Math Challenge - May 2020

[fresh_42]

Questions

1. (solved by @benorin ) Let $1<p<4$ and $f\in L^p((1,\infty))$ with the Lebesgue measure $\lambda$. We define $g\, : \,(1,\infty)\longrightarrow \mathbb{R}$ by
$$
g(x)=\dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\,d\lambda(t).
$$
Show that there exists a constant $C=C(p)$ which depends on $p$ but not on $f$ such that $\|g\|_2 \leq C\cdot \|f\|_p$ so $g\in L^2((1,\infty)).$ (FR)

2. We define $\mathbb{R}^\infty =\mathbb{R}^{(\mathbb{N})}=\{\,(x_1,x_2,\ldots)\,|\,x_i\stackrel{a.a.}{=}0\,\}$ and equip $\mathbb{R}^\infty$ with the Euclidean metric $d((x_1,x_2,\ldots),(y_1,y_2,\ldots)) = \sqrt{\sum_{i=1}^\infty \left|x_i-y_i\right|^2}.$ which defines a topology $\mathcal{S}:=\{\,U\subseteq \mathbb{R}^\infty\,|\, \forall\,p\in U\, \exists\, \varepsilon>0\,:\,B_\varepsilon(p)\subseteq U\,\}$ with the open ball $B_\varepsilon(p)=\{\,q\in\mathbb{R}^\infty\,|\,d(p,q)<\varepsilon\,\}.$ (FR)

a.) Show that the function
\begin{align*}
\alpha\, : \,(\mathbb{R}^\infty,\mathcal{S})&\longrightarrow (\mathbb{R},\mathcal{E})
(x_1,x_2,\ldots) &\longmapsto \sum_{i=1}^\infty 2^i\cdot x_i \end{align*}
is not continuous, where $\mathcal{E}$ is the usual Euclidean topology on $\mathbb{R}.$

b.) Let $B$ be the diagonal matrix where the diagonal entries are $2^i$ for $i=1,2,\ldots,$ i.e.
$$ B=\begin{bmatrix}2&0&0&\ldots\\0&4&0&\ldots\\0&0&8&\ldots\\ \vdots&\vdots&\vdots&\ddots \end{bmatrix}$$
Show that $\beta\, : \,(\mathbb{R}^\infty,\mathcal{S}) \longrightarrow (\mathbb{R}^\infty,\mathcal{S})$ defined by $\beta(x)=Bx$ is not continuous.

c.) Define a topology $\mathcal{T}$ on $\mathbb{R}^\infty$ such that the inclusion maps
\begin{align*}
\iota_n \, : \, (\mathbb{R}^n,\mathcal{E}) & \longrightarrow (\mathbb{R}^\infty,\mathcal{T}) \\
(x_1,\ldots,x_n)&\longmapsto (x_1,\ldots,x_n,0,\ldots) \end{align*}
are continuous for any $n\in \mathbb{N}_0.$

3. (solved by @cbarker1, @benorin , alternative solution possible) Calculate (FR) $$\int_{-\infty}^{+\infty}\dfrac{\cos(\alpha x)}{1+x^2}\,dx \quad (\alpha \geq 0).$$

4. Calculate $$\int_0^1 \sin(\pi x)\,x^x\,(1-x)^{1-x}\,dx.$$
Hint: You may use calculators to determine residues. (FR)

5. (FR) The $p$-Prüfer group is defined as
$$
G:=\mathbb{C}_{p^\infty} =\{\,\exp(2n\pi i/p^m )\,|\,n\in \mathbb{Z},m\in \mathbb{N}\,\}\cong\mathbb{Z}\left[\frac{1}{p}\right]/\mathbb{Z}
$$
Show that $G$ is isomorphic to the factor group $F/R$ of the free Abelian group over an countably infinite basis $\{\,a_1,a_2,\ldots ,a_n,\ldots \,\}$ with the subgroup of relations $R$ generated by $\{\,pa_1,a_1-pa_2,\ldots , a_n-pa_{n+1},\ldots\,\}$, so
$$
G = \langle x_1,x_2,\ldots\,|\,x_1^p=1,x_2^p=x_1,x_3^p=x_2,\ldots \rangle
$$

6. a.) (solved by @benorin ) (FR) Let $u_1,\ldots,u_n$ be solutions of the one dimensional heat equation $\dfrac{du}{dt}-\dfrac{d^2u}{dx^2}=0\;(x\in \mathbb{R},t>0).$ Show that $$u(x_1,\ldots,x_n,t):=\displaystyle{\prod_{k=1}^n}u_k(x_k,t)$$ is a solution of the $n$ dimensional heat equation $\dfrac{\partial u}{\partial t}-\Delta u=0.$
b.) (solved by @benorin ) (FR) Calculate a solution for
$$
\begin{cases}
\dfrac{\partial u}{\partial t}(x,t)-\Delta u(x,t)=0 &\text{ for } x\in \mathbb{R}^3,\,t>0 \\[6pt]
u(x,0)=x_1^2x_2^2x_3&\text{ for } x=(x_1,x_2,x_3)\in \mathbb{R}^3
\end{cases}
$$
Hint: use part a.)

