Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

Topics - Rhamel

Pages: [1]
Home Assignment 5 / Problem 1
« on: February 21, 2019, 05:54:56 PM »
I would like to know if my solution is correct.
for problem 1. 3

I am supposed to solve $$u_{t} = ku_{xx}, u(x, 0) = g(x)$$

where $$g(x) = \exp(-a |x|)$$

My solution:
 $$u(x, t) = \frac{1}{4\sqrt{kt\pi}} \int_{-\inf}^{\inf} \exp(\frac{-(x-y)^2}{4kt}) \exp(-a|y|) dy$$
(*) to get rid of the absolute value:
 $$u(x, t) = \frac{2}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{-(x-y)^2}{4kt}) \exp(-ay) dy$$
I then complete the square for: $\frac{-(x-y)^2 - 4ktay}{4kt}$
to get:
 $$u(x, t) = 2 \exp(ax-a^2kt) \frac{1}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{-(y+2kat-x)^2}{4kt}) dy$$

 I then use:  $$1 =  \frac{1}{4\sqrt{kt\pi}} \int_{-\inf}^{\inf} \exp(\frac{-(y+2kat-x)^2}{4kt}) dy$$
(**) to write:  $$\frac{1}{2} =  \frac{1}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{-(y+2kat-x)^2}{4kt}) dy$$
  $$u(x, t) = \frac{2 \exp(ax-a^2kt)}{2} = \exp(ax-a^2kt) $$

I feel like I might have taken a wrong turn at step (*) and/or (**); could someone let me know if this is right or wrong   

Home Assignment 3 / HW2.3 Problem 6
« on: January 30, 2019, 11:45:14 PM »
Part 2 of question 6 asks us to solve for v but then gives us the general form of v?

I am not sure what we are supposed to do here

Home Assignment 2 / Home Assignment 2 problem 1(a)
« on: January 15, 2019, 04:12:16 PM »
For problem 1(a) equation 8: $u_t + (x^2+1)u_x = 0$

From solving for the characteristic equation I get $x = \tan(t-k)$ where k is a constant.
My question is how would I draw the characteristics in this case? The periodic nature of tan would result in just an indistinguishable blob of lines if I were to plot it out? Should I just draw the lines for one value of k?

Pages: [1]