3.

$$f=[\cos (2x)+3]^4$$

$$g=\cos (2x)+3, f=g^4$$

$$(\cos (2x)+3)'.\Big((\cos (2x)+3)^4\Big)'$$

$$-\sin (2x).4(\cos (2x)+3)^3 = -4\sin (2x)(\cos (2x)+3)^3$$

aga - 24-5-2016 at 00:50

4.

$$f=\sqrt [5]{1-2x}$$

$$g=1-2x, f=\sqrt [5] g, f=g^{\frac15}$$

$$(1-2x)'.\Big((1-2x)^{\frac15}\Big)'$$

$$=-2. \frac 15 (1-2x) ^{-\frac 45}$$

$$=\frac {-2}{5} \Big( \frac {1}{\sqrt [5]{1-2x}}\Big)^4 $$

aga - 24-5-2016 at 00:57

5.

$$f=\sin\Big(\ln x+\frac{1}{x^2}\Big)$$

$$g=\ln x+\frac{1}{x^2} = \ln x +x^{-2}$$

$$\sin\Big(\ln x+\frac{1}{x^2}\Big)'.(\ln x +x^{-2})'$$

$$=\cos\Big(\ln x+\frac{1}{x^2}\Big). (\frac 1x -2x^{-3})$$

aga - 24-5-2016 at 01:19

6.

$$f=\sqrt [3] {2x^4-x^2-3}$$

$$g=2x^4-x^2-3$$

$$(\sqrt [3] {2x^4-x^2-3})'.(2x^4-x^2-3)'$$

$$= (({2x^4-x^2-3})^\frac 13)'.(8x^3-2x)$$

$$= \frac 13 (2x^4-x^2-3)^{-\frac 23}.(8x^3-2x)$$

$$= \frac {8x^3-2x}{3} \Big( \frac {1}{\sqrt [3] {(2x^4-x^2-3)}} \Big)^2 $$

$$= \frac {8x^3-2x}{3} \frac {1}{(2x^4-x^2-3)^{\frac 23}} $$

$$= \frac {8x^3-2x}{3(2x^4-x^2-3)^{\frac 23}} $$

[Edited on 24-5-2016 by aga]

blogfast25 - 24-5-2016 at 07:00

2.

$$\Big(\frac{1}{1+x}\Big)' = \Big((1+x)^{-1}\Big)' = -1(x+1)^{-2} = \frac {2}{-1(x+1)} = -\frac{2}{x+1}$$

... is correct up to and including the second identity. Then, for unknown reasons you turn the - 2 exponent into a coefficient!

Correct answer:

$$(x+1).-1(x+1)^{-2}=-(x+1)^{-1}=\frac{1}{1-x}$$

3.

$$-\sin (2x).4(\cos (2x)+3)^3 = -4\sin (2x)(\cos (2x)+3)^3$$

Correct apart from one minor mistake:

$$(\cos (2x)+3)'=(\cos(2x))'+0=-2\sin(2x)$$

The cos(2x) also needed chain ruling!

4. is correct.

5. is also correct.

6. is also correct and nicely reworked.

<hr>

Seems to me the chain gremlins have been dealt a lethal blow and apart from one or two still bleeding out, we won't hear from them again!

Remember also that for simple functions:

$$f(g)=\cos g, \sin g, e^g, \ln g, g^n$$

If (with

$$g(x)=ax\:\text{...OR...}\:g(x)=ax+b$$

Then:

$$\frac{dg}{dx}=a$$

So:

$$f'(x)=a\frac{df(g)}{dg}$$

e.g.:

$$f(x)=\sin(ax+b)$$

$$f'(x)=a\cos(ax+b)$$

These are the simplest but also most frequently encountered cases of Chain Rule application!

[Edited on 24-5-2016 by blogfast25]

aga - 24-5-2016 at 08:07

(blush)

Yes, it does seem a lot clearer now.

Basically, if something happens to x (e.g. x+1, x^2, 3x+5 etc) then something happens to the

Still not comfortable with dg/dx etc but it feels like it's coming together.

Thanks again for taking the time to check these answers !

blogfast25 - 24-5-2016 at 12:05

Quote: Originally posted by aga |

Not entirely sure what you mean there. The whole Chain Rule business can be proved from the limit theorem.

For a function:

$$f\big(g(x)\big)$$

Then:

$$[f\big(g(x)\big)]'=\lim_{\Delta x \to 0}\frac{f\big(g(x+\Delta x)\big)-f\big(g(x)\big)}{\Delta x}$$

How this leads to:

$$[f\big(g(x)\big)]'=\frac{df(g)}{dg(x)}\times \frac{dg(x)}{dx}$$

... can be found in this formal proof:

http://kruel.co/math/chainrule.pdf

Enjoy!

aga - 24-5-2016 at 12:38

if the same brain can 'hear' the sound of that brain imploding, i just heard it.

aga - 24-5-2016 at 13:13

OK. You've cited Limit Theorem about 3 times now.

Best come clean and unload.

What's Limit Theorem all about ?

blogfast25 - 24-5-2016 at 14:52

Quote: Originally posted by aga |

It would have been more correct for me to refer to the 'limit definition of derivatives' (it's not really a theorem ):

$$f'(x)=\frac{df}{dx}=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

All the rules of derivation (sum/difference, product, quotient, chain,...) can be derived from that simple principle.

Here are the actual computations of these limits (i.e. first derivatives) of a few simple functions:

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/defderd...

(most of them require rather boring and lengthy algebra but aren't really that hard)

But in real life we simply use tabled derivatives of simple functions and the derivation rules for more complicated ones, instead of using the harder limit taking route.

[Edited on 25-5-2016 by blogfast25]

blogfast25 - 30-5-2016 at 07:28

Ooopsie. Double post deleted.

[Edited on 30-5-2016 by blogfast25]

In the previous post we saw that for a second order DE of the type:

$$ay''+by'+cy=0\:\text{, where: }y=f(x)$$

... a Chracteristic Equation (CE) can be defined as:

$$a\lambda^2+b\lambda +cy=0$$

The roots of this quadratic equation will be used to determine a full solution of the DE.

The two roots of the CE are determined from:

$$\lambda_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

We can now distinguish several cases:

If:

$$b^2-4ac>0$$

Then both roots are Real and distinct and the general solution of the DE is given by:

$$y=c_1e^{\lambda_1 x}+c_2e^{\lambda_2 x}$$

Where c

An example was shown in the previous post.

