MHB Why Does the Minimum Value of u(x,y) Occur on the Boundary of the Unit Disk?

[evinda]

Hello! (Wave)

Let $u(x,y), x^2+y^2 \leq 1$, a solution of

$$u_{xx}(x,y)+2u_{yy}(x,y)+e^{u(x,y)}=0, x^2+y^2\leq 1$$

Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y) $.

We suppose that $\min_{x^2+y^2 \leq 1} u(x,y) \neq \min_{x^2+y^2=1} u(x,y) $.At the solution it is said that since $\{ (x,y) | x^2+y^2 \leq 1\} \supseteq \{ (x,y)| x^2+y^2 =1 \}$ it has to hold that $\min_{x^2+y^2 \leq 1} u(x,y)<\min_{x^2+y^2=1} u(x,y)$ .

I haven't understood how we deduce that the above inequality holds. Could you explain it to me? (Thinking)

 

[I like Serena]

Hello! (Wave)

Let $u(x,y), x^2+y^2 \leq 1$, a solution of

$$u_{xx}(x,y)+2u_{yy}(x,y)+e^{u(x,y)}=0, x^2+y^2\leq 1$$

Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y) $.

We suppose that $\min_{x^2+y^2 \leq 1} u(x,y) \neq \min_{x^2+y^2=1} u(x,y) $.At the solution it is said that since $\{ (x,y) | x^2+y^2 \leq 1\} \supseteq \{ (x,y)| x^2+y^2 =1 \}$ it has to hold that $\min_{x^2+y^2 \leq 1} u(x,y)<\min_{x^2+y^2=1} u(x,y)$ .

I haven't understood how we deduce that the above inequality holds. Could you explain it to me? (Thinking)

Hey! (Happy)We have that if set $A \supseteq B$, then $$\min_{x\in A} f(x) \le \min_{x\in B} f(x)$$.
That is, if a set is larger, the range of some function $f$ on it must be at least as big.If we combine that with $$\min_{x\in A} f(x) \ne \min_{x\in B} f(x)$$, we get that $$\min_{x\in A} f(x) < \min_{x\in B} f(x)$$. (Wasntme)

 

[evinda]

Hey! (Happy)We have that if set $A \supseteq B$, then $$\min_{x\in A} f(x) \le \min_{x\in B} f(x)$$.
That is, if a set is larger, the range of some function $f$ on it must be at least as big.If we combine that with $$\min_{x\in A} f(x) \ne \min_{x\in B} f(x)$$, we get that $$\min_{x\in A} f(x) < \min_{x\in B} f(x)$$. (Wasntme)

Could you give me an example of sets $A,B$ such that $A \supseteq B$ and $\min_{x \in A} f(x) < \min_{x \in B} f(x)$ ? (Thinking)

Also, they continue the proof as follows:We set $u(x,y)=U(x,Y), Y= \frac{1}{ \sqrt{2}} y$

So we have $u_{xx}(x,y)+2u_{yy}(x,y)+e^{u(x,y)}=0 , x^2+y^2 \leq 1 \Leftrightarrow U_{xx}(x,Y)+U_{YY}(x,Y)+e^{u(x,Y)}=0, x^2+2Y^2 \leq 1$.

Since $\min_{X^2+2Y^2 \leq 1} U(x,Y) < \min_{X^2+Y^2=1} U(x,y)$, we deduce that $U$ has an interior global minimum and furthermore if $(x_1, Y_1)$ is such a point we have

$$\nabla{U(x_1, Y_1)}=0$$

$$\text{ Hessian } U(x_1, Y_1) \geq 0 \Rightarrow \Delta U(x_1, Y_1) \geq 0$$

However $e^{U(x_1, Y_1)}>0 \Rightarrow U_{xx} (x_1, Y_1)+U_{YY}(x_1, Y_1)+e^{U(x_1, Y_1)}>0$, contradiction.Could you explain me th proof from the point "we deduce that $U$ has an interior global minimum ..."?

Couldn't we also do it as follows?

$U_{xx}(x,Y)+U_{YY}(x,Y)+e^{u(x,Y)}=0 \Rightarrow U_{xx}(x,Y)+U_{YY}(x,Y)=-e^{u(x,Y)} \leq 0$

So from the minimum principle we have that $\min_{X^2+2Y^2 \leq 1} U=\min_{X^2+2Y^2=1} U $, that is a contradiction.

Or am I wrong?

 

[I like Serena]

Could you give me an example of sets $A,B$ such that $A \supseteq B$ and $\min_{x \in A} f(x) < \min_{x \in B} f(x)$ ? (Thinking)

Sure.
Suppose $A=[0,3], B=[1,2], f(x)=x$, then $0 = \min_{x \in A} f(x) < \min_{x \in B} f(x) = 1$. (Wasntme)

Could you explain me th proof from the point "we deduce that $U$ has an interior global minimum ..."?

