MATH 251 Lecture 36

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Written Homework 8

	FTC:	$\int _{a}^{b}f'(x)\,\mathrm {d} x=f(b)-f(a)$
+	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:	$\int _{a}^{b}\left(u'v\right)\,\mathrm {d} x=\left.uv\right|_{a}^{b}-\int _{a}^{b}\left(uv'\right)\,\mathrm {d} x$

Using the divergence theorem, $\iiint _{R}\mathrm {div} {\vec {F}}\,\mathrm {d} V=\iint _{\partial R}{\vec {F}}\cdot {\vec {n}}\,\mathrm {d} S$

Problem 1

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

Part A

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

We need to verify this.

Let ${\vec {F}}=\left\langle P,Q,R\right\rangle$, so $u{\vec {F}}=\left\langle uP,uQ,uR\right\rangle$

${\begin{aligned}{\vec {\nabla }}\cdot \left\langle uP,uQ,uR\right\rangle &=(u_{x}P+uP_{x})+(u_{y}Q+uQ_{y})+(u_{z}R+uR_{z})\\&=(u_{x}P+u_{y}Q+u_{z}R)+(uP_{x}+uQ_{y}+uR_{z})\\&=({\vec {\nabla }}u\cdot {\vec {F}})+u({\vec {\nabla }}\cdot {\vec {F}})\end{aligned}}$

Part B

Find ${\vec {\nabla }}\cdot (u{\vec {\nabla }}v)$

Plug in ${\vec {F}}={\vec {\nabla }}v$:

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

Problem 2

Integration by Parts:

${\begin{aligned}\iiint _{R}{\vec {\nabla }}\cdot (u{\vec {F}})\,\mathrm {d} V&=\iiint _{R}\left({\vec {\nabla }}u\cdot {\vec {F}}+u{\vec {\nabla }}\cdot {\vec {F}}\right)\,\mathrm {d} V\\\iint _{\partial R}u{\vec {F}}\cdot {\vec {n}}\,\mathrm {d} S&=\\\iiint _{R}u{\vec {\nabla }}\cdot {\vec {F}}\,\mathrm {d} V&=\iint _{\partial R}u{\vec {F}}\cdot {\vec {n}}\,\mathrm {d} S-\iiint _{R}\left({\vec {\nabla }}u\cdot {\vec {F}}\right)\,\mathrm {d} V\\\end{aligned}}$

Substitute ${\vec {F}}={\vec {\nabla }}v$:

${\begin{aligned}\iiint _{R}u\Delta v\,\mathrm {d} V=\iint _{\partial R}u\left({\vec {\nabla }}v\cdot {\vec {n}}\right)\,\mathrm {d} S-\iiint _{R}\left({\vec {\nabla }}u\cdot {\vec {\nabla }}v\right)\,\mathrm {d} V\end{aligned}}$

Problem 3

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

$u_{t}-\alpha \Delta u=0$

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: $\mathrm {Heat\ flow} =-\alpha {\vec {\nabla }}u$, where $\alpha$ is a property of the material called heat diffusivity.

$u_{t}=-{\vec {\nabla }}\cdot (-\alpha {\vec {\nabla }}u)=\alpha \Delta u$

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

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

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

$\iiint _{R}u_{t}\,\mathrm {d} V=0$

Substitute $u_{t}=\alpha \Delta u$:

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

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

Problem 4

${\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} t}}\iiint _{R}{\frac {1}{2}}u^{2}\,\mathrm {d} V&=\iiint {\frac {\mathrm {d} }{\mathrm {d} t}}\,\mathrm {d} V\\&=\iiint _{R}u\,u_{t}\,\mathrm {d} V\\&=\alpha \iiint _{R}u\Delta u\,\mathrm {d} V\end{aligned}}$

The last step is from problem 3. Continue from there

Problem 6

Kinetic Energy and Potential Energy...

$E(t)=\iiint _{R}{\frac {1}{2}}u_{t}^{2}+{\frac {c^{2}}{2}}\left\|{\vec {\nabla }}u\right\|^{2}\,\mathrm {d} V$

Find $E'(t)=0$ to show that energy is conserved.

${\begin{aligned}E'(t)&=\iiint _{R}u_{t}\,u_{tt}+c^{2}{\vec {\nabla }}u\cdot {\vec {\nabla }}u_{t}\,\mathrm {d} V\\&=\iiint _{R}u_{t}c^{2}\Delta u+c^{2}{\vec {\nabla }}u\cdot {\vec {\nabla }}u_{t}\,\mathrm {d} V\\&=\iint _{\partial R}u_{t}{\vec {\nabla }}u\cdot {\vec {n}}\,\mathrm {d} S\\&=0\end{aligned}}$

This is true because $u=0$ along ∂R so $u_{t}=0$ as well.

Therefore energy is conserved.

Q.E.D.

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