7. (solved by @mathwonk ) Give an example of a quotient $R-$module $M/N$ which is Artinian although neither the ring $R$ nor the modules $M,N$ are. (FR)

8. Prove and give an example of a solvable group which is not supersolvable. (FR)

9. Let $\gamma$ be a non-zero limit ordinal. Consider the order topology on $[0, \gamma]$. Show that this topological space is compact. (MQ)

10. Can you completely cover a disk of diameter $10$ with nine $1\times 10$ rectangles? (IR)

High Schoolers only

11. (solved by @etotheipi ) For which natural numbers is $1!+\ldots + n!$ a square number? $n!=1\cdot 2\cdot \ldots \cdot n\,.$

12. (solved by @physion ) Determine $\{\,(x,y)\in \mathbb{N}_0\times \mathbb{N}_0\,|\,x^3+8x^2-6x+8-y^3=0\,\}\,.$

13. Given two different, coprime, positive natural numbers $a,b \in \mathbb{N}$. Then there are two natural numbers $x,y \in \mathbb{N}$ such that $ax-by=1\,.$

14. How many moves do the towers of Hanoi require to solve by an optimal strategy?

15. (solved by @etotheipi ) Among six people are always three who know each other or three who don't. Why?

 

[etotheipi]

Spoiler: Problem 15

There's likely a nicer way to do it, but this what I first thought of. In the set of 6 people we consider the different cases of the largest possible subset(s) of people who knows everyone else in the subset. This largest subset for any given set of 6 people can only have a size between 1 and 6 inclusive.

If the largest subset has a size between 4 and 6 then the statement is true. If the largest subset has size 3, then there is the possibility of another largest subset of size 3 but in any case the statement is still true.

If the largest subset has a size of 2, then there are a 3 different cases. If there are 3 such subsets in the group of 6, then we can choose one person from each subset to obtain 3 people who will not know each other. If there are 2 such subsets, then we can simply pick remaining other 2 people and any 1 further person from either of these 2 subsets to obtain 3 people who don't know each other. If there is 1 such subset, then the other 4 people don't know each other and the statement is also true.

Finally, if the largest subset has a size of 1, then no-one knows each other and the statement is true. This I think covers all cases.

 

[etotheipi]

Spoiler: Problem 11

Let $p = 10q + r$ with $r \in \mathbb{N}_0 \cap [0, 9]$ and $q \in \mathbb{N}_0$, then $p^2 = (10q+r)^2 = 10(10q^2 + 2qr) + r^2$. The squares of all 10 possible $r$ all have last digits in $\{0,1,4,5,6,9\}$, so the last digits of all possible $p^2$ are also members of that set.

For any value of $k \geq 5$, $k! \, \text{mod}\, 10 = 0$ since $k! = k(k-1)(k-2)\dots (5)(4)(3)(2)(1)$ has at least one pair of $5 \times 2$. It follows that the last digit of the sum $n! + (n-1)! + (n-2)! + \dots + 2! + 1!$ equals $(4! + 3! + 2! + 1!) \, \text{mod}\, 10 = 3$ if $n \geq 4$.

And $3 \notin \{0,1,4,5,6,9\}$, hence the last digit of the sum cannot be a perfect square if $n \geq 4$.

So we need only consider $n \in \{1,2,3\}$, in which case only $n =1$ and $n=3$ produce perfect squares.

 

[cbarker1]

Spoiler: Problem 3

Calculate the of this integral:
$$\int_{-\infty}^{\infty} \frac{\cos(\alpha x)}{x^2+1} \, dx\, (\alpha \geq 0)$$

Solution:
Let $f: \mathbb{C} \to \mathbb{C}$ is given by $f(z)=\frac{e^{kzi}}{z^2+1}$, where $k=\alpha$. We know $f$ is holomorphic everywhere except for $z=\pm i$. Now, we need to determine the region of this integral. The singularity $z=i$ lies in the interior of the semicircular region whose boundary consists of the segment $[-R,R]$ of the real axis. and the upper half $C_R$ of $\left |z\right |=R\, (R\gt 0)$ from $z=-R$ to $z=R$. Using the Cauchy Residue Theorem, the integration of $f$ is the following: $$\int_{-R}^{R} \frac{e^{ikx}}{x^2+1} \, dx=2 i \pi B_1-\oint_{C_R} f(z)\, dz \, (1)$$ where $B_1$ is the residue of $f$ at $z=i$. Now we need to determine the residue. The point $z=i$ is a simple pole of $f$. Let $f:=\frac{p(z)}{q(z)}$ where $p(z)=\frac{e^{kzi}}{z+i}$ and $q(z)=z-i$. We know that $q(i)=0$, $p(i)=\frac{e^{-k}}{2i} \neq 0$, and $q'(i)=1 \neq 0$. By a theorem (Churchill, pg 231), We have: $$B_1= \frac{p(i)}{q'(i)}=\frac{e^{-k}}{2i}$$ Plug in $B_1$ to (1), it yields:
$$\int_{-R}^{R} \frac{e^{ikx}}{x^2+1} \, dx=2 i \pi \frac{e^{-k}}{2i}-\oint_{C_R} f(z)\, dz $$
With some simplification, it yields,
$$\int_{-R}^{R} \frac{e^{ikx}}{x^2+1} \, dx= \pi e^{-k}-\oint_{C_R} f(z)\, dz $$
We take the Real part on both sides of the equation to yield:
$$\int_{-R}^{R} \frac{\cos(kx)}{x^2+1} \, dx= \pi e^{-k}-\Re{\oint_{C_R} f(z)\, dz}$$
We need to demonstrate the $\Re{\oint_{C_R} f(z)\, dz} \to 0$ as $R \to \infty$
Let $M_R=\frac 1 {R^2-1}$. So by the ML Theorem (Churchill, pg 136), $$\left|\Re{\oint_{C_R} f(z)\, dz}\right| \le \left |\oint_{C_R} f(z)\, dz \right | \le M_R \pi R=\pi \frac R {R^2-1} $$
Taking the limit on all sides of the inequalities as $R \to \infty$, the rational limit is zero because the polynomial in the denominator is higher degree than the numerator polynomial.
Thus, Both the real part of the integral and the contour integral tends to 0. Thus the integral is the following with the back substitution of $k$ to $\alpha$ :
$$\int_{-\infty}^{\infty} \frac{\cos(\alpha x)}{x^2+1} \, dx =\pi e^{-\alpha}$$Source on Some Theorems: Complex Variable and Applications by James Ward Brown and Ruel V. Churchill Ninth Edition McGraw Hill Education.

