Web Bonus Problem –– Week 1

[Victor Ivrii]

Solutions should be posted and discussed here (on the forum, in this topic) only. Do not post duplicate solutions. However significant improvements and corrections are different matter.

Find the general solution $u=u(x,y)$  of the overdetermined system (more equations than unknown functions. Usually it is a good idea to check such solution by the substitution):
\begin{align}
& u_{xx}=0,\label{A}\\
&u_{yy}=0.\label{B}
\end{align}

[Tristan Fraser]

I've given the problem a try, and I'm still trying to figure out how to reconcile my results.

What I did:

Given $ u_{xx} = 0 $ (1) and $ u _{yy} = 0 $ (2), then let's work on (1) first:

$ u_{xx} = v_{x} = 0 $ which implies that $ v = \phi (y) $, then $ u_{x} = \phi (y) $ and $ u(x,y) = x\phi(y) + \psi(y) $

but then there is (2)

$ u_{yy} = w_{y} = 0 , w = \alpha(x) $ then $ u_{y} = \alpha(x), u(x,y) = y \alpha(x) + \beta(x) $

So comparing our two solutions for u:
(I) : $ u (x,y)  = x\phi(y) + \psi(y) $
(II) : $ u(x,y) = y\alpha(x) + \beta(x) $

Could we simply conclude that in this family of solutions, $\phi(y) = ay$ and similarly $\alpha(x) = bx$. Furthermore, $\psi(y) = \beta(x)$ , and since we know those terms go to 0 when the first partial is taken (see above), can we just make them an arbitrary constant?

(Note: a,b are some arbitrary constant)

Am I on the right track, or am I making too many assumptions?

[Adam Gao]

Tristan,

I don't think you are making too many assumptions if your goal is to find one possible solution, but it is valid to be concerned if it there are leaps in logic for a general solution.

Your solutions (I) and (II) are interpreted to be the general solutions to u_xx=0 and u_yy=0, respectively. I believe your solution steps for u_x and u_y for respective constraints to be correct. However, I have modified the final steps for the general solutions:

(I) : u(x,y) =  xϕ(y)+ψ(y)+bx+C
(II) : u(x,y) = yα(x)+β(x)+cy+C

Note that if you take the derivative of (I) with respect to x, b gets absorbed into C. Analogous conditions apply to (II).

When constrained to variables x and y, C is a constant (and so are b and c).

Finally, note that the general solution which obeys both u_xx=0 and u_yy=0 is equal to both solution (I) and solution (II).

Thus, solve: xϕ(y)+ψ(y)+bx+C = yα(x)+β(x)+cy+C

Let's assume none of the terms are equal to 0. This will allow us to impose constraints that will deny trivial solutions. If any of ϕ(y), ψ(y), α(x), and β(x) are equal to 0, the general solution should still be valid.

Thus, xϕ(y) = yα(x).

Thus, ϕ(y)=ay and α(x)=ax. (This is pretty much what you said, except I imposed the extra constraint that the functions share the same coefficient). Check that xϕ(y) = yα(x) = axy.

So now we have axy + ψ(y) + bx + C = axy + β(x) + cy + C

Thus, ψ(y) = cy and  β(x) = bx.

Thus, the general solution is axy + bx + cy + C.

[Victor Ivrii]

Tristan started well, but then went wrong way in the place I could not expect. Adam made the same "tactical error" but then managed to solve (but it was  overcomplicated). The most obvious way after one solves the first equation (\ref{A})
\begin{equation}
u(x,y)= \phi (y) x +\psi(y),
\label{C}
\end{equation}
is not to solve the second one (\ref{B}) from the scratch, but to plug (\ref{C}) in it... Continue, please.
 I really want the simplest solution...

NB. Adam, please use $\LaTeX$ rather than html commands.

[Jaisen Kuhle]

Here is my attempt at a solution, perhaps someone can review it?

[Victor Ivrii]

Jaisen, if I did not know the solution, I would not be able to read your post. If you want to scan, write at least properly.

On Quiz, Test or Final such handwriting could lead to "Unreadable" and 0 mark.

And typing is much more preferable.

[Jaisen Kuhle]

Ok, typed here are my steps:

$ u_{x} = f(y) $
$ u(x,y) = xf(y) + g(y) $
$ u_{y} = x f_{y} + g_{y} $, but
$ u_{yy} = 0 $ which implies
$ f_{y} = c_{1} $ and $ g_{y} = c_{2} $ which in turn implies
$ f(y) = c_{1}y + c_{3} $ and $ g(y) = c_{2}y + c_{4} $, concluding:

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

[Victor Ivrii]

Ok, typed here are my steps:
$$
 u_{x} = f(y)  \implies  u(x,y) = xf(y) + g(y) \implies  u_{y} = x f_{y} + g_{y},$$
but
$$ u_{yy} = 0 \implies  f_{y} = c_{1} \ \ \text{and} \ \  g_{y} = c_{2},
$$
 which in turn implies
$$ f(y) = c_{1}y + c_{3} \ \ \text{and} \ \   g(y) = c_{2}y + c_{4} ,
$$
concluding:
$$ u(x,y) = c_{1}xy + c_{3}x + c_{2}y + c_{4}
 $$

Now this is solution I was looking for (I just made it more readable).