If:

$$b^2-4ac<0$$

Then note that the square root of that expression becomes Complex (imaginary):

$$\sqrt{b^2-4ac}=\sqrt{(4ac-b^2) \times -1}=\sqrt{4ac-b^2}\sqrt{-1}$$

Traditionally we call:

$$i=\sqrt{-1}$$

If in addition we set:

$$\beta =\frac{-b}{2a},\gamma = \frac{\sqrt{4ac-b^2}}{2a}$$

Then the Complex and distinct roots can be written as:

$$\lambda_{1,2} = \beta \pm \gamma i$$

We can then insert these roots into the general solution as above but that yields a complex function.

But using Euler's Formula and with some reworking (not shown here) we then obtain a Real solution:

$$y=c_1e^{\beta x}\cos \gamma x+c_2e^{\beta x}\sin \gamma x$$

Where c

<hr>

Two more cases to follow:

$$b^2-4ac=0\:\text{and }b=0$$

[Edited on 30-5-2016 by blogfast25]

aga - 30-5-2016 at 09:17

Gulp !

blogfast25 - 30-5-2016 at 10:30

Quote: Originally posted by aga |

Just try it for:

$$y''-4y'+5y=0\:\text{where: }y=f(x)$$

The CE has two complex roots. No boundary conditions have been given with the DE, so you don't need to determine values for the integration constants.

[Edited on 30-5-2016 by blogfast25]

For:

$$ay''+by'+cy=0$$

$$b^2-4ac=0$$

$$\lambda_{1,2}=\frac{-b}{2a}=\lambda$$

$$\implies y=c_1e^{\lambda x}+c_2xe^{\lambda x}$$

$$y''+2y'+y=0$$

$$b^2-4ac=2^2-4.1.1=0$$

$$\implies \lambda=-\frac{b}{2a}=-\frac22=-1$$

Solution:

$$y=c_1e^{-x}+c_2xe^{-x}$$

$$b=0 \implies \beta =0, \gamma = \sqrt{\frac{c}{a}}$$

a. Real roots:

$$\implies y=c_1e^{\lambda x}+c_2xe^{\lambda x}$$

b. Complex roots:

$$\implies y=c_1\cos \gamma x+c_2\sin \gamma x$$

<hr>

Ever wondered why a plucked guitar string sounds the way it does? How precisely it vibrates? It’s a surprisingly cool calculus problem. It’s also a big derivation, so we’ll do it in bite-size chunks.

Imagine a flexible string with linear density ρ (kg/m) clamped between x=0 and x=L, with a tension T:

The string is deformed at time t=0 as shown by the red line, then released.

Mathematically the motion the string will assume is governed by the wave equation, partial differential equation (PDE):

$$\frac{\partial^2y}{\partial t^2}=\frac{T}{\rho}\frac{\partial^2y}{\partial x^2}$$

Where y is the displacement of a mass element of the string, as a function of x and t:

$$y(x,t)$$

Also we call:

$$\frac{T}{\rho}=c^2$$

(A full derivation of the PDE can be found here. It’s quite simple)

So:

$$\frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}$$

We also need some initial conditions:

$$y(x,0)=f(x)\:\text{and }\frac{\partial y(x,0)}{\partial t}=0$$

The left one describes the string’s position at t=0 (don’t worry about f(x) just yet) and the right one the string’s initial velocity (in the y-direction), that is zero.

And some boundary conditions:

$$y(0,t)=0\:\text{and }y(L,t)=0$$

These represent the fact that the string is clamped at 0 and L and thus y=0 at all times in these points.

Ok, so far so good (and so little!) For solving this type of equations, as for DEs in general, no exact recipe exists. Instead a mixture of known techniques, useful theorems and a good dollop of guesswork are used. Here we’ll try separation of variables for a starter: that means separating all the xs from all the ts.

To do so, we make an important assumption:

$$y(x,t)=\phi(x)h(t)$$

Where φ(x) and h(t) are separate functions, resp. in x and t. Our sought after function y(x,t) is then presumed to be the product of the two, φ(x)h(t).

Firstly, we insert the original boundary conditions into the proposed solution, so we get:

$$y(0,t)=\phi(0)h(t)=0\:\text{and }y(L,t)=\phi(L)h(t)=0$$

And assuming:

$$h(t) \neq 0$$

We get the boundary conditions for φ(x):

$$\phi(0)=0\:\text{and }\phi(L)=0$$

Next we plug the proposed solution y=(x,t)= φ(x)h(t) into the original PDE.

$$\frac{\partial^2}{\partial t^2}\big(\phi(x)h(t)\big)=c^2\frac{\partial^2}{\partial x^2}\big(\phi(x)h(t)\big)$$

This look a lot scarier than it is because of course φ(x) is independent of t and h(t) is independent of x, so both can be multiplied out of the resp. derivatives. That also means we can dispense with the partial derivatives and we obtain:

$$\phi(x)\frac{d^2h}{dt^2}=c^2h(t)\frac{d^2\phi}{dx^2}$$

Minimal reworking yields separation of variables:

$$\frac{1}{c^2h}\frac{d^2h}{dt^2}=\frac{1}{\phi}\frac{d^2\phi}{dx^2}$$

Nice, huh? Hmmm... more like a ‘nice mess’, right now. And still no idea if this is even going to lead to a comprehensive solution of the PDE.

Stay tuned!

[Edited on 9-6-2016 by blogfast25]

aga - 9-6-2016 at 09:07

I'm starting to understand what they mean when they say a formula has a certain 'beauty'.

These hieroglyphs are incredible !

They look more like Art than Maths.

[Edited on 9-6-2016 by aga]

blogfast25 - 9-6-2016 at 10:01

Quote: Originally posted by aga |

Yes. Trust me, it gets more beautiful as we get on with this problem.

The 'hieroglyphs' could be avoided by a notation we used above, for example:

$$\frac{\partial^2u}{\partial x^2}=u_{xx}$$

The wave equation then becomes:

$$y_{tt}=c^2y_{xx}$$

But that doesn't have the same allure. It is much more practical though...

[Edited on 9-6-2016 by blogfast25]

aga - 9-6-2016 at 14:12

Aspersions cast i not, young 25 of blogfastingness.

The drawn glyphs admire i, and that muchly.

Of great beauty they are.

So far we obtained:

$$\frac{1}{c^2h}\frac{d^2h}{dt^2}=\frac{1}{\phi}\frac{d^2\phi}{dx^2}$$

This is really a bit of a strange animal. h(t) and φ(x) are obviously not the same (they’re not even dependent on the same variable!). This can only make sense mathematically if both derivative equations (each side of the equation) equals the same number:

$$\frac{1}{c^2h}\frac{d^2h}{dt^2}=\frac{1}{\phi}\frac{d^2\phi}{dx^2}=-\lambda$$

Here λ is called the separation constant. The negative sign is for convenience (in some other PDEs a positive sign is needed, it’s case dependent) Note that we don’t know its value yet though.

