# Partial Differential Equation

$\partial_t f=\sin (t)\,\partial_x f+\lambda \,\partial_{xx} f$

with initial condition $f(x,0)=1$ for all $x$

• Hint: there's a very trivial solution. Commented Dec 13, 2012 at 3:26
• separation of variables? i didn't get along with the initial conditions Commented Dec 13, 2012 at 3:28
• let me make it a bit harder: $\int_{-\infty}^{\infty} f(x,t)dx=1$ for all $t$ and $f$ is nonegative. this is obtained from kolmogorov forward equation Commented Dec 13, 2012 at 3:30

A slightly more interesting initial condition $f(x,0) = a + b x + c x^2$ has the solution $f(x,t) = a + b (x + 1 - \cos(t)) + c ((x+1-\cos(t))^2 + 2 \lambda t)$

EDIT: Oh, even better:

$f(x,t) = e^{i\omega (x + 1 - \cos(t)) - \lambda \omega^2 t}$ for initial condition $f(x,0) = e^{i\omega x}$.

And using Fourier transform we can get any $L^2$ initial condition $f(x,0) = g(x)$: if $G(k)$ is the inverse Fourier transform of $g$, so $g(x) = \int_{-\infty}^\infty G(k) e^{-2 \pi i k x}\ dk$, then $$f(x,t) = \int_{-\infty}^\infty G(k) e^{-2 \pi i k (x + 1 - \cos(t)) - 4 \pi^2 k^2 \lambda t}\ dk$$

EDIT: Hmmm: this is simpler than I thought. It's the heat equation in disguise. Namely, $g(x,t) = f(x+\cos(t),t)$ satisfies the heat equation $$\frac{\partial g}{\partial t} = \lambda \frac{\partial^2 g}{\partial x^2}$$

• interesting! can you find a solution that vanishes as x tends to infinity? i.e. f to be a PDF wrt x for all t Commented Dec 13, 2012 at 4:37
• i also see that your solution is not compatible with $f(x,0)=0$ Commented Dec 13, 2012 at 4:42
• $f(x,0) = 0$ would have the solution $f(x,t) = 0$. Commented Dec 13, 2012 at 4:47
• this is wonderful. really. so you claim that there isn't such a solution that vanishes when x tends to infinity? this will contradict kolmogorov forward equation Commented Dec 13, 2012 at 4:51
• When did I claim that? Commented Dec 13, 2012 at 6:18

Case $1$ : $\lambda=0$

Then $\partial_tf=\sin t~\partial_xf$

$\dfrac{1}{\sin t}\partial_tf=\partial_xf$

This belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde1104.pdf

$f(x,t)=C\biggl(e^{\int\frac{1}{\frac{1}{\sin t}}dt-\int-dx}\biggr)=C\left(e^{\int dx+\int\sin t~dt}\right)=C(x-\cos t)$

$f(x,0)=1$ :

$C(x-1)=1$

$C(x)=1$

$\therefore f(x,t)=1$

Case $2$ : $\lambda\neq0$

Then $\partial_tf=\sin t~\partial_xf+\lambda\partial_{xx}f$

Let $\begin{cases}u=x-\cos t\\v=t\end{cases}$ ,

Then $\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial x}=\dfrac{\partial f}{\partial u}$

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

$\dfrac{\partial f}{\partial t}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial t}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial t}=\sin t\dfrac{\partial f}{\partial u}+\dfrac{\partial f}{\partial v}$

$\therefore\sin t\dfrac{\partial f}{\partial u}+\dfrac{\partial f}{\partial v}=\sin t\dfrac{\partial f}{\partial u}+\lambda\dfrac{\partial^2f}{\partial u^2}$

$\dfrac{\partial f}{\partial v}=\lambda\dfrac{\partial^2f}{\partial u^2}$

Case $2a$ : $\text{Re}(\lambda v)\geq0$

Let $f(u,v)=U(u)V(v)$ ,

Then $U(u)V'(v)=\lambda U''(u)V(v)$

$\dfrac{V'(v)}{\lambda V(v)}=\dfrac{U''(u)}{U(u)}=-s^2$

$\begin{cases}\dfrac{V'(v)}{V(v)}=-\lambda s^2\\U''(u)+s^2U(u)=0\end{cases}$

$\begin{cases}V(v)=c_3(s)e^{-\lambda vs^2}\\U(u)=\begin{cases}c_1(s)\sin us+c_2(s)\cos us&\text{when}~s\neq0\\c_1u+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore f(u,v)=\int_0^\infty C_1(s)e^{-\lambda vs^2}\sin us~ds+\int_0^\infty C_2(s)e^{-\lambda vs^2}\cos us~ds$

Case $2b$ : $\text{Re}(\lambda v)\leq0$

Let $f(u,v)=U(u)V(v)$ ,

Then $U(u)V'(v)=\lambda U''(u)V(v)$

$\dfrac{V'(v)}{\lambda V(v)}=\dfrac{U''(u)}{U(u)}=s^2$

$\begin{cases}\dfrac{V'(v)}{V(v)}=\lambda s^2\\U''(u)-s^2U(u)=0\end{cases}$

$\begin{cases}V(v)=c_6(s)e^{\lambda vs^2}\\U(u)=\begin{cases}c_4(s)\sinh us+c_5(s)\cosh us&\text{when}~s\neq0\\c_1u+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore f(u,v)=\int_0^\infty C_3(s)e^{\lambda vs^2}\sinh us~ds+\int_0^\infty C_4(s)e^{\lambda vs^2}\cosh us~ds$

