### Author Topic: Web Bonus Problem –– Week 1  (Read 6262 times)

#### Victor Ivrii ##### Re: Web Bonus Problem –– Week 1
« Reply #15 on: January 06, 2018, 08:32:37 PM »
Jingxuan, I removed excessive quotation.

With (13)  you found the general solution. Great!

#### Jaisen Kuhle

• Full Member
•   • Posts: 20
• Karma: 2 ##### Re: Web Bonus Problem –– Week 1
« Reply #16 on: January 06, 2018, 10:16:07 PM »
For (14)

$$u_{xx}=y^2 \implies u=\frac{x^2y^2}{2} + xf(y)+g(y) \implies u_{y}=x^2y+xf_{y}+g_{y} \implies u_{yy}= x^2 + xf_{yy} + g_{yy} = -x^2$$

So we have:

$$2x^2 + xf_{yy} + g_{yy}= 0$$

Suppose x = 0, therefore

$$g_{yy} = 0 \implies g_{y} = c \implies g(y)=cy+d \implies u(x,y) = u(0,y) = cy + d$$

Suppose x not equal to 0

When we solve the quadratic we arrive at imaginary values for x. I'm not sure if I ought to continue.
« Last Edit: January 07, 2018, 08:39:28 AM by Victor Ivrii »

#### Jaisen Kuhle

• Full Member
•   • Posts: 20
• Karma: 2 ##### Re: Web Bonus Problem –– Week 1
« Reply #17 on: January 06, 2018, 10:24:00 PM »
oops.

#### Jaisen Kuhle

• Full Member
•   • Posts: 20
• Karma: 2 ##### Re: Web Bonus Problem –– Week 1
« Reply #18 on: January 06, 2018, 10:24:31 PM »
But the interesting question: Why this solution includes just 4 arbitrary constants rather than arbitrary functions of one variable?

Perhaps a hint? Are we supposed to know this from the solution or prior to arriving at the solution? The solution indicates this is a linear(?) PDE so perhaps it is related to that fact. Alternatively, perhaps I can ask, as this is a second order PDE, and our only constraints are related to the second partial derivatives, does that mean we need constants to determine the initial values and boundaries?
« Last Edit: January 06, 2018, 10:43:50 PM by Jaisen Kuhle »

#### Victor Ivrii ##### Re: Web Bonus Problem –– Week 1
« Reply #19 on: January 06, 2018, 11:13:31 PM »
Quote
Perhaps a hint? Are we supposed to know this from the solution or prior to arriving at the solution?
In this particular case we can arrive to the correct conclusion in either way: we can get it in the process of solving or we can get it without solving.

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: Web Bonus Problem –– Week 1
« Reply #20 on: January 07, 2018, 06:00:24 AM »
Quote from: Jaisen Kuhle
Suppose x not equal to 0

When we solve the quadratic we arrive at imaginary values for x. I'm not sure if I ought to continue.

I agree with your first part. Now suppose x is not identically $0$, what function of $f_{yy}(y)$ multiply to $x$ will give you such a quadratic in $x$?
Also the matter seems not to be with possible complex values, but that in this case the quadratic formula gives a relation $x=F(y).$

Quote from: Victor Ivrii
Is it ever possible?
No. Hence there is no common solution if $x$ is not identically zero. But if it is then upon substituting $u=u(y)=g(y)$ and so
$$u_{xx}=0=y^2 \implies y=0.$$
Thus $u$ can only be defined on the origin, and takes any constant value. (Though I doubt if derivatives are well defined then.)
« Last Edit: January 07, 2018, 10:34:06 AM by Jingxuan Zhang »

#### Victor Ivrii ##### Re: Web Bonus Problem –– Week 1
« Reply #21 on: January 07, 2018, 06:15:05 AM »
The answer to the question "why solution does not include arbitrary functions off one variable" the answer is simple: because it is overdetermined system. One can arrive to this conclusion by the following reasoning: solution to equation $u_{xx}=0$ includes arbitrary functions of $y$ but does not include arbitrary functions of $x$; solution to $u_{yy}=0$ does not include arbitrary functions of $y$. Then the common solution to both equations contains neither.

Pending: what is the general solution to (14) from the previous page

#### Jaisen Kuhle

• Full Member
•   • Posts: 20
• Karma: 2 ##### Re: Web Bonus Problem –– Week 1
« Reply #22 on: January 07, 2018, 08:01:51 AM »
Quote from: Jaisen Kuhle
Suppose x not equal to 0

When we solve the quadratic we arrive at imaginary values for x. I'm not sure if I ought to continue.

