Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 1.1

1. Show that if $z=u(x,y)$ is an integral surface of $V=\langle a,b,c\rangle$ containing a point $P$, then the surface contains the characteristic curve $\chi$ passing through $P$. (Assume the vector field $V$ is $C^1$.)
Solution
Let $\chi(t)=(x(t),y(t),z(t))$ be the characteristic curve passing through $P$. Since $P$ belongs to the integral surface, there exists $t_0\in\mathbb R$ satisfying $z(t_0)=u(x(t_0),y(t_0))$. Consider the function $\phi(t)$ is defined as follows:
$\phi(t)=z(t)-u(x(t),y(t))$ for $|t-t_0|<\delta$ and some $\delta>0$.
By the definition of $t_0$, we have $\phi(t_0)=z(t_0)-u(x(t_0),y(t_0))$. By the multivariable chain rule, we can obtain
$\begin{aligned}\phi'(t)&=z'(t)-u_x(x(t),y(t))x'(t)-u_y(x(t),y(t))y'(t)\\&=(-u_x(x(t),y(t)),-u_y(x(t),y(t)),1)\cdot(x'(t),y'(t),z'(t))\quad\text{for}~|t-t_0|<\delta.\end{aligned}$
According the definition of the characteristic curve, $(x'(t),y'(t),z'(t))=(a,b,c)$. On the other hand, we know that $(-u_x(x(t),y(t)),-u_y(x(t),y(t)),1)$ is the normal vector to the surface $z=u(x,y)$. Hence, $\phi'(t)=0$ which implies $\phi(t)$ is a zero function. Thus, $\chi$ is contained in the surface.


2. If $S_1$ and $S_2$ are two graphs [i.e., $S_i$ is given by $z=u_i(x,y)$, $i=1,2$] that are integral surfaces of $V=\left\langle a,b,c\right\rangle$ and intersect in a curve $\chi$, show that $\chi$ is a characteristic curve.
Solution
Let $P\in S_1\cap S_2$. Then by Exercise 1, there exist $\Gamma_1\subset S_1$ and $\Gamma_2\subset S_2$ passing through $P$. Since $a,b,c$ are $C^1$, we know the characteristic equation of the system of ordinary differential equations has an unique solution by the theory of ordinary differential equation. Hence the characteristic curve passing through $P$ is unique. Thus, $\Gamma_1=\Gamma_2$ is characteristic curve.


3. If $\Gamma$ is a characteristic curve of $V$, show that there is an infinite number of solutions $u$ of $(1)$ containing $\Gamma$ in the graph of $u$.
Solution
Let $\Gamma$ be the characteristic curve and parameterized as $(x(t),y(t),z(t))$. Assume that $\Gamma$ passes through $P(x_0,y_0,z_0)$ when $t=t_0$. Then we can consider the non-characteristic curve $\gamma:(x_0+\alpha(t-t_0),y_0+\beta(t-t_0),\bar{z}(t))$, where $\bar{z}(t)$ is an arbitrary $C^1$-function satisfying $\bar{z}(t_0)=z_0$ and the Cauchy problem
$\left\{\begin{aligned} &a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u),\\&u(x_0+\alpha(t-t_0),y_0+\beta(t-t_0))=\bar{z}(t).\end{aligned}\right.$
In addition, we have
$J:=\text{det}\begin{bmatrix}a\left(x_0,y_0,z_0\right)&\alpha\\b\left(x_0,y_0,z_0\right)&\beta\end{bmatrix}=a\left(x_0,y_0,z_0\right)\beta-b\left(x_0,y_0,z_0\right)\alpha$.
Since $a^2\left(x_0,y_0,z_0\right)+b^2\left(x_0,y_0,z_0\right)\neq0$,we can choose $\alpha$ and $\beta$ such that $J\neq0$. Under this choice, there exists an unique solution $\bar{u}$ containg $\Gamma$ and $\gamma$. Due to arbitrary of $\alpha$ and $\beta$, we have infinite number of solutions $u$ of $\left(1\right)$ containing $\Gamma$ in the graph of $u$.


