# existence of solution of a degenerate pde with change of variables

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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my original function was defined on $[0,\infty)$, so I want that function to exist there, but for the transformed function this value corresponds to the limit in $y$. However, with a change of variables you have provided it looks like we just set the value to be 0 there and thus obtain the existence for all $x$. So, I wonder then why there so much literature about degenerate equations and issues of existence, can't you always find some suitable change of variables to perform transformation to the "good" parabolic equation? What are possible consequences? – Medan Nov 9 '12 at 4:08
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? – Medan Nov 9 '12 at 4:13

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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thanks for the detailed answer, however I was interested in the theory behind possible change of variables. I expected the solution to exist, so I just considered a simple example. I wanted to know if it is true in general, that is if the existence for the function with changed variables is equivalent to the existence of the original function. – Medan Nov 9 '12 at 4:01
The aims of changing variables of the equations are transforming the equations to the different forms, so the properties of the transformed equations should be equivalent to that of the original equations, otherwise they should be unreasonable or contradictious. The only things between the original equations and the transformed equations are the easiness of determining the properties. – doraemonpaul Nov 18 '12 at 21:43
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? – Medan Nov 19 '12 at 0:21