 

[fresh_42]

There's likely a nicer way to do it, but this what I first thought of.

That's possibly a matter of taste. My solution uses graphs as language, but the reasoning is similar.

In the set of 6 people we consider the different cases of the largest possible subset(s) of people who knows everyone else in the subset. This largest subset for any given set of 6 people can only have a size between 1 and 6 inclusive.

If the largest subset has a size between 4 and 6 then the statement is true. If the largest subset has size 3, then there is the possibility of another largest subset of size 3 but in any case the statement is still true.

If the set has three elements, the statement trivially holds.

If the largest subset has a size of 2, then there are a 3 different cases. If there are 3 such subsets in the group of 6, then we can choose one person from each subset to obtain 3 people who will not know each other. If there are 2 such subsets, then we can simply pick remaining other 2 people and any 1 further person from either of these 2 subsets to obtain 3 people who don't know each other. If there is 1 such subset, then the other 4 people don't know each other and the statement is also true.

Finally, if the largest subset has a size of 1, then no-one knows each other and the statement is true. This I think covers all cases.

What if I have a 3 cycle, or 4-cycle: one set of your kind with two persons knowing each other, and all others know a different one? I don't really see that you covered all cases.

 

[fresh_42]

Spoiler: Problem 3

Calculate the of this integral:
$$\int_{-\infty}^{\infty} \frac{\cos(\alpha x)}{x^2+1} \, dx\, (\alpha \geq 0)$$

Solution:
Let $f: \mathbb{C} \to \mathbb{C}$ is given by $f(z)=\frac{e^{kzi}}{z^2+1}$, where $k=\alpha$. We know $f$ is holomorphic everywhere except for $z=\pm i$. Now, we need to determine the region of this integral. The singularity $z=i$ lies in the interior of the semicircular region whose boundary consists of the segment $[-R,R]$ of the real axis. and the upper half $C_R$ of $\left |z\right |=R\, (R\gt 0)$ from $z=-R$ to $z=R$. Using the Cauchy Residue Theorem, the integration of $f$ is the following: $$\int_{-R}^{R} \frac{e^{ikx}}{x^2+1} \, dx=2 i \pi B_1-\oint_{C_R} f(z)\, dz \, (1)$$ where $B_1$ is the residue of $f$ at $z=i$. Now we need to determine the residue. The point $z=i$ is a simple pole of $f$. Let $f:=\frac{p(z)}{q(z)}$ where $p(z)=\frac{e^{kzi}}{z+i}$ and $q(z)=z-i$. We know that $q(i)=0$, $p(i)=\frac{e^{-k}}{2i} \neq 0$, and $q'(i)=1 \neq 0$. By a theorem (Churchill, pg 231), We have: $$B_1= \frac{p(i)}{q'(i)}=\frac{e^{-k}}{2i}$$ Plug in $B_1$ to (1), it yields:
$$\int_{-R}^{R} \frac{e^{ikx}}{x^2+1} \, dx=2 i \pi \frac{e^{-k}}{2i}-\oint_{C_R} f(z)\, dz $$
With some simplification, it yields,
$$\int_{-R}^{R} \frac{e^{ikx}}{x^2+1} \, dx= \pi e^{-k}-\oint_{C_R} f(z)\, dz $$
We take the Real part on both sides of the equation to yield:
$$\int_{-R}^{R} \frac{\cos(kx)}{x^2+1} \, dx= \pi e^{-k}-\Re{\oint_{C_R} f(z)\, dz}$$
We need to demonstrate the $\Re{\oint_{C_R} f(z)\, dz} \to 0$ as $R \to \infty$
Let $M_R=\frac 1 {R^2-1}$. So by the ML Theorem (Churchill, pg 136), $$\left|\Re{\oint_{C_R} f(z)\, dz}\right| \le \left |\oint_{C_R} f(z)\, dz \right | \le M_R \pi R=\pi \frac R {R^2-1} $$
Taking the limit on all sides of the inequalities as $R \to \infty$, the rational limit is zero because the polynomial in the denominator is higher degree than the numerator polynomial.
Thus, Both the real part of the integral and the contour integral tends to 0. Thus the integral is the following with the back substitution of $k$ to $\alpha$ :
$$\int_{-\infty}^{\infty} \frac{\cos(\alpha x)}{x^2+1} \, dx =\pi e^{-\alpha}$$Source on Some Theorems: Complex Variable and Applications by James Ward Brown and Ruel V. Churchill Ninth Edition McGraw Hill Education.

Well done.

And to all others: There is also a (even shorter) real solution. You might want to try.

 

[etotheipi]

What if I have a 3 cycle, or 4-cycle: one set of your kind with two persons knowing each other, and all others know a different one? I don't really see that you covered all cases.