4. Solve the given initial value problem and determine the values of $x$ and $y$ for which it exists:
   1. $xu_x+u_y=y$, $u(x,0)=x^2$
   2. $u_x-2u_y=u$, $u\left(0,y\right)=y$
   3. $y^{-1}u_x+u_y=u^2$, $u\left(x,1\right)=x^2$
Solution

1. Consider the characteristic curve $x=x(t)$, $y=y(t)$, $z=z(t)=u\left(x(t),y(t)\right)$. We have the following ordinary differential equation

   $\displaystyle\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=x,\\&\frac{\mathrm dy}{\mathrm dt}=1,\\&\frac{\mathrm dz}{\mathrm dt}=y,\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned} &x(0)=x_0,\\&y(0)=0,\\&z(0)=u(x(0),y(0))=x_0^2.\end{aligned}\right.$

   We can solve these equations to obtain

   $\displaystyle\left\{\begin{aligned} &x(t)=x_0e^t,\\&y(t)=t,\\&z(t)=\frac{1}{2}t^2+x_0^2.\end{aligned}\right.$

   By $z(t)=u\left(x(t),y(t)\right)$, we have

   $\displaystyle u\left(x_0e^t,t\right)=x_0^2+\frac{t^2}{2}$.

   By taking $t=y$ and $x_0=xe^{-y}$, we can deduce that

   $\displaystyle u(x,y)=x^2e^{-2y}+\frac{y^2}{2}~~\text{for}~(x,y)\in\mathbb R^2$.

2. Consider the characteristic curve $x=x(t)$, $y=y(t)$, $z=z(t)=u\left(x(t),y(t)\right)$. We have the following ordinary differential equation

   $\displaystyle\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=1,\\&\frac{\mathrm dy}{\mathrm dt}=-2,\\&\frac{\mathrm dz}{\mathrm dt}=z,\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned} &x(0)=0,\\&y(0)=y_0,\\&z(0)=u(x(0),y(0))=y_0.\end{aligned}\right.$

   We can solve these equations to obtain

   $\left\{\begin{aligned} &x(t)=t,\\&y(t)=-2t+y_0,\\&z(t)=y_0e^t.\end{aligned}\right.$

   By $z(t)=u(x(t),y(t))$, we have

   $u\left(t,-2t+y_0\right)=y_0e^t$.

   By taking $t=x$ and $y_0=2x+y$, we can deduce that

   $u(x,y)=(2x+y)e^x~~\text{for}~(x,y)\in\mathbb R^2$.

3. Consider the characteristic curve $x=x(t)$, $y=y(t)$, $z=z(t)=u(x(t),y(t))$. We have the following ordinary differential equation

   $\displaystyle\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=\frac{1}{y},\\&\frac{\mathrm dy}{\mathrm dt}=1,\\&\frac{\mathrm dz}{\mathrm dt}=z^2,\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned} &x(0)=x_0,\\&y(0)=1,\\&z(0)=u\left(x(0),y(0)\right)=x_0^2.\end{aligned}\right.$

   We can solve these equations to obtain

   $\displaystyle\left\{\begin{aligned} &x(t)=x_0+\ln\left(t+1\right),\\&y(t)=t+1,\\&z(t)=\frac{x_0^2}{1-tx_0^2}.\end{aligned}\right.$

   By $z(t)=u\left(x(t),y(t)\right)$, we have

   $\displaystyle u\left(x_0+\ln\left(t+1\right),t+1\right)=\frac{x_0^2}{1-tx_0^2}$.

   By taking $t=y-1$ and $x_0=x-\ln y$, we can deduce that

   $\displaystyle u(x,y)=\frac{\left(x-\ln y\right)^2}{1-\left(y-1\right)\left(x-\ln y\right)^2},~~\text{for}~x\in\mathbb R,~y>0$



5. Solve the given initial value problem and determine the values of $x,y$ and $z$ for which it exists:
   1. $xu_x+yu_y+u_z=u$, $u\left(x,y,0\right)=h(x,y)$
   2. $u_x+u_y+zu_z=u^3$, $u\left(x,y,1\right)=h(x,y)$
Solution

1. By rewriting variable, the equation becomes

   $x_1u_{x_1}+x_2u_{x_2}+u_{x_3}=u,~u\left(x_1,x_2,0\right)=h\left(x_1,x_2\right)$.

