Solution
The cylindrical and spherical coordinates are defined as given in the textbook figures shown below
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.
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 \(H\).
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*}
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}
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\[ y\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}0 & & x\geq at\\ f\left ( t-\frac{x}{a}\right ) & & x<at \end{array} \right . \]
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\[ y\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}0 & & x\geq at\\ \sin \left ( \pi \left ( t-\frac{x}{a}\right ) \right ) & & a\left ( t-1\right ) <x<at \end{array} \right . \]
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\[ y_{uu}\left ( A+B\alpha +C\alpha ^{2}\right ) +y_{uv}\left ( 2A+B\left ( \alpha +\beta \right ) +2C\alpha \beta \right ) +y_{vv}\left ( A+B\beta +C\beta ^{2}\right ) =0 \]
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 \[ y_{uv}=0 \]
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
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.
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\[ y_{uu}+y_{vv}=0 \]
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\[ y_{uu}=0 \]