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2.4 HW 4

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2.4.1 Section 27, Problem 8

Figure 2.48:Problem statement

Solution

The cylindrical and spherical coordinates are defined as given in the textbook figures shown below

Figure 2.49:Cylinderical coordinates

Figure 2.50:Spherical coordinates

The relation between these is given by (13) in the book

$$\begin{align} z & =r\cos \theta \tag{1}\\ \rho & =r\sin \theta \tag{2}\\ \phi & =\phi \tag{3} \end{align}$$

To obtain the required formula, we will use the chain rule. Since in spherical we have $u\equiv u\left ( r,\theta \right )$ and in cylindrical we have $u\equiv u\left ( \rho ,z\right )$, then by chain rule

$$\frac{\partial u}{\partial \theta }=\frac{\partial u}{\partial \rho }\frac{\partial \rho }{\partial \theta }+\frac{\partial u}{\partial z}\frac{\partial z}{\partial \theta }$$ But from (2) $\frac{\partial \rho }{\partial \theta }=r\cos \theta$ and from (1) $\frac{\partial z}{\partial \theta }=-r\sin \theta$, hence the above becomes $$\frac{\partial u}{\partial \theta }=\frac{\partial u}{\partial \rho }\left ( r\cos \theta \right ) +\frac{\partial u}{\partial z}\left ( -r\sin \theta \right )$$ But $r\cos \theta =z$ and $-r\sin \theta =\rho$, hence the above simplifies to $$\begin{equation} \frac{\partial u}{\partial \theta }=z\frac{\partial u}{\partial \rho }-\rho \frac{\partial u}{\partial z} \tag{4} \end{equation}$$ Which is the result required to show. Now we need to show that $\frac{\partial u}{\partial \theta }$ evaluated at boundary $r=1,\theta =\frac{\pi }{2}$ is zero. But $\theta =\frac{\pi }{2}$ implies that $z=0$, since $z=r\cos \theta$. Hence (4) now reduces to $$\begin{equation} \frac{\partial u}{\partial \theta }=-\rho \frac{\partial u}{\partial z} \tag{4} \end{equation}$$ Since $\theta =\frac{\pi }{2}$, then $\frac{\partial u}{\partial z}$ is the directional derivative normal to the base surface. But we are told it is insulated. This implies that $\frac{\partial u}{\partial z}=0$, since by definition this is what insulated means. Therefore $\frac{\partial u}{\partial \theta }=0$ at $r=1,\theta =\frac{\pi }{2}$, which is what we are asked to show.

2.4.2 Section 28, Problem 1

Figure 2.51:Problem statement

Eq (6) in section 28 is

$$y_{tt}\left ( x,t\right ) =a^{2}y_{xx}\left ( x,t\right ) -g$$ At static displacement, by definition, there is no time dependency, hence $y_{tt}=0$ and the above becomes $$0=a^{2}y_{xx}\left ( x,t\right ) -g$$ Therefore now this becomes an ODE instead of a PDE since it does not depend on time, and we can write the above as $$\begin{equation} a^{2}y^{\prime \prime }\left ( x\right ) =g \tag{1} \end{equation}$$ The boundary conditions $y\left ( 0,t\right ) =0$ and $y\left ( 2x,t\right ) =0$ now become $y\left ( 0\right ) =0,y\left ( 2x\right ) =0$. Now we need to solve (1) with these boundary conditions. This is an boundary value ODE. $$y^{\prime \prime }\left ( x\right ) =\frac{g}{a^{2}}$$ The RHS is constant. The solution to the homogeneous ODE $y^{\prime \prime }=0$ is $y_{h}=Ax+B$. Let the particular solution be $y_{p}=C_{3}x^{2}$, then $y_{p}^{\prime }=2C_{3}x$ and $y_{p}^{\prime \prime }=2C_{3}$. Substituting this in the above ODE gives $$\begin{align*} 2C_{3} & =\frac{g}{a^{2}}\\ C_{3} & =\frac{g}{2a^{2}} \end{align*}$$

Hence $y_{p}\left ( x\right ) =\frac{g}{2a^{2}}x^{2}$. Therefore the general solution is

$$\begin{align} y & =y_{h}+y_{p}\nonumber \\ & =Ax+B+\frac{g}{2a^{2}}x^{2} \tag{2} \end{align}$$