So, here we have a general solution, depending on 4 arbitrary constants, rather than a number of arbitrary functions of one variable. Why so?

[Jingxuan Zhang]

I am trying to follow the quick way as in lecture below.
$$\text{ something wrong }.$$
Hence upon substitution,
$$\text{ some other thing wrong}. $$
Thus I have the unknown functions. I think the constants and explicit forms shows up as a result of not presume these functions beforehand.


I hope what follows is no longer nonsense. We indeed have the not-so-arbitrary functions as desired,
$$c_{1}xy, c_{3}x,  c_{2}y, c_{4}.$$
They lost their arbitrariness in form by constrain $(2)$, since upon substitution
$$x f_{y}(y) + g_{y}(y)=h(x)$$
the particular form of $f,g$ is dictated to be polynomial in $y$ of degree less than one.

[Victor Ivrii]

Jingxuan,

please read carefully what other students posted, you are on the wrong pass from the beginning.

[Lingyun Xu]

Based on Tristan's work,
\begin{eqnarray}
u(x,y)&=&\phi(y)x+\psi(y)\\
u_{yy}&=&0
\end{eqnarray}
Consider derive $u(x,y)$ with respect to y:
\begin{eqnarray}
u_y&=&\phi'(y)x+\psi'(y)\\
u_{yy}&=&\phi''(y)x+\psi''(y)
\end{eqnarray}
Then plug (5) into (7):
\begin{eqnarray}
\phi''(y)x=-\psi''(y)
\end{eqnarray}
Integrating both sides with respect to y:
\begin{eqnarray}
x(\phi'(y)+g(x))=-\psi'(y)+f(x)
\end{eqnarray}
Let me rearrange it into:
\begin{eqnarray}
x\phi'(y)+\psi'(y)=f(x)-xg(x)
\end{eqnarray}
Then plug (10) into (6):
\begin{eqnarray}
u_y=f(x)-xg(x)
\end{eqnarray}
Integrate it with respect to y, we get:
\begin{eqnarray}
u(x,y)=-xyg(x)+yf(x)+p(x)
\end{eqnarray}

[Victor Ivrii]

Lingyun, think: after you got $\phi''(y)x=-\psi''(y)$; this is an identity, it must be satisfied for all $x$ and $y$, which is possible only if $\psi''=\phi''=0$.

Jaisen did everything. I just want a discussion, not another solution (espacially the wrong one)

PS Lingyun, two $\LaTeX$ remarks.
1) eqnarray should not be used, it is old, buggy and has been deprecated long ago
2) MathJax supports autonumbering amd \label / \ref mechanism

[Ruite Xu]

Here is my solution to this question:

Given $u_{xx} = 0 \text{ , } u_{yy} = 0$ and $u=u(x,y)$

$u_{xx}=0$ implies $u_x = \phi(y)$

so we have $u(x,y) = \int u_x dx = \phi(y) x + \psi(y) $ (*)

And from $u_{yy}=0$, we have $u_y = \varphi(x)$

With (*) implies $u(x,y)_y = {(\phi(y) x + \psi(y))}_y  = \varphi(x)$

Partial derivative on both sides with respect to y we have: 

$\phi_{yy}(y)x + \psi_{yy}(y) = 0 $

which implies that:

$\phi_y(y) = c$ and $\psi_y(y) = d$ where both of c and d are constant

Thus:

$\phi(y) = cy + a$ and $\psi(y) = dy + b$ (**) where both of a and b are constant

Combine (*) and (**): we have $u(x,y) = cyx + ax + dy +b$ as our general solution to this pde question

[Victor Ivrii]

Guys, there is already a perfect solution by Jaisen, and Tristan made the first part correctly. There is no point to post more solutions! But the interesting question: Why this solution includes just 4 arbitrary constants rather than arbitrary functions of one variable?

And please find the general solutions to two other overdetermined systems
\begin{equation}
\left\{\begin{aligned}
&u_{xx}=y^2,\\
&u_{yy}=x^2
\end{aligned}\right.
\label{Q}
\end{equation}
and
\begin{equation}
\left\{\begin{aligned}
&u_{xx}=y^2,\\
&u_{yy}=-x^2
\end{aligned}\right.
\label{R}
\end{equation}

[Jingxuan Zhang]

As for $(\ref{Q})$,
$$u_{xx}=y^2 \implies u=\frac{x^2y^2}{2} + xf(y)+g(y); u_{yy}=x^2 \implies f_{yy}(y)=g_{yy}(y)=0 \implies u = \frac{x^2y^2}{2} + x(ay+b) + (cy+d).$$
Soppose $u$ solves $(\ref{R})$, we would quickly arrive at what seems to me a contradiction
$$-2x^2=xf(y)+g(y).$$