Hence $f(u,v)=\begin{cases}\int_0^\infty C_1(s)e^{-\lambda vs^2}\sin us~ds+\int_0^\infty C_2(s)e^{-\lambda vs^2}\cos us~ds&\text{when Re}(\lambda v)\geq0\\\int_0^\infty C_3(s)e^{\lambda vs^2}\sinh us~ds+\int_0^\infty C_4(s)e^{\lambda vs^2}\cosh us~ds&\text{when Re}(\lambda v)\leq0\end{cases}$

$f(x,t)=\begin{cases}\int_0^\infty C_1(s)e^{-\lambda ts^2}\sin((x-\cos t)s)~ds+\int_0^\infty C_2(s)e^{-\lambda ts^2}\cos((x-\cos t)s)~ds&\text{when Re}(\lambda t)\geq0\\\int_0^\infty C_3(s)e^{\lambda ts^2}\sinh((x-\cos t)s)~ds+\int_0^\infty C_4(s)e^{\lambda ts^2}\cosh((x-\cos t)s)~ds&\text{when Re}(\lambda t)\leq0\end{cases}$

$f(x,0)=1$ :

$\begin{cases}\int_0^\infty C_1(s)\sin((x-1)s)~ds+\int_0^\infty C_2(s)\cos((x-1)s)~ds=1&\text{when Re}(\lambda t)\geq0\\\int_0^\infty C_3(s)\sinh((x-1)s)~ds+\int_0^\infty C_4(s)\cosh((x-1)s)~ds=1&\text{when Re}(\lambda t)\leq0\end{cases}$

$\begin{cases}\int_0^\infty C_2(s)\cos((x-1)s)~ds=1-\int_0^\infty C_1(s)\sin((x-1)s)~ds&\text{when Re}(\lambda t)\geq0\\\int_0^\infty C_4(s)\cosh((x-1)s)~ds=1-\int_0^\infty C_3(s)\sinh((x-1)s)~ds&\text{when Re}(\lambda t)\leq0\end{cases}$

$\begin{cases}\int_0^\infty C_2(s)\cos((x-1)s)~ds=1-\int_0^\infty C_1(s)\sin((x-1)s)~ds&\text{when Re}(\lambda t)\geq0\\\int_0^\infty C_4(s)\cos(i(x-1)s)~ds=1+i\int_0^\infty C_3(s)\sin(i(x-1)s)~ds&\text{when Re}(\lambda t)\leq0\end{cases}$

$\begin{cases}\int_0^\infty C_2(s)\cos xs~ds=1-\int_0^\infty C_1(s)\sin xs~ds&\text{when Re}(\lambda t)\geq0\\\int_0^\infty C_4(s)\cos xs~ds=1+i\int_0^\infty C_3(s)\sin xs~ds&\text{when Re}(\lambda t)\leq0\end{cases}$

$\begin{cases}\mathcal{F}_{c,s\to x}\{C_2(s)\}=1-\mathcal{F}_{s,s\to x}\{C_1(s)\}&\text{when Re}(\lambda t)\geq0\\\mathcal{F}_{c,s\to x}\{C_4(s)\}=1+i\mathcal{F}_{s,s\to x}\{C_3(s)\}&\text{when Re}(\lambda t)\leq0\end{cases}$

$\begin{cases}C_2(s)=\delta(s)-\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}&\text{when Re}(\lambda t)\geq0\\C_4(s)=\delta(s)+i\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}&\text{when Re}(\lambda t)\leq0\end{cases}$

$\therefore f(x,t)=\begin{cases}\int_0^\infty C_1(s)e^{-\lambda ts^2}\sin((x-\cos t)s)~ds+\int_0^\infty\delta(s)e^{-\lambda ts^2}\cos((x-\cos t)s)~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}e^{-\lambda ts^2}\cos((x-\cos t)s)~ds&\text{when Re}(\lambda t)\geq0\\\int_0^\infty C_3(s)e^{\lambda ts^2}\sinh((x-\cos t)s)~ds+\int_0^\infty\delta(s)e^{\lambda ts^2}\cosh((x-\cos t)s)~ds+i\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{\lambda ts^2}\cosh((x-\cos t)s)~ds&\text{when Re}(\lambda t)\leq0\end{cases}$

$f(x,t)=\begin{cases}1+\int_0^\infty C_1(s)e^{-\lambda ts^2}\sin((x-\cos t)s)~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}e^{-\lambda ts^2}\cos((x-\cos t)s)~ds&\text{when Re}(\lambda t)\geq0\\1+\int_0^\infty C_3(s)e^{\lambda ts^2}\sinh((x-\cos t)s)~ds+i\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{\lambda ts^2}\cosh((x-\cos t)s)~ds&\text{when Re}(\lambda t)\leq0\end{cases}$