HW 14 Mathematics 503, Mathematical Modeling, CSUF , August 6, 2007

Nasser M. Abbasi

June 15, 2014

1 Problem 9 page 346 section 6.2 (PDE's)
2 Problem 3 page 365 section 6.2 (PDE's)
3 Problem 5 page 365 section 6.2 Conservation laws

1 Problem 9 page 346 section 6.2 (PDE's)

problem:

Find all solutions to the heat equation $ut = κuxx$ of the form $u(x,t) = U (z)$ where $x√-- z= κt$

answer:

We have that $z(x,t) = √xκt$ , hence $∂∂zx = √1κt$ and $2 ∂∂zx2-= 0$ and $3 − 3 ∂z∂t = −2x(κt)−2 κ = −2xt√2k-$

Now

and

Plug in the above expressions into the PDE we obtain

But $√x- z = κt$ , hence the above becomes

1 ′ ′′ − 2z U (z) = U (z)

|---------------------| | ′′ | | U (z)+ 12z U ′(z)= 0 | -----------------------

Let $U ′(z)= y(z),$ hence the above becomes

Integrate both sides

1 lny = − -z2+ C 4

Hence

−1z2 y(z) = Ae 4

But since $′ U (z)= y(z)$ , then

I think now I need to write the above in terms of $x,t$ again. Fix time, and change $x$ and so we have $∂z 1-- dz = ∂xdx= √κtdx$ and the above integral becomes

∫ −(x−4κξt)2√-1- u (x,t;ξ)= A (ξ)e κtd ξ + B (ξ )

for any $ξ$ location along the space dimension $x$ , where $A(ξ),B (ξ )$ are functions that depend on the value $ξ$

2 Problem 3 page 365 section 6.2 (PDE's)

problem:

Use the energy method to prove the uniqueness for the problem

Solution

First note that $2 2 2 ∇2u ≡ ∂∂ux2-+ ∂∂ux2+ ⋅⋅⋅+ ∂∂x2nu 1 2$ i.e. the Laplacian.

Proof by contradiction. Assume there is no unique solution. Let $u1(x,t)$ and $u2(x,t)$ be 2 diﬀerent solutions to the above PDE. Let $w (x,t)$ be the diﬀerence between these 2 solutions. i.e. $w (x,t)= u1 (x,t)− u2(x,t)$ , hence $w (x,t)$ must satisfy the following conditions: it must be zero at the boundaries $x∈ ∂ Ω$ for all time, and also it must be zero inside $Ω$ initially. Hence

Now if we can show that $w(x,t)= 0$ for $t > 0$ inside $Ω$ , then this would imply that $u1 (x,t)= u2(x,t)$ , showing a contradiction, hence completing the proof.

i.e. we need to show that $wt(x,t) = ∇2w (x,t)$ yields a solution $w (x,t)= 0$ for $x ∈ Ω,t > 0$

Using the energy argument, we write

∫ E (t)= w2 (x,t)dx Ω

First we note that $E (0)= 0$ since $w (x,0)= 0$ from the initial conditions above.

But $∂∂tw (x,t)= ∇2w(x,t)$ from the PDE itself, hence the above becomes

∫ E ′(t)= 2 w (x,t) ∇2w(x,t) d x Ω

(1)

But from Green ﬁrst identity which states the following

∫ ( ) ∫ u ∇2w+ ∇u ⋅∇w dx = udw-dA Ω ∂Ω dn

Replace $u$ by $w$ in the above, we obtain

Comparing (1) and (2) we see that LHS of (2) is $1E′(t) 2$ Hence the above become

∫ ∫ 1E ′(t)= wdw-dA − ∇w ⋅∇w dx 2 ∂Ω dn Ω

But $∇w ⋅∇w = ∥∇w ∥2$ , so the above becomes

∫ ∫ 1E′(t)= w dw-dA− ∥∇w ∥2 dx 2 ∂Ω dn Ω

But $w(x,t)= 0$ on $∂Ω$ for $t > 0$ , since this is the boundary conditions. Hence the above becomes

∫ E′(t) = − 2 ∥∇w ∥2 dx Ω

Therefore we showed that $E′(t)$ is $≤ 0$ since $∫ Ω∥∇w ∥2 dx≥ 0$

So energy inside $Ω$ is nonincreasing with time. But since $E (0 )= 0$ then $E (t) = 0$ (since energy can not be negative, this is the only choice left).

Therefore, from $∫ 2 E(t)= Ω w (x,t)dx$ , we conclude that $w(x,t)= 0$ everywhere in $Ω$ for $t > 0$ since $w (x,t)$ is continuous in both its arguments.

Hence we conclude since $w (x,t) = u1(x,t)− u2(x,t) = 0$ then $u1(x,t) = u2(x,t)$ , then the PDE solution is unique.

3 Problem 5 page 365 section 6.2 Conservation laws

problem:

In absence of sources derive the diﬀusion equation for radial motion in the plane $D ut = r (rur)r$ from ﬁrst principles. That is, take an arbitrary domain between circles $r = a,r = b$ and apply conservation law for the density $u = u(r,t)$ assuming the ﬂux is $J(r,t)= − Dur$ . Assume no sources.

Answer:

First note that the density $u (r,t)$ is measured in quantity per unit volume.

Consider a cross sectional area through circle $ra = a$ . This area is $2πhra$ where $h$ is the width of the strip.

Let $J(r,t)$ be the ﬂux at $r$ at time $t$ , measured in quantity per unit area per unit time.

Hence amount $u$ that passes though cross sectional area at $ra$ , per unit time, is $A (ra)J (ra,t)$ where $A (ra)= 2πhra$

Similarly, amount $u$ that passes though cross sectional area at $rb$ , per unit time, is $A (rb)J (ra,t)$ where $A (rb)= 2πhrb$

Hence the net amount that ﬂows, per unit time, between $rb$ and $ra$ is $A(ra)J(ra,t)− A(rb)J(rb,t)$

Since there is no source nor sink inside this region, then the above equal the rate at which the amount $u$ itself changes between $rb$ and $ra$ , which is $d- dt (u (r,t)× volume between ra and rb)$ .