Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 2.1

1. Consider the initial value problem $u_{zz}=u^2u_x+(u_{xy})^2$, $u(x,y,0)=x-y$, $u_z(x,y,0)=\sin x$. Find the values of $u_{xz}$, $u_{yz}$ and $u_{zz}$ when $z=0$.
Solution

Suppose that $u\in\mathcal C^2$. From $u(x,y,0)=x-y$, we have $u_x(x,y,0)=1$ and $u_y(x,y,0)=-1$, and hence $u_{xy}(x,y,0)=0$. On the other hand, from $u_z(x,y,0)=\sin x$, we get

$u_{xz}(x,y,0)=\cos x,\quad u_{yz}(x,y,0)=0$.

Now, we use the partial differential equation to obtain

$u_{zz}(x,y,0)=u^2(x,y,0)u_x(x,y,0)+u_{xy}^2(x,y,0)=(x-y)^2\cdot1+0^2=(x-y)^2$.



2. Is the heat equation $u_t=ku_{xx}$ in normal form for Cauchy data on the $x$-axis? On the $t$-axis? What form would the Cauchy data (3) take?
Solution

Note that $u_{xx}$ is the second order partial derivative in $x$ and $u_t$ is first order partial derivative in $t$.

The heat equation is not normal form when the Cauchy data is given on the $x$-axis because the order of the right hand side of equation $u_t=ku_{xx}$ is greater than that of the left hand side of the equation.

The heat equation is in normal form when the Cauchy data is given on $t$-axis, and the equation can be represented as

$\displaystyle u_{xx}=\frac1ku_t$

with the Cauchy data $u(0,t)=g_1(t)$ and $u_x(0,t)=g_2(t)$. Using the matematical induction, it is clear that $\displaystyle\frac{\partial^{n+m}u}{\partial x^n\partial t^m}(0,t)$ can be determined by the heat equation and these Cauchy data, provided $u$ is smooth.



3. Find the solution of the initial value problem $u_{yy}=u_{xx}+u$, $u(x,0)=e^x$, $u_y(x,0)=0$ in the form of power series expansion with respect to $y$ [i.e., $\displaystyle\sum_{n=0}^\infty a_n(x)y^n$]. (Note: This is not a Taylor series.)
Solution

Suppose $u$ is smooth and $\displaystyle u(x,y)=\sum_{n=0}^\infty a_n(x)y^n$. From $u(x,0)=e^x$ and $u_y(x,0)=0$, we get $a_0(x)=e^x$ and $a_1(x)\equiv0$. By the partial differential equation, we have

$\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}(x)y^n\sum_{n=2}^\infty n(n-1)a_n(x)y^{n-2}=u_{yy}=u_{xx}+u=\sum_{n=0}^\infty(a_n''(x)+a_n(x))y^n$

Thus, by comparing the coefficient of $y^n$, it follows that

$\displaystyle a_{n+2}(x)=\frac{a_n''(x)+a_n(x)}{(n+2)(n+1)}\quad\text{for}~n\in\mathbb N\cup\{0\}$.

Since $a_1(x)\equiv0$, it is easy to see that $a_{2k-1}(x)\equiv0$ for $k\in\mathbb N$. On the other hand, from $a_0(x)=e^x$, we find that

$\displaystyle a_2(x)=e^x,\quad a_4(x)=\frac{e^x}{6},\quad a_6(x)=\frac{e^x}{90},\quad a_8(x)=\frac{e^x}{2520},\quad\dots$.

By the mathematical induction, it is easy to show that $\displaystyle a_{2n}(x)=\frac{2^n}{(2n)!}e^x$ for $n\in\mathbb N\cup\{0\}$. Therefore, we obtain

$\displaystyle u(x,y)=\sum_{n=0}^\infty\frac{2^n}{(2n)!}e^xy^{2n}=e^x\sum_{n=0}^\infty\frac{(\sqrt 2y)^{2n}}{(2n)!}=e^x\cosh(\sqrt2y)$.



4. Find the Taylor series solution about $x,y=0$ of the initial value problem $u_y=\sin u_x$, $u(x,0)=\pi x/4$.
Solution

Suppose that $u$ is smooth. It is clear that $u_x(x,0)=\pi/4$ and $\partial_x^nu(x,0)=0$ for $n\geq2$. From the partial differential equation, we have $u_y(x,0)=\sin(u_x(x,0))=\sqrt2/2$, which implies $\partial_x^nu_y(x,0)=0$ for $n\in\mathbb N$ by differentiating it with respect to $x$. Differentiating the partial differential equation with respect to $y$, we obtain

$u_{yy}(x,y)=\cos(u_x(x,y))u_{xy}(x,y)$,

which gives $\partial_x^nu_{yy}(x,0)=0$ for $n\in\mathbb N\cup\{0\}$. By chain rule and mathematical induction, we arrive at $\partial_x^n\partial_y^mu(x,0)=0$ for $n\in\mathbb N\cup\{0\}$ and $m\geq2$. Therefore, the Taylor series for $u$ about $(x,y)=(0,0)$ is

$\begin{aligned}T(x,y)&=u(0,0)+u_x(0,0)x+u_y(0,0)y+\frac{u_{xx}(0,0)x^2+2u_{xy}(0,0)xy+u_{yy}(0,0)y^2}{2}+\sum_{n+m\geq3}\frac{\partial_x^n\partial_y^mu(0,0)}{(n+m)!}x^ny^m\\&=0+\frac\pi4x+\frac{\sqrt2}2y+\frac{0x^2+0xy+0y^2}{2}+0=\frac\pi4x+\frac{\sqrt2}2y.\end{aligned}$



