one of the most beautiful problems of all time

[logistic_guy]

$\displaystyle \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2} + 2\beta \frac{\partial u}{\partial t}$

 

[khansaheb]

$\displaystyle \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2} + 2\beta \frac{\partial u}{\partial t}$

There is NO QUESTION to answer. And,

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[logistic_guy]

Nine months ago a very little lizard sneaked into my room!

Please show us what you have tried and exactly where you are stuck.

I was stuck on how to catch him, so I ignored him.

During these $\displaystyle 9$ months, it was very rare to see him because he could hide himself pretty good. But I knew that he was still in my room because I heard his sounds from time to time. Now he became very big but he is still afraid from me.

I named him khan!

 

[khansaheb]

still afraid from me

You need to go back to "grammar" school.

 

[logistic_guy]

You need to go back to "grammar" school.

I fed you, sheltered you, and raised you for $\displaystyle 9$ months. And this is how you repay me? Betrayal. Get out of my room. Now!

 

[jonah2.0]

Beer soaked non sequitur ramblings follow.

Nine months ago a very little lizard sneaked into my room!


I was stuck on how to catch him, so I ignored him.

During these $\displaystyle 9$ months, it was very rare to see him because he could hide himself pretty good. But I knew that he was still in my room because I heard his sounds from time to time. Now he became very big but he is still afraid from me.

I named him khan!

The other day, shortly after turning off the hot water shower, I stood awhile in uffish thought as I placed my left hand in my right armpit and my right hand on my chin. I then began to wonder if I should just dry off and spend the whole day in bed by drinking or drugging myself back to sleep. I decided to urinate while gazing at the bathroom wall clock and noted that it took about 30 seconds to partially empty my bladder. Feeling that I could still expel more urine, I lathered some shampoo on them good old balls (and d.i.c.k.) to massage them balls a little bit (about 5 minutes) in the hopes that I could still convince me bladder to release what felt like trapped urine down there because of an enlarged prostate. Them trapped urine eventually came out. Decided to take some pills to sleep some more.

 

[logistic_guy]

Beer soaked non sequitur ramblings follow.


The other day, shortly after turning off the hot water shower, I stood awhile in uffish thought as I placed my left hand in my right armpit and my right hand on my chin. I then began to wonder if I should just dry off and spend the whole day in bed by drinking or drugging myself back to sleep. I decided to urinate while gazing at the bathroom wall clock and noted that it took about 30 seconds to partially empty my bladder. Feeling that I could still expel more urine, I lathered some shampoo on them good old balls (and d.i.c.k.) to massage them balls a little bit (about 5 minutes) in the hopes that I could still convince me bladder to release what felt like trapped urine down there because of an enlarged prostate. Them trapped urine eventually came out. Decided to take some pills to sleep some more.

I am really glad that you were able to empty your bladder in a short time. But if you have not yet heard there is a new technology on the market: a diaper that stays effective for three days straight and lets you relieve yourself without needing to go to the bathroom. It is mostly intended for the elderly, but I think that you are still in the prime of your youth.

What really puzzled me, though, is how someone your age has old balls? Did you undergo some kind of replacement surgery at some point in your life?

 

[jonah2.0]

Beer induced non sequitur ramblings follow.

I am really glad that you were able to empty your bladder in a short time. But if you have not yet heard there is a new technology on the market: a diaper that stays effective for three days straight and lets you relieve yourself without needing to go to the bathroom. It is mostly intended for the elderly, but I think that you are still in the prime of your youth.

What really puzzled me, though, is how someone your age has old balls? Did you undergo some kind of replacement surgery at some point in your life?

After taking a shower, I decided to dry off in front of a powerful electric fan. It was quite surprising to discover after I dried off that my left and right armpits smelled differently from one another.

 

[logistic_guy]

$\displaystyle u(x,t) = X(x)T(t)$

$\displaystyle T\frac{\partial^2 X}{\partial x^2} = X\frac{\partial^2 T}{\partial t^2} + 2\beta X \frac{\partial T}{\partial t}$

 

[logistic_guy]

$\displaystyle T\frac{\partial^2 X}{\partial x^2} = X\frac{\partial^2 T}{\partial t^2} + 2\beta X \frac{\partial T}{\partial t}$

$\displaystyle \frac{1}{X}\frac{\partial^2 X}{\partial x^2} = \frac{1}{T}\left(\frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t}\right)$


$\displaystyle \frac{1}{X}\frac{\partial^2 X}{\partial x^2} = \frac{1}{T}\left(\frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t}\right) = -\lambda$


$\displaystyle \frac{\partial^2 X}{\partial x^2} + \lambda X = 0$


$\displaystyle \frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t} + \lambda T = 0$

 

[logistic_guy]

$\displaystyle \frac{\partial^2 X}{\partial x^2} + \lambda X = 0$

Let us focus on the first equation first.

