Sarahisme

its cool, got it! don't worry guys

Hello peoples,

I think this is a trick question... well sort of

for part (a) i get that the cosine Fourier Series for f(x) = cos(x) to be:

i hope that is ok, but its part (b) that is troubling me...

is all that happens as is that the cosine Fourier series of cos(x) goes to 0?

i am guessing i am missing some trick to this question?

Cheers!

Sarah

its cool everyone, got the question, its all good!

Yes; I was still looking at the general solution. The constraints would seem to eliminate the exponential (real or imaginary) solution.

ah ok, yeah that all makes sense now.

since you people have been so helpful with this PDE question, if any of you guys have time, i would really love some help with some of my other questions

thanks guys!

Sarah

but don't we also need to satify the initial conditions?

hmmm i have no idea where to even start with this problem, i cannot find any examples that are similar or anything like that anywhere!

anyone got an idea as to a good first step to take?

thanks

sarah

edit: i tryed something wild and came up with [math] A(u) = e^{u(x,y)} [/math] but i didnt use any knowlage of PDE's to work that out....hmmm.... :S

hi all, sorry again for the thousands of stupid questions! but heres another one...

ok i said for part (a) that the PDE is a hyperbolic type and so has two sets of real characteristics. and that the characteristics are

then for part (b) i used the coordinates:

which leads to the equation:

but then i am stuck... how can i get the 'general solution' from this, i think i am heading in the right direction, but i am not sure.

any help would be greatly appreciated!

thanks

Sarah

but then how do you find a singularity when you have an answer of 1 and 0? :S

ok i got it now , thanks Bignose

lol my head is about to explode!

i think this is similar to a previous question i asked but i can't quite get it none the less...

now what i did was to following this :

then using equations (31) & (32) from that i just plugged in the values and got:

u(x,t) = 1 for x > ct and u(x,t) = 0 for 0 < x < ct

how does this look to you intelligent mathematically inclined people? :S

Sarah

You can add e^x+ct to that and still satisfy the solution, which means that it must be included in the answer.

how do you know to add that? i see that it works, but at what point do you figure out to add it? that is it doesnt come out in any of the steps i take to solve the problem...

does this mean for this problem: that i have to do a similar thing? (i.e. add [math] e^{x+ct} [/math] ?

i get for that problem that u(x,t) = 1 for x > ct and u(x,t) = 0 for 0 < x < ct but that seems to simple, am i forgetting something completely obvious swansont? :S

its ok i think its correct, its some other questions that i am having problems with now!

You need to write down the most general expression which could give the last term, and then work out the coefficients. For example, for (a) write down a polynomial in x and t up to four powers (in x or t) with arbitrary coefficients. Stick it in and solve for the coefficients. That gives you one solution, but then you can add anything to that solution which satisfies [math]u_{xx}=c^2u_{tt}[/math] and it will still be a solution. So you also need to find the solution to that inhomogenous equation and add it in.

ok, i think i did something like that...i said that part (a) has a solution of:

[math] u(x,t) = f(x+ct)+g(x-ct)-\frac{x^3t}{6c^2} [/math]

then using the initial conditions i solved for f and g... and came up with my answer of [math] u(x,t) = \frac{1}{6}xt^3 [/math]

ok. well i plugged my answer for part (a) back in and it works out, although i think what i have got as an answer could/should be simplified... leaving the answer as an integral looks a bit messy.

however for part (b) i am totally stumped... how do you work with the BC that u(0,t) = 1 ?? any suggestions?

well, u(x,t)=0 cannot be right, since the equation in a becomes 0=x*t

 

Have you tried separation of variables? That method is usually the first bullet out of my gun when given problems like that.

