# Problem 1

We have solved this problem by ourselves.

## Given

 ${\displaystyle \displaystyle \left.f(S,t)\right|_{S=Y^{1}(t)}=f(Y^{1}(t),t)}$ (1.1)

## Find

 ${\displaystyle {\frac {\mathrm {d^{2}} f}{\mathrm {d} t^{2}}}=f_{,S}(Y^{1},t){\ddot {Y}}^{1}+f_{,SS}(Y^{1},t)({\dot {Y}}^{1})^{2}+2f_{,St}(Y^{1},t){\dot {Y^{1}}}+f_{,tt}(Y^{1},t)}$, where ${\displaystyle \displaystyle \ f_{,S}(Y^{1},t):={\frac {\partial f(Y^{1}(t),t)}{\partial S}}}$ and ${\displaystyle \displaystyle \ f_{,St}(Y^{1},t):={\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S\partial t}}}$ (1.2)

## Solution

First, we get the first derivative of ${\displaystyle \displaystyle f}$ with respect to time,
 ${\displaystyle \displaystyle {\frac {\mathrm {d} f(Y^{1}(t),t)}{\mathrm {d} t}}={\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\dot {Y}}^{1}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}}}$, with ${\displaystyle \displaystyle {\dot {Y}}^{1}:={\frac {\mathrm {d} Y^{1}(t)}{\mathrm {d} t}}}$ (1.3)
Then we get,
 ${\displaystyle {\frac {\mathrm {d^{2}} f}{\mathrm {d} t^{2}}}=\underbrace {{\frac {\mathrm {d} }{\mathrm {d} t}}[{\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\dot {Y}}^{1}]} _{\color {blue}{RHS1}}+\underbrace {{\frac {\mathrm {d} }{\mathrm {d} t}}[{\frac {\partial f(Y^{1}(t),t)}{\partial t}}]} _{\color {blue}{RHS2}}}$ (1.4)
 ${\displaystyle RHS1={\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\ddot {Y}}^{1}+{\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S^{2}}}({\dot {Y}}^{1})^{2}+{\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S\partial t}}{\dot {Y}}^{1}}$ (1.5)
 ${\displaystyle RHS2={\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S\partial t}}{\dot {Y}}^{1}+{\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial t^{2}}}}$ (1.6)
Bring (1.5) and (1.6) into (1.4), we can get,
 ${\displaystyle {\frac {\mathrm {d^{2}} f}{\mathrm {d} t^{2}}}={\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\ddot {Y}}^{1}+{\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S^{2}}}({\dot {Y}}^{1})^{2}+2{\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S\partial t}}{\dot {Y^{1}}}+{\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial t^{2}}}}$ (1.7)
i.e.
 ${\displaystyle {\frac {\mathrm {d^{2}} f}{\mathrm {d} t^{2}}}=f_{,S}(Y^{1},t){\ddot {Y}}^{1}+f_{,SS}(Y^{1},t)({\dot {Y}}^{1})^{2}+2f_{,St}(Y^{1},t){\dot {Y^{1}}}+f_{,tt}(Y^{1},t)}$, where ${\displaystyle \displaystyle \ f_{,S}(Y^{1},t):={\frac {\partial f(Y^{1}(t),t)}{\partial S}}}$ and ${\displaystyle \displaystyle \ f_{,St}(Y^{1},t):={\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S\partial t}}}$ (1.8)

# Problem 2

We have solved this problem by ourselves.

## Given

The Equation of Motion (EOM) for the meglev train is modeled by:

 ${\displaystyle \displaystyle \left.f(s,t)\right|_{s=Y^{1}(t)}=f(Y^{1}(t),t)}$ (2.1)

## Find

• Get the derivation of the 1st total time derivative.
• Get the derivation of the 2nd total time derivative.
• Show the similarity with the derivation of the Coriolis force.