Couldn't we also do it as follows?

$U_{xx}(x,Y)+U_{YY}(x,Y)+e^{u(x,Y)}=0 \Rightarrow U_{xx}(x,Y)+U_{YY}(x,Y)=-e^{u(x,Y)} \leq 0$

So from the minimum principle we have that $\min_{X^2+2Y^2 \leq 1} U=\min_{X^2+2Y^2=1} U $, that is a contradiction.

Or am I wrong?

It's the same thing.
Basically they are proving the minimum principle for this particular case. (Thinking)

That is, if a function would have a global minimum in its interior, the derivatives have to be zero at that point, and the Hessian has to be $\ge 0$.
Since the differential equation shows that cannot be the case, there is no global minimum in the interior. (Emo)

However $e^{U(x_1, Y_1)}>0 \Rightarrow U_{xx} (x_1, Y_1)+U_{YY}(x_1, Y_1)+e^{U(x_1, Y_1)}>0$, contradiction.

Hmm... I think that should be $e^{U(x_1, Y_1)}>0 \wedge U_{xx} (x_1, Y_1)+U_{YY}(x_1, Y_1)+e^{U(x_1, Y_1)}=0 \Rightarrow \Delta U(x_1, Y_1) < 0$, contradiction. (Thinking)

 

[evinda]

Suppose that we din't know that we have to set $Y=\frac{1}{\sqrt{2}}y$, how could we find $U$ so that we have the known Laplace's equation?

 

[I like Serena]

Suppose that we din't know that we have to set $Y=\frac{1}{\sqrt{2}}y$, how could we find $U$ so that we have the known Laplace's equation?

Well, we want to find an Y such that $U_{YY}=2u_{yy}$. (Thinking)

If we substitute $y=aY$ where $a$ is an as yet unknown constant, we get:
$$\d U Y = \d u y \d y Y = \d u y a \Rightarrow U_Y = a u_y$$
The second derivative brings us:
$$U_{YY} = a^2 u_{yy}$$
So $a=\sqrt 2$. (Mmm)

 

[evinda]

Well, we want to find an Y such that $U_{YY}=2u_{yy}$. (Thinking)

If we substitute $y=aY$ where $a$ is an as yet unknown constant, we get:
$$\d U Y = \d u y \d y Y = \d u y a \Rightarrow U_Y = a u_y$$
The second derivative brings us:
$$U_{YY} = a^2 u_{yy}$$
So $a=\sqrt 2$. (Mmm)

I am a little confused now... (Worried)

Why do we want that the equality $\frac{dU}{dY}=\frac{du}{dy} \frac{dy}{dY}$ holds? (Thinking)

 

[evinda]

Sure.
Suppose $A=[0,3], B=[1,2], f(x)=x$, then $0 = \min_{x \in A} f(x) < \min_{x \in B} f(x) = 1$. (Wasntme)

Ah, I see...

It's the same thing.
Basically they are proving the minimum principle for this particular case. (Thinking)

That is, if a function would have a global minimum in its interior, the derivatives have to be zero at that point, and the Hessian has to be $\ge 0$.
Since the differential equation shows that cannot be the case, there is no global minimum in the interior. (Emo)
Hmm... I think that should be $e^{U(x_1, Y_1)}>0 \wedge U_{xx} (x_1, Y_1)+U_{YY}(x_1, Y_1)+e^{U(x_1, Y_1)}=0 \Rightarrow \Delta U(x_1, Y_1) < 0$, contradiction. (Thinking)

Could you explain me further how we get the contradiction?
Would the following justification also suffice?

$U_{xx}(x,Y)+U_{YY}(x,Y)+e^{u(x,Y)}=0 \Rightarrow U_{xx}(x,Y)+U_{YY}(x,Y)=-e^{u(x,Y)} \leq 0$

So from the minimum principle we have that $\min_{X^2+2Y^2 \leq 1} U=\min_{X^2+2Y^2=1} U $, that is a contradiction.

 

[I like Serena]

I am a little confused now... (Worried)

Why do we want that the equality $\frac{dU}{dY}=\frac{du}{dy} \frac{dy}{dY}$ holds? (Thinking)

It's not so much something we want - it's an application of the chain rule.

We have that $u(x,y) = U(x,Y(y))$.
And actually, it should be:
$$\pd U Y=\pd u x \d y x + \pd u y \d y Y = \pd u y \d y Y$$
(Blush)

Ah, I see...

Would the following justification also suffice?

$U_{xx}(x,Y)+U_{YY}(x,Y)+e^{u(x,Y)}=0 \Rightarrow U_{xx}(x,Y)+U_{YY}(x,Y)=-e^{u(x,Y)} \leq 0$

So from the minimum principle we have that $\min_{X^2+2Y^2 \leq 1} U=\min_{X^2+2Y^2=1} U $, that is a contradiction.

Yes, that is fine. (Nod)