   Consider the characteristic plane $x_1=x_1(t)$, $x_2=x_2(t)$, $x_3=x_3(t)$, $z=z(t)=u\left(x_1(t),x_2(t),x_3(t)\right)$. We have the following ordinary differential equation

   $\displaystyle\left\{\begin{aligned} &\frac{\mathrm dx_1}{\mathrm dt}=x_1,\\&\frac{\mathrm dx_2}{\mathrm dt}=x_2,\\&\frac{\mathrm dx_3}{\mathrm dt}=1,\\&\frac{\mathrm dz}{\mathrm dt}=z,\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned} &x_1(0)=x_1^0,\\&x_2(0)=x_2^0,\\&x_3(0)=0,\\&z(0)=u\left(x_1(0),x_2(0),x_3(0)\right)=h\left(x_1^0,x_2^0\right).\end{aligned}\right.$

   We can solve these equations to obtain

   $\left\{\begin{aligned} &x_1(t)=x_1^0e^t,\\&x_2(t)=x_2^0e^t,\\&x_3(t)=t,\\&z(t)=h\left(x_1^0,x_2^0\right)e^t.\end{aligned}\right.$

   By $z(t)=u\left(x_1(t),x_2(t),x_3(t)\right)$, we have

   $u\left(x_1^0e^t,x_2^0e^t,t\right)=h\left(x_1^0,x_2^0\right)e^t$.

   By taking $t=x_3$, $x_1^0=x_1e^{-x_3}$ and $x_2^0=x_2e^{-x_3}$, we can deduce that

   $u\left(x_1,x_2,x_3\right)=h\left(x_1e^{-x_3},x_2e^{-x_3}\right)e^{x_3},~~\text{for}~(x_1,x_2,x_3)\in\mathbb R^3$.

   Finally, we use original variables to express the solution:

   $u\left(x,y,z\right)=h\left(xe^{-z},ye^{-z}\right)e^z,~~\text{for}~(x,y,z)\in\mathbb R^3$.

2. By rewriting variable, the equation becomes

   $u_{x_1}+u_{x_2}+x_3u_{x_3}=u^3$, $u\left(x_1,x_2,1\right)=h\left(x_1,x_2\right)$.

   Consider the characteristic plane $x_1=x_1(t)$, $x_2=x_2(t)$, $x_3=x_3(t)$, $z=z(t)=u\left(x_1(t),x_2(t),x_3(t)\right)$. We have the following ordinary differential equation

   $\displaystyle\left\{\begin{aligned} &\frac{\mathrm dx_1}{\mathrm dt}=1,\\&\frac{\mathrm dx_2}{\mathrm dt}=1,\\&\frac{\mathrm dx_3}{\mathrm dt}=x_3,\\&\frac{\mathrm dz}{\mathrm dt}=z^3,\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned} &x_1(0)=x_1^0,\\&x_2(0)=x_2^0\\&x_3(0)=1,\\&z(0)=u\left(x_1(0),x_2(0),x_3(0)\right)=h\left(x_1^0,x_2^0\right).\end{aligned}\right.$

   We can solve these equations to obtain

   $\displaystyle\left\{\begin{aligned} &x_1(t)=x_1^0+t,\\&x_2(t)=x_2^0+t,\\&x_3(t)=e^t,\\&z(t)=\frac{h\left(x_1^0,x_2^0\right)}{\sqrt{1-2th^2\left(x_1^0,x_2^0\right)}}.\end{aligned}\right.$

   By $z(t)=u\left(x_1(t),x_2(t),x_3(t)\right)$, we have

   $\displaystyle u\left(x_1^0+t,x_2^0+t,e^t\right)=\frac{h\left(x_1^0,x_2^0\right)}{\sqrt{1-2th^2\left(x_1^0,x_2^0\right)}}$.

   By taking $t=\ln x_3$, $x_1^0=x_1-\ln x_3$, and $x_2^0=x_2-\ln x_3$, we can deduce that

   $\displaystyle u\left(x_1,x_2,x_3\right)=\frac{h\left(x_1-\ln x_3,x_2-\ln x_3\right)}{\sqrt{1-2h^2\left(x_1-\ln x_3,x_2-\ln x_3\right)\ln x_3}},~~\text{for}~(x_1,x_2)\in\mathbb R^2,~x_3>0$.

