3

I want to solve this equation $$\frac{{\partial u}}{{\partial t}} = 2ct\frac{{\partial u}}{{\partial x}} + \frac{1}{2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}$$ with initial data $u(x,0) = \varphi (x)$ and $u(0,t) = 0$ where $x \in [0, + \infty )$

At first, I wanted to use Fourier Transformation and I got that in the case $x \in ( - \infty , + \infty )$ the solution is $$u(x,t) = \frac{1}{{\sqrt {2\pi t} }}\int\limits_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x + c{t^2} - \xi )}^2}}}{{2t}}}}\varphi (\xi )d\xi } $$ but in this case we can't use the method of images to get a solution in a semi-interval from all-interval solution...

Another way - using a substitution $y = x + c{t^2}$. We get the equation ${u_t} = \frac{1}{2}{u_{xx}}$ but with moving boundary condition $$u(c{t^2},t) = 0$$ and I think it's not easier to solve than the original equation.

Any ideas? Thanks for any help.

P.S.Sorry for my terrible English

Dylan
  • 16,575
Msdos4
  • 63
  • You can extend the function $\phi$ to whole $\mathbb{R}$ with same regularity if $\phi$ is sufficiently regular. See for instance page 13 of the book "Elliptic Boundary Value Problems in Domains with Point Singularities". – Student Nov 07 '17 at 14:48
  • Yes, but the standart method - method of images $\varphi (x) \to - \varphi ( - x)$ (for $x < 0$) does not work here. – Msdos4 Nov 07 '17 at 19:33
  • $\phi(x)\to -\phi(-x)$ is in no meaning standard, since such extension can not guarantee the differentiablity of $\phi$ at zero. For your case, $\phi$ should have $H^2$-regularity, so its extension is not a trivial one, since it should keep differentiability up to second order. Please have a careful look at the book. – Student Nov 07 '17 at 20:00

1 Answers1

1

Hint:

Let $\begin{cases}p=x+ct^2\\q=t\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial t}=2ct\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$

$\therefore2ct\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}=2ct\dfrac{\partial u}{\partial p}+\dfrac{1}{2}\dfrac{\partial^2u}{\partial p^2}$

$\dfrac{\partial u}{\partial q}=\dfrac{1}{2}\dfrac{\partial^2u}{\partial p^2}$

Let $u(p,q)=P(p)Q(q)$ ,

Then $P(p)Q'(q)=\dfrac{1}{2}P''(p)Q(q)$

$\dfrac{2Q'(q)}{Q(q)}=\dfrac{P''(p)}{Q'(q)}=-s^2$

$\begin{cases}\dfrac{Q'(q)}{Q(q)}=-\dfrac{s^2}{2}\\P''(p)+s^2P(p)=0\end{cases}$

$\begin{cases}Q(q)=c_3(s)e^{-\frac{qs^2}{2}}\\P(p)=\begin{cases}c_1(s)\sin ps+c_2(s)\cos ps&\text{when}~s\neq0\\c_1p+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-\frac{qs^2}{2}}\sin ps~ds+\int_0^\infty C_2(s)e^{-\frac{qs^2}{2}}\cos ps~ds=\int_0^\infty C_1(s)e^{-\frac{ts^2}{2}}\sin((x+ct^2)s)~ds+\int_0^\infty C_2(s)e^{-\frac{ts^2}{2}}\cos((x+ct^2)s)~ds$

doraemonpaul
  • 16,178
  • 3
  • 31
  • 75