## Solution

### * Get the derivation of the 1st total time derivative

Apply the chain rule to ${\displaystyle \displaystyle f(s,t)}$, we have,

 ${\displaystyle \displaystyle {\frac {d}{dt}}f(s,t)={\frac {\partial f}{\partial s}}{\frac {ds}{dt}}+{\frac {\partial f}{\partial t}}{\cancelto {1}{\frac {dt}{dt}}}}$ (2.2)

Given that for ${\displaystyle \displaystyle s}$ is evaluated at ${\displaystyle \displaystyle Y^{1}(t)}$ in the Eq. (2-1). Then the equation can be rewritten as,

 ${\displaystyle \displaystyle {\frac {df(Y^{1}(t),t)}{dt}}={\frac {\partial f(Y^{1}(t),t)}{\partial Y^{1}(t)}}\underbrace {\frac {\partial Y^{1}(t)}{\partial t}} _{{\dot {Y}}^{1}}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}}{\cancelto {1}{\frac {\partial t}{\partial t}}}}$ (2.3)

Define the notation as below,

 ${\displaystyle \displaystyle {\dot {Y}}^{1}(t)={\frac {\partial Y^{1}(t)}{\partial t}}}$ (2.4)

As the result, Eq. (2-1) can be simplified to,

 ${\displaystyle \displaystyle {\frac {d}{dt}}f(Y^{1}(t),t)={\frac {\partial f(Y^{1}(t),t)}{\partial Y^{1}(t)}}{\dot {Y}}^{1}(t)+{\frac {\partial f(Y^{1}(t),t)}{\partial t}}}$ (2.5)

### * Get the derivation of the 2nd total time derivative

Take the 2nd derivative of the Eq. (2.2) with respect to ${\displaystyle \displaystyle t}$, and apply the chain rule yields. We have,

 {\displaystyle \displaystyle {\begin{aligned}{\frac {d^{2}}{dt^{2}}}f\left(s,t\right)&={\frac {d}{dt}}\left({\frac {\partial f}{\partial s}}{\frac {\partial s}{\partial t}}+{\frac {\partial f}{\partial t}}\right)\\&={\frac {\partial ^{2}f}{\partial s^{2}}}\left({\frac {\partial s}{\partial t}}\right)^{2}+{\frac {\partial ^{2}f}{\partial s\partial t}}{\frac {\partial s}{\partial t}}{\cancelto {1}{\frac {\partial t}{\partial t}}}+{\frac {\partial f}{\partial s}}{\cancelto {0}{\frac {\partial ^{2}s}{\partial s\partial t}}}{\frac {\partial s}{\partial t}}+{\frac {\partial f}{\partial s}}{\frac {\partial ^{2}s}{d\partial ^{2}}}{\cancelto {1}{\frac {\partial t}{\partial t}}}+{\frac {\partial ^{2}f}{\partial s\partial t}}{\frac {\partial s}{\partial t}}+{\frac {\partial ^{2}f}{\partial t^{2}}}{\cancelto {1}{\frac {\partial t}{\partial t}}}\\&=\underbrace {\frac {\partial ^{2}f}{\partial s^{2}}} _{f_{,ss}}\left(\underbrace {\frac {\partial s}{\partial t}} _{{\dot {Y}}^{1}}\right)^{2}+2\underbrace {\frac {\partial ^{2}f}{\partial s\partial t}} _{f_{,st}}\underbrace {\frac {\partial s}{\partial t}} _{{\dot {Y}}^{1}}+\underbrace {\frac {\partial f}{\partial s}} _{f_{,s}}\underbrace {\frac {\partial ^{2}s}{\partial t^{2}}} _{{\ddot {Y}}^{1}}+\underbrace {\frac {\partial ^{2}f}{\partial t^{2}}} _{f_{,tt}}\\\end{aligned}}} (2.6)

Define the notations as below,

 {\displaystyle \displaystyle {\begin{aligned}f_{,ss}&={\frac {\partial ^{2}f}{\partial s^{2}}}\\f_{,s}&={\frac {\partial f}{\partial s}}\\f_{,st}&={\frac {\partial ^{2}f}{\partial s\partial t}}\\f_{,tt}&={\frac {\partial ^{2}f}{\partial t^{2}}}\\{\dot {Y}}^{1}&={\frac {\partial s}{\partial t}}\\{\ddot {Y}}^{1}&={\frac {\partial ^{2}s}{\partial t^{2}}}\\\end{aligned}}} (2.7)