   Finally, we use original variables to express the solution:

   $\displaystyle u\left(x,y,z\right)=\frac{h\left(x-\ln z,y-\ln z\right)}{\sqrt{1-2h^2\left(x-\ln z,y-\ln z\right)\ln z}},~~\text{for}~(x,y)\in\mathbb R^2,~z>0$.



6. Solve the initial value problem and determine the values of $x$ and $y$ for which it exists:
   1. $u_x+u^2u_y=1$, $u(x,0)=1$
   2. $u_x+\sqrt{u}u_y=0$, $u(x,0)=x^2+1$
Solution

1. Consider the characteristic curve $x=x(t)$, $y=y(t)$, $z=z(t)=u\left(x(t),y(t)\right)$. We have the following ordinary differential equation

   $\displaystyle\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=1,\\&\frac{\mathrm dy}{\mathrm dt}=z^2,\\&\frac{\mathrm dz}{\mathrm dt}=1,\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned} &x(0)=x_0,\\&y(0)=0,\\&z(0)=u\left(x(0),y(0)\right)=1.\end{aligned}\right.$

   We can solve these equations to obtain

   $\left\{\begin{aligned} &x(t)=x_0+t,\\&y(t)=\frac{\left(t+1\right)^3-1}{3},\\&z(t)=t+1.\end{aligned}\right.$

   By $z(t)=u\left(x(t),y(t)\right)$, we have

   $\displaystyle u\left(x_0+t,\frac{\left(t+1\right)^3-1}{3}\right)=t+1$.

   By taking $x_0=x+1-\sqrt[3]{1+3y}$ and $t=\sqrt[3]{1+3y}-1$, we can deduce that

   $u(x,y)=\sqrt[3]{1+3y},~\text{for}~(x,y)\in\mathbb R^2$.

2. Consider the characteristic curve $x=x(t)$, $y=y(t)$, $z=z(t)=u\left(x(t),y(t)\right)$. We have the following ordinary differential equation

   $\displaystyle\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=1,\\&\frac{\mathrm dy}{\mathrm dt}=\sqrt{z},\\&\frac{\mathrm dz}{\mathrm dt}=0,\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned} &x(0)=x_0,\\&y(0)=0,\\&z(0)=u(x(0),y(0))=x_0^2+1.\end{aligned}\right.$

   We can solve these equations to obtain

   $\left\{\begin{aligned} &x(t)=x_0+t,\\&y(t)=t\sqrt{x_0^2+1},\\&z(t)=x_0^2+1.\end{aligned}\right.$

   By $z(t)=u\left(x(t),y(t)\right)$, we have

   $u\left(x_0+t,t\sqrt{x_0^2+1}\right)=x_0^2+1$.

   It seems impossible to solve $x_0$ and $t$ in terms of $x$ and $y$ unless we use the formula of the quartic equation. Instead of solving $x_0$ and $t$, we can express implictly the solution:

   $\displaystyle u(x,y)=\left(x-\frac{y}{\sqrt{u(x,y)}}\right)^2+1$

   In order to determine the domain of solution, we compute the Jacobian matrix

   $\displaystyle J=\text{det}\begin{bmatrix}x_{x_0}&y_{x_0}\\x_t&y_t\end{bmatrix}=\text{det}\begin{bmatrix}1&\displaystyle\frac{x_0t}{\sqrt{x_0^2+1}}\\1&\sqrt{x_0^2+1}\end{bmatrix}=\sqrt{x_0^2+1}-\frac{x_0t}{\sqrt{x_0^2+1}}$.

   It is clear that $J=0$ implies $\displaystyle t=s+\frac{1}{s}$. Thus, if $J\neq0$, we can solve $x_0$ and $t$ in terms of $x$ and $y$ implictly. The solution can be defined on $\mathbb{R}^2$ expcet the curve

   $\left\{\begin{aligned} &x=2s+\frac{1}{s}\\&y=\sqrt{s^2+1}\left(s+\frac{1}{s}\right)\end{aligned}\right.,~~\text{for}~s\in\mathbb R$.