Now we will use the boundary conditions to find $A,B$ above. At $x=0$, (2) becomes

$$0=B$$ Hence solution (2) reduces to $$\begin{equation} y\left ( x\right ) =Ax+\frac{g}{2a^{2}}x^{2} \tag{3} \end{equation}$$ At $x=2c$, the second boundary condition gives $$\begin{align*} 0 & =2cA+\frac{g}{2a^{2}}\left ( 4c^{2}\right ) \\ A & =\frac{-g}{2a^{2}}\frac{\left ( 4c^{2}\right ) }{2c}\\ & =\frac{-gc}{a^{2}} \end{align*}$$

Hence the solution (3) becomes

$$\begin{align} y & =\frac{-gc}{a^{2}}x+\frac{g}{2a^{2}}x^{2}\nonumber \\ y & =\frac{gx^{2}-2gcx}{2a^{2}} \tag{4} \end{align}$$

To get the result needed, we can manipulate this more as follows. From (4)

$$\begin{align} 2a^{2}y & =gx^{2}-2gcx\nonumber \\ & =g\left ( x^{2}-2cx\right ) \nonumber \\ & =g\left ( x-c\right ) ^{2}-gc^{2}\nonumber \end{align}$$

Hence

$$\begin{align*} g\left ( x-c\right ) ^{2} & =2a^{2}y+gc^{2}\\ \left ( x-c\right ) ^{2} & =\frac{2a^{2}y}{g}+c^{2}\\ & =\frac{2a^{2}}{g}\left ( y+\frac{gc^{2}}{2a^{2}}\right ) \end{align*}$$

Now since $a^{2}=\frac{H}{\delta }$ then the above becomes

$$\begin{align*} \frac{g}{2a^{2}}\left ( x-c\right ) ^{2} & =y+\frac{gc^{2}}{2a^{2}}\\ y & =\frac{1}{2a^{2}}\left ( g\left ( x-c\right ) ^{2}-gc^{2}\right ) \\ & =\frac{g}{2\frac{H}{\delta }}\left ( \left ( x-c\right ) ^{2}-c^{2}\right ) \\ & =\frac{\delta }{H}\frac{g}{2}\left ( \left ( x-c\right ) ^{2}-c^{2}\right ) \end{align*}$$

We see now that $y$ is directly proportional to $\delta$ and $c^{2}$ and inversely proportional to $H$.

2.4.3 Section 28, Problem 5  

Figure 2.52:Problem statement

solution

The wave PDE in 1D is given by

$$\begin{equation} y_{tt}\left ( x,t\right ) =a^{2}y_{xx}\left ( x,t\right ) \tag{1} \end{equation}$$ Where $$a^{2}=\frac{H}{\delta }$$ Where $H$ is the tension in the strand and $\delta$ is the mass per unit length of the strand. But $weight=\left ( mass\right ) g$. hence $\delta =\frac{weight}{g}$. We are given that $weight=0.032$ lb, and that $g=32$ ft/s$^{2}$. This implies that $$\delta =\frac{0.032}{32}=\frac{1}{1000}$$ Hence $$a^{2}=\frac{10}{\frac{1}{1000}}=10^{4}$$ Therefore (1) becomes $$\begin{equation} y_{tt}\left ( x,t\right ) =10^{4}y_{xx}\left ( x,t\right ) \tag{2} \end{equation}$$ Since at $t=0$ we are told that strand lies along the $x-axis$, then $y\left ( x,0\right ) =0$ and problem says $y_{t}\left ( x,0\right ) =1$. For boundary conditions, since strand fixed at $x=0$ and $x=1$, then this implies $y\left ( 0,t\right ) =0$ and $y\left ( 1,t\right ) =0$. Therefore the PDE is $$\begin{align*} y_{tt}\left ( x,t\right ) & =10^{4}y_{xx}\left ( x,t\right ) \qquad 0<x<1,t>0\\ y\left ( x,0\right ) & =0\\ y_{t}\left ( x,0\right ) & =1\\ y\left ( 0,t\right ) & =0\\ y\left ( 1,t\right ) & =0 \end{align*}$$