<hr>

We can now split the PDE into two parts (two DEs). First part:

$$\frac{1}{c^2h}\frac{d^2h}{dt^2}=-\lambda$$

Which reworks to:

$$\frac{d^2h}{dt^2}+c^2\lambda h=0$$

Second part:

$$\frac{1}{\phi}\frac{d^2\phi}{dx^2}=-\lambda$$

Which reworks to:

$$\frac{d^2\phi}{dx^2}+\lambda \phi=0$$

For the latter we also have the boundary conditions:

$$\phi(0)=0\:\text{and }\phi(L)=0$$

The DE for φ(x) is a very simple one of the type we’ve already solved higher up (see e.g. the Pi1DB problem) and the solutions are:

$$\lambda_n=\Big(\frac{n\pi}{L}\Big)^2\:\text{so }\phi_n(x)=A\sin\Big(\frac{n\pi x}{L}\Big)\:\text{for: }n=1,2,3,...$$

The λ<sub>n</sub> are called

<hr>

And of course that means we can use these eigenvalues in the DE for h(t):

$$\frac{d^2h}{dt^2}+\Big(\frac{n\pi c}{L}\Big)^2h=0$$

This is a straightforward 2nd order linear DE and its CE has two repeat complex roots and general solution:

$$h(t)=c_1\cos\Big(\frac{n\pi ct}{L}\Big)+c_2\sin\Big(\frac{n\pi ct}{L}\Big)$$

Here we have a useful initial condition:

$$\frac{\partial y(x,0)}{\partial t}=0 \implies \frac{dh(0)}{dt}=0$$

$$\frac{dh}{dt}=-c_1\frac{n\pi c}{L}\sin\Big(\frac{n\pi ct}{L}\Big)+c_2\frac{n\pi c}{L}\cos\Big(\frac{n\pi ct}{L}\Big)$$

$$\frac{dh(0)}{dt}=0 \implies c_2=0$$

So:

$$h(t)=c_1\cos\Big(\frac{n\pi ct}{L}\Big)$$

<hr>

Now remember that we assumed (in German mathspeak: our 'Ansatz'):

$$y(x,t)=\phi(x)h(t)$$

Which gives us a set of solutions:

$$y_n(x,t)=A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)\:\text{with }A_n=Ac_1$$

$$\text{For: }n=1,2,3,...$$

To put together a comprehensive solution we need to use the Superposition Principle (SP). There’s nothing new about it and we’ve used the SP several times before when solving second order linear DEs.

Formally the principle is:

<hr>

$$\text{If }(y_1, y_2,...,y_i,...,y_n)\:\text{are solutions to a linear, homogeneous DE, then }\displaystyle\sum_{i=1}^{n}(c_iy_i)\:\text{,where }c_i\:\text{are arbitrary constants, is also a solution to the DE.}$$

<hr>

Looking at the expression for y<sub>n</sub>(x,t), for each value of n we have a particular solution of the PDE, so in accordance with the SP these have to be combined to:

$$y(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$

Scary as it looks, we’ll soon see what it means with some function analysis!

[Edited on 10-6-2016 by blogfast25]

aga - 10-6-2016 at 12:04

Scary ? Terrifying more like !

There's gonna be chain rules everywhere, probably whips and gimp suits too.

blogfast25 - 10-6-2016 at 13:54

Quote: Originally posted by aga |

In fact there's almost nothing we've done here that we haven't seen in action before in the thread.

That's the beauty of

The solution of the Schrödinger equation applied to the Hydrogen atom also uses SoV. The assumed solution is then:

$$\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)$$

... with

aga - 10-6-2016 at 14:05

Erm, are you

blogfast25 - 10-6-2016 at 14:41

Quote: Originally posted by aga |

I oppose the Culture of Low Expectations

Seriously, that depends on ones level of dedication, of course. At the very least you should be able to

There won't be any exercises on PDEs, just one more example of a simple PDE governing Diffusion. And maybe a chemical example, if I can find a suitable one...

The main purpose of a vibrating string is of course to generate sound. The string ‘bangs’ against the air, an elastic medium, and sound waves are generated. These our ears receive, turn the waves back to oscillations and the resulting electrical impulses are perceived by our brain as sound (here a music note).

We’ve yet to determine the coefficients A<sub>n</sub>but let’s first look and first time-dependent term in the series, cos(nπct/L).

In wave/oscillation mechanics we write:

$$\cos\Big(\frac{n\pi ct}{L}\Big)=\cos\omega t$$

Where ω is the angular velocity and f the frequency of oscillation:

$$\omega=2\pi f=\frac{n\pi c}{L}$$

$$f=\frac{nc}{2L}$$

$$\frac{T}{\rho}=c^2$$

$$f=\frac{n}{2L}\sqrt{\frac{T}{\rho}}$$

We can see that the vibrating string doesn’t have one single frequency but a frequency spectrum. If for n=1 we call:

$$f_1=\frac{1}{2L}\sqrt{\frac{T}{\rho}}$$

... the fundamental frequency (which defines the note), then the frequency spectrum is:

$$f_1, 2f_1, 3f_1,...$$

... where f<sub>2</sub>, f<sub>3</sub>,... are the so-called

Note also:

• Shorter string length gives higher pitch

• Higher string tension gives higher pitch

• Higher density strings give lower pitch

And next we’ll see not all the frequencies are equally represented in the spectrum.

We now need to derive values for A<sub>n</sub> in:

$$y(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$

We do this by using the initial condition (see first page of the full derivation):

$$y(x,0)=f(x)$$

Since as t=0, all the time dependent cosine terms drop out, which leaves us with:

$$y(x,0)=f(x)=\displaystyle\sum_{n=1}^{\infty}A_n\sin\Big(\frac{n\pi x}{L}\Big)$$

In material not covered in this thread it’s proved that:

$$A_n=\frac{2}{L}\int_0^{L}f(x)\sin\Big(\frac{n\pi x}{L}\Big)dx$$

Here’s f(x):

