z+ k_z^2 u_z=0$$ where the solutions are

$$u_x(x)= A_x\sin(k_x x) + B_x\cos(k_x x)$$

$$u_y(y)= A_y\sin(k_y y) + B_y\cos(k_y y)$$

$$u_z(z)= A_z\sin(k_z z) + B_z\cos(k_z z)$$

The given boundary conditions are ##u(x,y,0)=\sin(\pi x)\cos(\pi y)## and ##u(x,y,1)=0## in the region ##0<=x,y<=1##

So I have

$$\sin(\pi x)\cos(\pi y)=(A_x\sin(k_x x) + B_x\cos(k_x x))(A_y\sin(k_y y) + B_y\cos(k_y y))(A_z\sin(k_z 0) + B_z\cos(k_z 0))$$ setting ##B_x,A_y=0## we have

$$sin(\pi x)\cos(\pi y)=A_x\sin(k_x x)B_y\cos(k_y y)B_z$$ forcing the product of the constants=1 and ##k_x,k_y=\pi## Making ##u_x=A_x\sin(\pi x)## and ##u_y=B_y\cos(\pi y)## Then ##u(x,y,1)=0## we have

$$0=A_x\sin(\pi x)B_y\cos(\pi y)(A_z\sin(k_z) +B_z\cos(k_z))$$ but here is where I see several solutions because there aren't enough conditions to completely specify the solution. One solution would be setting ##A_z=0## and ##k_z=\frac{n\pi}{2}## giving ##u_z=B_z\cos(\frac{n\pi z}{2})## then there are infinite solutions of the form

$$u(x,y,z)_n= C_n\sin(\pi x)\cos(\pi y) \cos(\frac{n\pi z}{2})$$ and the most general solution is an infinite sum.

But also, if we set $$A_z=-B_z\frac{\cos(k_z)}{\sin(k_z)}$$ making

$$u_z=-B_z\frac{\cos(k_z)}{\sin(k_z)}\sin(k_z z) + B_z\cos(k_z z)$$ where ##k_z\ne n\pi## also meets the boundary conditions.

$$u_x(x)= A_x\sin(k_x x) + B_x\cos(k_x x)$$

$$u_y(y)= A_y\sin(k_y y) + B_y\cos(k_y y)$$

$$u_z(z)= A_z\sin(k_z z) + B_z\cos(k_z z)$$

The given boundary conditions are ##u(x,y,0)=\sin(\pi x)\cos(\pi y)## and ##u(x,y,1)=0## in the region ##0<=x,y<=1##

So I have

$$\sin(\pi x)\cos(\pi y)=(A_x\sin(k_x x) + B_x\cos(k_x x))(A_y\sin(k_y y) + B_y\cos(k_y y))(A_z\sin(k_z 0) + B_z\cos(k_z 0))$$ setting ##B_x,A_y=0## we have

$$sin(\pi x)\cos(\pi y)=A_x\sin(k_x x)B_y\cos(k_y y)B_z$$ forcing the product of the constants=1 and ##k_x,k_y=\pi## Making ##u_x=A_x\sin(\pi x)## and ##u_y=B_y\cos(\pi y)## Then ##u(x,y,1)=0## we have

$$0=A_x\sin(\pi x)B_y\cos(\pi y)(A_z\sin(k_z) +B_z\cos(k_z))$$ but here is where I see several solutions because there aren't enough conditions to completely specify the solution. One solution would be setting ##A_z=0## and ##k_z=\frac{n\pi}{2}## giving ##u_z=B_z\cos(\frac{n\pi z}{2})## then there are infinite solutions of the form

$$u(x,y,z)_n= C_n\sin(\pi x)\cos(\pi y) \cos(\frac{n\pi z}{2})$$ and the most general solution is an infinite sum.

But also, if we set $$A_z=-B_z\frac{\cos(k_z)}{\sin(k_z)}$$ making

$$u_z=-B_z\frac{\cos(k_z)}{\sin(k_z)}\sin(k_z z) + B_z\cos(k_z z)$$ where ##k_z\ne n\pi## also meets the boundary conditions.

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