Then the Eq. (2.6) can be rewritten as below,

 ${\displaystyle \displaystyle {\frac {d^{2}f}{dt^{2}}}=f_{,s}\left({Y^{1},t}\right){\ddot {Y}}^{1}+f_{,ss}(Y^{1},t)({\dot {Y}}^{1})^{2}+2f_{,st}(Y^{1},t){\dot {Y}}^{1}+f_{,tt}(Y^{1},t)}$ (2.8)

### * Show the similarity with the derivation of the Coriolis force

Define ${\displaystyle \displaystyle {\mathbf {r}}}$ as the position vector indicating the position of the origin of the reference frame, and define ${\displaystyle \displaystyle A}$ is a point fixed on an object rotating with an angular velocity of ${\displaystyle \displaystyle {}^{N}{\boldsymbol {\omega }}^{A}}$ with respect to inertial reference frame ${\displaystyle \displaystyle N}$ . And define the three mutual perpendicular unit vectors ${\displaystyle \displaystyle {\mathbf {i}},{\mathbf {j}},{\mathbf {k}}}$. Then, ${\displaystyle \displaystyle {\mathbf {r}}}$ can be defined as,

 ${\displaystyle \displaystyle {\mathbf {r}}=x{\mathbf {i}}+y{\mathbf {j}}+z{\mathbf {k}}}$ (2.9)

The velocity of the object as viewed by an observer fixed to the inertial reference frame ${\displaystyle \displaystyle N}$ is defined as below,

 ${\displaystyle \displaystyle {}^{N}{\mathbf {v}}={\frac {{}^{N}d{\mathbf {r}}}{dt}}=\underbrace {\frac {{}^{A}d{\mathbf {r}}}{dt}} _{{}^{A}{\mathbf {v}}}+{}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}}$ (2.10)

The acceleration of the object as viewed by an observer fixed to the inertial reference frame ${\displaystyle \displaystyle N}$ is defined as below,

 ${\displaystyle \displaystyle {}^{N}{\mathbf {a}}={\frac {{}^{N}d}{dt}}{}^{N}{\mathbf {v}}={\frac {{}^{A}d}{dt}}{}^{N}{\mathbf {v}}+{}^{N}{\boldsymbol {\omega }}^{A}\times {}^{N}{\mathbf {v}}}$ (2.11)

Substitute ${\displaystyle \displaystyle {}^{N}{\mathbf {v}}}$ from the Eq. (2.10) into the Eq. (2.11),and we have,

 ${\displaystyle \displaystyle {}^{N}{\mathbf {a}}={\frac {{}^{A}d}{dt}}\left({}^{A}{\mathbf {v}}+{}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}\right)+{}^{N}{\boldsymbol {\omega }}^{A}\times \left({}^{A}{\mathbf {v}}+{}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}\right)}$ (2.12)

Expand the Eq. (2.12) as below.

{\displaystyle \displaystyle {\begin{aligned}{}^{N}{\mathbf {a}}&={\frac {{}^{A}d}{dt}}{}^{A}{\mathbf {v}}+{\frac {{}^{A}d}{dt}}\left({}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}\right)+{}^{N}{\boldsymbol {\omega }}^{A}\times {}^{A}{\mathbf {v}}+{}^{N}{\boldsymbol {\omega }}^{A}\times \left({}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}\right)\\&={}^{A}{\mathbf {a}}+{\frac {{}^{A}d}{dt}}{}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}+{}^{N}{\boldsymbol {\omega }}^{A}\times {\frac {{}^{A}d{\mathbf {r}}}{dt}}+{}^{N}{\boldsymbol {\omega }}^{A}\times {}^{A}{\mathbf {v}}+{}^{N}{\boldsymbol {\omega }}^{A}\times \left({}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}\right)\\&={}^{A}{\mathbf {a}}+{}^{N}{\boldsymbol {\alpha }}^{A}\times {\mathbf {r}}+2{}^{N}{\boldsymbol {\omega }}^{A}\times {}^{A}{\mathbf {v}}+{}^{N}{\boldsymbol {\omega }}^{A}\times \left({}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}\right)\\\end{aligned}}}

