Problem $3.($Please don't use AI$,$ I can tell and it doesnt give the right answer$)$ Given a bounded region $D$ in $R^3,$ we can consider the vector space

$L^2(D)=\{F$:$DC|{}_D|F{|}^{2}dV<mi\; id="EditorElement\_557233"></mi><mi\; id="EditorElement\_557235">\}</mi>$

with its standard bracket

$($:$F|G$:$)=\frac{{}_D}{b}ar(F(x,y,z))G(x,y,z)dV$

a$.$ Show that for functions $G$ and $F$ we have

Gvec$(grad{)}^{2}F=-(vec(grad)G)*(vec(grad)F)+$vec$(grad)*(Gvec(grad)F)$

and conclude that

$(vec(grad{)}^{2}G)F-G(vec(grad{)}^{2}F)=-$vec$(grad)*(Gvec(grad)F-$Fvec$(grad)G)$

Note: We used this identity to show that the Laplace operator is formally self$-$adjoint.

Hint: Just expand both sides term by term. Make sure you use the correct definition of the gradient

$(vec(grad)),$ divergence $(vec(grad)*),$ and Laplace $(vec(grad{)}^2)$ operators. The first dot on the right hand side is a dot

product, and the second dot is part of the divergence operator.

b$.$ Show that

$($:$F|vec(grad{)}^{2}F$:$)=-{}_D||vec(grad)F|{|}^{2}dV+{}_S$vec$(F)vec(grad)F*hat(n)dA$

where $S$ is the boundary of $D,$hat$(n)$ is a unit normal vector to $S,$ and

$||vec(grad)F|{|}^2=\frac{\frac{?}{\text{b}\text{a}\text{r}}(delF)}{\text{d}\text{e}\text{l}\text{x}}\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{x}}+\frac{\frac{?}{\text{b}\text{a}\text{r}}(delF)}{\text{d}\text{e}\text{l}\text{y}}\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{y}}\frac{+}{b}ar(\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{z}})\frac{\text{d}\text{e}\text{l}\text{F}}{\text{d}\text{e}\text{l}\text{z}}$

is the magnitude of vec$(grad)F$ as a $3$D complex vector.

Hint : Integrate the identity in a$,$ in the case $G\frac{?}{b}=ar(F).$ Apply the divergence theorem.

c$.$ Suppose that $F$ is an eigenfunction of the Laplace operator, i$.$e$.$

vec$(gra{d}^2)F=F$

for some scalar $.$ If in addition $F$ satisfies the Neumann boundary condition

vec$(grad)F*hat(n)=0onS$

or the Dirichlet boundary condition

$F=0onS$

show that $$ is a real number, and $0.$

Hint: Apply the identity in $b,$ solve for $,$ and explain why it is a non$-$positive real number.

Note: This explains why one usually considers the eigenfunction equation for the negative Laplace

operator,

$-$vec$(grad{)}^{2}F=F$

which leads to positive values for $.$