2.4.4 Section 30, Problem 3

Figure 2.53:Problem statement

Part a

Applying the first initial conditions $y\left ( x,0\right ) =0$ to the solution

$$\begin{equation} y\left ( x,t\right ) =\phi \left ( x+at\right ) +\psi \left ( x-at\right ) \tag{1} \end{equation}$$ Gives $$\begin{equation} 0=\phi \left ( x\right ) +\psi \left ( x\right ) \tag{2} \end{equation}$$ But $y_{t}=a\phi ^{\prime }-a\psi ^{\prime }$. Hence the second initial conditions at $t=0$ gives $$\begin{equation} 0=a\phi ^{\prime }\left ( x\right ) -a\psi ^{\prime }\left ( x\right ) \tag{3} \end{equation}$$ Taking derivative of (2) and multiplying the resulting equation by $a$ gives $$\begin{equation} 0=a\phi ^{\prime }\left ( x\right ) +a\psi ^{\prime }\left ( x\right ) \tag{2A} \end{equation}$$ Adding (3,2A) gives $$\begin{align*} 2a\phi ^{\prime }\left ( x\right ) & =0\\ \phi ^{\prime }\left ( x\right ) & =0 \end{align*}$$

Therefore

$$\begin{equation} \phi \left ( x\right ) =C \tag{4} \end{equation}$$ Where $C$ is an arbitrary constant. Substituting the above result back in (2) gives $$\begin{align} 0 & =C+\psi \left ( x\right ) \nonumber \\ \psi \left ( x\right ) & =-C \tag{5} \end{align}$$

From (4,5) we see that

$$\begin{align*} \phi \left ( x\right ) & =C\\ \psi \left ( x\right ) & =-C \end{align*}$$

Now applying boundary condition $y\left ( 0,t\right ) =f\left ( t\right )$ to (1) gives

$$f\left ( t\right ) =\phi \left ( at\right ) +\psi \left ( -at\right )$$ But $a$ is the speed of the wave given by $a=\frac{x}{t}$ or $t=\frac{x}{a}$. Hence the above becomes $$\begin{align*} f\left ( \frac{x}{a}\right ) & =\phi \left ( x\right ) +\psi \left ( -x\right ) \\ \psi \left ( -x\right ) & =f\left ( \frac{x}{a}\right ) -\phi \left ( x\right ) \end{align*}$$

Since $\phi \left ( x\right ) =C$ from equation (4), then the final result is obtained

$$\begin{equation} \psi \left ( -x\right ) =f\left ( \frac{x}{a}\right ) -C\qquad x\geq 0 \tag{6} \end{equation}$$

Part b

Since the part to the right of $x=at$ is unaffected by the movement of the right, then

$$\begin{equation} y\left ( x,t\right ) =0\qquad x\geq at \tag{1} \end{equation}$$ So now we need to find the solution for $x<at$ and $x\geq 0$. From $$y\left ( x,t\right ) =\phi \left ( x+at\right ) +\psi \left ( x-at\right )$$ And using (6) in part (a), we see that $\psi \left ( x-at\right ) =f\left ( \frac{-\left ( x-at\right ) }{a}\right ) -C$. Therefore the above becomes $$y\left ( x,t\right ) =\phi \left ( x+at\right ) +f\left ( \frac{-\left ( x-at\right ) }{a}\right ) -C$$ But also from part (a) $\phi \left ( x+at\right ) =C$. Hence the above simplifies to $$\begin{align} y\left ( x,t\right ) & =c+f\left ( \frac{-\left ( x-at\right ) }{a}\right ) -C\nonumber \\ & =f\left ( \frac{-x+at}{a}\right ) \nonumber \\ & =f\left ( t-\frac{x}{a}\right ) \qquad x<at \tag{2} \end{align}$$

Combining (1) and (2) shows that

$$y\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}0 & & x\geq at\\ f\left ( t-\frac{x}{a}\right ) & & x<at \end{array} \right .$$

2.4.5 Section 30, Problem 4

Figure 2.54:Problem statement

This requires just substitution of the function $f\left ( t\right )$ given into the solution found above which is

$$\begin{equation} y\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}0 & & x\geq at\\ f\left ( t-\frac{x}{a}\right ) & & x<at \end{array} \right . \tag{1} \end{equation}$$ But $$\begin{equation} f\left ( t\right ) =\left \{ \begin{array} [c]{ccc}\sin \pi t & & 0\leq t\leq 1\\ 0 & & t>1 \end{array} \right . \tag{2} \end{equation}$$ Substituting (2) into (1) gives, after replacing each $t$ in (2) by $t-\frac{x}{a}$ the result needed $$y\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}0 & & x\geq at\\ \sin \left ( \pi \left ( t-\frac{x}{a}\right ) \right ) & & a\left ( t-1\right ) <x<at \end{array} \right .$$