As the result, the Eq.(2.12) can be rewritten as,

 ${\displaystyle \displaystyle {}^{N}{\mathbf {a}}={}^{A}{\mathbf {a}}+{}^{N}{\boldsymbol {\alpha }}^{A}\times {\mathbf {r}}+2{}^{N}{\boldsymbol {\omega }}^{A}\times {}^{A}{\mathbf {v}}+{}^{N}{\boldsymbol {\omega }}^{A}\times \left({}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}\right)}$ (2.13)

by the definition,the expression on RHS can also be written as,

 {\displaystyle \displaystyle {\begin{aligned}{}^{A}{\mathbf {a}}&={\frac {{}^{A}d^{2}{\mathbf {r}}}{dt^{2}}}\\2{}^{N}{\boldsymbol {\omega }}^{A}\times {}^{A}{\mathbf {v}}&=2{}^{N}{\boldsymbol {\omega }}^{A}\times {\frac {{}^{A}d{\mathbf {r}}}{dt}}\\{}^{N}{\boldsymbol {\alpha }}^{A}\times {\mathbf {r}}&={\frac {{}^{A}d}{dt}}\left({}^{N}{\boldsymbol {\omega }}^{A}\right)\times {\mathbf {r}}\\\end{aligned}}} (2.14)

The Eq. (2-8) and the Eq. (2-13) are related as below,

 Eq. (2-8) Eq. (2-13) LHS ${\displaystyle \displaystyle {\frac {d^{2}f{Y^{1}{t},t}}{dt^{2}}}}$ ${\displaystyle \displaystyle {}^{N}{\mathbf {a}}}$ Expression1 on RHS ${\displaystyle \displaystyle f_{,tt}\left(Y^{1}(t),t\right)}$ ${\displaystyle \displaystyle {\frac {{}^{A}d^{2}{\mathbf {r} }}{dt^{2}}}}$ Expression 2 on RHS ${\displaystyle \displaystyle 2{\dot {Y}}^{1}f_{,st}\left(Y^{1}(t),t\right)}$ ${\displaystyle \displaystyle 2{}^{N}{\boldsymbol {\omega }}^{A}\times {\frac {{}^{A}d{\mathbf {r}}}{dt}}}$ Expression 3 on RHS ${\displaystyle \displaystyle {\ddot {Y}}^{1}f_{,s}\left(Y^{1}(t),t\right)}$ ${\displaystyle \displaystyle {\frac {{}^{A}d}{dt}}\left({}^{N}{\boldsymbol {\omega }}^{A}\right)\times {\mathbf {r}}}$ Expression 4 on RHS ${\displaystyle \displaystyle ({\dot {Y}}^{1})^{2}f_{,ss}\left(Y^{1}(t),t\right)}$ ${\displaystyle \displaystyle {}^{N}{\boldsymbol {\omega }}^{A}\times \left({}^{N}{\boldsymbol {\omega }}^{A}\times {\mathbf {r}}\right)}$

# Problem 3

We have solved this problem by ourselves.

## Given

The expression of ${\displaystyle c_{0}(Y^{1},t)}$ is given by,

${\displaystyle \displaystyle c_{0}(Y^{1},t)=-F^{1}(1-{\bar {R}}u_{,SS}^{2})-F^{2}u_{,S}^{2}-{\frac {T}{R}}+M[(1-{\bar {R}}u_{,SS}^{2})(u_{,tt}^{1}-{\bar {R}}u_{,Stt}^{2})+u_{,S}^{2}u_{tt}^{2}]}$

 (3.1)

## Find

Analyze the dimension of each term and provide the physical meaning.

## Solution

### *Dimensional analysis of each term

Left hand side:

${\displaystyle \displaystyle [c_{0}(Y^{1},t)]=F}$

 (3.2)

Right hand side:

• First term

${\displaystyle \displaystyle [F^{1}]=F}$

 (3.3)

${\displaystyle \displaystyle [1]=1}$

 (3.4)

${\displaystyle \displaystyle [{\bar {R}}]=L}$

 (3.5)

${\displaystyle \displaystyle [u^{2}]=L\Rightarrow [u_{,SS}^{2}]={\frac {L}{L^{2}}}=L^{-1}}$

 (3.6)

