4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ What did you try? $\endgroup$ – Romeo Sep 9 '13 at 21:09
  • $\begingroup$ Im a bit lost on where to start. Maybe guessing an exponential since we have du/dx = 2u? u = e^2x? $\endgroup$ – yellowsmoke Sep 9 '13 at 21:10
  • $\begingroup$ 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 $u_0$. $\endgroup$ – Pocho la pantera Sep 9 '13 at 21:17
  • $\begingroup$ The first equation seems to be an ordinary differential equation. Is that supposed to be the case? Is $u$ supposed to have parameters besides $x$? $\endgroup$ – Omnomnomnom Sep 9 '13 at 21:18
  • $\begingroup$ The existence of two parameters is implied by the initial condition. $\endgroup$ – J. W. Perry Sep 9 '13 at 21:31
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ No, that is not a solution. You want $\exp(2\theta/\lambda)$, not $\exp(\theta^2/\lambda^2)$. $\endgroup$ – Robert Israel Sep 18 '13 at 5:36
  • 1
    $\begingroup$ But in any case, these are certainly not the only solutions. For example, you could take linear combinations... $\endgroup$ – Robert Israel Sep 18 '13 at 5:37
  • $\begingroup$ @RobertIsrael: Thanks for the comment. It was a typo. It is corrected, $\endgroup$ – Mhenni Benghorbal Sep 18 '13 at 5:55
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.