2.4.6 Section 31, Problem 2

Figure 2.55:Problem Statement

Part  a

We want to do the transformation from $y\left ( x,t\right )$ to $y\left ( u,v\right )$. Therefore

$$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial x}$$ But $\frac{\partial u}{\partial x}=1$ and $\frac{\partial v}{\partial x}=1$, hence the above becomes $$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}+\frac{\partial y}{\partial v}$$ And $$\begin{align*} \frac{\partial ^{2}y}{\partial x^{2}} & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial x}\right ) \\ & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial u}+\frac{\partial y}{\partial v}\right ) \\ & =\frac{\partial }{\partial x}\frac{\partial y}{\partial u}+\frac{\partial }{\partial x}\frac{\partial y}{\partial v}\\ & =\left ( \frac{\partial ^{2}y}{\partial u^{2}}\frac{\partial u}{\partial x}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial v}{\partial x}\right ) +\left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial x}+\frac{\partial ^{2}y}{\partial vu}\frac{\partial u}{\partial x}\right ) \end{align*}$$

But $\frac{\partial u}{\partial x}=1,\frac{\partial v}{\partial x}=1$, hence the above becomes

$$\begin{align} \frac{\partial ^{2}y}{\partial x^{2}} & =\frac{\partial ^{2}y}{\partial u^{2}}+2\frac{\partial ^{2}y}{\partial uv}+\frac{\partial ^{2}y}{\partial v^{2}}\nonumber \\ y_{xx} & =y_{uu}+y_{vv}+2y_{uv} \tag{1} \end{align}$$

Similarly,

$$\frac{\partial y}{\partial t}=\frac{\partial y}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial t}$$ But $\frac{\partial u}{\partial t}=\alpha$ and $\frac{\partial v}{\partial t}=\beta$, hence the above becomes $$\frac{\partial y}{\partial t}=\alpha \frac{\partial y}{\partial u}+\beta \frac{\partial y}{\partial v}$$ And

$$\begin{align*} \frac{\partial ^{2}y}{\partial t^{2}} & =\frac{\partial }{\partial t}\left ( \frac{\partial y}{\partial t}\right ) \\ & =\frac{\partial }{\partial t}\left ( \alpha \frac{\partial y}{\partial u}+\beta \frac{\partial y}{\partial v}\right ) \\ & =\alpha \frac{\partial }{\partial t}\left ( \frac{\partial y}{\partial u}\right ) +\beta \frac{\partial }{\partial t}\left ( \frac{\partial y}{\partial v}\right ) \\ & =\alpha \left ( \frac{\partial ^{2}y}{\partial u^{2}}\frac{\partial u}{\partial t}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial v}{\partial t}\right ) +\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial t}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial u}{\partial t}\right ) \end{align*}$$

But $\frac{\partial u}{\partial t}=\alpha$ and $\frac{\partial v}{\partial t}=\beta$, hence the above becomes

$$\begin{align} \frac{\partial ^{2}y}{\partial t^{2}} & =\alpha \left ( \alpha \frac{\partial ^{2}y}{\partial u^{2}}+\beta \frac{\partial ^{2}y}{\partial uv}\right ) +\beta \left ( \beta \frac{\partial ^{2}y}{\partial v^{2}}+\alpha \frac{\partial ^{2}y}{\partial uv}\right ) \nonumber \\ & =\alpha ^{2}\frac{\partial ^{2}y}{\partial u^{2}}+\alpha \beta \frac{\partial ^{2}y}{\partial uv}+\beta ^{2}\frac{\partial ^{2}y}{\partial v^{2}}+\alpha \beta \frac{\partial ^{2}y}{\partial uv}\nonumber \\ y_{tt} & =\alpha ^{2}y_{uu}+\beta ^{2}y_{vv}+2\alpha \beta y_{uv} \tag{2} \end{align}$$

