Challenge Math Challenge - November 2020

[martinbn]

About 1: It is a very standard maximum principle for the heat equation.

About 8: Using the hint, where $s$ looks more like energy density and it should be integrated over $\Omega$ to get the total energy $E=\int_\Omega s$, doing the usual pde thing, namely differentiate, use the Green's formula, the boundary conditions one gets that

$\frac {dE}{dt}=-aD\int_\Omega \frac1u^2|\nabla u|^2-D\int_\Omega \frac1v^2|\nabla v|^2$

So it is decreasing (well, non-increasing). Easy to see that $E$ is bounded, so the above goes to zero.

One comment: either $\Omega$ should accully be bounded, or if it is just of finite volume, one need sufficient decay of $u$ and $v$ along with the boundary conditions.

 

[martinbn]

For 14: The observations that help is that any real zero has to be negative and that if $a$ is a zero of $f_n(x)$, then $f'_n(a)=-\frac{a^n}{n!}$. In fact for even $n$ there are no zeros, for odd exactly one.

 

[songoku]

Correct. You could have shortened it by investigating the function $L(x)=\dfrac{\sin^4x-\cos^4x}{\tan^4x-\cot^4x}$ on $(0,\pi/4)$.

That can be simplified to $L(x)=\dfrac{(\sin x . \cos x)^4}{\sin^4x+\cos^4x}$

Then how to continue? How to approach it differently from the one I did before? Thanks

 

[songoku]

Another mess

Spoiler: Question 14

$$f_n(x)=1+x+...+\frac{x^n}{n!}$$
$$f'_n(x)=1+x+...+\frac{x^{n-1}}{(n-1)!}=f_n(x)-\frac{x^n}{n!}$$

Let $x=a$ is the zero of $f_n(x)$ so $f_n(a)=0$, then $f'_n(a)=-\dfrac{a^n}{n!}$

Only negative value of $a$ can satisfy $f_n(a)=0$ so the sign of $f'_n(a)$ will depend on whether $n$ is odd or even.1) For odd value of $n$, the sign of $f'_n(a)$ will be positive and for even value of $n$, the sign of $f'_n(a)$ will be negative2) Checking the end behavior of $f_n(x)$
(i) for even value of $n$
$$\lim_{x \to \pm \infty} f_n(x)=+ \infty$$

(ii) for odd value of $n$
$$\lim_{x \to \infty} f_n(x)=+ \infty~\text{and}~\lim_{x \to -\infty} f_n(x) = -\infty$$3) If I "trace" the graph of $f_n(x)$ on interval $(-\infty, \infty)$:

(i) for even value of $n$
There must be at least one turning point to satisfy the end behavior of $f_n(x)$ so there should be at least two zeroes (let say the zeroes are $a$ and $b$ where $a<b<0$).

When the graph crosses x-axis at $x=a$, the slope of $f_n(x)$ will be negative since the graph will be decreasing on interval $(-\infty,a)$ and when the graph crosses the x-axis again at $x=b$, the slope should be positive, which contradicts the sign of $f'_n(x)$ in part (1).

Therefore, for even value of $n$, the graph can not have zero(ii) for odd value of $n$
When the graph crosses x-axis at $x=a$, the slope will be positive since the graph will be increasing on interval $(-\infty,a)$.

Assume $f_n(x)$ has another zero at $x=b$. To have another zero at $x=b$, there must be a turning point and when the graph crosses x-axis at $x=b$, the slope will be negative, which contradicts the sign of $f'_n(x)$ in part (1).

Since a turning point is not a must to satisfy the end behavior of $f_n(x)$, we can rule out the case where $f_n(x)$ has other zero besides $a$

Therefore, for odd value of $n$, the graph only has one zero

--------------------------------------------------------------------------------------------------------------------------------------

Conclusion:
Since $f_n (x)$ has no zero when $n$ is even and only has one zero when $n$ is odd, it is proven that $f_n (x)$ has at most one real zero

 

[fresh_42]

That can be simplified to $L(x)=\dfrac{(\sin x . \cos x)^4}{\sin^4x+\cos^4x}$

Then how to continue? How to approach it differently from the one I did before? Thanks

You first observe via symmetry considerations why it's sufficient to consider the interval $(0,\pi/4)$. Then you note that $L(x)$ is strictly monotone increasing on that interval, by writing $L(x)=F(\sin(2x))$ as function of the double angle.

 

[fresh_42]

Another mess "Question 14"
$$f_n(x)=1+x+...+\frac{x^n}{n!}$$
$$f'_n(x)=1+x+...+\frac{x^{n-1}}{(n-1)!}=f_n(x)-\frac{x^n}{n!}$$

Let $x=a$ is the zero of $f_n(x)$ so $f_n(a)=0$, then $f'_n(a)=-\dfrac{a^n}{n!}$

Only negative value of $a$ can satisfy $f_n(a)=0$ so the sign of $f'_n(a)$ will depend on whether $n$ is odd or even.1) For odd value of $n$, the sign of $f'_n(a)$ will be positive and for even value of $n$, the sign of $f'_n(a)$ will be negative2) Checking the end behavior of $f_n(x)$
(i) for even value of $n$
$$\lim_{x \to \pm \infty} f_n(x)=+ \infty$$

(ii) for odd value of $n$
$$\lim_{x \to \infty} f_n(x)=+ \infty~\text{and}~\lim_{x \to -\infty} f_n(x) = -\infty$$3) If I "trace" the graph of $f_n(x)$ on interval $(-\infty, \infty)$:

(i) for even value of $n$
There must be at least one turning point to satisfy the end behavior of $f_n(x)$ so there should be at least two zeroes (let say the zeroes are $a$ and $b$ where $a<b<0$).