5. Consider the initial value problem $u_t=u_{xx}$, $u(x,0)=g(x)$, where $g(x)=a_nx^n+\cdots+a_0$ is a polynomial. Find a Taylor series solution about $(0,0)$. Where does it converge?
Solution
By the mathematical induction, it is clear that $\partial_t^ku=\partial_x^{2k}u$ for $k\in\mathbb N$, which implies $\partial_t^k\partial_x^mu(x,0)=\partial_x^{2k+m}u(x,0)=g^{(2k+m)}(x)$ for $k,m\in\mathbb N\cup\{0\}$. Hence, we get $\partial_t^k\partial_x^mu(0,0)=g^{(2k+m)}(0)=(2k+m)!a_{2k+m}$ for $0\leq 2k+m\leq n$ and $\partial_t^k\partial_x^mu(0,0)=0$ for $2k+m>n$. Thus, the Taylor series for $u$ about $(0,0)$ is
$\displaystyle T(x,t)=\sum_{k,m=0}^\infty\frac{\partial_t^k\partial_x^mu(0,0)}{(k+m)!}t^kx^m=\sum_{k,m=0}^\infty\frac{g^{(2k+m)}(0)}{(k+m)!}t^kx^m=\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{m=0}^{n-2k}\frac{(2k+m)!a_{2k+m}}{(k+m)!}t^kx^m.$
It is clear that this Taylor series $T$ converges on whole $\mathbb R^2$.


6. Consider the same initial problem as in the preceding exercise, but with $g(x)=(1-ix)^{-1}$, which is real analytic for $-\infty<x<\infty$. Derive the formal Taylor series solution $u(x,t)$, but show that it fails to converge for any $x,t$ with $t\neq0$. Why does this not violate the Cauchy-Kovalevski theorem?
Solution
By the mathematical induction, it is clear that $\partial_t^ku=\partial_x^{2k}u$ for $k\in\mathbb N$, which implies
$\partial_t^k\partial_x^mu(x,0)=\partial_x^{2k+m}u(x,0)=g^{(2k+m)}(x)=(-1)^ki^m(2k+m)!(1-ix)^{-(2k+m+1)}\quad\text{for}~k,m\in\mathbb N\cup\{0\}$.
Then the Taylor series for $u$ about $(0,0)$ is
$\displaystyle T(x,t)=\sum_{k,m=0}^\infty\frac{\partial_t^k\partial_x^mu(0,0)}{(k+m)!}t^kx^m=\frac{(-1)^ki^m(2k+m)!}{(k+m)!}t^kx^m$.
It is clear that the series does not converge when $t\neq0$ because $\displaystyle\frac{(2k+2m)!}{k!m!}\geq1$ for $k,m\in\mathbb N\cup\{0\}$. This does not violate the Cauchy-Kovalevski theorem because the heat equation $u_t=u_{xx}$ with the Cauchy data $u(x,0)=(1-ix)^{-1}$ is not in normal form.


7. Consider the Cauchy problem for Laplace's equation $u_{xx}+u_{yy}=0$, $u(x,0)=0$, $u_y(x,0)=k^{-1}\sin kx$, where $k>0$. Use separation of variables to find the solution explicitly. If we let $k\to\infty$, notice that the Cauchy data tends uniformly to zero, but the solution does not converge to zero for any $y\neq0$. Therefore, a small change from zero Cauchy data [which has the solution $u(x,y)\equiv0$ induces more than a small change in the solution; this means that the Cauchy problem for the Laplace equation is not well posed.
Solution
To apply the separation of variables, we suppose that $u(x,y)=X(x)Y(y)$, where $X$ and $Y$ are not identically zero. From the Cauchy data, we have $X(x)Y(0)=0$ and $X(x)Y'(0)=k^{-1}\sin(kx)$, which gives $Y(0)=0$, $Y'(0)\neq0$ and $X(x)=k^{-1}\sin(kx)/Y'(0)$. Now by Laplace's equation, we have $X''(x)Y(y)+X(x)Y''(y)=0$, which implies
$-k\sin(kx)Y(y)+k^{-1}\sin(kx)Y''(y)=0$.
Thus, $Y''(y)-k^2Y(y)=0$, and hence $Y(y)=2c_1e^{ky}+2c_2e^{-ky}$. Using $Y(0)=0$, we get $c_2=-c_1\neq0$, and hence $Y(y)=c_1\sinh(ky)$. Clearly, $Y'(0)=kc_1$, which gives $X(x)=k^{-2}\sin(kx)/c_1$. Therefore, we solve $u$ by
$\displaystyle u(x,y)=X(x)Y(y)=\frac{k^{-2}\sin(kx)}{c_1}\cdot c_1\sinh(ky)=k^{-2}\sin(kx)\sinh(ky)$.
As $k\to\infty$, the Cauchy data are convergent to zero uniformly on the whole real line. However, the solution does not converge to zero uniformly on the whole plane $\mathbb R^2$. For instance, we observe that
$\displaystyle\lim_{k\to\infty}u\left(\frac{\pi}{2k},y\right)=\lim_{k\to\infty}\frac{\sinh(ky)}{k^2}\to\text{sgn}(y)\infty$.
This means the solution $u$ cannot depend on the Cauchy data continuously, and hence the problem for Laplace's equation with the Cauchy data is not well posed.