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**APM346--Lectures & Home Assignments / Re: Exam Practice: Prob 69-72**

« **on:**April 09, 2019, 10:09:58 PM »

yes

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yes

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When we are asked to "find the Euler-Lagrange of [some functional T[v]]" are we supposed to :

a) differentiate T[v+$\epsilon$ w] w.r.t to w and evaluate at $\epsilon = 0$ to get the EL, or

b) just plug the Lagrangian into

$$\frac{\partial}{\partial x_j} \frac{\partial \mathcal{L}}{\partial v_{x_j}} = \frac{\partial \mathcal{L}}{\partial v}$$

a) differentiate T[v+$\epsilon$ w] w.r.t to w and evaluate at $\epsilon = 0$ to get the EL, or

b) just plug the Lagrangian into

$$\frac{\partial}{\partial x_j} \frac{\partial \mathcal{L}}{\partial v_{x_j}} = \frac{\partial \mathcal{L}}{\partial v}$$

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Yes

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Could someone explain why u(x,t) =sin(x+3t) for x>t, when in the line above its says $\phi(x0 = \psi(x) = 0$ \\

Which would imply $$u = 0 \\for x > 0?$$

Which would imply $$u = 0 \\for x > 0?$$

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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 **

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

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

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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

I am not sure what we are supposed to do here

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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?

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?

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