And to obtain $y_{xt}$, then starting from above result obtained

$$\frac{\partial y}{\partial t}=\alpha \frac{\partial y}{\partial u}+\beta \frac{\partial y}{\partial v}$$ Now taking partial derivative w.r.t. $x$ gives $$\begin{align*} \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial t}\right ) & =\frac{\partial }{\partial x}\left ( \alpha \frac{\partial y}{\partial u}+\beta \frac{\partial y}{\partial v}\right ) \\ & =\alpha \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial u}\right ) +\beta \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial v}\right ) \\ & =\alpha \left ( \frac{\partial ^{2}y}{\partial u^{2}}\frac{\partial u}{\partial x}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial v}{\partial x}\right ) +\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial x}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial u}{\partial x}\right ) \end{align*}$$

But $\frac{\partial u}{\partial x}=1,\frac{\partial v}{\partial x}=1$, hence the above becomes

$$\begin{align} \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial t}\right ) & =\alpha \left ( \frac{\partial ^{2}y}{\partial u^{2}}+\frac{\partial ^{2}y}{\partial uv}\right ) +\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}+\frac{\partial ^{2}y}{\partial uv}\right ) \nonumber \\ y_{xt} & =\alpha y_{uu}+\left ( \alpha +\beta \right ) y_{vu}+\beta y_{vv} \tag{3} \end{align}$$

Substituting (1,2,3) into $Ay_{xx}+By_{xt}+Cy_{tt}=0$ results in

$$A\left ( y_{uu}+y_{vv}+2y_{uv}\right ) +B\left ( \alpha y_{uu}+\left ( \alpha +\beta \right ) y_{vu}+\beta y_{vv}\right ) +C\left ( \alpha ^{2}y_{uu}+\beta ^{2}y_{vv}+2\alpha \beta y_{uv}\right ) =0$$ Or $$y_{uu}\left ( A+B\alpha +C\alpha ^{2}\right ) +y_{uv}\left ( 2A+B\left ( \alpha +\beta \right ) +2C\alpha \beta \right ) +y_{vv}\left ( A+B\beta +C\beta ^{2}\right ) =0$$

Part b

Looking at the term above for $y_{uu}$ we see it is $A+B\alpha +C\alpha ^{2}$ which has the root

$$\begin{align*} \alpha & =-\frac{b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}\\ & =-\frac{B}{2C}\pm \frac{1}{2C}\sqrt{B^{2}-4AC} \end{align*}$$

Hence if we pick the root $\alpha =\alpha _{0}=-\frac{B}{2C}+\frac{1}{2C}\sqrt{B^{2}-4AC}$ then the term $y_{uu}$ vanishes. Similarly for the term multiplied by $y_{vv}$ which is $A+B\beta +C\beta ^{2}$. The root is

$$\beta =-\frac{B}{2C}\pm \frac{1}{2C}\sqrt{B^{2}-4AC}$$ And if we pick $\beta =\beta _{0}=-\frac{B}{2C}-\frac{1}{2C}\sqrt{B^{2}-4AC}$ then the term $y_{vv}$ vanishes also in the PDE obtained in part (a), and now the PDE becomes $$y_{uv}\left ( 2A+B\left ( \alpha +\beta \right ) +2C\alpha \beta \right ) =0$$ Substituting the above selected roots $\alpha _{0},\beta _{0}$ into the above in place of $\alpha ,\beta$ since these are the values we picked, then the above becomes $$\begin{align*} y_{uv}\left ( 2A+B\left ( -\frac{B}{2C}+\frac{1}{2C}\sqrt{B^{2}-4AC}-\frac{B}{2C}-\frac{1}{2C}\sqrt{B^{2}-4AC}\right ) +2C\alpha \beta \right ) & =0\\ y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+2C\alpha \beta \right ) & =0 \end{align*}$$

And again replacing $\alpha \beta$ above with $\alpha _{0},\beta _{0}$ results in

$$\begin{align*} y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+2C\left ( -\frac{B}{2C}+\frac{1}{2C}\sqrt{B^{2}-4AC}\right ) \left ( -\frac{B}{2C}-\frac{1}{2C}\sqrt{B^{2}-4AC}\right ) \right ) & =0\\ y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+2C\left ( \frac{B^{2}}{4C^{2}}+\frac{1}{4C^{2}}\left ( B^{2}-4AC\right ) \right ) \right ) & =0\\ y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+\frac{B^{2}}{2C}+\frac{1}{2C}\left ( B^{2}-4AC\right ) \right ) & =0\\ y_{uv}\left ( 2A-\frac{2B^{2}}{2C}+\frac{B^{2}}{2C}+\frac{B^{2}}{2C}-2A\right ) & =0\\ \frac{B^{2}}{2C}y_{uv} & =0 \end{align*}$$

Since $B\neq 0,C\neq 0$ then the above simplifies to

$$y_{uv}=0$$

Part c

Since

$$y_{uv}=0$$ Or $$\frac{\partial }{\partial v}\left ( \frac{\partial y}{\partial u}\right ) =0$$ The implies that $$\frac{\partial y}{\partial u}=\Phi \left ( u\right )$$ Integrating w.r.t. $u$ gives $$y\left ( u,v\right ) =\int \Phi \left ( u\right ) du+\psi \left ( v\right )$$ Where $\psi \left ( v\right )$ is the constant of integration which is a function.