7. Find a general solution:
   1. $\left(x+u\right)u_x+\left(y+u\right)u_y=0$
   2. $\left(x^2+3y^2+3u^2\right)u_x-2xyu_y+2xu=0$
Solution
1. In order to find the general solution, we need to find two independent functions $\phi$ and $\psi$ satisfying

   $\left\{\begin{aligned} &(x+z)\phi_x\left(x,y,z\right)+\left(y+z\right)\phi_y\left(x,y,z\right)+0\phi_z\left(x,y,z\right)=0,\\&\left(x+z\right)\psi_x\left(x,y,z\right)+\left(y+z\right)\psi_y\left(x,y,z\right)+0\psi_z\left(x,y,z\right)=0.\end{aligned}\right.$

   Thus, we can choose $\phi\left(x,y,z\right)=z$ and $\displaystyle\psi\left(x,y,z\right)=\frac{x+z}{y+z}$ from the following nonparameterized form

   $\displaystyle\frac{\mathrm dx}{x+z}=\frac{\mathrm dy}{y+z}=\frac{\mathrm dz}{0}$.

   Here the last term implies that $z$ is constant along the characteristic, so we take $\phi\left(x,y,z\right)=z$. Treating $z$ as a constant in the other terms, we have

   $\displaystyle\frac{\mathrm dx}{x+C_1}=\frac{\mathrm dy}{y+C_1}$.

   By integrating, we obtain

   $\ln|x+C_1|=\ln|y+C_1|+\ln|C_2|$,

   or

   $\displaystyle\frac{x+C_1}{y+C_1}=C_2$.

   Hence, we take $\psi\left(x,y,z\right)=\left(x+z\right)/\left(y+z\right)$. So the general solution is

   $\displaystyle F\left(u(x,y),\frac{x+u(x,y)}{y+u(x,y)}\right)=0$,

   where $F$ is an arbitrary $C^1$ function satisfying $F_\phi^2+F_\psi^2\neq0$. If $F_\phi\neq0$, we can use the implicit function theorem to get

   $\displaystyle u(x,y)=f\left(\frac{x+u(x,y)}{y+u(x,y)}\right)$,

   where $f$ is an arbitrary $C^1$ function.
2. In order to find the general solution, we need to find two independent functions $\phi$ and $\psi$ satisfying

   $\left\{\begin{aligned} &\left(x^2+3y^2+3z^2\right)\phi_x\left(x,y,z\right)-2xy\phi_y\left(x,y,z\right)-2xz\phi_z\left(x,y,z\right)=0,\\&\left(x^2+3y^2+3z^2\right)\psi_x\left(x,y,z\right)-2xy\psi_y\left(x,y,z\right)-2xz\psi_z\left(x,y,z\right)=0.\end{aligned}\right.$

   Thus, we can choose $\displaystyle\phi\left(x,y,z\right)=\frac{z}{y}$ and $\psi\left(x,y,z\right)=y\left(x^2+y^2+z^2\right)$ from the following nonparameterized form

   $\displaystyle\frac{\mathrm dx}{x^2+3y^2+3z^2}=\frac{\mathrm dy}{-2xy}=\frac{\mathrm dz}{-2xz}$.

   Notice that the second eqality implies $\mathrm dy/y=\mathrm dz/z$, which gives us $z/y=C_1$ by integrating. So we take $\phi(x,y,z)=z/y$. Treating $z/y$ as a constant in first two terms, we have

   $\displaystyle\frac{\mathrm dx}{x^2+3\left(1+C_1^2\right)y^2}=\frac{\mathrm dy}{-2xy}$,

   or

   $2xy\,\mathrm dx+x^2\,\mathrm dy+3\left(1+C_1^2\right)y^2\,\mathrm dy=0$.

   We can solve this exact differential equation to obtain

   $x^2y+(1+C_1^2)y^3=C_2$.

   Thus, we take $\psi(x,y,z)=y(x^2+y^2+z^2)$. So the general solution is

   $\displaystyle F\left(\frac{u(x,y)}{y},y\left(x^2+y^2+u^2(x,y)\right)\right)=0$,

   where $F$ is an arbitrary $C^1$ function satisfying $F_\phi^2+F_\psi^2\neq0$. If $F_\phi\neq0$, we can use the implict function theorem to get

   $u(x,y)=yf\left(y\left(x^2+y^2+u^2(x,y)\right)\right)$,

   where $f$ is an arbitrary $C^1$ function.