We consider the case where the subset is of size 2, and now suppose that we have either a 3-cycle (with one person left out ) or a 4-cycle of connections. In both of these cycles, any given vertex is connected to two other vertices. This is a contradiction, since this would mean that we have present a subset larger than 2. So wouldn't the 3-cycles and 4-cycles already be covered by the previous casework?]

Edit: Hang on, that's not right, I see your point... sorry!

In the case of a subset of 2 and a 3-cycle, we may pick one from the subset, one from the 3-cycle and the leftover person. In the case of a subset of 2 and the 4-cycle, we may pick one from the subset, and two from opposite vertices of the 4-cycle.

 

[fresh_42]

Edit: Hang on, that's not right, I see your point... sorry!

I hate those questions, so it's my fault. It is hard to keep all possibilities in mind without a strategy. You started well, but the smaller your set is, the more unknowns are in the remaining sample.

Maybe to color a graph of 6 vertices by two colors isn't that bad after all.

 

[physion]

Spoiler: Problem 12

We set $y = x + k$, where k is some integer. I initially attempted this problem by substituting this in and restricting k's range with the discriminant but was confronted with many integer solutions and so thus attempted to restrict it further.

For the case $k = 0, x = y$, one side is more than the other: $x^3 = x^3 + 8x^2 - 6x + 8$.

For the case $k=3$, we similarly find that $(x+3)^3$ is more than the other side.

We are left with $k = 1$ and $k = 2$. For the prior, we find $5x^2 - 9x + 7 = 0$, with no solutions. We are left with $k = 2$, yielding $2x^2 - 18x = 0$, with solutions $x = 0, x = 9$. $y = x + 2$, and so thus we have two pairs of solutions:

\begin{equation}
(0,2) \
(9,11)
\end{equation}

 

[PeroK]

Spoiler: Problem 15

There's likely a nicer way to do it, but this what I first thought of. In the set of 6 people we consider the different cases of the largest possible subset(s) of people who knows everyone else in the subset. This largest subset for any given set of 6 people can only have a size between 1 and 6 inclusive.

If the largest subset has a size between 4 and 6 then the statement is true. If the largest subset has size 3, then there is the possibility of another largest subset of size 3 but in any case the statement is still true.

If the largest subset has a size of 2, then there are a 3 different cases. If there are 3 such subsets in the group of 6, then we can choose one person from each subset to obtain 3 people who will not know each other. If there are 2 such subsets, then we can simply pick remaining other 2 people and any 1 further person from either of these 2 subsets to obtain 3 people who don't know each other. If there is 1 such subset, then the other 4 people don't know each other and the statement is also true.

Finally, if the largest subset has a size of 1, then no-one knows each other and the statement is true. This I think covers all cases.

For the key case where there the largest subset has size two, these subsets may not be disjoint. E.g. you could have $AB, CD, EF$ as pairs who know each other. Now, $C$, for example, might know $A$ but not know $B$. And, if you pick $A, C, E$ you may not have three people who do not know each other.

 

[fresh_42]

Spoiler: Problem 12

We set $y = x + k$, where k is some integer. I initially attempted this problem by substituting this in and restricting k's range with the discriminant but was confronted with many integer solutions and so thus attempted to restrict it further.

For the case $k = 0, x = y$, one side is more than the other: $x^3 = x^3 + 8x^2 - 6x + 8$.

For the case $k=3$, we similarly find that $(x+3)^3$ is more than the other side.

We are left with $k = 1$ and $k = 2$. For the prior, we find $5x^2 - 9x + 7 = 0$, with no solutions. We are left with $k = 2$, yielding $2x^2 - 18x = 0$, with solutions $x = 0, x = 9$. $y = x + 2$, and so thus we have two pairs of solutions:

\begin{equation}
(0,2) \
(9,11)
\end{equation}

What about the cases with negative $k$ or $k>3$?

 

[etotheipi]

For the key case where there the largest subset has size two, these subsets may not be disjoint. E.g. you could have $AB, CD, EF$ as pairs who know each other. Now, $C$, for example, might know $A$ but not know $B$. And, if you pick $A, C, E$ you may not have three people who do not know each other.

But say you were to list out everyone who A knows, e.g. B, C and F. If any of B, C, or F know each other also (e.g. B and F), then you have three people (i.e. A, B, F) who all know each other. If none of them know each other, the statement holds when the set $\{B,C,F\}$ has a size greater or equal to 3. But if that set has a size less than 3, then you could just as easily look at the set of people that A doesn't know and just flip the analysis (i.e. if any of pair of these people also don't know each other, you also get a valid group of 3 people including A). So I think you do always have the group of 3, but the case where the largest subset is of size 2 there are so many different cases that it just becomes annoying... I might go and have breakfast and then come back to it!

 

[physion]

What about the cases with negative $k$ or $k>3$?

As shown above, $x^3 < y^3$ (i.e. the case where k = 0) --> $x < y$ , and thus $k > 0$
Similarly, $y^3 < (x+3)^3$ --> $y < x + 3$
We are left with $k = 1, k = 2.$

 

[fresh_42]

But say you were to list out everyone who A knows, e.g. B, C and F. If any of B, C, or F know each other also (e.g. B and F), then you have three people (i.e. A, B, F) who all know each other. If none of them know each other, the statement holds when the set $\{B,C,F\}$ has a size greater or equal to 3. But if that set has a size less than 3, then you could just as easily look at the set of people that A doesn't know and just flip the analysis (i.e. if any of pair of these people also don't know each other, you also get a valid group of 3 people including A). So I think you do always have the group of 3, but the case where the largest subset is of size 2 there are so many different cases that it just becomes annoying... I might go and have breakfast and then come back to it!