 

Let u(x,t)=X(x)*T(t), then you should be able to write 2 equations, one for X(x) (and only a function of x) and T(t) for only a function of t. Then solve the two ODEs.

i can't see how to apply seperation of variables (at least for part (a)).... i get this after plugging u(x,t) = X(x)T(t) into the orginal PDE:

XT'' = (c^2)X''T + xt

but you can't seperate the x's and t's ...? lol i am probably just completely wrong!

hmm ok , its just that my textbook says nothing about seperation of variables, all it talks about is how to get an explicit solution (such as my one above), which works in maple...

i am little confused.... is my answer incorrect or...?

i don't think seperation of variables works for part (a) because i end up getting that X(x) = 0 (because X(+inifinity) = exp(-x))

I don´t have the time thinking about that atm, but since I suppose others might have the same question: Are the small subscripts at the u derivatives with respect to that parameter?

yes they are

hmm ok i tryed some more and i come up with this answer:

however i don't know how to simplify it from here, i have looked up integral tables and still no luck. any suggestions guys? lol assuming the answer is right in the first place!

cheers

-sarah

Hi all,

I am stuggling with this question ...

so far i have only tried part (a), but since i can't see how to do that so far...

ok so what to do...

do we first look at an 'associated problem' ? ... something like

lol, this stuff is all quite confusing

-Sarah

ok, i tried again and this is what i get...

i included a non-homogenous term...

for (a)

for (b)

what do you guys think?

Hi everyone,

I'm having a bit of trouble with this pde problem:

i get the answer to be u(x,t)=0 but i am guessing thats not right.

is the general solution to this problem: u(x,t) = f(x+ct) + g(x-ct) ??

thanks

sarah

Hmmm' date=' interesting to see that formula you give. I've never seen this before. Indeed, it works out to be the same as what I found, but I do not see the general underlying principle behind this formula. This formula only is valid for 2D surfaces in 3D space. The formalism, I outlined in the previous post, is general and works for any manifold. Also the mechanism with the determinant of the Jacobian is general, but it is somewhat more cumbersome.

 

But it is good to see that our answers now agree and that this issue is settled for you.[/quote']

 

yep its all good now thanks for all your help and especially your time

 

-Sarah

yes, and you should avoid going into these things for a while. there is no need to be at all concerned with manifolds, wedge products and the like, since you're only doing this with surfaces in 3 space. The motivation behind that is because you're doing stuff that is 'real life'. What woelen is talking about is the generalization to arbitrary dimensions, and not relevant to you.

ah ok, that would explain why i have never heard my lecturer mention them

nm, got it!

Hi all,

would anyone be able to give me a little bit of help with this question.

for part (a) i did this....

our sphere is given by [math] x^2 + y^2 +z^2 = 25 [/math]

putting in z = 3 to find out part of the sphere gives:

[math] x^2 + y^2 = 16 [/math]

so we change to the parametric representation of the part of the sphere we are interested in sphere which is

[math]\vec{r}(\theta,\psi) = (4sin\psi cos\theta) \vec{i} + (4sin\psi sin\theta) \vec{j} +(4cos\psi) \vec{k} [/math]

then

[math]\vec{r_\theta}(\theta,\psi) = (-4sin\psi sin\theta) \vec{i} + (4sin\psi cos\theta) \vec{j}[/math]

[math]\vec{r_\psi}(\theta,\psi) = (4cos\psi cos\theta) \vec{i} + (4cos\psi sin\theta) \vec{j} +(-4sin\psi) \vec{k} [/math]

so we get that

[math] \vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi) = -16sin^2\psi cos\theta)\vec{i} - 16sin^2\psi sin\theta \vec{j} -16sin\psi cos\psi \vec{k} [/math]

our unit normal vector is given by:

[math] \vec{\hat{n}} = -\frac{\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)}{|\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)|} = \frac{16sin^2\psi cos\theta)\vec{i} + 16sin^2\psi sin\theta \vec{j} + 16sin\psi cos\psi \vec{k}}{|\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)|} [/math]

does this look like i am heading in the right direction for part (a)?

cheers!

-Sarah