Let $\int \Phi \left ( u\right ) du=\phi \left ( u\right )$ then the above can be written as

$$y\left ( u,v\right ) =\phi \left ( u\right ) +\psi \left ( v\right )$$ Or in terms of $x,t$, since $u=x+\alpha t$ and $v=x+\beta t$ the above solution becomes $$y\left ( x,t\right ) =\phi \left ( x+\alpha t\right ) +\psi \left ( x+\beta t\right )$$ Where $\phi ,\psi$ are arbitrary functions twice differentiable.  When $\alpha =+a,\beta =-a$, then the above becomes $$y\left ( x,t\right ) =\phi \left ( x+at\right ) +\psi \left ( x-at\right )$$ Which is the general solution (7) in section (30). QED

2.4.7 Section 31, Problem 3

Figure 2.56:Problem Statement

The differential equation in problem 2 is

$$Ay_{xx}+By_{xt}+Cy_{tt}=0$$ We want to do the transformation from $y\left ( x,t\right )$ to $y\left ( u,v\right )$ with $$\begin{align*} u & =x\\ v & =\alpha x+\beta t \end{align*}$$

Now

$$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial x}$$ But $\frac{\partial u}{\partial x}=1$ and $\frac{\partial v}{\partial x}=\alpha$, hence the above becomes $$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}+\alpha \frac{\partial y}{\partial v}$$ And $$\frac{\partial y}{\partial t}=\frac{\partial y}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial t}$$ But $\frac{\partial u}{\partial t}=0$ and $\frac{\partial v}{\partial t}=\beta$, hence the above becomes $$\frac{\partial y}{\partial t}=\beta \frac{\partial y}{\partial v}$$ Therefore $$\begin{align} \frac{\partial ^{2}y}{\partial x^{2}} & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial x}\right ) \nonumber \\ & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial u}+\alpha \frac{\partial y}{\partial v}\right ) \nonumber \\ & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial u}\right ) +\alpha \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial v}\right ) \nonumber \\ & =\left ( \frac{\partial ^{2}y}{\partial u^{2}}\frac{\partial u}{\partial x}+\frac{\partial ^{2}y}{\partial uv}\frac{\partial v}{\partial x}\right ) +\alpha \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial x}+\frac{\partial ^{2}y}{\partial vu}\frac{\partial u}{\partial x}\right ) \nonumber \\ & =\left ( \frac{\partial ^{2}y}{\partial u^{2}}+\alpha \frac{\partial ^{2}y}{\partial uv}\right ) +\alpha \left ( \alpha \frac{\partial ^{2}y}{\partial v^{2}}+\frac{\partial ^{2}y}{\partial vu}\right ) \nonumber \\ & =\frac{\partial ^{2}y}{\partial u^{2}}+\alpha \frac{\partial ^{2}y}{\partial uv}+\alpha ^{2}\frac{\partial ^{2}y}{\partial v^{2}}+\alpha \frac{\partial ^{2}y}{\partial vu}\nonumber \\ y_{xx} & =y_{uu}+\alpha ^{2}y_{vv}+2\alpha y_{uv} \tag{1} \end{align}$$

Similarly,

$$\begin{align} \frac{\partial ^{2}y}{\partial t^{2}} & =\frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial t}\right ) \nonumber \\ & =\frac{\partial }{\partial x}\left ( \beta \frac{\partial y}{\partial v}\right ) \nonumber \\ & =\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial t}+\frac{\partial ^{2}y}{\partial vu}\frac{\partial u}{\partial t}\right ) \nonumber \\ & =\beta \left ( \beta \frac{\partial ^{2}y}{\partial v^{2}}\right ) \nonumber \\ y_{tt} & =\beta ^{2}y_{vv} \tag{2} \end{align}$$