Try my suggestion: use two colors on a graph with six persons. It will become a lot easier to reason!

 

[fresh_42]

As shown above, $x^3 < y^3$ (i.e. the case where k = 0) --> $x < y$ , and thus $k > 0$
Similarly, $y^3 < (x+3)^3$ --> $y < x + 3$
We are left with $k = 1, k = 2.$

You haven't shown this. If you set $k=0$, how can you argue about cases with a different value for $k$? You dealt with $k\in \{\,0,1,2,3\,\}$, so it remains to show that these are the only possible cases.

Your proof goes:

Assume we have a solution $(x,y)$.
Then we can write $y=x+k$.
(At this point you have only the information $k\in \mathbb{Z}$.)
Then you deal with $k\in \{\,0,1,2,3\,\}$.

Where have you ruled out other possibilities?
Hint: Consider $y^3-(x+1)^3$ and $(x+1)^3-y^3$

 

[physion]

Ah, thanks! My reasoning didn't really make much sense looking back on it now.

Let's say that, for negative k, we use $k = -n$ where n > 0, so we have $(x-n)^3 = x^3 + 8x^2 - 6x + 8$

Importantly, $8x^2 - 6x + 8 > 0$ for all positive x.

However, upon expanding, and canceling x^3 on both sides, on the LHS we have

$-3nx^2 + 3n^2x - n^3$, which evidenced by its discriminant $-3n^4$, will only be positive if n < 0, a contradiction. $k > 0$

Alternatively, we look at where we ruled out $k > 3$.

We show that $(x+3)^3 > x^3 + 8x^2 - 6x + 8$,

$x^3 + 9x^2 + 27x + 27 = x^3 + 8x^2 - 6x + 8$

We have a difference $x^2 + 33x + 19$, which is positive for all $x > 0$. (it has negative roots).

Thus, $(x+3)^3 > x^3 + 8x^2 - 6x + 8$

As $x+k > x+3$ where $k > 3$, $(x+k)^3 > (x+3)^3$. If $k > 3$, the LHS would still always be more than the RHS, for $x > 0$.

Again, thanks for your help!

 

[fresh_42]

Ah, thanks! My reasoning didn't really make much sense looking back on it now.

Let's say that, for negative k, we use $k = -n$ where n > 0, so we have $(x-n)^3 = x^3 + 8x^2 - 6x + 8$

Importantly, $8x^2 - 6x + 8 > 0$ for all positive x.

However, upon expanding, and canceling x^3 on both sides, on the LHS we have

$-3nx^2 + 3n^2x - n^3$, which evidenced by its discriminant $-3n^4$, will only be positive if n < 0, a contradiction. $k > 0$

Alternatively, we look at where we ruled out $k > 3$.

We show that $(x+3)^3 > x^3 + 8x^2 - 6x + 8$,

$x^3 + 9x^2 + 27x + 27 = x^3 + 8x^2 - 6x + 8$

We have a difference $x^2 + 33x + 19$, which is positive for all $x > 0$. (it has negative roots).

Thus, $(x+3)^3 > x^3 + 8x^2 - 6x + 8$

As $x+k > x+3$ where $k > 3$, $(x+k)^3 > (x+3)^3$. If $k > 3$, the LHS would still always be more than the RHS, for $x > 0$.

Again, thanks for your help!

Now you ruled out everything except $x=0$ and $k>3$. Bad this is due to bad notation, as you can assume $x\geq 0$ instead of $x>0$.

 

[physion]

Apologies, doesn't my line of reasoning rule out all k > 3, not everything except it? Also, my argument should indeed instead specify x 'more than or equal' to 0 and 'non-negative' instead of 'positive', but the line of reasoning should still hold.

 

[fresh_42]

Apologies, doesn't my line of reasoning rule out all k > 3, not everything except it? Also, my argument should indeed instead specify x 'more than or equal' to 0 and 'non-negative' instead of 'positive', but the line of reasoning should still hold.

Yes, as written: a bad notation. Your cases are written a bit confusing, so it's hard to keep track. A more elegant way is to show that $y^3-(x+1)^3 > 0$ and $(x+3)^3-y^3>0$ so only $y=x+2$ is left.

 

[benorin]

6. a.) (FR) Let $u_1,\ldots,u_n$ be solutions of the one dimensional heat equation $\dfrac{du}{dt}-\dfrac{d^2u}{dx^2}=0\;(x\in \mathbb{R},t>0).$ Show that $$u(x_1,\ldots,x_n,t):=\displaystyle{\prod_{k=1}^n}u_k(x_k,t)$$ is a solution of the $n$ dimensional heat equation $\dfrac{\partial u}{\partial t}-\Delta u=0.$