If $\displaystyle \lambda = 0$, then we have:

$\displaystyle \frac{\partial^2 X}{\partial x^2} = 0$

The solution to this differential equation is:

$\displaystyle X(x) = c_1x + c_2$

 

[logistic_guy]

$\displaystyle \frac{\partial^2 X}{\partial x^2} + \lambda X = 0$

If $\displaystyle \lambda = \mu^2 > 0$, then

$\displaystyle \frac{\partial^2 X}{\partial x^2} + \mu^2 X = 0$

The solution to this differential equation is:

$\displaystyle X(x) = c_3\cos \mu x + c_4\sin \mu x$

 

[logistic_guy]

If $\displaystyle \lambda = -\mu^2 < 0$, then

$\displaystyle \frac{\partial^2 X}{\partial x^2} - \mu^2 X = 0$

The solution to this differential equation is:

$\displaystyle X(x) = c_5\cosh \mu x + c_6\sinh \mu x$

 

[logistic_guy]

$\displaystyle \frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t} + \lambda T = 0$

Now we go to the second differential equation. And it seems that solving this equation will be more fun!

If $\displaystyle \lambda = 0$, then we have:

$\displaystyle \frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t} = 0$

Let $\displaystyle R'(t) = T''(t)$, then we have:

$\displaystyle R' + 2\beta R = 0$

It is easy to solve this differential equation.

$\displaystyle R' = -2\beta R$

Or

$\displaystyle \frac{dR}{dt} = -2\beta R$

$\displaystyle \frac{1}{R} \ dR = -2\beta \ dt$

$\displaystyle \int \frac{1}{R} \ dR = \int -2\beta \ dt$

$\displaystyle \ln R = -2\beta t + c$

$\displaystyle R = e^{-2\beta t + c} = Ae^{-2\beta t}$

We know that $\displaystyle T'(t) = R(t)$, then

$\displaystyle \frac{dT}{dt} = Ae^{-2\beta t}$

$\displaystyle dT = Ae^{-2\beta t} \ dt$

$\displaystyle \int dT = \int Ae^{-2\beta t} \ dt$

$\displaystyle T = \frac{1}{-2\beta}Ae^{-2\beta t} + B$

Or

$\displaystyle T(t) = \frac{c_7}{\beta}e^{-2\beta t} + c_8$

 

[logistic_guy]

$\displaystyle \frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t} + \lambda T = 0$

If $\displaystyle \lambda = \mu^2 > 0$

then we have:

$\displaystyle \frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t} + \mu^2 T = 0$

The characteristic equation gives:

$\displaystyle r = \frac{-2\beta \pm \sqrt{4\beta^2 - 4(1)\mu^2}}{2(1)} = -\beta \pm \sqrt{\beta^2 - \mu^2}$

Now to find $\displaystyle r_1$ and $\displaystyle r_2$, we need to know more about the values of $\displaystyle \beta$.

If we assume that $\displaystyle 0 < \beta < 1$, then the value inside the square root will be negative. This gives us:

$\displaystyle r = -\beta \pm \sqrt{-(\mu^2 - \beta^2)}$

Let $\displaystyle \alpha^2 = \mu^2 - \beta^2$

$\displaystyle r = -\beta \pm \sqrt{-\alpha^2} = -\beta \pm i\alpha$

Then, the solution to the differential equation is:

$\displaystyle T(t) = c_9e^{-\beta t}\cos \alpha t + c_{10}e^{-\beta t}\sin \alpha t$

 

[logistic_guy]

$\displaystyle \frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t} + \lambda T = 0$

If $\displaystyle \lambda = -\mu^2 < 0$

then, we have:

$\displaystyle \frac{\partial^2 T}{\partial t^2} + 2\beta \frac{\partial T}{\partial t} - \mu^2 T = 0$

The characteristic equation gives:

$\displaystyle r = \frac{-2\beta \pm \sqrt{4\beta^2 - 4(1)(-\mu^2)}}{2(1)} = -\beta \pm \sqrt{\beta^2 + \mu^2}$

Let $\displaystyle \alpha^2 = \beta^2 + \mu^2$

$\displaystyle r = -\beta \pm \sqrt{\alpha^2} = -\beta \pm \alpha$

Then, the solution to this differential equation is:

$\displaystyle T(t) = c_{11}e^{-\beta t}\cosh \alpha t + c_{12}e^{-\beta t}\sinh \alpha t$