8. Consider the equation $u_x+u_y=\sqrt{u}$. Derive the general solution $u(x,y)=\left(x+f\left(x-y\right)\right)^2/$4. Observe that the trivial solution $u(x,y)\equiv0$ is not covered by the general solution.
Solution
In order to find the general solution, we need to find two independent functions $\phi$ and $\psi$ satisfying
$\left\{\begin{aligned} &\phi_x\left(x,y,z\right)+\phi_y\left(x,y,z\right)+\sqrt{z}\phi_z\left(x,y,z\right)=0,\\&\psi_x\left(x,y,z\right)+\psi_y\left(x,y,z\right)+\sqrt{z}\psi_z\left(x,y,z\right)=0.\end{aligned}\right.$
Thus, we can choose $\phi\left(x,y,z\right)=x-y$ and $\psi\left(x,y,z\right)=2\sqrt{z}-x$ from the following nonparamterized form
$\displaystyle\frac{\mathrm dx}{1}=\frac{\mathrm dy}{1}=\frac{\mathrm dz}{\sqrt{z}}$.
Therefore, the general solution is
$F(x-y,2\sqrt{z}-x)=0$,
where $F$ is an arbitrary $C^1$ function satisfying $F_\phi^2+F_\psi^2\neq0$. If $F_\psi\neq0$, we can use the implicit function theorem to get
$\displaystyle u(x,y)=\frac{(x+f(x-y))^2}{4}$,
where $f$ is an arbitrary $C^1$ function. We can check that $u\equiv0$ is also a solution, but zero function cannot be obtained from the general solution.


9. Consider the equation $y^2u_x+xu_y=\sin\left(u^2\right)$.
   1. Describe all projected characteristic curves in the $xy$-plane.
   2. For the solution $u$ of the initial value problem with $u(x,0)=x$, determine the values of $u_x, u_y, u_{xx}, u_{xy}, u_{yy}$ on the $x$-axis.
Solution
1. The characteristic curves $\left(x(t),y(t),z(t)\right)$ satisfies that

   $\displaystyle\frac{\mathrm dx}{\mathrm dt}=y^2,~~\frac{\mathrm dy}{\mathrm dt}=x,~~\frac{\mathrm dz}{\mathrm dt}=\sin(z^2)$.

   Since we want to find the projected characteristic curves in the $xy$-plane, we foucs on the first two equations and write them

   $\displaystyle\mathrm dt=\frac{\mathrm dx}{y^2}=\frac{\mathrm dy}{x}$,

   which implies $\displaystyle\frac{y^3}{3}=\frac{x^2}{2}+\frac{C}{6}$, or $2y^3=3x^2+C$.
2. By differentiating $u$ with respect to $x$ on $x$-axis, we have

   $u_x(1,0)=1$, $u_xx(x,0)=0$.

   On the other hand, by putting $y=0$ into the equation, we can get

   $xu_y(x,0)=\sin(u^2(x,0))=\sin(x^2)$,

   or $\displaystyle u_y(x,0)=\frac{\sin(x^2)}{x}$ for $x\neq0$. Note that we may define $u_y\left(0,0\right)=0$ to make $u_y$ has continuity. Then we differentiate $u_y$ with respect to $x$ on $x$-axis, we have

   $\displaystyle u_{xy}(x,0)=2\cos\left(x^2\right)-\frac{\sin\left(x^2\right)}{x^2}$ for $x\neq0$.

   Note that we may define $u_{xy}\left(0,0\right)=1$ to make $u_{xy}$ has continuity. Finally, we differentiate the equation with respect to $y$ to obtain

   $2yu_x+y^2u_{xy}+xu_{yy}=2uu_y\cos\left(u^2\right)$.

   Then by plugging $y=0$ into this equation, we get

   $\displaystyle xu_{yy}(x,0)=2u(x,0)u_y(x,0)\cos(u^2(x,0))=2x\cdot\frac{\sin(x^2)}{x}\cos\left(x^2\right)=\sin\left(2x^2\right)$,

   or $\displaystyle u_{yy}(x,0)=\frac{\sin(2x^2)}{x}$ for $x\neq0$. Note that we may define $u_{yy}(0,0)=0$ to make $u_y$ has continuity.