Help solving a partial differential equation Find all $u_0(x)$ for which
$$
\frac{\partial^2z}{\partial x\partial\theta}-2z=0~;\qquad z(x,0)=u_0(x)
$$
has a solution, and for each such function, find all solutions.
 A: Formally, there are solutions of the form
$$ z(x,\theta) = \int T(\lambda)\; \exp(\lambda x - 2 \theta/\lambda)\; d\lambda$$
where $T$ is a distribution on $\mathbb C$, defined on all $(x,\theta)$ for which
the integral converges and the mixed partial derivative exists.
  At $\theta = 0$ you want
$$ u_0(x) = \int T(\lambda) \exp(\lambda x)\; d\lambda $$
For example, you might express $u_0$ as a Fourier transform as Felix did.  But there are many more possibilities, e.g. $u_0(x) = P(\lambda) \exp(\lambda x)$ for some polynomial
$P$ with $$z(x,\theta) = P\left(\frac{\partial}{\partial \lambda}\right) \exp(\lambda x - 2 \theta/\lambda)$$
A: First, note that, the pde can be solved using the separation of variables techniques. Assuming 

$$ z(x,\theta) = F(x)G(\theta) $$

gives rise to the two ordinary differential equations

$$ F'(x) = \lambda F(x), \quad G'(\theta)=\frac{2}{\lambda}G(\theta).$$

The above yields the solution

$$ z( x,\theta ) = A e^{\lambda x } e^{\frac{2\theta}{\lambda}}\longrightarrow (1) $$

Now, apply the initial condition to get 

$$ A = u_0(x) e^{-\lambda x} \implies u_{0}(x) = Ae^{\lambda x}. $$

which gives the general form of $u_{0}(x)$. Now, subsbtituting back in $(1)$ gives the solution

