Q. 70 - Calculus

Question

Alex is modeling traffic patterns at a bottleneck on a freeway as it leaves Denver. He uses a well-known equation $u_x+u_t=0$, where $u=u(x, t)$ is a scaled traffic density at a point x on the highway at time t.

(a) Show that $u_n(x,t)=\sin \left(n\right(x-t\left)\right)$ is a solution of the equation for any integer n.

(b) Alex finds that in heavy traffic at the bottleneck the solution of his equation/ can look like

$24+\sum _{n=1}^N\frac{\sin \left(n\right(x-t\left)\right)}{n}$ for some integer $N .$

Show that this is a solution, and plot it for $N=8$. Can you tell what happens as N gets large?

Step-by-Step Solution

Answer

Ans: 

Part (a). $$\begin{gathered} u_n(x,t)=24+\sum _{n=1}^N\frac{\sin \left(n\right(x-t\left)\right)}{n} \\ {u}_x=\sum _{n=1}^N\cos \left(n\right(x-t\left)\right) \end{gathered}$$

Part (b).$$\begin{gathered} u_n(x,t)=24+\sum _{n=1}^8\frac{\sin \left(n\right(x-t\left)\right)}{n} \\ =24+\sin (x-t)+\frac{\sin \left(2\right(x-t\left)\right)}{2}+\frac{\sin \left(3\right(x-t\left)\right)}{3}+\dots +\frac{\sin \left(8\right(x-t\left)\right)}{8} \end{gathered}$$

Step 1. Given information:

The differential equation for the modeling traffic patterns is given to be

$u_x+u_t=0$

Step 2. Differentiating the function:

The partial derivative of a function is determined by differentiating it with respect to the involved variable, keeping the other variable as a constant.

Differentiate the function 'f ' partially with respect to ' x ' treating ' t ' as constant.

$$\begin{gathered} u_n(x,t)=\sin (nx-nt) \\ {u}_x=n\cos (nx-nt) \end{gathered}$$

Differentiate the function 'f' partially with respect to ' t ' treating ' x ' as constant.

$$\begin{gathered} u_n(x,t)=\sin (nx-nt) \\ {u}_t=-n\cos (nx-nt) \end{gathered}$$

Step 3. Checking:

Substitute these double partial derivatives in the differential equation to check if they satisfy the equation or now.

$$\begin{gathered} u_x+u_t=0 \\ \left(n\cos \right(nx-nt\left)\right)+(-n\cos (nx-nt\left)\right)=0 \\ n\cos (nx-nt)-n\cos (nx-nt)=0 \\ 0=0 \end{gathered}$$

Thus, this function satisfied the differential equation very well.

Hence, these functions do satisfy the given differential equation, $u_x+u_t=0$

Step 4. Solving part (a):

(a) The objective is to show that $u_n(x,t)=\sin \left(n\right(x-t\left)\right)$ is a of this equation for any numbers n

Consider the solution given as $u_n(x,t)=24+\sum _{n=1}^N\frac{\sin \left(n\right(x-t\left)\right)}{n}$.

The partial derivative of a function is determined by differentiating it with respect to the involved variable, keeping the other variable as a constant.

Differentiate the function ' f ' partially with respect to 'x' treating 't' as constant.

$$\begin{gathered} u_n(x,t)=24+\sum _{n=1}^N\frac{\sin \left(n\right(x-t\left)\right)}{n} \\ {u}_x=\sum _{n=1}^N\frac{n\cos \left(n\right(x-t\left)\right)}{n} \\ {u}_x=\sum _{n=1}^N\cos \left(n\right(x-t\left)\right) \end{gathered}$$

Step 5. Differentiating the function:

Differentiate the function 'f' partially with respect to 't' treating 'x' as constant.

$$\begin{gathered} u_n(x,t)=24+\sum _{n=1}^N\frac{\sin \left(n\right(x-t\left)\right)}{n} \\ {u}_x=\sum _{n=1}^N\frac{-n\cos \left(n\right(x-t\left)\right)}{n} \\ {u}_x=\sum _{n=1}^N\cos \left(n\right(x-t\left)\right) \end{gathered}$$

Step 6. Checking:

Substitute these double partial derivatives in the differential equation to check if they satisfy the equation or now.

$$\begin{gathered} u_x+u_t=0 \\ \left(\sum _{n=1}^N\cos \left(n\right(x-t\left)\right)\right)+\left(-\sum _{n=1}^N\cos \left(n\right(x-t\left)\right)\right)=0 \\ \sum _{n=1}^N\cos \left(n\right(x-t\left)\right)-\sum _{n=1}^N\cos \left(n\right(x-t\left)\right)=0 \\ 0=0 \end{gathered}$$

Thus, this function satisfied the differential equation very well.

Hence, these functions do satisfy the given differential equation, $u_x+u_t=0$

Step 7. Plotting the graph:

Rewrite the solution function, using $N=8$.

$$\begin{gathered} u_n(x,t)=24+\sum _{n=1}^8\frac{\sin \left(n\right(x-t\left)\right)}{n} \\ =24+\sin (x-t)+\frac{\sin \left(2\right(x-t\left)\right)}{2}+\frac{\sin \left(3\right(x-t\left)\right)}{3}+\dots +\frac{\sin \left(8\right(x-t\left)\right)}{8} \end{gathered}$$

Use computer software to plot the graph of the above function.