Obviously, f(x) is not a continuous function, as it’s gradient changes abruptly changes at L/a but we can construct two linear function that act resp. on 0 < x < L/a and L/a < x < L:

$$f_1(x)=\frac{ay_0}{L}x\:\text{ and }f_2(x)=\frac{ay_0}{(a-1)L}(L-x)$$

A<sub>n</sub> then becomes:

$$A_n=\frac{2}{L}\Big[\int_0^{L/a}f_1(x)\sin\Big(\frac{n\pi x}{L}\Big)dx+\int_{L/a}^{L}f_2(x)\sin\Big(\frac{n\pi x}{L}\Big)dx\Big]$$

$$A_n=\frac{2ay_0}{L^2}\Big[\int_0^{L/a}x\sin\Big(\frac{n\pi x}{L}\Big)dx+\frac{1}{a-1}\int_{L/a}^{L}(L-x)\sin\Big(\frac{n\pi x}{L}\Big)dx\Big]$$

$$A_n=\frac{2ay_0}{L^2}\Big[\int_0^{L/a}x\sin\Big(\frac{n\pi x}{L}\Big)dx+\frac{L}{a-1}\int_{L/a}^{L}\sin\Big(\frac{n\pi x}{L}\Big)dx\Big]-\frac{1}{a-1}\int_{L/a}^{L}x\sin\Big(\frac{n\pi x}{L}\Big)dx\Big]$$

This isn’t a particularly difficult integral but it is tedious and I’ll only present the final result:

$$A_n=\frac{4ay_0}{n^2\pi^2}\sin\Big(\frac{n\pi }{a}\Big)$$

The A<sub>n</sub> should be seen as amplitudes, associated with each frequency and they vary

$$f_1,2f_1,3f_2,4f_1,...$$

$$A_1,\frac{A_1}{4},\frac{A_1}{9},\frac{A_1}{16},...$$

So the amplitudes diminish very quickly with increasing values of n.

We can also see that the A<sub>n</sub> increase with y<sub>0</sub> but are independent of L.

The spectrum:

$$A_n\cos\Big(\frac{n\pi ct}{L}\Big)$$

... determines the timbre (as do other factors like instrument shape) of the note.

[Edited on 12-6-2016 by blogfast25]

aga - 12-6-2016 at 12:14

Quote: Originally posted by blogfast25 |

Looks good to me : an integration problem that Can be solved and has an answer that can be used to check if i did it right.

Deffo gonna attempt it.

Might need more paper for this one ...

blogfast25 - 12-6-2016 at 13:08

Quote: Originally posted by aga |

Lots of paper, yes. You end up with like 6 individual pieces you need to put back together again! But it requires only listed integrals (see cheat sheet). More lots of algebra than calculus, really...

So far our vibrating string time-dependent equation is:

$$y(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$

Where the amplitude coefficients A<sub>n</sub> have been determined above.

Now look at an individual term:

$$\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$

Using the well-known trigonometric identity:

$$\cos \alpha \sin \beta=\frac12 \Big(\sin(\alpha+\beta)+\sin(\alpha-\beta)\Big)$$

... y(x,t) can then be reworked to:

$$y(x,t)=\frac12\displaystyle\sum_{n=1}^{\infty} A_n\Big(\sin\frac{n\pi(x+ct)}{L}+\sin\frac{n\pi(x-ct)}{L}\Big)$$

In:

$$\sin\frac{n\pi(x+ct)}{L}+\sin\frac{n\pi(x-ct)}{L}$$

... the left hand term represents a wave travelling from 0 to L and the right hand term an identical wave travelling from L to 0. The superposition (sum) of both is a

Standing waves (amplitudes not to scale) in a string:

(Source of pic)

<hr>

[Edited on 13-6-2016 by blogfast25]

Diffusion is an important mass transport phenomenon and a part of many chemical, biological and physical processes. There are also some really interesting and colourful experiments that illustrate it (I’ll suggest one further down).

On the left hand side concentration of a species (e.g. a dye) is high. Diffusion causes concentration to level out over time. The rate of diffusion is proportional to the concentration gradient of the solute, acc. to Fick’s first law, here in one spatial dimension only:

$$J=-D\frac{\partial u}{\partial x}$$

Where

From it, Fick’s second law, aka the diffusion equation, can be derived:

$$\frac{\partial u}{\partial t}=D\frac{\partial^2 u}{\partial x^2}\:\text{for: } 0\leq x \leq L$$

In shorthand:

$$u_t=Du_{xx}$$

We’ll apply this to a long, narrow tube, filled with a dye solution and closed off at both ends:

As initial condition we set:

$$u(x,0)=f(x)$$

Where f(x) is the initial distribution of dye at t=0.

Now for some boundary conditions. As the tube is closed off at both ends, no mass transfer can take place through these boundaries. With Fick’s first law this means:

$$0=D\frac{\partial u}{\partial x} \implies \frac{\partial u}{\partial x}=0\:\text{at } x=0, x=L$$

... at all times. In shorthand:

$$u_x(0,t)=0\:\text{and: }u_x(L,t)=0$$

Now for our Ansatz, we assume that:

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

Where X(x) depends only on x and T(t) only on t. Inserting into the diffusion equation we get:

$$X(x)T'(t)=DT(t)X''(x)$$

Divide both sides by X(x)T(t), rearrange slightly and dispense with the (x) and (t) notation and we get:

$$\frac{1}{D}\frac{T'}{T}=\frac{X''}{X}$$

Introducing the separation constant λ:

$$\frac{1}{D}\frac{T'}{T}=\frac{X''}{X}=-\lambda$$

Splitting then gives us two ordinary DEs:

$$X''+\lambda X=0\:\text{and: }T'+\lambda DT=0$$

<hr>

Next: solving the separated DEs.

[Edited on 15-6-2016 by blogfast25]

So far we obtained the fully separated DEs:

$$X''+\lambda X=0\:\text{and: }T'+\lambda DT=0$$

$$X''+\lambda X=0$$

$$X=c_1\sin(\sqrt{\lambda}x)+c_2\cos(\sqrt{\lambda}x)$$

With the first boundary condition:

$$u_x(0,t)=X'(0)T(t)=0 \implies X'(0)=0$$

$$X'=c_1\sqrt{\lambda}\cos(\sqrt{\lambda}x)-c_2\sqrt{\lambda}\sin(\sqrt{\lambda}x)=0$$

$$x=0 \implies c_1=0$$

Which gives us:

$$X=c_2\cos(\sqrt{\lambda}x)$$

With the second boundary condition:

$$u_x(L,t)=X'(L)T(t)=0 \implies X'(L)=0$$

$$X’(L)=-c_2\sqrt{\lambda}\sin(\sqrt{\lambda}L)=0$$

$$\implies \sqrt{\lambda}L=n\pi$$

$$\lambda=\Big(\frac{n\pi}{L}\Big)^2\:\text{for: }n=0,1,2,3,...$$

$$X=c_2\cos\Big(\frac{n\pi x}{L}\Big)$$

Note that the series n=0,1,2,3,... includes the zeroth term because X is of the cosine form (a consequence of the boundary condition). Had X been of the sine form then n=0 would be a trivial solution, making u(x,t) zero over the entire x-domain. But cos 0 is not zero (it’s one). For that reason the n=0 term has to be included in the cosine series.

$$T'+\lambda DT=0$$

$$T=Ae^{-\lambda Dt}=Ae^{-\Big(\frac{n\pi}{L}\Big)^2Dt}\:\text{for: }n=0,1,2,3,...$$

So with u(x,t)=T(t)X(x):

$$u_n=A_ne^{-\Big(\frac{n\pi}{L}\Big)^2Dt}\cos\Big(\frac{n\pi x}{L}\Big)$$

And with the Superposition Principle:

$$u(x,t)=A_0+\displaystyle \sum_{n=1}^{\infty}A_ne^{-\Big(\frac{n\pi}{L}\Big)^2Dt}\cos\Big(\frac{n\pi x}{L}\Big)$$

The term A<sub>0</sub> stems from the fact that for n=0 both the exponential factor and the cosine factor of the zeroth term equal one (1). That way the series can be started at n=1.