$$z( x,\theta ) = u_{0}(x) e^{\frac{2\theta }{\lambda}}  .$$

A: $\displaystyle{%
\frac{\partial^{2}\,{\rm z}\left(x,\theta\right)}{\partial x\,\partial\theta}
-
2{\rm z}\left(x,\theta\right) = 0;
\qquad 
{\rm z}\left(x, 0\right) = {\rm u_{0}}\left(x\right)}$.
$$
{\rm z}\left(x,\theta\right)
\equiv
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{{\rm d}k_{x}\,{\rm d}\theta \over \left(2\pi\right)^{2}}\,
\tilde{\rm z}\left(k_{x}, k_{\theta}\right)
{\rm e}^{{\rm i}\left(k_{x}\,x\ +\ k_{\theta}\,\theta\right)}
\quad\Longrightarrow\quad
\left(-k_{x}k_{\theta} - 2\right)\tilde{\rm z}\left(k_{x}, k_{\theta}\right)
=
0
$$
Then, $k_{x}$ and $k_{\theta}$ are not independent: $k_{\theta} = -2/k_{x}$ and we just need a one dimensional Fourier technique:
$$
{\rm z}\left(x,\theta\right)
\equiv
\int_{-\infty}^{\infty}
{{\rm d}k_{x} \over 2\pi}\,
{\rm A}\left(k_{x}\right)
{\rm e}^{{\rm i}\left\lbrack k_{x}\,x\ +\ \left(-2/k_{x}\right)\,\theta\right\rbrack}
$$
$$
{\rm u_{0}}\left(x\right)
=
{\rm z}\left(x,0\right)
=
\int_{-\infty}^{\infty}{{\rm d}k_{x} \over 2\pi}\,{\rm A}\left(k_{x}\right)
{\rm e}^{{\rm i}k_{x}\,x}
\quad\Longrightarrow\quad
{\rm A}\left(k_{x}\right)
=
\int_{-\infty}^{\infty}{\rm d}x\,
{\rm u_{0}}\left(x\right){\rm e}^{-{\rm i}k_{x}\,x}
$$
\begin{align}
{\rm z}\left(x,\theta\right)
&=
\int_{-\infty}^{\infty}
{{\rm d}k_{x} \over 2\pi}\,\left\lbrack%
\int_{-\infty}^{\infty}{\rm d}x'\,
{\rm u_{0}}\left(x'\right){\rm e}^{-{\rm i}k_{x}\,x'}
\right\rbrack
{\rm e}^{{\rm i}\left\lbrack k_{x}\,x\ -2\theta/k_{x}\right\rbrack}
\\[3mm]&=
\int_{-\infty}^{\infty}
{\rm u_{0}}\left(x'\right)\left\lbrace%
\int_{-\infty}^{\infty}{{\rm d}k_{x} \over 2\pi}\,
{\rm e}^
{{\rm i}\left\lbrack k_{x}\left(x - x'\right)\ -\ 2\theta/k_{x}\right\rbrack}
\right\rbrace\,{\rm d}x'
\end{align}
$$
\begin{array}{|c|}\hline\\
{\rm z}\left(x,\theta\right)
=
\int_{-\infty}^{\infty}{\rm K}\left(x - x', \theta\right)
{\rm u_{0}}\left(x'\right)\,{\rm d}x'
\\[3mm]
{\rm K}\left(x,\theta\right)
\equiv
\int_{-\infty}^{\infty}{{\rm d}k_{x} \over 2\pi}\,
{\rm e}^{{\rm i}\left(k_{x}x - 2\theta/k_{x}\right)}
=
{1 \over \left\vert x\right\vert}\,\overline{\rm K}\left(2x\theta\right)
\\[3mm]
\overline{\rm K}\left(\mu\right)
\equiv
\int_{-\infty}^{\infty}{{\rm d}k_{x} \over 2\pi}\,
{\rm e}^{{\rm i}\left(k_{x} - \mu/k_{x}\right)}
=
\overline{\rm K}^{\,*}\left(\mu\right)
\\ \\ \hline
\end{array}
$$
Once $\overline{\rm K}\left(\mu\right)$ is evaluated,
${\rm z}\left(x, \theta\right)$ is completely known. However, it would be useful to have some information about $x$ and $\theta$ domains.
A: Let $\begin{cases}p=\sqrt2(x+\theta)\\q=\sqrt2(x-\theta)\end{cases}$ ,
Then $\dfrac{\partial z}{\partial\theta}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial\theta}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial\theta}=\sqrt2\dfrac{\partial z}{\partial p}-\sqrt2\dfrac{\partial z}{\partial q}$
$\dfrac{\partial^2z}{\partial x\partial\theta}=\dfrac{\partial}{\partial x}\left(\sqrt2\dfrac{\partial z}{\partial p}-\sqrt2\dfrac{\partial z}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\sqrt2\dfrac{\partial z}{\partial p}-\sqrt2\dfrac{\partial z}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\sqrt2\dfrac{\partial z}{\partial p}-\sqrt2\dfrac{\partial z}{\partial q}\right)\dfrac{\partial q}{\partial x}=\sqrt2\left(\sqrt2\dfrac{\partial^2z}{\partial p^2}-\sqrt2\dfrac{\partial^2z}{\partial p\partial q}\right)+\sqrt2\left(\sqrt2\dfrac{\partial^2z}{\partial p\partial q}-\sqrt2\dfrac{\partial^2z}{\partial q^2}\right)=2\dfrac{\partial^2z}{\partial p^2}-2\dfrac{\partial^2z}{\partial q^2}$
$\therefore2\dfrac{\partial^2z(p,q)}{\partial p^2}-2\dfrac{\partial^2z(p,q)}{\partial q^2}-2z(p,q)=0$
$\dfrac{\partial^2z(p,q)}{\partial p^2}=z(p,q)+\dfrac{\partial^2z(p,q)}{\partial q^2}$
Similar to PDE - solution with power series:
Consider $z(q,q)=f(q)$ and $z_p(q,q)=g(q)$ ,
Let $z(p,q)=\sum\limits_{n=0}^\infty\dfrac{(p-q)^n}{n!}\dfrac{\partial^nz(q,q)}{\partial p^n}$ ,
Then $z(p,q)=\sum\limits_{n=0}^\infty\dfrac{(p-q)^{2n}}{(2n)!}\dfrac{\partial^{2n}z(q,q)}{\partial p^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(p-q)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}z(q,q)}{\partial p^{2n+1}}$
$\dfrac{\partial^4z(p,q)}{\partial p^4}=\dfrac{\partial^2z(p,q)}{\partial p^2}+\dfrac{\partial^4z(p,q)}{\partial q^2\partial p^2}=z(p,q)+\dfrac{\partial^2z(p,q)}{\partial q^2}+\dfrac{\partial^2z(p,q)}{\partial q^2}+\dfrac{\partial^4z(p,q)}{\partial q^4}=z(p,q)+2\dfrac{\partial^2z(p,q)}{\partial q^2}+\dfrac{\partial^4z(p,q)}{\partial q^4}$
Similarly, $\dfrac{\partial^{2n}z(p,q)}{\partial p^{2n}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k}z(p,q)}{\partial q^{2k}}$
$\dfrac{\partial^3z(p,q)}{\partial p^3}=\dfrac{\partial z(p,q)}{\partial p}+\dfrac{\partial^3z(p,q)}{\partial q^2\partial p}$
$\dfrac{\partial^5z(p,q)}{\partial p^5}=\dfrac{\partial^3z(p,q)}{\partial p^3}+\dfrac{\partial^5z(p,q)}{\partial q^2\partial p^3}=\dfrac{\partial z(p,q)}{\partial p}+\dfrac{\partial^3z(p,q)}{\partial q^2\partial p}+\dfrac{\partial^3z(p,q)}{\partial q^2\partial p}+\dfrac{\partial^5z(p,q)}{\partial q^4\partial p}=\dfrac{\partial z(p,q)}{\partial p}+2\dfrac{\partial^3z(p,q)}{\partial q^2\partial p}+\dfrac{\partial^5z(p,q)}{\partial q^4\partial p}$
Similarly, $\dfrac{\partial^{2n+1}z(p,q)}{\partial p^{2n+1}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k+1}z(p,q)}{\partial q^{2k}\partial p}$
$\therefore z(p,q)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(q)(p-q)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(q)(p-q)^{2n+1}}{(2n+1)!}$
$z(x,\theta)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(\sqrt2(x-\theta))\theta^{2n}}{8^n(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(\sqrt2(x-\theta))\theta^{2n+1}}{2\sqrt28^n(2n+1)!}$
This is the most general solution of $\dfrac{\partial^2z}{\partial x\partial\theta}-2z=0$ .
Now $z(x,0)=u_0(x)$ :
$f(\sqrt2x)=u_0(x)$
$f(x)=u_0\left(\dfrac{x}{\sqrt2}\right)$
$\therefore z(x,\theta)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nu_0^{(2k)}(x-\theta)\theta^{2n}}{8^n(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(\sqrt2(x-\theta))\theta^{2n+1}}{2\sqrt28^n(2n+1)!}$
