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I was looking at the pde $u_t=u_{xx}+1_{\{x\geq c\}}u_x$ for some constant $c$. I see that in order even weak solution to exist I need to insure that the coefficients are Hölder continuous. Which doesn't hold here because of the indicator function. Does it imply there is no solution here?

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What is the $1_{\{x\geq c\}}$ meaning? –  doraemonpaul Jul 18 '12 at 21:01
    
that is an indicator function, i.e. for those $x$ s.t. $x\geq c$ function is equal to $1$, otherwise $0$ –  Medan Jul 18 '12 at 22:57
    
One needs to define "solution" before deciding whether it exists or not. Given $u$, how would you tell if it's a solution? –  user31373 Jul 21 '12 at 4:51
    
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. –  doraemonpaul Sep 10 '12 at 1:19
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1 Answer 1

up vote 2 down vote accepted

In fact $u_t=u_{xx}+1_{\{x\geq c\}}u_x$ should have solutions, but you should find them piecewisely.

First consider $u_t=u_{xx}$ alone:

Of course we use separation of variables:

Case 1: $t\geq 0$

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

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

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

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

$\begin{cases}T(t)=c_3(s^2)e^{-ts^2}\\X(x)=\begin{cases}c_1(s^2)\sin((x-c)s)+c_2(s^2)\cos((x-c)s)&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s^2)e^{-ts^2}\sin((x-c)s)~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos((x-c)s)~ds$

Case 2: $t\leq 0$

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

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

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

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

$\begin{cases}T(t)=c_3(s^2)e^{ts^2}\\X(x)=\begin{cases}c_1(s^2)\sinh((x-c)s)+c_2(s^2)\cosh((x-c)s)&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s^2)e^{ts^2}\sinh((x-c)s)~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh((x-c)s)~ds$

Hence $u(x,t)=\begin{cases}C_1x+C_2+\int_0^\infty C_3(s^2)e^{-ts^2}\sin((x-c)s)~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos((x-c)s)~ds&\text{when}~t\geq0\\C_1x+C_2+\int_0^\infty C_3(s^2)e^{ts^2}\sinh((x-c)s)~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh((x-c)s)~ds&\text{when}~t\leq0\end{cases}$

Now consider $u_t=u_{xx}+u_x$ alone:

Of course we use separation of variables:

Case 3: $t\geq 0$

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

Then $X(x)T'(t)=X''(x)T(t)+X'(x)T(t)=(X''(x)+X'(x))T(t)$

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

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

$\begin{cases}T(t)=c_3'(s^2)e^{-\frac{t(4s^2+1)}{4}}\\X(x)=\begin{cases}c_1'(s^2)e^{-\frac{x}{2}}\sin((x-c)s)+c_2'(s^2)e^{-\frac{x}{2}}\cos((x-c)s)&\text{when}~s\neq0\\c_1'xe^{-\frac{x}{2}}+c_2'e^{-\frac{x}{2}}&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\sin((x-c)s)~ds+\int_0^\infty C_4'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\cos((x-c)s)~ds$

Case 4: $t\leq 0$

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

Then $X(x)T'(t)=X''(x)T(t)+X'(x)T(t)=(X''(x)+X'(x))T(t)$

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

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

$\begin{cases}T(t)=c_3'(s^2)e^{\frac{t(4s^2-1)}{4}}\\X(x)=\begin{cases}c_1'(s^2)e^{-\frac{x}{2}}\sinh((x-c)s)+c_2'(s^2)e^{-\frac{x}{2}}\cosh((x-c)s)&\text{when}~s\neq0\\c_1'xe^{-\frac{x}{2}}+c_2'e^{-\frac{x}{2}}&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\sinh((x-c)s)~ds+\int_0^\infty C_4'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\cosh((x-c)s)~ds$

Hence $u(x,t)=\begin{cases}C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\sin((x-c)s)~ds+\int_0^\infty C_4'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\cos((x-c)s)~ds&\text{when}~t\geq0\\C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\sinh((x-c)s)~ds+\int_0^\infty C_4'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\cosh((x-c)s)~ds&\text{when}~t\leq0\end{cases}$

Therefore the general solution of $u_t=u_{xx}+1_{\{x\geq c\}}u_x$ , without considering the continuity at $x=c$ , should be

$u(x,t)=\begin{cases}C_1x+C_2+\int_0^\infty C_3(s^2)e^{-ts^2}\sin((x-c)s)~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos((x-c)s)~ds&\text{when}~t\geq0~\text{and}~x\leq c\\C_1x+C_2+\int_0^\infty C_3(s^2)e^{ts^2}\sinh((x-c)s)~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh((x-c)s)~ds&\text{when}~t\leq0~\text{and}~x\leq c\\C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\sin((x-c)s)~ds+\int_0^\infty C_4'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\cos((x-c)s)~ds&\text{when}~t\geq0~\text{and}~x\geq c\\C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\sinh((x-c)s)~ds+\int_0^\infty C_4'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\cosh((x-c)s)~ds&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

But if you want $u(x,t)$ should continuous but don't care about the smoothness at $x=c$ , you should play further trick.