In the so far unused initial condition u(x,0)=f(x), the exponential factors all become 1 (e

$$u(x,0)=f(x)=A_0+\displaystyle \sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi x}{L}\Big)$$

As in the vibrating spring problem, the coefficients A<sub>n</sub> are determined from:

$$A_n=\frac{2}{L}\int_0^Lf(x)\cos\Big(\frac{n\pi x}{L}\Big)dx$$

<hr>

Next we’ll use two realistic initial dye distributions (u(x,0)=f(x)) and do some plotting of u(x,t)!

[Edited on 17-6-2016 by blogfast25]

Richard3050 - 16-6-2016 at 17:26

Chemistry! For beginners, with a ‘no elements’ approach!

blogfast25 - 16-6-2016 at 17:46

Quote: Originally posted by Richard3050 |

Welcome to the thread!

To approximately evaluate the diffusion equation graphically, let’s try two different

We have a constant concentration u<sub>0</sub> over a narrow length a, centred about L/2, like this:

Now we calculate the coefficients A<sub>n</sub>:

$$A_n=\frac{2}{L}\int_0^Lf(x)\cos\Big(\frac{n\pi x}{L}\Big)dx$$

$$A_n=\frac{2}{L}\int_{\frac{L-a}{2}}^{\frac{L+a}{2}}u_0\cos\Big(\frac{n\pi x}{L}\Big)dx$$

$$A_n=\frac{2u_0}{n\pi}\big[\sin\Big(\frac{n\pi x}{L}\Big)\Big]_{\frac{L-a}{2}}^{\frac{L+a}{2}}$$

$$A_n=\frac{2u_0}{n\pi}\Big[\sin\Big(\frac{n\pi}{2}+\frac{n\pi a}{2L}\Big)-\sin\Big(\frac{n\pi}{2}-\frac{n\pi a}{2L}\Big)\Big]$$

Using the well-known trigonometric identity:

$$\sin(\alpha + \beta)-\sin(\alpha - \beta)=2\cos\alpha \sin \beta$$

$$A_n=\frac{4u_0}{n\pi}\cos \frac{n\pi}{2} \sin \frac{n\pi a}{2L}$$

Now we apply the small angle approximation for sine functions:

$$2L \gg n\pi a \implies \sin\frac{n\pi a}{2L} \approx \frac{n\pi a}{2L}$$

$$A_n\approx \frac{2u_0a}{L}\cos \frac{n\pi}{2}$$

Which gives:

$$A_0=\frac{2u_0a}{L}$$

$$A_1=0$$

$$A_2=-\frac{2u_0a}{L}$$

With:

$$u(x,t)=A_0+\displaystyle \sum_{n=1}^{\infty}A_ne^{-\Big(\frac{n\pi}{L}\Big)^2Dt}\cos\Big(\frac{n\pi x}{L}\Big)$$

Applied to the first three terms only (n=0, n=1, n=2):

$$u(x,t) \approx \frac{2u_0a}{L}-\frac{2u_0a}{L}e^{-4\Big(\frac{\pi}{L}\Big)^2Dt}\cos\Big(\frac{2\pi x}{L}\Big)$$

$$u(x,t) \approx \frac{2u_0a}{L}\Big[1-e^{-4\Big(\frac{\pi}{L}\Big)^2Dt}\cos\Big(\frac{2\pi x}{L}\Big)\Big]$$

Before we do some graphical plotting, a word about time scales. To allow dimensionless plotting we’ll use dimensionless time t<sub>r</sub>:

$$t_r=\frac{D}{L^2}t$$

$$\implies t=\frac{L^2}{D}t_r$$

The following resource:

http://webserver.dmt.upm.es/~isidoro/dat1/Mass%20diffusivity...

... lists some values for D for common chemicals in water. They are typically in the order of 10<sup>-9</sup> m<sup>2</sup> s<sup>-1</sup>!

Assuming a tube length of 10 cm (0.1 m), then t= 1 x 10<sup>7</sup> x t<sub>r</sub> = 16 weeks x t<sub>r</sub>! Diffusion in liquids is a decidedly slow process.

Here’s the result using the first three terms approximation. All parameters are relative values.

Vertical axis: relative concentration, horizontal axis: relative time t<sub>r</sub>.

If a narrow region 0 < x < a has an initial concentration of u<sub>0</sub>, then calculation of A<sub>n</sub> yields:

$$A_n \approx \frac{2u_0a}{L}$$

We’ve used the following 4 term approximation for u(x,t):

$$u(x,t) \approx \frac{2u_0a}{L}\Big[1+e^{-\frac{\pi^2}{L^2}Dt}\cos\Big(\frac{\pi x}{L}\Big)+e^{-\frac{4\pi^2}{L^2}Dt}\cos\Big(\frac{2\pi x}{L}\Big)+e^{-\frac{9\pi^2}{L^2}Dt}\cos\Big(\frac{3\pi x}{L}\Big)\Big]$$

Vertical axis: relative concentration, horizontal axis: relative time t<sub>r</sub>.

Place a small amount of potassium permanganate crystals at the bottom of a test tube. Add about 0.5 ml of water and swirl gently to dissolve some of the crystals. Now carefully (so as not to disturb the permanganate layer) pipette some water into the test tube, filling it to about half way. Stopper and place in an undisturbed place, preferably of constant temperature (convection currents will promote advective diffusion!)

Be patient and watch the permanganate diffuse through the water.

[Edited on 18-6-2016 by blogfast25]

We previously derived the wave functions of a particle in 1D box, here. The wave functions for 2D rectangles and 3D cuboids can be found easily by means of Separation of Variables of the Schrödinger equation.

The solutions lead to some interesting plots, like these:

The probability density

But things tend to get a whole lot more interesting when they’re circular (or cylindrical or spherical – see hydrogenic atoms!).