And to obtain $y_{xt}$, then starting from above result obtained

$$\frac{\partial y}{\partial t}=\beta \frac{\partial y}{\partial v}$$ Now taking partial derivative w.r.t. $x$ gives $$\begin{align} \frac{\partial }{\partial x}\left ( \frac{\partial y}{\partial t}\right ) & =\frac{\partial }{\partial x}\left ( \beta \frac{\partial y}{\partial v}\right ) \nonumber \\ & =\beta \left ( \frac{\partial ^{2}y}{\partial v^{2}}\frac{\partial v}{\partial x}+\frac{\partial ^{2}y}{\partial vu}\frac{\partial u}{\partial x}\right ) \nonumber \\ & =\beta \left ( \alpha \frac{\partial ^{2}y}{\partial v^{2}}+\frac{\partial ^{2}y}{\partial vu}\right ) \nonumber \\ y_{xt} & =\alpha \beta y_{vv}+\beta y_{vu} \tag{3} \end{align}$$

Substituting (1,2,3) into $Ay_{xx}+By_{xt}+Cy_{tt}=0$ results in

$$A\left ( y_{uu}+\alpha ^{2}y_{vv}+2\alpha y_{uv}\right ) +B\left ( \alpha \beta y_{vv}+\beta y_{vu}\right ) +C\left ( \beta ^{2}y_{vv}\right ) =0$$ Or $$\begin{equation} Ay_{uu}+y_{uv}\left ( 2A\alpha +B\beta \right ) +y_{vv}\left ( A\alpha ^{2}+B\alpha \beta +C\beta ^{2}\right ) =0 \tag{4} \end{equation}$$ Which is what asked to show.

Part a

Setting $\alpha =\frac{-B}{\sqrt{4AC-B^{2}}},\beta =\frac{2A}{\sqrt{4AC-B^{2}}}$ in (4) above results in

$$\begin{align*} Ay_{uu}+y_{uv}\left ( 2A\left ( \frac{-B}{\sqrt{4AC-B^{2}}}\right ) +B\left ( \frac{2A}{\sqrt{4AC-B^{2}}}\right ) \right ) +y_{vv}\left ( A\alpha ^{2}+B\alpha \beta +C\beta ^{2}\right ) & =0\\ Ay_{uu}+y_{vv}\left ( A\alpha ^{2}+B\alpha \beta +C\beta ^{2}\right ) & =0 \end{align*}$$

And the above now becomes

$$\begin{align*} Ay_{uu}+y_{vv}\left ( A\left ( \frac{-B}{\sqrt{4AC-B^{2}}}\right ) ^{2}+B\left ( \frac{-B}{\sqrt{4AC-B^{2}}}\right ) \left ( \frac{2A}{\sqrt{4AC-B^{2}}}\right ) +C\left ( \frac{2A}{\sqrt{4AC-B^{2}}}\right ) ^{2}\right ) & =0\\ Ay_{uu}+y_{vv}\left ( \frac{AB^{2}}{4AC-B^{2}}-\frac{2B^{2}A}{4AC-B^{2}}+\frac{4CA^{2}}{4AC-B^{2}}\right ) & =0\\ Ay_{uu}+y_{vv}\left ( \frac{AB^{2}-2B^{2}A+4CA^{2}}{4AC-B^{2}}\right ) & =0\\ Ay_{uu}+Ay_{vv}\left ( \frac{-B^{2}+4CA}{4AC-B^{2}}\right ) & =0\\ Ay_{uu}+Ay_{vv} & =0\\ A\left ( y_{uu}+y_{vv}\right ) & =0 \end{align*}$$

Therefore, since $A\neq 0$ the above becomes

$$y_{uu}+y_{vv}=0$$

Part b

Setting $\alpha =-B,\beta =2A$ in (4) above results in

$$\begin{align*} Ay_{uu}+y_{uv}\left ( -2AB+2AB\right ) +y_{vv}\left ( AB^{2}-2B^{2}A+4CA^{2}\right ) & =0\\ Ay_{uu}+y_{vv}\left ( 4CA^{2}-B^{2}A\right ) & =0\\ Ay_{uu}-Ay_{vv}\left ( B^{2}-4CA\right ) & =0 \end{align*}$$

But $B^{2}-4CA=0$, therefore the above becomes

$$y_{uu}=0$$