${\displaystyle \displaystyle [{\bar {R}}u_{,SS}^{2}]=L\cdot L^{-1}=1}$

 (3.7)

${\displaystyle \displaystyle [F^{1}(1-{\bar {R}}u_{,SS}^{2})]=F\cdot 1=F}$

 (3.8)
• Second term

${\displaystyle \displaystyle [F^{2}]=F}$

 (3.9)

${\displaystyle \displaystyle [u^{2}]=L\Rightarrow [u_{,S}^{2}]={\frac {L}{L}}=1}$

 (3.10)

${\displaystyle \displaystyle [F^{2}u_{,S}^{2}]=F\cdot {1}=F}$

 (3.11)
• Third term

${\displaystyle \displaystyle [T]=F\cdot {L}}$

 (3.12)

${\displaystyle \displaystyle [R]=L}$

 (3.13)

${\displaystyle \displaystyle [{\frac {T}{R}}]={\frac {F\cdot {L}}{L}}=F}$

 (3.14)
• Fourth term

${\displaystyle \displaystyle [M]=M}$

 (3.15)

${\displaystyle \displaystyle [1-{\bar {R}}u_{,SS}^{2}]=1}$

 (3.16)

${\displaystyle \displaystyle [u_{,tt}^{1}]={\frac {L}{T^{2}}}=LT^{-2}}$

 (3.17)

${\displaystyle \displaystyle [u_{,Stt}^{2}]={\frac {L}{L\cdot {T^{2}}}}=T^{-2}\Rightarrow [{\bar {R}}u_{,Stt}^{2}]=LT^{-2}}$

 (3.18)

${\displaystyle \displaystyle [(1-{\bar {R}}u_{,SS}^{2})(u_{,tt}^{1}-{\bar {R}}u_{,Stt}^{2})]=1\cdot {L}\cdot {T^{-2}}=LT^{-2}}$

 (3.19)

${\displaystyle \displaystyle [u_{,S}^{2}]={\frac {L}{L}}=1}$

 (3.20)

${\displaystyle \displaystyle [u_{,tt}^{2}]={\frac {L}{T^{2}}}=LT^{-2}}$

 (3.21)

${\displaystyle \displaystyle [u_{,S}^{2}u_{,tt}^{2}]=1\cdot {L}\cdot {T^{-2}}=LT^{-2}}$

 (3.22)

${\displaystyle \displaystyle [M[(1-{\bar {R}}u_{,SS}^{2})(u_{,tt}^{1}-{\bar {R}}u_{,Stt}^{2})+u_{,S}^{2}u_{tt}^{2}]]=M\cdot {L}\cdot {T^{-2}}=F}$

 (3.23)

### *Physical meaning

${\displaystyle c_{0}}$ is the horizontal force which act on wheel/magnet of the train. Other terms are forces from different source.

# Problem 4

We have solved this problem by ourselves.

## Given

 The polar coordinate ${\displaystyle \displaystyle (\xi _{1},\xi _{2})=(r,\theta )}$

## Find

 Draw the polar coordinate lines in a 2-D plane emanating from a point, not at the origin.

## Solution

 Suppose a non-origin point ${\displaystyle \displaystyle (C_{1},C_{2})}$.
 Draw the polar coordinate lines in XY coordinate system from the point ${\displaystyle \displaystyle (C_{1},C_{2})}$, then we could derive:
 ${\displaystyle \displaystyle \left\{{\begin{matrix}x=r\cos \theta +C_{1}\\y=r\sin \theta +C_{2}\end{matrix}}\right.}$ (4.1)
 ${\displaystyle \displaystyle \Rightarrow \left\{{\begin{matrix}(x-C_{1})^{2}+(x-C_{2})^{2}=r^{2}\\(y-C_{1})/(x-C_{2})=\tan \theta \end{matrix}}\right.}$ (4.2)
 According to (4.2), as ${\displaystyle \displaystyle (r,\theta )}$ changes, two sets of lines could be draw as shown in the following figure(4-1):

# Problem 5

We have solved this problem by ourselves.

