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**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: $$\frac{1}{2} = \frac{1}{4\sqrt{kt\pi}} \int_{0}^{\inf} \exp(\frac{-(y+2kat-x)^2}{4kt}) dy$$

$$u(x, t) = \frac{2 \exp(ax-a^2kt)}{2} = \exp(ax-a^2kt) $$

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**