So let’s try and determine the wave functions for a circular 2D box of radius

In Cartesian coordinates (x,y), a 2D, one particle quantum system is described by the Schrödinger equation (here time-independent):

$$- \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} \bigg) \psi(x,y) = [E-V(x,y)] \psi(x,y)$$

In our case we have:

$$V(x,y)=0$$

$$\implies - \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} \bigg) \psi(x,y) = E\psi(x,y)$$

But this problem, because of the circular geometry, is easier to solve in polar coordinates:

$$x=r\cos \theta\:\text{and }y=r\sin \theta\:\text{and }r^2=x^2+y^2$$

So we need to convert the Cartesian coordinates based equation to a polar coordinates based equation. I’ll also switch to u instead of ψ, because of easier notation:

$$\psi(x,y) \to u(r,\theta)$$

This transformation of coordinates is tedious and finicky but is also well known (it can be found on plenty web pages), so I can just present the polar coordinates Schrödinger equation, here:

$$-\frac{\hbar^2}{2m}\Big(\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}\Big)=Eu$$

Slight reworking:

$$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}+\frac{2mE}{\hbar^2}u=0$$

For shortness of notation we set:

$$k^2=\frac{2mE}{\hbar^2}$$

$$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}+k^2u=0$$

Due to:

$$u=0\:\text{for: }V=+\infty\:\text{,that is for: }r \geq R$$

... we have a boundary condition:

$$u(R,\theta)=0$$

<hr>

Next: separating the variables r and theta.

[Edited on 24-6-2016 by blogfast25]

aga - 24-6-2016 at 12:02

Polar co-ordinates.

Now we're talking business ...

blogfast25 - 24-6-2016 at 13:12

Quote: Originally posted by aga |

You haven't lived if you haven't used polar coordinates!

aga - 24-6-2016 at 13:29

Astonishingly, i have !

I was introduced to them as a convenient way to plot shapes on a computer screen, then rotate those shapes.

blogfast25 - 24-6-2016 at 14:10

Grab a piece of chalk, the blackboard is yours!

aga - 24-6-2016 at 14:56

In 'normal' Cartesian co-ordinates, every point is defined as having an X,Y value relative to the Zero point where the X and Y axes cross.

Polar co-ordinates can define exactly the same spatial position using Distance (d) from a zero-point on the x-axis, and an Angle relative to that x-axis.

If you think about it, 'where the thing is' in Cartesian (normal X,Y) co-ordinates generates an Angle from the zero-point anyway, so you could use trigonometry to work out what the Polar co-ordinates would be and vice versa.

For some problems, Polar co-ordinates simplify the maths enormously.

For example, in the one and only case where i used them, i defined a shape on a computer screen in Polar co-ordinates, each point desribed by a Distance from zero and an Angle.

Simply by adding 1 to the Angle of every point, the entire shape rotated.

Changing the Zero point alters the position of the entire shape.

Adding/subtracting 1 from the Distance of each point alters the size of the entire share, retaining the overall structure.

In a Cartesian system this would be Much more difficult (take more time to solve) for each point in the plotted image.

Yes, i'm Pedagogically disabled.

You can take the chalk back now, Please.

[Edited on 25-6-2016 by aga]

blogfast25 - 24-6-2016 at 16:24

That wasn't a bad lecture, actually.

I like these plots: cycloids:

http://curvebank.calstatela.edu/cycloidmaple/cycloid.htm

We assume u can be written as the product of two functions, one only dependent on r, the other only dependent on θ:

$$u(r,\theta)=R(r)Y(\theta)$$

Then we insert the assumed form of u into the PDE. This looks frightening (?) but because of the nature of u, it’s very easy. The first term e.g. becomes:

$$\frac{\partial^2}{\partial r^2}u(r,\theta)=\frac{\partial^2}{\partial r^2}(R(r)Y(\theta))=Y(\theta)\frac{d^2R}{dr^2}=Y(\theta)R''$$

Ploughing on with the other terms we get:

$$YR''+\frac{1}{r}YR'+\frac{1}{r^2}RY''+k^2RY=0$$

Divide both sides by RY:

$$\frac{R''}{R}+\frac{1}{r}\frac{R'}{R}+\frac{1}{r^2}\frac{Y''}{Y}+k^2=0$$

Multiply both sides with r<sup>2</sup>:

$$r^2\frac{R''}{R}+r\frac{R'}{R}+\frac{Y''}{Y}+r^2k^2=0$$

Bringing the Y based term to the right and introducing a separation constant m<sup>2</sup>:

$$r^2\frac{R''}{R}+r\frac{R'}{R}+r^2k^2=-\frac{Y''}{Y}=m^2$$

So we obtain two ODEs, one in R(r):

$$r^2R''+rR'+\Big(r^2k^2-m^2\Big)R= m^2$$

... and one in Y(θ):

$$\frac{Y''}{Y}=-m^2$$

The value of the separation constant m<sup>2</sup> is yet to be determined.

[Edited on 25-6-2016 by blogfast25]

aga - 25-6-2016 at 13:40

Quote: Originally posted by blogfast25 |

Ever the Master of the Cliff-Hanger

As an aside, when polar co-ordinates are used in n-dimensions, is there any hard-and-fast rule dictating which axis the zero point lies on ?

I.e. in 3-D space it could be x,y,z or time ... or anything.

[Edited on 25-6-2016 by aga]

blogfast25 - 25-6-2016 at 14:31

Quote: Originally posted by aga |

The choice or origin and axis is usually yours and made out of mathematical convenience (there's no

In Cartesian coordinates we can describe a system of n dimensions, e.g. x<sub>1</sub>,x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, etc. Of course you can't represent that

For spherical coordinates there's an unwritten convention:

But really it's up to the user.

wg48 - 27-6-2016 at 09:20

Quote: Originally posted by blogfast25 |

Then you haven't lived WELL if you haven't used Euler angles plus range their first and second derivatives their estimates and time.

But if you want to live really well you need to get in to metrics of relativity and black holes.

morganbw - 27-6-2016 at 10:41

Sorry gentlemen, I do like math and even calculus, I am not extremely well versed and I usually have to do some research to use it meaningfully. I have not read all post in this thread and I am drinking just a little bit but it seems to me that this is a mastrubation post.

Self love screams from these post. Am I wrong? Think about it just a little. Calculus was developed between three and four hundred years ago, this is not new shit. Difficult shit for sure, for most of us mortals, but entirely doable.

And no, I am not able to give the theorem of calculus in its entirety without hitting the literature, but I am able to hit the literature and do pretty damn well. I am no doubt slower that a true mathematician, but tortoise or rabbit the end is the same place.

