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I am looking at the pde $u_t=x^2u_{xx},\; x\in [0,\infty) ,\; t\in (0,T], \; u(0,x)=u_0(x)$ This is a degenerate pde with a diffusion coefficient which is not bounded from 0, so I can't apply the classic theory of existence and uniqueness since the operator is not uniformly parabolic. However, I can do change of variables as $y=ln(x)$ and arrive at another pde: $v_t=v_{yy}-v_y,\; y \in (-\infty, +\infty),\; t\in (0,T],\; v_0(y)=u_0(e^y)$ Now there is no degeneracy and I can state the existence of function $v=v(t,y)$. Thus, I claim that $u(t,x)$ exists as well, however only for $x\in (0,\infty)$(Question 1: What about $x=0$? Does the function exist there, can we additionally define it?).

Question 2: Is that a right line of thought? So, as long as I can find change of variables s.t. it produces non degenerate pde I obtain the existence for the degenerate pde? What should I be careful about?

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$x=0$ corresponds to the limit $y \to -\infty$. Does your solution $v(y,t)$ have a limit as $y \to -\infty$? That's a bit of a delicate question. However, note that if $v(t,y) = e^{y/2 - t/4} V(t,y)$ the equation for $V$ is the classical heat equation: $\dfrac{\partial V}{\partial t} = \dfrac{\partial^2 V}{\partial x^2}$. If, for example, $V(0,y)$ is bounded, then $V(t,y)$ has the same bound for all $t > 0$, and $v(t,y) \to 0$ as $y \to -\infty$ for any fixed $t > 0$.

EDIT: actually it would have been better, I think, to use a different transformation: $v(t,y) = w(t,z)$ where $y = z + t$ so you get $ \frac{\partial w}{\partial t} = \frac{\partial^2 w}{\partial z^2}$ and $\lim_{y \to -\infty} v(t,y) = \lim_{z \to -\infty} w(t,z)$.

But I doubt that there will always be a transformation that will act so nicely.

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    In fact, if I understand correctly, for the equation $a(x)u_{xx}=u_t$ boundedness from $0$ of the diffusion coefficient is very important as the formula of the fundamental solution has the lower bound in the denominator. So, if there is no lower boundary the fundamental solution blows up? Does it imply anything about the solution itself? As pointed out above with change of variables we can find a perfectly fine solution. So, if the fundamental solution behaves "bad", is that a sign of issues on existence or it doesn't really matter much?2012-11-09
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Let $u(t,x)=T(t)X(x)$ ,

Then $T'(t)X(x)=x^2T(t)X''(x)$

$\dfrac{T'(t)}{T(t)}=\dfrac{x^2X''(x)}{X(x)}=-\dfrac{4s^2+1}{4}$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-\dfrac{4s^2+1}{4}\\x^2X''(x)+\dfrac{4s^2+1}{4}X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-\frac{t(4s^2+1)}{4}}\\X(x)=\begin{cases}c_1(s)\sqrt{x}\sin(s\ln x)+c_2(s)\sqrt{x}\cos(s\ln x)&\text{when}~s\neq0\\c_1\sqrt{x}\ln x+c_2\sqrt{x}&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty C_4(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds$

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

$C_1\sqrt{x}\ln x+C_2\sqrt{x}+\int_0^\infty C_3(s)\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty C_4(s)\sqrt{x}\cos(s\ln x)~ds=u_0(x)$

$C_1\ln x+C_2+\int_0^\infty C_3(s)\sin(s\ln x)~ds+\int_0^\infty C_4(s)\cos(s\ln x)~ds=\dfrac{u_0(x)}{\sqrt{x}}$

$\int_0^\infty C_4(s)\cos(s\ln x)~ds=\dfrac{u_0(x)}{\sqrt{x}}-C_1\ln x-C_2-\int_0^\infty C_3(s)\sin(s\ln x)~ds$

$\int_0^\infty C_4(s)\cos xs~ds=e^{\frac{x}{2}}u_0(e^x)-C_1x-C_2-\int_0^\infty C_3(s)\sin xs~ds$

$\mathcal{F}_{c,s\to x}\{C_4(s)\}=e^{\frac{x}{2}}u_0(e^x)-C_1x-C_2-\mathcal{F}_{s,s\to x}\{C_3(s)\}$

$C_4(s)=\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}+C_1\delta'(s)-C_2\delta(s)-\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}$

$\therefore u(x,t)=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds+\int_0^\infty C_1\delta'(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-\int_0^\infty C_2\delta(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-C_1e^{-\frac{t}{4}}\sqrt{x}\ln x-C_2e^{-\frac{t}{4}}\sqrt{x}-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds=\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds$

The function not only exist for at least $x\in(0,\infty)$ , but also has one arbitrary function left.

To determine whether the function exist at $x=0$ or not, we just need to consider $\lim\limits_{x\to0}\sqrt{x}\sin(s\ln x)$ and $\lim\limits_{x\to0}\sqrt{x}\cos(s\ln x)$ .

Since both $\sin(s\ln x)$ and $\cos(s\ln x)$ are bounded,

$\lim\limits_{x\to0}\sqrt{x}\sin(s\ln x)=0$ and $\lim\limits_{x\to0}\sqrt{x}\cos(s\ln x)=0$

So the function not only also exist at $x=0$ , but also equals to $0$ when $x=0$ .

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    even for the regularity properties? Because when I think of the first derivative, for example, $u_x=v_y*y_x$, so even if new function $v$ is perfectly smooth, $y_x$ might destroy the smoothness of the original problem, or am I wrong?2012-11-19