Work:
$$\dfrac{\partial u}{\partial t} = \displaystyle{\sum_{j=1}^n} \dfrac{\partial u_j}{\partial t} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t)$$
and
$$\Delta u = \displaystyle{\sum_{j=1}^n} \dfrac{\partial ^2 u_j}{\partial x_j ^2} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t)$$
Hence
$$\begin{gathered} \dfrac{\partial u}{\partial t}-\Delta u = \displaystyle{\sum_{j=1}^n} \dfrac{\partial u_j}{\partial t} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t) - \displaystyle{\sum_{j=1}^n} \dfrac{\partial ^2 u_j}{\partial x_j ^2} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t) \\ = \displaystyle{\sum_{j=1}^n}\underbrace{\left( \dfrac{\partial u_j}{\partial t} - \dfrac{\partial ^2 u_j}{\partial x_j ^2}\right)}_{=0\forall j} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t)=0 \\ \end{gathered}$$

b.) (FR) Calculate a solution for
$$
\begin{cases}
\dfrac{\partial u}{\partial t}(x,t)-\Delta u(x,t)=0 &\text{ for } x\in \mathbb{R}^3,\,t>0 \\[6pt]
u(x,0)=x_1^2x_2^2x_3&\text{ for } x=(x_1,x_2,x_3)\in \mathbb{R}^3
\end{cases}
$$

Note: My PDEs are 2 decades rusty, I tried to look for separable solutions and got

$$u(x,t)=c_0e^{kt}\prod_{j=1}^3\left( c_j e^{\sqrt{k_j}x_j}+c_j^{\prime} e^{-\sqrt{k_j}x_j}\right)$$

where $k_3:=k-|k_1|-|k_2|$ and here's how I know that the separable solution didn't work, because the boundary condition becomes

$$u(x,0)=c_0\prod_{j=1}^3\left( c_j e^{\sqrt{k_j}x_j}+c_j^{\prime} e^{-\sqrt{k_j}x_j}\right) =x_1^2x_2^2x_3$$

which... I give.

 

[fresh_42]

6. a.) (FR) Let $u_1,\ldots,u_n$ be solutions of the one dimensional heat equation $\dfrac{du}{dt}-\dfrac{d^2u}{dx^2}=0\;(x\in \mathbb{R},t>0).$ Show that $$u(x_1,\ldots,x_n,t):=\displaystyle{\prod_{k=1}^n}u_k(x_k,t)$$ is a solution of the $n$ dimensional heat equation $\dfrac{\partial u}{\partial t}-\Delta u=0.$

Work:
$$\dfrac{\partial u}{\partial t} = \displaystyle{\sum_{j=1}^n} \dfrac{\partial u_j}{\partial t} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t)$$
and
$$\Delta u = \displaystyle{\sum_{j=1}^n} \dfrac{\partial ^2 u_j}{\partial x_j ^2} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t)$$
Hence
$$\begin{gathered} \dfrac{\partial u}{\partial t}-\Delta u = \displaystyle{\sum_{j=1}^n} \dfrac{\partial u_j}{\partial t} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t) - \displaystyle{\sum_{j=1}^n} \dfrac{\partial ^2 u_j}{\partial x_j ^2} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t) \\ = \displaystyle{\sum_{j=1}^n}\underbrace{\left( \dfrac{\partial u_j}{\partial t} - \dfrac{\partial ^2 u_j}{\partial x_j ^2}\right)}_{=0\forall j} \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t)=0 \\ \end{gathered}$$

b.) (FR) Calculate a solution for
$$
\begin{cases}
\dfrac{\partial u}{\partial t}(x,t)-\Delta u(x,t)=0 &\text{ for } x\in \mathbb{R}^3,\,t>0 \\[6pt]
u(x,0)=x_1^2x_2^2x_3&\text{ for } x=(x_1,x_2,x_3)\in \mathbb{R}^3
\end{cases}
$$

Note: My PDEs are 2 decades rusty, I tried to look for separable solutions and got

$$u(x,t)=c_0e^{kt}\prod_{j=1}^3\left( c_j e^{\sqrt{k_j}x_j}+c_j^{\prime} e^{-\sqrt{k_j}x_j}\right)$$

where $k_3:=k-|k_1|-|k_2|$ and here's how I know that the separable solution didn't work, because the boundary condition becomes

$$u(x,0)=c_0\prod_{j=1}^3\left( c_j e^{\sqrt{k_j}x_j}+c_j^{\prime} e^{-\sqrt{k_j}x_j}\right) =x_1^2x_2^2x_3$$

which... I give.

The first part is correct, but I don't really see what you did in part b). However, you can use the "separation" suggested in the first part.

 

[benorin]