Please gentlemen, you are just who you are, please inform all of us when you are able to expand on ( an example, Newton ) his math and thoughts are hundreds of years old. ( Can you do it without referencing perhaps Einstein )

No gentlemen, you cannot, because you are not the genius you wished to be. We play with the thoughts of those who came before us, (effing A, I do as well )wishing to show something different/better, but it is not there yet, I do believe that it will be someday, but I am certain that it does not exist in this thread.

blogfast25 - 27-6-2016 at 11:34

@morganbw:

Despite your unnecessary reference to 'mastrubation' [sic], your multiple spelling, grammar and punctuation mistakes, not to mention your conflation between 'thread' and 'post', I'll keep this short and polite.

This is an educational thread for those who wish to learn something 'in house' about calculus and its applications in science. As such nothing new occurs here, which of course is also true for 99 % of what goes on in other SM threads. That's the nature of something educational.

Maybe morgan wants to point to whatever it is that he has done here that hasn't already been done elsewhere and possibly a million times already? Perhaps the many excellent but far, far from new (or unique) syntheses that are performed here by

Finally, if you don't like the subject of this thread or how it's being presented, then don't read it. Inane comments like yours don't help one iota.

As regards:

Quote: |

Again that's offensive (but it's your right to be so) and wholly incorrect. I'm pretty sure know my limitations better than you know yours.

[Edited on 27-6-2016 by blogfast25]

aga - 27-6-2016 at 11:39

Erm, in this thread I learnt at least a bit of Calculus from bloggers, and for Free !

For that reason alone, i am Very Glad this thread exists.

However the knowledge cums, i'm glad it does, no matter if it is from the enlightening words of morganbw or from Dr Flexi Jerkoff himself.

As it is a Maths thread, erm, perhaps you'd care to share some of your own pearls of wisdom ?

Edit:

I'll swap for my secret formula for

[Edited on 27-6-2016 by aga]

blogfast25 - 27-6-2016 at 11:53

Quote: Originally posted by aga |

Very slow, controlled evaporation of an Al sulphate solution, with careful nucleation should yield a defined hydrate as first crystals.

$$\frac{Y''}{Y}=-m^2$$

$$Y''+m^2 Y=0$$

With two complex roots, +im and –im this has the general solution:

$$Y(\theta)=c_1\cos m\theta + c_2\sin m\theta$$

θ is of course an angle, in radians, and an angle ‘repeats’ itself every 2π. For that reason we can choose either c<sub>1</sub> or<sub>2</sub> to be zero. We choose the latter:

$$c_2=0 \implies Y=c_1\cos m\theta$$

Again, because θ is an angle the following condition must also be met:

$$Y_m(\theta)=Y_m(\theta+2\pi ) \implies \cos(m\theta)=\cos(m(\theta+2\pi))=\cos(m\theta+2m\pi)$$

$$\implies m=0, \pm 1, \pm 2, \pm 3,...$$

The negative values are trivial because:

$$\cos(|m|\theta)=\cos(-|m|\theta)$$

The function needs to be normalised, with the normalisation requirement:

$$\int_0^{2\pi}Y^*Yd\theta=1$$

Y<sup>*</sup> is the complex conjugate of Y. But Y is a Real function, so:

$$Y^*=Y \implies Y^*Y=Y^2$$

$$\int_0^{2\pi}Y^2d\theta=\int_0^{2\pi}c_1^2\cos^2(m\theta) d\theta=1$$

$$\implies c_1=\frac{1}{\sqrt{\pi}}$$

$$Y_m=\frac{1}{\sqrt{\pi}}\cos m\theta$$

There’s one exception, for m=0:

$$m=0 \implies Y_0=c_1$$

$$\int_0^{2\pi}c_1^2d\theta=1 \implies c_1=\frac{1}{\sqrt{2\pi}}$$

$$Y_0=\frac{1}{\sqrt{2\pi}}$$

So we have a fully defined solution for Y(θ), as well as values for m:

$$m=0,1,2,3,...$$

[Edited on 27-6-2016 by blogfast25]

aga - 27-6-2016 at 12:03

Yay ! morganbw's post caused the next page to appear.

it was taking ages for page 18 to render.

Edit:

Hold on hoss.

So we're in Polar co-ordinates now and we got a Derivative, then we got a derivative of a derivative.

In plain english, what's the first derivative mean ?

The rate of change of the angular rotation ?

[Edited on 27-6-2016 by aga]

blogfast25 - 28-6-2016 at 06:41

Quote: Originally posted by aga |

Sorry, only just saw your edit.

Firstly, both polar coordinates and Cartesian coordinates are completely valid ways to define the position of a point (the position does not change in function of the coordinate system used).

Derivatives (first, second or whatnot) can be used in both systems, the same rules of derivation apply.

A derivative like:

$$\frac{\partial u(r,\theta)}{\partial r}$$

... is the rate of change of the wave function (in polar coordinates) vis-a-vis the r-component of the position (r,θ).

Quote: |

That's a perfectly legitimate question but hard to answer. Mathematical statements like the Schrödinger equation are jam packed full of logic. But that makes 'unpacking' the parcel into intelligible spoken or written language quite hard.

Incidentally, the Schrödinger equation used here is the time-independent one. It is free of time and therefore cannot be used to determine velocities or angular velocities (both are derivatives to time). For these we need to apply quantum operators on the wave function.

I'll wait for your response before solving the radial part of the SE and then we'll be able to start plotting some functions!

aga - 28-6-2016 at 09:26

OK. It was a silly question really now i'm sober enough to think about it.

It's a bit like asking what a 5-dimension co-ordinate set actually means ...

Please, continue.

So far we have:

$$r^2R''+rR'+(k^2 r^2-m^2)R=0$$

With boundary condition:

$$u(R,\theta)=R(R)Y(\theta)=0 \implies R(R)=0$$

We make a simple substitution:

$$\rho=kr$$

Minimal reworking then yields:

$$\frac{d^2R(\rho)}{d\rho^2}+\frac{1}{\rho}\frac{dR(\rho)}{d\rho}+\Big(1-\frac{m^2}{\rho^2}\Big)R(\rho)=0$$

This is a Bessel Differential Equation, a type of DE that also pops up in the solution of the hydrogen SE, the Fourier (heat) equation and other important PDEs.

It has a solution for every

$$m \to J_m(\rho)=J_m(kr)$$

Where:

$$R(\rho)=J_m(\rho)=\displaystyle \sum_{n=0}^{+\infty}\frac{(-1)^n}{n!\Gamma(n+m+1)}\Big(\frac{\rho}{2}\Big)^{2n+m}$$

The

Later we’ll make use of these resemblances to define some quasi-solutions, because the actual Bessel functions really don’t lend themselves easily to graphical 2D plotting.