Introduce a dummy I.C. $u(c,t)=f(t)$ :

$\begin{cases}C_1c+C_2+\int_0^\infty C_4(s^2)e^{-ts^2}~ds=f(t)&\text{when}~t\geq0~\text{and}~x\leq c\\C_1c+C_2+\int_0^\infty C_4(s^2)e^{ts^2}~ds=f(t)&\text{when}~t\leq0~\text{and}~x\leq c\\C_1'ce^{-\frac{t+2c}{4}}+C_2'e^{-\frac{t+2c}{4}}+\int_0^\infty C_4'(s^2)e^{-\frac{t(4s^2+1)+2c}{4}}~ds=f(t)&\text{when}~t\geq0~\text{and}~x\geq c\\C_1'ce^{-\frac{t+2c}{4}}+C_2'e^{-\frac{t+2c}{4}}+\int_0^\infty C_4'(s^2)e^{\frac{t(4s^2-1)-2c}{4}}~ds=f(t)&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

$\begin{cases}\int_0^\infty C_4(s^2)e^{-ts^2}~ds=f(t)-C_1c-C_2&\text{when}~t\geq0~\text{and}~x\leq c\\\int_0^\infty C_4(s^2)e^{ts^2}~ds=f(t)-C_1c-C_2&\text{when}~t\leq0~\text{and}~x\leq c\\e^{-\frac{t+2c}{4}}\int_0^\infty C_4'(s^2)e^{-ts^2}~ds=f(t)-C_1'ce^{-\frac{t+2c}{4}}-C_2'e^{-\frac{t+2c}{4}}&\text{when}~t\geq0~\text{and}~x\geq c\\e^{-\frac{t+2c}{4}}\int_0^\infty C_4'(s^2)e^{ts^2}~ds=f(t)-C_1'ce^{-\frac{t+2c}{4}}-C_2'e^{-\frac{t+2c}{4}}&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

$\begin{cases}\int_0^\infty\dfrac{C_4(s^2)e^{-ts^2}}{2s}d(s^2)=f(t)-C_1c-C_2&\text{when}~t\geq0~\text{and}~x\leq c\\\int_0^\infty\dfrac{C_4(s^2)e^{ts^2}}{2s}d(s^2)=f(t)-C_1c-C_2&\text{when}~t\leq0~\text{and}~x\leq c\\\int_0^\infty\dfrac{C_4(s^2)e^{-ts^2}}{2s}d(s^2)=f(t)e^{\frac{t+2c}{4}}-C_1'c-C_2'&\text{when}~t\geq0~\text{and}~x\geq c\\\int_0^\infty\dfrac{C_4(s^2)e^{ts^2}}{2s}d(s^2)=f(t)e^{\frac{t+2c}{4}}-C_1'c-C_2'&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

$\begin{cases}\int_0^\infty\dfrac{C_4(s)e^{-ts}}{2\sqrt{s}}ds=f(t)-C_1c-C_2&\text{when}~t\geq0~\text{and}~x\leq c\\\int_0^\infty\dfrac{C_4(s)e^{ts}}{2\sqrt{s}}ds=f(t)-C_1c-C_2&\text{when}~t\leq0~\text{and}~x\leq c\\\int_0^\infty\dfrac{C_4(s)e^{-ts}}{2\sqrt{s}}ds=f(t)e^{\frac{t+2c}{4}}-C_1'c-C_2'&\text{when}~t\geq0~\text{and}~x\geq c\\\int_0^\infty\dfrac{C_4(s)e^{ts}}{2\sqrt{s}}ds=f(t)e^{\frac{t+2c}{4}}-C_1'c-C_2'&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_4(s)}{2\sqrt{s}}\biggr\}=\begin{cases}f(t)-C_1c-C_2&\text{when}~t\geq0~\text{and}~x\leq c\\f(-t)-C_1c-C_2&\text{when}~t\leq0~\text{and}~x\leq c\\f(t)e^{\frac{t+2c}{4}}-C_1'c-C_2'&\text{when}~t\geq0~\text{and}~x\geq c\\f(-t)e^{-\frac{t-2c}{4}}-C_1'c-C_2'&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

$C_4(s)=\begin{cases}2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{f(t)\}-(C_1c+C_2)\delta(\sqrt{s})&\text{when}~t\geq0~\text{and}~x\leq c\\2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{f(-t)\}-(C_1c+C_2)\delta(\sqrt{s})&\text{when}~t\leq0~\text{and}~x\leq c\\2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{f(t)e^{\frac{t+2c}{4}}\}-(C_1'c+C_2')\delta(\sqrt{s})&\text{when}~t\geq0~\text{and}~x\geq c\\2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{f(-t)e^{-\frac{t-2c}{4}}\}-(C_1'c+C_2')\delta(\sqrt{s})&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