6 b.) (FR) Calculate a solution for
$$
\begin{cases}
\dfrac{\partial u}{\partial t}(x,t)-\Delta u(x,t)=0 &\text{ for } x\in \mathbb{R}^3,\,t>0 \\[6pt]
u(x,0)=x_1^2x_2^2x_3&\text{ for } x=(x_1,x_2,x_3)\in \mathbb{R}^3
\end{cases}
$$
Work:
$$u(x_1,x_2,x_3,t):=\displaystyle{\prod_{k=1}^n}u_k(x_k,t)$$
From part (a), the given heat equation becomes
$$\dfrac{\partial u}{\partial t}-\Delta u = \displaystyle{\sum_{j=1}^n}\left( \dfrac{\partial u_j}{\partial t} - \dfrac{\partial ^2 u_j}{\partial x_j ^2}\right) \displaystyle{\prod_{k=1,k\neq j}^n}u_k(x_k,t)=0$$
divide through by $u_1u_2u_3$ to get
$$ \displaystyle{\sum_{j=1}^n}\dfrac{1}{u_j}\left( \dfrac{\partial u_j}{\partial t} - \dfrac{\partial ^2 u_j}{\partial x_j ^2}\right) =0$$
the summand, for each fixed $j$, represents a function solely of $x_j$ and $t$ whence it reduces to at most a constant $\lambda _j$ such that $\sum_j \lambda _j =0$ hence we have the 3 PDEs
$$\dfrac{1}{u_j}\left( \dfrac{\partial u_j}{\partial t} - \dfrac{\partial ^2 u_j}{\partial x_j ^2}\right) =\lambda _j ,\quad \forall 1\leq j\leq 3$$
$$\Rightarrow \dfrac{\partial u_j}{\partial t} - \dfrac{\partial ^2 u_j}{\partial x_j ^2} -\lambda _j u_j =0,\quad \forall 1\leq j\leq 3$$
Assume that there exists a power series solution for each $u_j(x_j,t)$, namely
$$u_j(x_j,t):=\sum_{n=0}^\infty\sum_{m=0}^\infty a_n^{(j)}b_m^{(j)}x_j^{n}t^{m}$$
The initial condition implies
$$u(x,0)=x_1^2x_2^2x_3\Rightarrow u_1(x_1,0)=x_1^2\wedge u_2(x_2,0)=x_2^2\wedge u_3(x_3,0)=x_3$$
hence
$$\boxed{\text{ for }j=1,2,} \, \, u_j(x_j,0):=\sum_{n=0}^\infty a_n^{(j)}b_0^{(j)}x_j^{n}=x_j^2\Rightarrow a_2^{(j)}b_0^{(j)}=1\wedge a_n^{(j)}=0\forall n\neq 2$$
and
$$u_3(x_3,0):=\sum_{n=0}^\infty a_n^{(3)}b_0^{(3)}x_3^{n}=x_3\Rightarrow a_1^{(3)}b_0^{(3)}=1\wedge a_n^{(3)}=0\forall n\neq 1$$
whence
$$\begin{gathered} \boxed{\text{ for }j=1,2,} \, \, u_j(x_j,t):=x_j^{2}\left( 1+a_2^{(j)}\sum_{m=0}^\infty b_{m+1}^{(j)}t^{m+1}\right) \\ \Rightarrow \dfrac{\partial u_j}{\partial t}=x_j^{2}a_2^{(j)}\sum_{m=0}^\infty (m +1)b_{m+1}^{(j)}t^{m}= x_j^{2}a_2^{(j)}b_{1}^{(j)}+x_j^{2}a_2^{(j)}\sum_{m=0}^\infty (m+2) b_{m+2}^{(j)}t^{m+1}\\ \Rightarrow \dfrac{\partial ^2 u_j}{\partial x_j ^2}=2+2a_2^{(j)}\sum_{m=0}^\infty b_{m+1}^{(j)}t^{m+1}\\ \end{gathered}$$
and
$$\begin{gathered} u_3(x_3,t):=x_3\left( 1+a_1^{(3)}\sum_{m=0}^\infty b_{m+1}^{(3)}t^{m+1}\right) \\ \Rightarrow \dfrac{\partial u_3}{\partial t}=x_3a_1^{(3)}\sum_{m=0}^\infty (m+1) b_{m+1}^{(3)}t^{m} = x_3a_1^{(3)}b_1^{(3)}+x_3a_1^{(3)}\sum_{m=0}^\infty (m+2) b_{m+2}^{(3)}t^{m+1}\\ \Rightarrow \dfrac{\partial ^2 u_3}{\partial x_3 ^2}=0 \\ \end{gathered}$$
Thus
$$\begin{gathered} \boxed{\text{ for }j=1,2,} \, \,\dfrac{\partial u_j}{\partial t}-\dfrac{\partial ^2 u_j}{\partial x_j ^2}-\lambda _j u_j(x_j,t)=0\\ \Rightarrow \left( x_j^{2}a_2^{(j)}b_{1}^{(j)} \, \, \boxed{\boxed{-2}} \, \, -\lambda _j x_j^{2}\right) t^0 +\sum_{m=0}^\infty\left[ x_j^2 a_2^{(j)}(m+2) b_{m+2}^{(j)}-2a_2^{(j)}b_{m+1}^{(j)}-\lambda _j x_j^{2}a_2^{(j)}b_{m+1}^{(j)}\right] t^{m+1} =0\\ \end{gathered}$$

@fresh_42 at this point in the problem I know I've made an error because of that $\boxed{\boxed{-2}}$ term (the boxes were to just highlight the presence of the $-2$), it shouldn't be there as the only other terms in the coefficient of the $t^0$ term have a factor of $x_j^2$ in them and hence cannot cancel the $-2$ term effectively saying that $-2=0$. What was my error? Maybe I can't assume series solution exists? I've redone this problem several times and as my PDEs are kinda shaky I'm arriving at different dead-ends each time I work it through. Time for me to get some help. Thank you for your time!

 

[benorin]

@fresh_42 read this first: I had an idea right after posting that^, I think this part of the previous post "the summand, for each fixed $j$, represents a function solely of $x_j$ and $t$ whence it reduces to at most a constant $\lambda _j$ such that $\sum_j \lambda _j =0$..." part might be incorrect since instead of reducing to constant functions they should have been reduced to at most functions of $t$ that sum to zero?

 

[fresh_42]

@fresh_42 read this first: I had an idea right after posting that^, I think this part of the previous post "the summand, for each fixed $j$, represents a function solely of $x_j$ and $t$ whence it reduces to at most a constant $\lambda _j$ such that $\sum_j \lambda _j =0$..." part might be incorrect since instead of reducing to constant functions they should have been reduced to at most functions of $t$ that sum to zero?

