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For pde $$u_t=x^2u_{xx},\; x\in [0,\infty) ,\; t\in (0,T], \; u(0,x)=u_0(x)$$ I can do change of variables as $y=ln(x)$ and arrive at another pde: $$v_t=v_{yy}-v_y,\; y \in (-\infty, +\infty),\; t\in (0,T],\; v_0(y)=u_0(e^y)$$ I can show the existence and uniqueness of the transformed pde and therefore imply the existence and uniqueness of the original pde. Does the results of the regularity transfer as well? My new equation might be infinitely smooth but I guess the statement of the smoothness would depend on change of variables via chain rule? Is that correct or there are some regularity that I can obtain for the original equation for free from the transformed equation?

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2 Answers 2

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Since $u(x,t) = v(\log x, t)$ for all $x, \ t$ and $v$ is $C^\infty$ smooth on $\mathbb{R} \times (0,\infty)$, so is $u$ on $(0,\infty) \times (0, \infty)$. However, there might be some problems for $x = 0, \, t > 0$ (where you can't prescribe data) and very likely at $x = t = 0$.

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When you apply separation of variables on the original equation,

Let $u(t,x)=T(t)X(x)$ ,

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

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

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

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

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

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

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

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

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

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

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

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

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

This can already obviously to see that $u(t,x)$ is regular on at least $x\in[0,\infty)$ , so as $v(t,y)$ is regular on at least $y\in(-\infty,\infty)$ .

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I am sorry to say it so bluntly. While your argument may be correct, it is also singularly uninformative. –  Hans Engler Nov 30 '12 at 1:56

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