# 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.

• What did you try? – Romeo Sep 9 '13 at 21:09
• Im a bit lost on where to start. Maybe guessing an exponential since we have du/dx = 2u? u = e^2x? – yellowsmoke Sep 9 '13 at 21:10
• You are on the right path. The solutions of the pde are $u(x,y)=B(y)e^{2x}$, where B is an arbitrary function. From here, you can find $u_0$. – Pocho la pantera Sep 9 '13 at 21:17
• The first equation seems to be an ordinary differential equation. Is that supposed to be the case? Is $u$ supposed to have parameters besides $x$? – Omnomnomnom Sep 9 '13 at 21:18
• The existence of two parameters is implied by the initial condition. – J. W. Perry Sep 9 '13 at 21:31

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 $$z(x,\theta) = P\left(\frac{\partial}{\partial \lambda}\right) \exp(\lambda x - 2 \theta/\lambda)$$

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

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

• No, that is not a solution. You want $\exp(2\theta/\lambda)$, not $\exp(\theta^2/\lambda^2)$. – Robert Israel Sep 18 '13 at 5:36
• But in any case, these are certainly not the only solutions. For example, you could take linear combinations... – Robert Israel Sep 18 '13 at 5:37
• @RobertIsrael: Thanks for the comment. It was a typo. It is corrected, – Mhenni Benghorbal Sep 18 '13 at 5:55

$\displaystyle{% \frac{\partial^{2}\,{\rm z}\left(x,\theta\right)}{\partial x\,\partial\theta} - 2{\rm z}\left(x,\theta\right) = 0; \qquad {\rm z}\left(x, 0\right) = {\rm u_{0}}\left(x\right)}$.

$${\rm z}\left(x,\theta\right) \equiv \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} {{\rm d}k_{x}\,{\rm d}\theta \over \left(2\pi\right)^{2}}\, \tilde{\rm z}\left(k_{x}, k_{\theta}\right) {\rm e}^{{\rm i}\left(k_{x}\,x\ +\ k_{\theta}\,\theta\right)} \quad\Longrightarrow\quad \left(-k_{x}k_{\theta} - 2\right)\tilde{\rm z}\left(k_{x}, k_{\theta}\right) = 0$$ 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.

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 $z(x,0)=u_0(x)$ :

$f(\sqrt2x)=u_0(x)$

$f(x)=u_0\left(\dfrac{x}{\sqrt2}\right)$

$\therefore z(x,\theta)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nu_0^{(2k)}(x-\theta)\theta^{2n}}{8^n(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(\sqrt2(x-\theta))\theta^{2n+1}}{2\sqrt28^n(2n+1)!}$