I agree with your first part. Now suppose x is not identically $0$, what function of $f_{yy}(y)$ multiply to $x$ will give you such a quadratic in $x$?
Also the matter seems not to be with possible complex values, but that in this case the quadratic formula gives a relation $x=F(y).$

I'm not sure I follow. If x is not equal to zero, the quadratic would be satisfied if $f_{yy}$ and $g_{yy}$ are constant. So we end up with:

$$u(x,y)=\frac{x^2y^2}{2} + \frac{c_{1}xy^2}{2}+c_{3}xy+c_{5}x+\frac{c_{2}y^2}{2}+{c_{4}y}+c_{6}$$
« Last Edit: January 07, 2018, 08:05:08 AM by Jaisen Kuhle »

#### Victor Ivrii ##### Re: Web Bonus Problem –– Week 1
« Reply #23 on: January 07, 2018, 08:44:15 AM »
Jaisen, you arrived to the equality I maked by red. And it is not just an equality, it must be an identity, that is, fulfilled for all $x$ and $y$.

What are $f$ and $g$ here by  your definition? And which of these functions make that equation to be an identity?

#### Jaisen Kuhle

• Full Member
•   • Posts: 20
• Karma: 2 ##### Re: Web Bonus Problem –– Week 1
« Reply #24 on: January 07, 2018, 09:38:09 AM »
Jaisen, you arrived to the equality I maked by red. And it is not just an equality, it must be an identity, that is, fulfilled for all $x$ and $y$.

What are $f$ and $g$ here by  your definition? And which of these functions make that equation to be an identity?

Ok, picking up from where I erred:

$$x = f_{yy} = g_{yy} = 0 \implies f_{y} = c_{1}$$ and $$g_{y} = c_{2}$$

Concluding:

$$u(x,y)=\frac{x^2y^2}{2} + {c_{1}xy}+c_{3}x+{c_{2}y}+c_{4}$$

#### Victor Ivrii ##### Re: Web Bonus Problem –– Week 1
« Reply #25 on: January 07, 2018, 09:43:14 AM »
Wrong again. What are $f$ and $g$?

#### Jaisen Kuhle

• Full Member
•   • Posts: 20
• Karma: 2 ##### Re: Web Bonus Problem –– Week 1
« Reply #26 on: January 07, 2018, 09:55:35 AM »
Wrong again. What are $f$ and $g$?

Functions dependent on the variable y but constant with respect to x. Other than that, I'm unsure.

#### Victor Ivrii ##### Re: Web Bonus Problem –– Week 1
« Reply #27 on: January 07, 2018, 10:03:05 AM »
So $f=f(y)$, $g=g(y)$ and
$$2x^2 + xg(y)+ h(y)=0$$
is an identity. Is it ever possible?

• Full Member
•   • Posts: 15
• Karma: 4 ##### Re: Web Bonus Problem –– Week 1
« Reply #28 on: January 07, 2018, 12:12:41 PM »
The identity is impossible. In other words, there does not exist any $g(y)$ or $h(y)$ such that the equation $2x^2 + xg(y) + h(y) = 0$ is true for all $x$ and $y$ on the domain of $u(x,y)$.

Proof:

Denote $v(x,y) = 2x^2 + xg(y) + h(y)$. Say we find some value $x=x_0$ and $y=y_0$ such that $v(x_0, y_0) = 0$. If $g(y_0)$ is positive or 0, let $x_1 > x_0$. Now closely examine the expression $v(x_1, y_0)$ and compare to $v(x_0, y_0)$. $2x_{1}^2 > 2x_{0}^2$ and $x_{1}g(y_0) > x_{0}g(y_0)$. Then $v(x_1, y_0) \ne v(x_0, y_0) = 0$.

If $g(y_0)$ is negative or 0, let $x_2 < x_0$. We would similarly find $v(x_2, y_0) \ne v(x_0, y_0) = 0$.

Thus, there is always some set of points $x$ and $y$ where $v(x,y) \ne 0$.

How this applies to the problem (14):

The fact that the identity is never possible means that there does not exist a general solution for the constraints $u_{xx} = y^2$ and $u_{yy} = -x^2$.

#### Victor Ivrii ##### Re: Web Bonus Problem –– Week 1
« Reply #29 on: January 07, 2018, 12:35:25 PM »
Adam nailed it: solution does not exist. It is the general situation for overdetermined systems: solution does not exist unless compatibility conditions are satisfied. If compatibility conditions are satisfied, then there exist fewer solutions.

You studied in Calculus II and ODE such system
\begin{equation}
\left\{\begin{aligned}
&u_x = f,\\
&u_y=g
\end{aligned}\right.
\end{equation}
In other words, when $f\,dx+g\,dy$ is an exact differential. And the necessary (and for simple-connected domains, sufficient) condition was $f_y=g_x$.

For the system
\begin{equation}
\left\{\begin{aligned}
&u_{xx} = f,\\
&u_{yy}=g
\end{aligned}\right.
\end{equation}
such condition is $f_{yy}=g_{xx}$. Prove the necessity!
« Last Edit: January 11, 2018, 04:52:23 AM by Victor Ivrii »