I just made an attempt at answering the following problem:
Suppose $D$ is the unit square $(0,1) \times (0,1)$. Then,
- verify the function $u(x,y)=xy$ satisfies the following boundary-value problem (BVP):
$u_{xx}(x,y)+u_{yy}(x,y)=0$ in $D$
$u(x,0)=0,\,$ $u(x,1)=x,\,$ $0<x<1$
$u(0,y)=0,\,$ $u(1,y)=y,\,$ $0<y<1$
- At what point $D$ do the max and min values of $u(x,y)$ occur and what are they?
Here is my attempt at 1.:
$u_{x}(x,y)=y$, therefore $u_{xx}(x,y)=0$ and $u_{y}(x,y)=x$, therefore $u_{yy}(x,y)=0$. Thus, we have that $u_{xx}(x,y)+u_{yy}(x,y)=0 +0 = 0$. Is that all there is to this part? I'm concerned because I didn't really use the fact that $u$ is satisfying that equation in $D$...
Same idea for the boundary conditions on $0<x<0$: $u(x,0)=x\cdot0=0$; $u(x,1)=x\cdot 1=x$.
Same question as well: did I use the fact that $0<x<1$ here, or do I need to?
Also, for the boundary conditions on $0<x<1$: $u(0,y)=0\cdot y=0$; $u(1,y)=1\cdot y=y$.
Again: did I use the fact that $0<y<1$ here, or do I need to?
Now, here is my attempt at 2.:
$D$ consists of all possible points $(x,y)$ such that $0<x<1$ and $0<y<1$.
The function $u(x,y)=xy$; therefore, has a minimum of $0$ at $(0,0)$, and a maximum of $1$ at $(1,1)$. However, $\mathbf{D}$ does not contain either of these points, and so $\mathbf{u}$ does not attain its max or min in $\mathbf{D}$.
The best we can hope for is a minimum function value of $\epsilon^{2}$, which would occur at the point $(\epsilon,\epsilon)$; and a maximum of $1-2\epsilon+ \epsilon^{2}$, which would occur at the point $(1-\epsilon,1-\epsilon)$, both for small $\epsilon >0$.
Now, I am not sure if this is what j was supposed to do for 2. If not, please let me know what I should do instead. (Also, in any hints/solutions that you provide, please pretend like I've never seen the maximum/minimum principle before - because, for all intents and purposes, in this course, I have not yet.)