# PDE problem : diffusion equation! help!

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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

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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

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Sarah, do you know what the method of separation of variables is? (I mentioned it in your wave equation post.)

Let me show you

First, we let u(x,t) equal the product of two functions, one only a function of x and the other only a function of t

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

Now, plug this into your equation (I'm sorry, but latex doesn't appear right on my screen so I'm gonna try to do this with normal text. And d means partial in this case)

du/dt becomes d(XT)/dt which is equal to XdT/dt since X is only a function of x which I'm going to shorthand even further to XT' where the prime means differentiation only on that functions variable.

kd^2u/dx^2 becomes kTd^2X/dx^2 or kTX''

XT' = kTX'' or rewritten

T'/T = kX''/X

Now, look at this, the lefthandside is only a function of t, and the righthandside is only a function of x. How can this be? If and only if both the lefthand and righthand sides are equal to a constant, which I am going to denote R.

T'/T = kX''/X = R

this can be written as two ODEs now.

T'/T = R

&

kX''/X = R

This two ODEs can be solved.

The boundary conditions can also be converted using the separation of variables, and should be applied to the appropriate ODE.

For example, using the BCs of problem b, it becomes:

T'/T = R & T(0)=0

kX''/X = R & X(0)=1 & X(+infinity)=0

p.s. I assume that as x goes to infinity, u goes to zero right? Otherwise your problem statement is missing a boundary condition.

After solving the ODEs, you put the solution back together, u=X*T and you have an answer.

p.p.s. Like I mentioned in your wave equation post, separation of variables is the very first thing I always try on a PDE that is supposed to have an analytical solution. There are few ODEs that have analytical solutions, and even fewer PDEs, so the PDE homework problems that are out there are almost all separable. There are some out there that require other tricks (like Fourier or Laplace or other transforms), but it rarely hurts to try separation first. You can usually see pretty quickly if it is going to work or not.

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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))

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Well, I didnt calculate an answer, so I don't know if it is right or wrong

You can put your answer back into the PDE and see if the two sides are equal or not. You can always check, and probably should whenever possible.

There are many different ways to solve this kind of problem. I know from your posts over the last few months you've been going through orthogonal functions, and these problems from 0 to infinity do lend themselves to being solved via the method of Laplace transforms. Some of the other ones can be solved via the method of Fourier transforms.

But, whatever way you solve it, you should get the same answer, provided the problem fulfills the uniqueness theorems (which pretty much any homework problem will). It may be expressed a little differently depending on the method you used, but the answer from any valid method you performed correctly will be unique. So, whether you use separation of variables or any other method, it should be the same answer. BTW, almost any book on PDEs you get from the library will have lots of info on separation of variables, if you want to read up on it. Like I said, it is the first method I always try, since it only takes a few seconds and you can see quickly whether the method will work or not.

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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? ##### Share on other sites

ok i got it now , thanks Bignose ## Create an account

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