## Given

From the lecture note, we have

 ${\displaystyle \displaystyle {\frac {1}{g_{i}(\xi _{i})}}{\frac {d}{d\xi _{i}}}\left[g_{i}(\xi _{i}){\frac {dX_{i}(\xi _{i})}{d\xi _{i}}}\right]+f_{i}(\xi _{i})X_{i}(\xi _{i})=0}$ (5.1)

## Find

Show that equation(5.1) becomes

 ${\displaystyle \displaystyle y''+\underbrace {\frac {g'(x)}{g(x)}} _{a_{1}(x)}y'+a_{0}(x)y=0}$ (5.2)

## Solution

According to the Chain Rule, equation (5.1) yields

 ${\displaystyle \displaystyle {\frac {1}{g_{i}(\xi _{i})}}\left[g'_{i}(\xi _{i}){\frac {dX_{i}(\xi _{i})}{d\xi _{i}}}+g_{i}(\xi _{i}){\frac {d^{2}X_{i}(\xi _{i})}{d\xi _{i}^{2}}}\right]+f_{i}(\xi _{i})X_{i}(\xi _{i})=0.}$ (5.3)

After simplification, it becomes

 ${\displaystyle \displaystyle X''_{i}(\xi _{i})+{\frac {g'_{i}(\xi _{i})}{g_{i}(\xi _{i})}}X'_{i}(\xi _{i})+f_{i}(\xi _{i})X_{i}(\xi _{i})=0}$ (5.4)

Obviously, When ${\displaystyle \displaystyle y=X_{i}(\xi _{i})}$, equation(5.4) and equation(5.2) are the same.

# Problem 6

We have solved this problem by ourselves.

## Given

The Equation of Motion of Wheel/Magnet has four terms, which are given by,

 ${\displaystyle \displaystyle C_{3}(Y^{1},t){\ddot {Y}}^{1}+C_{2}(Y^{1},t)({\dot {Y}}^{1})^{2}+C_{1}(Y^{1},t){\dot {Y}}^{1}+C_{0}(Y^{1},t)=0}$ (6.1)

The general expression of ${\displaystyle \displaystyle C_{3}(Y^{1},t){\ddot {Y}}^{1}}$ is given by,

 ${\displaystyle \displaystyle c_{3}(Y^{1},t){\ddot {Y}}^{1}=[1-{\bar {R}}u_{,ss}(Y^{1},t)]{\ddot {Y}}^{1}}$ (6.2)

From the lecture note[}, we can Consider ${\displaystyle \displaystyle F(\cdot )}$ as an operator, ${\displaystyle \displaystyle F(\cdot )}$ is linear if and only if,

 ${\displaystyle \displaystyle F(\alpha f+\beta g)=\alpha F(f)+\beta F(g)}$ (6.3)

Where${\displaystyle \displaystyle f}$ and ${\displaystyle \displaystyle g}$ are two possible arguments for ${\displaystyle \displaystyle F(\cdot )}$, and ${\displaystyle \displaystyle \forall \alpha ,\beta \in R.}$

The derivative operator is a linear operator, which is given by,

 ${\displaystyle \displaystyle F(\cdot )={\frac {(\cdot )}{dx}}}$ (6.4)

## Find

${\displaystyle \displaystyle C_{3}(Y^{1},t){\ddot {Y}}^{1}}$ is nonlinear with repect ${\displaystyle \displaystyle Y^{1}.}$

## Solution

Suppose we have,

 ${\displaystyle \displaystyle Y^{1}:[t_{0},+\infty )\rightarrow R}$ ${\displaystyle \displaystyle \forall u,v:[t_{0},+\infty )\rightarrow R}$ ${\displaystyle \displaystyle \forall \alpha ,\beta \in R}$ (6.5)

For the right-hand side, we have,

 ${\displaystyle \displaystyle RHS:=\alpha C_{3}(u,t){\frac {\alpha d^{2}u}{dt^{2}}}+\beta C_{3}(v,t){\frac {\beta d^{2}v}{dt^{2}}}}$ (6.6)

The left-hand side is given by,

 ${\displaystyle \displaystyle LHS:=C_{3}(\alpha u+\beta v,t){\frac {d^{2}(\alpha u+\beta v)}{dt^{2}}}}$ (6.7)

