Integration

[Yllanes]

Which step do you not understand?

By the way, I get the same answers as Zygar.

 

[andyandy]

ok, maybe i'm just being really dense here...it's been a while since i've done this....

step by step

solving

acosx + bsinx = 0

bsinx = -acosx

btanx = -a

so

subbing a = -btanx into equation (2) gives

btanxsinx + bcosx = 1+tanx

transforming the left hand side to the form Kcos(x+@)

[latex] ({{b^2}{tanx}{tanx} + b^2)^{0.5} {cos({x+{{arctan}{(-cotx)}}}) [/latex]

and [latex] {arctan}{(-cotx)} = x - pi/2[/latex]

so

this simplifies to

[latex] cos(2x-pi/2) [/latex]

but then i'm stuck...i can square both sides and get rid of the sqrt on the left, but it still seems rather messy....
have i gone wrong already or can this be solved?

 

[Yllanes]

subbing a = -btanx into equation (2) gives

btanxsinx + bcosx = 1+tanx

Yes.

transforming the left hand side to the form Kcos(x+@)

[latex] $$(b^2\tan^2 x + b^2)^{1/2} \cos\bigl(x+\arctan(-\cot x)\bigr) $$[/latex]

This makes the expression even more complicated. A simpler way:

[latex]
\begin{align*}
1+\tan x &= b \tan x \sin x + b \cos x \\
&= b\left[ \frac{\sin^2 x}{\cos x} + \cos x\right] \\
&= b \frac{1}{\cos x} \left[\sin^2 x +\cos^2 x\right]= \frac{b}{\cos x}
\end{align*}
[/latex]

And you get b = cos x (1+ tan x ) = cos x + sin x. Multiplying by - tan x we recover a = - tan x cos x (1+tan x) = - sin x (1 + tan x) .

Your first step with these equations is to try and find sin² x + cos² = 1 somewhere. If there is no way to get it you can try the formulas for half angles; double angles; sin, cos or tan of the sum or difference etc. But try the easier stuff first.

 

[69dodge]

Multiply the first equation by sin x, the second by cos x, and add them to get an equation in b but not a.

Multiply the first equation by cos x, the second by sin x, and subtract them to get an equation in a but not b.

Factor out sin² x + cos² x from the left side of each of the resulting equations, and replace it by 1.

And you're done.

 

[andyandy]

many thanks guys - that's been annoying me all morning

 

[andyandy]

PDEs.....

First, the question asks to find the ODE satisfied by [latex] $$ f(r) [/latex]

if [latex]$ {\frac{1}{r}} f(r)coswt [/latex] is a solution of the PDE;



[latex] $$ {\frac{\partial^2 u}{\partial r^2}} + {\frac{2}{r}} {\frac {\partial u}{\partial r}} = {\frac{1}{c^2}}{\frac{\partial^2 u}{\partial t^2}}



[/latex]

(where w and c and constants)

With some subbing, you get the ODE

[latex]$ {f^''(r)} + {\frac{w^2}{c^2}}f(r) = 0 [/latex]

as such a solution of this PDE is

[latex] $ u={{\frac{1}{r}}(Acosnr + Bsinnr)coswt}

n=${\frac{w}{c}} [/latex]

where A and B are constants

so far so good....but,

next we're told that
For a particular solution the following conditions all hold;
u is finite at r=0 for all t, [latex]$ {\frac{\partial u}{\partial r}}[/latex] = 0 at r=a for all t, and u is not identically zero.

and then have to show that A=0 and obtain an equation which must be satisfied by w.

The solution states that for u to be finite at r=0 requires A=0
I don't understand why. Something to do with w/c being the b in the complex number b i perhaps? lol. Just a guess

I also don't understand the significance of introducing a, or how one arrives at the given solution
[latex]$ {tan{\frac{wa}{c}}} = {\frac{wa}{c}} [/latex]

any explanations? Or a pointer as to what area to read up on?

thanks

ETA

having thought about this,
obviously one can't divide by zero - so if you multiply both sides by r (hmm...i think that's ok...)
then the left side equals zero, and the right side A(coswt)
so for all values of t, A must be zero......
is that right?

now, the rest of it......

 

[Jekyll]

The solution states that for u to be finite at r=0 requires A=0
I don't understand why.

You're overlooking the silly little things.

cos n0 is 1.
a/0 is always non-finite unless a is 0.

Whoops missed a step:

1/r*sin nr = n( 1/nr *sin nr ) ->n as nr ->0. So Bn is finite.

 

[andyandy]

You're overlooking the silly little things.

cos n0 is 1.
a/0 is always non-finite unless a is 0.

Whoops missed a step:

1/r*sin nr = n( 1/nr *sin nr ) ->n as nr ->0. So Bn is finite.

ok i like that reasoning better

 

[Yllanes]

PDEs.....

[latex] $ u={{\frac{1}{r}}(A\cos nr + B\sin nr)\cos wt}$
[/latex]

where A and B are constants

so far so good....but,

next we're told that
For a particular solution the following conditions all hold;
u is finite at r=0 for all t, [latex]$ {\frac{\partial u}{\partial r}}$[/latex] = 0 at r=a for all t, and u is not identically zero.

and then have to show that A=0 and obtain an equation which must be satisfied by w.

[...]

(Forgive me for fixing the LaTeX, but I am a huge LaTeX nerd, you have to enclose the formula in $...$, not only put one $ at the beginning)

You guessed the first part: A/r cos n r -> oo as r -> 0, so A must be zero to take care of this. On the other hand B/r sen nr -> Bn as r -> 0, so that part is Ok.