The particle in a 2D circular box is a quantum system and its wave functions and energy levels are quantised. Quantisation is derived from the boundary condition:

$$J_m(kR)=0$$

Basically this means that for the roots (zero-points)

$$k_{m,n}R=z_{m,n}$$

The z values can be found here:

(Source).

$$k_{m,n}=\frac{z_{m,n}}{R}$$

Much higher up we stated:

$$k^2=\frac{2mE}{\hbar^2}$$

We can now use that expression to derive the energy spectrum of the system:

$$E_{m,n}=\frac{\hbar^2}{2mR^2}z_{m,n}^2$$

And the ground state energy:

$$E_{0,1}=\frac{\hbar^2}{2mR^2}(2.4048)^2$$

<hr>

Next we'll replace the Bessel functions with appropriate sines/cosines and the plotting can begin!

[Edited on 29-6-2016 by blogfast25]

Two cases need to be considered:

1.

$$R(r)=c_1\cos(kr)$$

$$R(R)=c_1\cos(kR)=0 \implies kR=\frac{n\pi}{2}$$

$$n=1,3,5,7,...$$

$$k=\frac{n\pi}{2R}$$

$$R_0(r)=c_1\cos\Big(\frac{n\pi r}{2R}\Big)$$

Normalisation condition to obtain c<sub>1</sub>:

$$\int_0^Rc_1^2\cos^2\Big(\frac{n\pi r}{2R}\Big)rdr \implies c_1=\frac{2}{R}$$

$$R_0(r)=\frac{2}{R}\cos\Big(\frac{n\pi r}{2R}\Big)$$

2.

$$R(r)=c_2\sin(kr)$$

$$R(R)=c_2\sin(kR)=0 \implies kR=n\pi$$

$$n=1,2,3,...$$

$$k=\frac{n\pi r}{R}$$

$$R_m(r)=c_2\sin\Big(\frac{n\pi r}{R}\Big)$$

Normalisation condition to obtain c<sub>2</sub>:

$$\int_0^Rc_2^2\sin^2\Big(\frac{n\pi r}{R}\Big)rdr \implies c_2=\frac{2}{R}$$

$$R(r)=\frac{2}{R}\sin\Big(\frac{n\pi r}{R}\Big)$$

$$u(r,\theta)=R(r)Y(\theta)$$

m = 0:

$$u_0(r,\theta)\approx \frac{2}{R\sqrt{2\pi}}\cos\Big(\frac{n\pi r}{2R}\Big)$$

$$n=1,3,5,...$$

m = 1,2,3,...

$$u_m(r,\theta)\approx \frac{2}{R\sqrt{\pi}}\sin\Big(\frac{n\pi r}{R}\Big)\cos(m\theta)$$

$$n=1,2,3,...$$

Instead of plotting the values of the wave functions u, the probability density functions will be plotted, for a circular well of R=1 m. As u is a Real function, the probability density function is simply:

$$\rho=u_m^2(r,\theta)$$

(not to be confused with the previously used rho)

Of course, R=1 m is not a realistic value for a quantum system. For a more realistic size of say 1 nm (1x10<sup>-9</sup> m) the vertical axis values would have to be multiplied by 1x10<sup>18</sup> m<sup>-2</sup>.

Left (next post) are the 3D plots, right the contour plots, with the actual boundary indicated as a thick black line. In the contour plots, areas of high probability density are marked as light colours and values close to zero as red.

m=0, n=1

m=0, n=3

m=0, n=5

m=1, n=1

m=1, n=2

To come: m=2, m=3

[Edited on 30-6-2016 by blogfast25]

m=2, n=1

m=2, n=2

m=3, n=1

m=3, n=2

wg48 - 26-7-2016 at 16:56

Blogfast25:

After titillating me with your color rendered 3D graphs, I am waiting for your orbital derivations. I have even fired up a cobbled together xp pc to run my old copy of mathcad so I might give them a try.

[Edited on 27-7-2016 by wg48]

blogfast25 - 27-7-2016 at 05:42

Quote: Originally posted by wg48 |

What specific orbital derivations are you looking for?

Recently I tried 3D plotting of the H 2p

For now I leave you with the contour plot of the actual wave function (not squared), which also shows the zero-lines (nodal lines), for the system above (m=3, n=2). All straight lines and circles are nodes where the wave function changes sign:

And here's a fascinating tidbit. The energy quantisation of the particle in a circular well with infinite walls is essentially chaotic. And this explains why a circular flexible membrane, like a drum for musical purposes, has no fundamental tone or well defined harmonics. It's basically noise.

[Edited on 27-7-2016 by blogfast25]

wg48 - 27-7-2016 at 10:52

Quote: Originally posted by blogfast25 |

I had no particular orbital in mind. My interest was just general curiosity. In part motivated by how an object with some symmetry can have modes with a different symmetry. I have now sorted that out now.

I thought drums do have a distinct modes but I suspect you mean something different than that. I will reread your particle in box to see if I can understand what you mean or even better play with it in mathcad.

blogfast25 - 27-7-2016 at 12:59

Quote: Originally posted by wg48 |

Look here:

http://www.sciencemadness.org/talk/viewthread.php?tid=65532&...

$$k_{m,n}R=z_{m,n}$$

The roots of the Bessel function (z

The PDE for a circular, elastic membrane is, bar the coefficients, the SAME as for the particle in a circular box. In that derivation the z

Now look at the vibrating modes of the membrane:

https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membr...

Ring a bell?

Let me know what I can do to help with orbital plotting: it would be a first here at SM!

<hr>

Some excellent sources of orbital representations:

http://www.st-andrews.ac.uk/physics/quvis/simulations_chem/c...

And:

http://winter.group.shef.ac.uk/orbitron/

[Edited on 27-7-2016 by blogfast25]

wg48 - 27-7-2016 at 14:17

Ok I think I understand what you mean now.

Personally I would not use the term noise ( or chaotic) to describe the spectrum of a drum in this technical context.

In practice the modes of a drum are excited preferentially by where you hit the drum and by the resilience of the drum stick.

As for ringing a bell: the whealing sounds of bells is caused by the anharmonic relationship between its fundamental and overtones. The fundamental decays faster than some overtones producing a sound that chances frequency, the walling. LOL

Thanks for the offer of help. I may get round to it some time. I am presently improving my attenuation graph and trying to work out how I can get mathcad to calculate masses from a chemical equation. I am stuck at inputting the atomic weights and abbreviations at the moment.

blogfast25 - 27-7-2016 at 14:24

Quote: Originally posted by wg48 |

Chaotic and noise in the sense of inharmonic.

The amplitude spectrum (relative weight of the various frequencies) is indeed determined by the initial condition:

$$u(r,\theta,0)=f(x,y)$$

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