I cannot say whether my solution is unique, so I'm not sure whether or what was wrong with yours. I simply used part one to separate dimensions, and made an ansatz with separated variables: $u_k(x_k,t)=v_k(x_k)+w_k(t)$ for the rest. At least this keeps powers low.

 

[benorin]

...and made an ansatz with separated variables: $u_k(x_k,t)=v_k(x_k)+w_k(t)$ for the rest.

I was looking for separated solutions too, just in the form of a product instead of a sum (no joy). But wouldn't such a possibility (sum) be included in my power series solution though? I used double power series thinking that the only restriction I imposed was that I wouldn't be able to identify solutions not having a power series expansion

 

[fresh_42]

I was looking for separated solutions too, just in the form of a product instead of a sum (no joy). But wouldn't such a possibility (sum) be included in my power series solution though? I used double power series thinking that the only restriction I imposed was that I wouldn't be able to identify solutions not having a power series expansion

Yes, they should be included. The trick is to use the conditions such that the series break down early. We have something like $f'+g''=0$ and if the variables are separated, we can simply integrate.

 

[wrobel]

Spoiler: 2c

inductive limit topology:)

 

[fresh_42]

Spoiler: 2c

inductive limit topology:)

Would you mind to define the open sets explicitly?

 

[benorin]

6 b) Answer derived using separable sum for each $u_i$ I got is
$$u(x_1,x_2,x_3,t)=(x_1^2+2t)(x_2^2+2t)x_3$$
and I checked by hand that this indeed solves both the initial conditions and the heat equation. @fresh_42 your ansatz made it really, really easy.

 

[benorin]

I have a Q regarding the alternate solution of #3: is $\int_{-\infty}^{+\infty}\dfrac{| x\sin(\alpha x) |}{1+x^2}\,dx$ (absolutely) convergent? Because I don't think it is, tell you why after some quick work. (Btw, using the upper bound of $1$ for sine yields a divergent integral).
I was attempting the differentiation under the integral sign solution:
Let $$I(\alpha ) :=\int_{-\infty}^{+\infty}\dfrac{\cos(\alpha x)}{1+x^2}\,dx \quad (\alpha \geq 0)$$
Then $$I(\alpha )-\dfrac{d^2I}{d\alpha ^2}=\int_{-\infty}^{+\infty}\dfrac{\cos (\alpha x)}{1+x^2}\,dx+\int_{-\infty}^{+\infty}\dfrac{x^2\cos(\alpha x)}{1+x^2}\,dx = \int_{-\infty}^{+\infty}\cos (\alpha x)\, dx$$
where the latter integral is obviously divergent. But @cbarker1 's solution stated that
$$I(\alpha )=\pi e^{-\alpha}\Rightarrow I(\alpha )-\dfrac{d^2I}{d\alpha ^2}=0$$
so I'm thinking that the second application of differentiation under the integral sign was not justified? The first application was clearly justified as $I(\alpha )$ is absolutely convergent. Also note that attempted DE's with the first derivative lead to some not so easy integrals, ugh.

 

[fresh_42]

I solved it by foot, only using symmetry properties, substitutions and integration by parts, and Fubini. My result is the same.

 

[mathwonk]

Question on #7: did you want a non - zero quotient module?

 

[fresh_42]

Question on #7: did you want a non - zero quotient module?

That would be nice of course.

 

[benorin]

1. Let $1<p<4$ and $f\in L^p((1,\infty))$ with the Lebesgue measure $\lambda$. We define $g\, : \,(1,\infty)\longrightarrow \mathbb{R}$ by
$$
g(x)=\dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\,d\lambda(t).
$$
Show that there exists a constant $C=C(p)$ which depends on $p$ but not on $f$ such that $\|g\|_2 \leq C\cdot \|f\|_p$ so $g\in L^2((1,\infty)).$ (FR)

Work: Let $\tfrac{1}{p}+\tfrac{1}{q}=1$ and note that $1<p<4\Rightarrow \tfrac{4}{3}<q<\infty$.
$$\begin{gathered} \| g \| _2^2 =\int_1^\infty\left| \dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\, d\lambda(t)\right| ^2\, d\lambda(x) \\ \leq \int_1^\infty \dfrac{1}{x^2} \left\{ \int_x^{10x} | f | ^p\, d\lambda(t)\right\} ^{\tfrac{2}{p}}\left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ \end{gathered} $$
by Holder's inequality, also note that $\forall x\in (1,\infty ),\quad \left\{ \int_x^{10x} | f | ^p\, d\lambda(t)\right\} ^{\tfrac{2}{p}}\leq \| f \| _p^2$ hence it follows that
$$\begin{gathered} \| g \| _2^2 \leq \| f \| _p^2 \int_1^\infty \dfrac{1}{x^2} \left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ = \tfrac{q}{4-q}\left( 10^\tfrac{4-q}{4}-1\right) \| f \| _p^2 \int_1^\infty x^{\tfrac{2}{q}-3}\, d\lambda(x) \\ \end{gathered} \\ =\tfrac{q^2}{2(4-q)(1-q)}\left( 1-10^\tfrac{4-q}{4}\right) \| f \| _p^2 $$

and the desired result comes upon taking square roots of both sides a remembering that $q=q(p)$ (q may be thought of as a function of $p$).
Edit: post contains an error!

 

[fresh_42]

I think you made an integration error, but I'm not sure since you haven't shown what you've done. And what if $q=4$?