Partial Differential Equations

[Hadrian]

I need help with Diffusion on the Whole Line.

For instance, my first homework problem is:

Solve the diffusion equation Ut = Uxx with the initial value conditions phi(x) = 1 for |x| > l, phi(x) = 0 for |x| < l. I don't know what to do, and the book I'm using isn't the least bit enlightening.

 

[Tide]

Have you tried solving the diffusion equation using a Laplace transform wrt to the time variable?

 

[HallsofIvy]

I need help with Diffusion on the Whole Line.
For instance, my first homework problem is:
Solve the diffusion equation Ut = Uxx with the initial value conditions phi(x) = 1 for |x| > l, phi(x) = 0 for |x| < l. I don't know what to do, and the book I'm using isn't the least bit enlightening.

Your book won't be enlightening if you don't take more care in reading it. No, the problem does NOT say " phi(x) = 1 for |x| > l, phi(x) = 0 for |x| < l". That doesn't make sense- phi is a function of both x and t. I suspect that the problem says: phi(x,0)= 1 for |x|> 1, ph(x,0)= |x|< 1. Also, there will have to be boundary conditions on x: probabably that the solution stays bounded as x goes to both infinity and -infinity. Looks to me like you will need a Fourier Transform solution: write phi(x,t) in the form
$$\phi(x,t)= \frac{1}{\sqrt{2\pi}\int_{-\infty}^{\infty}f(t)e^{ixs}ds$$.
Put that into the differential equation to get an ordinary differential equations for f(t).

 

[Hadrian]

Phi is not a function of x and t, only x. The solution equation u(x,t) is a function of x and t... maybe you could picture phi(x) as being u(x,0), but I don't think that's quite right either. Also, it's |x| < the letter l, not 1. Here are exact words:

"Solve the diffusion equation with the initial condition

phi(x) = 1 for |x| < L and phi(x) = 0 for |x| > L.

Write your answer in terms of Erf(x) <== the error function of statistics."

The only thing I'm left to go on is the equation the guy derived in the section:

1/(4*pi*k*t)^(1/2)*[integral from -infinity to infinity] {exp((-(x-y)^2)/4kt) * phi(y) * dy}

In one of the examples, an initial condition was given without constraints, and it was in terms of x. He then proceeded to plug it into the equation given above as the same function, in terms of y.

I say the book sucks because it dedicates all of two pages to this topic, and it's very very veryyyyy ambiguous with the examples; it doesn't really let me know how I should go about solving a problem of this nature.

Also, that image didn't work. I think I need software?

Thanks for telling me what my book says, though :)

 

[saltydog]

Is this the problem:


$$\text{DE:}\quad \frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial t^2};\quad -\infty<x<\infty,\quad t>0$$

$$\text{IC:}\quad u(x,0)=\Phi(x)=<br /> \left\{<br /> \begin{array}{rcl}<br /> 1 & \mbox{for} & |x|\le L \\<br /> 0 & \mbox{for} & |x|>L<br /> \end{array}<br /> \right$$

As you so stated, the solution is given by the heat kernel:

$$u(x,t)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty} e^{-(x-y)^2/4t} \Phi(y)dy$$

So there you go . . . now, plug in [itex]\Phi(x)[/tex] (use any value of L for starters like L=1) and then express the integral as you so indicated.<br /> <br /> Try and find in the library, "Basic Partial Differential Equations" by D. Bleecker and G. Csordas.<br /> <br /> Edit: Oh yea, a plot wouldn't hurt neither you know.[/itex]

 

[Hadrian]

I don't understand. I'm supposed to plug in zero for phi? doesn't that just give me zero? I'm so confused by this material that my brain is imploding.

 

[saltydog]

I don't understand. I'm supposed to plug in zero for phi? doesn't that just give me zero? I'm so confused by this material that my brain is imploding.

Well, let's say for example that L=1. So for the integral:

$$u(x,t)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty} e^{-(x-y)^2/4t} \Phi(y)dy$$

Phi is non-zero only in [-1,1], everywhere else it's zero so the integral is zero there too. So isn't that the same as:

$$u(x,t)=\frac{1}{\sqrt{4\pi t}}\int_{-1}^{1} e^{-(x-y)^2/4t} (1)dy$$

So what's the value at x=15 and t=100? Wouldn't that be:

$$u(15,100)=\frac{1}{\sqrt{4\pi (100)}}\int_{-1}^{1} e^{-(15-y))^2/(400)} (1)dy$$