$\therefore u(x,t)=\begin{cases}C_1x+C_2+\int_0^\infty C_3(s^2)e^{-ts^2}\sin((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)\}e^{-ts^2}\cos((x-c)s)~ds-(C_1c+C_2)\int_0^\infty\delta(s)e^{-ts^2}\cos((x-c)s)~ds&\text{when}~t\geq0~\text{and}~x\leq c\\C_1x+C_2+\int_0^\infty C_3(s^2)e^{ts^2}\sinh((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)\}e^{ts^2}\cosh((x-c)s)~ds-(C_1c+C_2)\int_0^\infty\delta(s)e^{ts^2}\cosh((x-c)s)~ds&\text{when}~t\leq0~\text{and}~x\leq c\\C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\sin((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)e^{\frac{t+2c}{4}}\}e^{-\frac{t(4s^2+1)+2x}{4}}\cos((x-c)s)~ds-(C_1'c+C_2')\int_0^\infty\delta(s)e^{-\frac{t(4s^2+1)+2x}{4}}\cos((x-c)s)~ds&\text{when}~t\geq0~\text{and}~x\geq c\\C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\sinh((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)e^{-\frac{t-2c}{4}}\}e^{\frac{t(4s^2-1)-2x}{4}}\cosh((x-c)s)~ds-(C_1'c+C_2')\int_0^\infty\delta(s)e^{\frac{t(4s^2-1)-2x}{4}}\cosh((x-c)s)~ds&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

$u(x,t)=\begin{cases}C_1x+C_2+\int_0^\infty C_3(s^2)e^{-ts^2}\sin((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)\}e^{-ts^2}\cos((x-c)s)~ds-(C_1c+C_2)&\text{when}~t\geq0~\text{and}~x\leq c\\C_1x+C_2+\int_0^\infty C_3(s^2)e^{ts^2}\sinh((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)\}e^{ts^2}\cosh((x-c)s)~ds-(C_1c+C_2)&\text{when}~t\leq0~\text{and}~x\leq c\\C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\sin((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)e^{\frac{t+2c}{4}}\}e^{-\frac{t(4s^2+1)+2x}{4}}\cos((x-c)s)~ds-(C_1'c+C_2')e^{-\frac{t+2x}{4}}&\text{when}~t\geq0~\text{and}~x\geq c\\C_1'xe^{-\frac{t+2x}{4}}+C_2'e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\sinh((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)e^{-\frac{t-2c}{4}}\}e^{\frac{t(4s^2-1)-2x}{4}}\cosh((x-c)s)~ds-(C_1'c+C_2')e^{-\frac{t+2x}{4}}&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

$u(x,t)=\begin{cases}C_1(x-c)+\int_0^\infty C_3(s^2)e^{-ts^2}\sin((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)\}e^{-ts^2}\cos((x-c)s)~ds&\text{when}~t\geq0~\text{and}~x\leq c\\C_1(x-c)+\int_0^\infty C_3(s^2)e^{ts^2}\sinh((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)\}e^{ts^2}\cosh((x-c)s)~ds&\text{when}~t\leq0~\text{and}~x\leq c\\C_1'(x-c)e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{-\frac{t(4s^2+1)+2x}{4}}\sin((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(t)e^{\frac{t+2c}{4}}\}e^{-\frac{t(4s^2+1)+2x}{4}}\cos((x-c)s)~ds&\text{when}~t\geq0~\text{and}~x\geq c\\C_1'(x-c)e^{-\frac{t+2x}{4}}+\int_0^\infty C_3'(s^2)e^{\frac{t(4s^2-1)-2x}{4}}\sinh((x-c)s)~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{f(-t)e^{-\frac{t-2c}{4}}\}e^{\frac{t(4s^2-1)-2x}{4}}\cosh((x-c)s)~ds&\text{when}~t\leq0~\text{and}~x\geq c\end{cases}$

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+1 for effort, but it remains unclear to me in what sense this piecewise function solves the PDE. –  user31373 Aug 20 '12 at 1:01
    
@LVK: Since this PDE has the coefficient $1_{\{x\geq c\}}$ for the $u_x$ term, that means when $x\leq c$ this will follow the behaviour of $u_t=u_{xx}$ while $x\geq c$ this will follow the behaviour of $u_t=u_{xx}+u_x$ . So the solution should be at least classified into $x\leq c$ and $x\geq c$ . This is just the fundamental concept of piecewise function. –  doraemonpaul Aug 20 '12 at 1:33
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