As for the second part, it is only a matter of imposing the condition.

[latex]
$$
\frac{\partial u}{\partial r} = \underbrace{B\cos \omega t}_{\substack{\text{irrelevant, does not}\\ \text{ depend on $r$}}} \left( \frac{n\cos nr}{r} - \frac{\sin nr}{r^2}\right)
$$
[/latex]

Now you subsitute r=a and set the resulting expression equal to zero, to get an equation for a.

With this very simple equations you only have to substitute r= boundary value (or the relevant variable) and solve for the required parameter.

 

[boooeee]

The significance of introducing a is that it in order to get a unique solution to a differential equation, you need to specify your boundary conditions.

They could have said the 1st derivative = 0 at 2.5 and have you solve for that. However, then your solution is only good for that particular boundary condition. By leaving a as a variable, you can now solve for a whole host of potential boundary conditions. If somebody tells you that the 1st derivative = 0 at 4.95, then you can just plug that value of a into your final expression.

 

[andyandy]

ok that's helpful

solving for du/dr = 0

you get 1/na*tan(na) = 0

so does the "u not identically zero" bit mean that you discount u(0) and look at what 1/na*tan(na) tends to as a -> 0 ?

which therefore gives the required

1 = 1/na*tan(na)
na=tan(na)

(n=w/c)

 

[boooeee]

you get 1/na*tan(na) = 0

You may want to double check that step.

 

[andyandy]

You may want to double check that step.

oops....i mean

(n/a)tan(na) = 0

 

[Yllanes]

so does the "u not identically zero" bit mean that you discount u(0) and look at what 1/na*tan(na) tends to as a -> 0 ?

The sentence 'u identically zero' means u(r,t)=0 for all values of r and t. Now, the problem asks you to find a particular solution to an equation. Clearly u(r,t)=0 works but it would be too easy... That sentence is only there to rpevent you from giving this trivial answer, it has nothing to do with the value of u at r=0.

 

[andyandy]

phew, partial differentiation is tricky.....

Given the error function as

[latex]$ erf(x) = {\frac{2}{\sqrt{\pi}}}{{\int_0^x}e^{-u^2}du} $ [/latex]

and told to show that

[latex] $ u(x,t)=A(1-erf(\frac{x}{2\sqrt{kt}}) $ [/latex]

is a solution of the equation (A and k constants)

[latex] $ \frac {\partial^2u}{\partial x^2} = \frac {1}{k} \frac {\partial u}{\partial t} $[/latex]

what's the best way to find [latex] $\frac {\partial u}{\partial t}$ [/latex]?

I've subbed so that

[latex] $ u(x,t)=A(1+\frac{2}{\sqrt{\pi}x2\sqrt{kt}} {e^\frac{-x^2}{4kt})}

[/latex]

and then differentiated - but upon simplifying i don't get the given answer of

[latex] $ \frac{-2A}{\sqrt{\pi}}e^\frac{-x^2}{4kt}\frac{x}{2\sqrt{k}}\frac{-1}{2t\sqrt{t}}$[/latex]

i'm wondering if it's just my working out that's wrong, or if my initial sub is wrong,

and also if there's an easier way (such as chain rule....
i've already found du/dx and d^2u/dx^2 so is it easier to use these and dx/dt?)

 

[andyandy]

That sentence is only there to prevent you from giving this trivial answer.

fair enough

 

[Yllanes]

I've subbed so that

[latex] $ u(x,t)=A(1+\frac{2}{\sqrt{\pi}x2\sqrt{kt}} {e^\frac{-x^2}{4kt})}

[/latex]

An integral is missing there.

and also if there's an easier way (such as chain rule.... i've already found du/dx and d^2u/dx^2 so is it easier to use these and dx/dt?)

You can't do that, x and t are independent variables.

You have to differentiate [latex]$-A\,\mathrm{erf}\left(\frac{x}{2\sqrt{kt}}\right)$[/latex] with respect to t. The way to do this is to write z = x /2 sqrt(kt) and calculate

[latex]
$$
\frac{\partial u}{\partial t} = \frac{\partial u}{\partial z}\frac{\partial z}{\partial t}
$$
[/latex]

This shouldn't be difficult. The first step is only the fundamental theorem of calculus, the second is a simple potential function.

 

[boooeee]

oops....i mean

(n/a)tan(na) = 0

Actually, I think that's still incorrect. Applying the differentiation rules and the algebra correctly should lead you directly to the given solution: tan(wa/c) = wa/c.

Hint: Look at Yllanes expression for du/dr and set that to zero.

 

[andyandy]

An integral is missing there.


You can't do that, x and t are independent variables.

You have to differentiate [latex]$-A\,\mathrm{erf}\left(\frac{x}{2\sqrt{kt}}\right)$[/latex] with respect to t. The way to do this is to write z = x /2 sqrt(kt) and calculate

[latex]
$$
\frac{\partial u}{\partial t} = \frac{\partial u}{\partial z}\frac{\partial z}{\partial t}
$$
[/latex]

This shouldn't be difficult. The first step is only the fundamental theorem of calculus, the second is a simple potential function.

doh! You're right....doing it that way isn't too difficult at all....

with my sub - i integrated the erf(x) with limits 0 and (x/(2sqrt(kt)))
to get u
can't you do this?

 

[andyandy]

Actually, I think that's still incorrect. Applying the differentiation rules and the algebra correctly should lead you directly to the given solution: tan(wa/c) = wa/c.

Hint: Look at Yllanes expression for du/dr and set that to zero.

thanks - you're right.

stupid maths