MATH 251 Lecture 36

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Written Homework 8

	FTC:	$\underset{a}{\overset{b}{\int }}{f}^{\prime }(x)\mathrm{d}x=f(b)-f(a)$
+	Product Rule:	$\frac{\mathrm{d}}{\mathrm{d}x}(uv)=\left(\frac{\mathrm{d}}{\mathrm{d}x}u\right)v+u\frac{\mathrm{d}}{\mathrm{d}x}v$
=	Integration by Parts:	$\underset{a}{\overset{b}{\int }}\left(u^{\prime }v\right)\mathrm{d}x={uv|}_a^b-\underset{a}{\overset{b}{\int }}\left(u{v}^{\prime }\right)\mathrm{d}x$

Using the divergence theorem, $\underset{R}{\iiint }\mathrm{d}\mathrm{i}\mathrm{v}\overrightarrow{F}\mathrm{d}V=\underset{\partial R}{\iint }\overrightarrow{F}\cdot \overrightarrow{n}\mathrm{d}S$

Problem 1

$u$ is a function, $v$ is a function, and $\overrightarrow{F}$ is a vector field.

Part A

$\overrightarrow{\mathrm{\nabla }}\cdot (u\overrightarrow{F})=(\overrightarrow{\mathrm{\nabla }}u)\cdot \overrightarrow{F}+u\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{F}$

We need to verify this.

Let $\overrightarrow{F}=⟨P,Q,R⟩$, so $u\overrightarrow{F}=⟨uP,uQ,uR⟩$

$\begin{array}{rl}\overrightarrow{\mathrm{\nabla }}\cdot ⟨uP,uQ,uR⟩& =(u_{x}P+u{P}_x)+(u_{y}Q+u{Q}_y)+(u_{z}R+u{R}_z)\\ & =(u_{x}P+u_{y}Q+u_{z}R)+(u{P}_x+u{Q}_y+u{R}_z)\\ & =(\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{F})+u(\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{F})\end{array}$

Part B

Find $\overrightarrow{\mathrm{\nabla }}\cdot (u\overrightarrow{\mathrm{\nabla }}v)$

Plug in $\overrightarrow{F}=\overrightarrow{\mathrm{\nabla }}v$:

$\overrightarrow{\mathrm{\nabla }}\cdot (u\overrightarrow{\mathrm{\nabla }}v)=(\overrightarrow{\mathrm{\nabla }}u)\cdot \overrightarrow{\mathrm{\nabla }}v+u\mathrm{\Delta }v$, where $\mathrm{\Delta }v$ is the Laplace operator ${\overrightarrow{\mathrm{\nabla }}}^{2}v=\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{\mathrm{\nabla }}v$

Problem 2

Integration by Parts:

$\begin{array}{rl}\underset{R}{\iiint }\overrightarrow{\mathrm{\nabla }}\cdot (u\overrightarrow{F})\mathrm{d}V& =\underset{R}{\iiint }(\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{F}+u\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{F})\mathrm{d}V\\ \underset{\partial R}{\iint }u\overrightarrow{F}\cdot \overrightarrow{n}\mathrm{d}S& =\\ \underset{R}{\iiint }u\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{F}\mathrm{d}V& =\underset{\partial R}{\iint }u\overrightarrow{F}\cdot \overrightarrow{n}\mathrm{d}S-\underset{R}{\iiint }(\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{F})\mathrm{d}V\\ \end{array}$

Substitute $\overrightarrow{F}=\overrightarrow{\mathrm{\nabla }}v$:

$\begin{array}{r}\underset{R}{\iiint }u\mathrm{\Delta }v\mathrm{d}V=\underset{\partial R}{\iint }u(\overrightarrow{\mathrm{\nabla }}v\cdot \overrightarrow{n})\mathrm{d}S-\underset{R}{\iiint }(\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{\mathrm{\nabla }}v)\mathrm{d}V\end{array}$

Problem 3

Application of this stuff: temperature at a point $(x,y,z)i$ at time $t$

$u_t-\alpha \mathrm{\Delta }u=0$

Newton's law of cooling states that heat flows against the gradient of the temperature and is proportional to the difference in temperature between two objects: $\mathrm{H}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{f}\mathrm{l}\mathrm{o}\mathrm{w}=-\alpha \overrightarrow{\mathrm{\nabla }}u$, where $\alpha$ is a property of the material called heat diffusivity.

$u_t=-\overrightarrow{\mathrm{\nabla }}\cdot (-\alpha \overrightarrow{\mathrm{\nabla }}u)=\alpha \mathrm{\Delta }u$

For an insulated region $R$, $\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{n}=0$ on a boundary. There is no gradient of heat leaving (pointing outward) along the boundary.

Show that $\frac{\mathrm{d}}{\mathrm{d}t}\underset{R}{\iiint }u(x,y,z,t)\mathrm{d}V=0$

Since we are not integrating with respect to $t$, so we can differentiate the inside:

$\underset{R}{\iiint }{u}_t\mathrm{d}V=0$

Substitute $u_t=\alpha \mathrm{\Delta }u$:

$\alpha \underset{R}{\iiint }\mathrm{\Delta }u=\alpha \underset{\partial R}{\iint }(\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{n})\mathrm{d}S=0$

Since $u$ does not change over time, temperature enclosed must be constant.

Problem 4

$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\underset{R}{\iiint }\frac{1}{2}{u}^2\mathrm{d}V& =\iiint \frac{\mathrm{d}}{\mathrm{d}t}\mathrm{d}V\\ & =\underset{R}{\iiint }u{u}_t\mathrm{d}V\\ & =\alpha \underset{R}{\iiint }u\mathrm{\Delta }u\mathrm{d}V\end{array}$

The last step is from problem 3. Continue from there

Problem 6

Kinetic Energy and Potential Energy...

$E(t)=\underset{R}{\iiint }\frac{1}{2}{u}_t^2+\frac{c^2}{2}{\Vert \overrightarrow{\mathrm{\nabla }}u\Vert }^2\mathrm{d}V$

Find $E^{\prime }(t)=0$ to show that energy is conserved.

$\begin{array}{rl}{E}^{\prime }(t)& =\underset{R}{\iiint }{u}_t{u}_{tt}+c^2\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{\mathrm{\nabla }}{u}_t\mathrm{d}V\\ & =\underset{R}{\iiint }{u}_t{c}^2\mathrm{\Delta }u+c^2\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{\mathrm{\nabla }}{u}_t\mathrm{d}V\\ & =\underset{\partial R}{\iint }{u}_t\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{n}\mathrm{d}S\\ & =0\end{array}$

This is true because $u=0$ along ∂R so $u_t=0$ as well.

Therefore energy is conserved.

Q.E.D.

Note: $\overrightarrow{\mathrm{\nabla }}u\cdot \overrightarrow{n}$ is often written as $\frac{\partial u}{\partial n}$, called the "normal derivative of $u$".