Why should there be a zero at all? You have to make two cases here: no zero and we are done, so second case, we assume a zero. If there is one, there have to be at least two with a turning point in between.

When the graph crosses x-axis at $x=a$, the slope of $f_n(x)$ will be negative since the graph will be decreasing on interval $(-\infty,a)$ and when the graph crosses the x-axis again at $x=b$, the slope should be positive, which contradicts the sign of $f'_n(x)$ in part (1).

Therefore, for even value of $n$, the graph can not have zero

It would have been far easier, if you only concentrated on $b$. $a$ is irrelevant and confusing, the more as there could be many more zeroes in between. You only have at least two, and did not specify which of the possibly many you are referring to. Just take the rightmost zero $b$ and forget about the rest.

(ii) for odd value of $n$
When the graph crosses x-axis at $x=a$, the slope will be positive since the graph will be increasing on interval $(-\infty,a)$.

Assume $f_n(x)$ has another zero at $x=b$. To have another zero at $x=b$, there must be a turning point and when the graph crosses x-axis at $x=b$, the slope will be negative, which contradicts the sign of $f'_n(x)$ in part (1).

Same as before. Just consider the second to the leftmost zero $b$.

Since a turning point is not a must to satisfy the end behavior of $f_n(x)$, we can rule out the case where $f_n(x)$ has other zero besides $a$

I neither understand this nor do I see where you need it. The turning points are irrelevant.

Therefore, for odd value of $n$, the graph only has one zero

Your idea is the same in all cases:

Whenever the function crosses the line from negative to positive, the slope has to be positive. This rules out even $n$ immediately. Now for odd $n$ it means, we can only cross the line once, since positive to negative would imply a negative slope, again contradicting (1).

--------------------------------------------------------------------------------------------------------------------------------------

Conclusion:
Since $f_n (x)$ has no zero when $n$ is even and only has one zero when $n$ is odd, it is proven that $f_n (x)$ has at most one real zero

Conclusion:
You should concentrate on your basic idea and omit all the confusing irrelevant rest. Once you had (1) you were basically done. Of course purists might perform an induction instead, but this is not necessary.

 

[benorin]

1. Let $u(x,t)$ satisfy the one dimensional diffusion equation $u_t=Du_{xx}$ in a space-time rectangle $R=\{0\leq x\leq l,0\leq t\leq T\}$, then the maximum value of $u(x,t)$ is assumed either on the initial line $(t=0)$ or on the boundary lines $(x=0 \,or\,x=l )$. $D > 0.$

Spoiler: #1) hand-wavy proof

Well it seems obvious that the maximum of $u(x,t)$ would either occur on the boundary lines $t=0$ or $x=0$ or $x=l$ since suppose it were the one dimensional heat equation of a rod of length $l$, either the temperature $u(x,t)$ is maximized at the radiating point source to begin with or one end of the rod inevitably becomes the hottest point. But I suppose you want a math proof? Well, I'm still thinking...
I found this nifty 1-D heat equation slides for a worked out example problem.

 

[fresh_42]

Spoiler: #1) hand-wavy proof

Well it seems obvious that the maximum of $u(x,t)$ would either occur on the boundary lines $t=0$ or $x=0$ or $x=l$ since suppose it were the one dimensional heat equation of a rod of length $l$, either the temperature $u(x,t)$ is maximized at the radiating point source to begin with or one end of the rod inevitably becomes the hottest point. But I suppose you want a math proof? Well, I'm still thinking...
I found this nifty 1-D heat equation slides for a worked out example problem.

You have to rule out the possibility that the maximum is somewhere in the inner of the spacetime diagram. It could heat up and cool down at some point. If we knew $u_{xx}<0$ at the maximum, we would have a contradiction for inner points. But $u_{xx}=0$ cannot be ruled out.

My solution uses a standard technique: Assume a maximum at the three boundaries, vary that value, and show that it is impossible to get a lower value. Good old calculus of variations.

 

[songoku]

Why should there be a zero at all? You have to make two cases here: no zero and we are done, so second case, we assume a zero. If there is one, there have to be at least two with a turning point in between.

It would have been far easier, if you only concentrated on $b$. $a$ is irrelevant and confusing, the more as there could be many more zeroes in between. You only have at least two, and did not specify which of the possibly many you are referring to. Just take the rightmost zero $b$ and forget about the rest.

Same as before. Just consider the second to the leftmost zero $b$.

I neither understand this nor do I see where you need it. The turning points are irrelevant.

Your idea is the same in all cases:

Whenever the function crosses the line from negative to positive, the slope has to be positive. This rules out even $n$ immediately. Now for odd $n$ it means, we can only cross the line once, since positive to negative would imply a negative slope, again contradicting (1).

Conclusion:
You should concentrate on your basic idea and omit all the confusing irrelevant rest. Once you had (1) you were basically done. Of course purists might perform an induction instead, but this is not necessary.

Ah I see. Thank you very much