Since the derivative operator is linear operator, we can get,

 ${\displaystyle \displaystyle {\frac {d^{2}(\alpha u+\beta v)}{dt^{2}}}={\frac {\alpha d^{2}u}{dt^{2}}}+{\frac {\beta d^{2}v}{dt^{2}}}}$ (6.8)

So (6.7) turns into,

 ${\displaystyle \displaystyle C_{3}(\alpha u+\beta v,t){\frac {d^{2}(\alpha u+\beta v)}{dt^{2}}}=C_{3}(\alpha u+\beta v,t){\frac {\alpha d^{2}u}{dt^{2}}}+C_{3}(\alpha u+\beta v,t){\frac {\beta d^{2}v}{dt^{2}}}}$ (6.9)

If ${\displaystyle \displaystyle C_{3}(Y^{1},t)}$ is nonlinear with respect to ${\displaystyle \displaystyle Y^{1}}$, by comparision between (6.5) and (6.8) we can find that,

 ${\displaystyle \displaystyle LHS\neq RHS}$ (6.10)

Even if ${\displaystyle \displaystyle C_{3}(Y^{1},t)}$ is linear with respect to ${\displaystyle \displaystyle Y^{1}}$, on the left-hsnd side we have,

 ${\displaystyle \displaystyle LHS:=[\alpha C_{3}(u,t)+\beta C_{3}(v,t)]{\frac {\alpha d^{2}u}{dt^{2}}}+[\alpha C_{3}(u,t)+\beta C_{3}(v,t)]{\frac {\beta d^{2}v}{dt^{2}}}}$ (6.11)

Obviously,

 ${\displaystyle \displaystyle LHS\neq RHS}$ (6.12)

# Problem 7

We have solved this problem by ourselves.

## Given

${\displaystyle \displaystyle L_{2}(\cdot )={\frac {d^{2}(\cdot )}{dx^{2}}}+a_{1}(x){\frac {d(\cdot )}{dx}}+a_{0}(x)(\cdot )}$

## Find

Show that ${\displaystyle \displaystyle L_{2}(\cdot )}$ is linear.

## Solution

If ${\displaystyle \displaystyle L_{2}(\cdot )}$ is linear, then the following must be true:

${\displaystyle \displaystyle L_{2}(\alpha u+\beta v)=\alpha L_{2}(u)+\beta L_{2}(v)}$

${\displaystyle \displaystyle \forall \alpha ,\beta \in \mathbb {R} }$

${\displaystyle \displaystyle L_{2}(\alpha u+\beta v)={\frac {d^{2}(\alpha u+\beta v)}{dx^{2}}}+a_{1}(x){\frac {d(\alpha u+\beta v)}{dx}}+a_{0}(x)(\alpha u+\beta v)}$

${\displaystyle \displaystyle L_{2}(\alpha u+\beta v)=\alpha u''+\beta v''+a_{1}(x)[\alpha u'+\beta v']+a_{0}(x)[\alpha u+\beta v]}$

${\displaystyle \displaystyle L_{2}(\alpha u+\beta v)=[\alpha u''+a_{1}(x)\alpha u'+a_{0}(x)\alpha u]+[\beta v''+a_{1}(x)\beta v'+a_{0}(x)\beta v]}$

${\displaystyle \displaystyle L_{2}(\alpha u+\beta v)=\alpha [u''+a_{1}(x)u'+a_{0}(x)u]+\beta [v''+a_{1}(x)v'+a_{0}(x)v]}$

${\displaystyle \displaystyle L_{2}(\alpha u+\beta v)=\alpha L_{2}(u)+\beta L_{2}(v)}$

Thus ${\displaystyle \displaystyle L_{2}(\cdot )}$ is a linear operator.

# Contribution

Problem Assignments
Problem # Solved&Typed by Reviewed by
1 Goran Marjanovic, Zixiang Liu all
2 Xiaolong Wu, Lin,Hao all
3 Zhou Zhe, Rui Kong, Anup Parikh all
4 Goran Marjanovic, Zixiang Liu all
5 Xiaolong Wu, Lin,Hao all
6 Zhou Zhe, Rui Kong, Anup Parikh all
7 Goran Marjanovic, Zixiang Liu all