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I am exposed to a PDE in the following form:

$\frac{\partial f}{\partial t}=\alpha \frac{\partial^2 f}{\partial x^2}-\beta \frac{\partial f}{\partial x} + \mu_a P_a(t) \delta(x-1)+ \mu_b P_b(t) \delta(x-N+1)$

I'm trying to find two linearly independent solutions. I know that the time derivative could be treated with Laplace transform and that I should divide the solution interval according to dirac arguments. Although, I don't get the expected result.

Could someone help me to proceed in solving this?

It is about applying Feller results on a modified Fokker-planck (Chapman-Kolmogovo) which has the same form of the given equation.

Edit:

$P_a$ and $P_b$ are defined by their differential equations as follows: \begin{equation} \frac{dP_a}{dt}=-\mu_a P_a + \lim _{x\rightarrow 0} \left( \alpha \frac{\partial^2 f}{\partial x^2}-\beta \frac{\partial f}{\partial x}\right) \end{equation} \begin{equation} \frac{dP_b}{dt}=-\mu_b P_b + \lim _{x\rightarrow N} \left( \alpha \frac{\partial^2 f}{\partial x^2}-\beta \frac{\partial f}{\partial x}\right) \end{equation}

Actually I am only interested in fundamental solutions of the PDE, I am not sure if the presence of dirac function affects the two fundamental solutions $e^{\xi_i}$ given that $\xi_i$ are roots of the associated characteristic equation.

Imane_07
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    Since the support of the Dirac Delta is ${0}$, one can find homogeneous solutions in the regions $x<1$, $1<x<N-1$ and $N-1<x$. Then, apply conditions on continuity and discontinuity of the first partial derivatives across these sub-regions. – Mark Viola Jul 15 '16 at 20:07
  • I suppose that applying conditions on continuity would help only to find constants and I don't know if the presence of dirac functions affects fundamental solutions. – Imane_07 Jul 15 '16 at 20:45
  • Yes, the presence of the Deltas certainly affects the answer as it prescribes the discontinuity of the first partial with respect to $x$. – Mark Viola Jul 15 '16 at 21:36
  • @Dr.MV : !!!!! the support of $\delta$ is any neighborhood $]-\epsilon,\epsilon[$ of $0$ – reuns Jul 15 '16 at 21:57
  • this is a convolution equation : $(\partial_t\delta - \alpha\partial_x\delta-\beta \partial_x^2\delta) \ast f = g$ and as every convolution equation you need to find the inverse filter $h$, such that $h \ast (\partial_t\delta - \alpha\partial_x\delta-\beta \partial_x^2\delta) = \delta$, then the solution is $f = h \ast (\partial_t\delta - \alpha\partial_x\delta-\beta \partial_x^2\delta) \ast f = h \ast g$. For doing this, you can compute the Fourier transform of the (tempered) distribution $\partial_t\delta - \alpha\partial_x\delta-\beta \partial_x^2\delta$ and do what AlexM said. – reuns Jul 15 '16 at 22:00
  • @user1952009: !!!!! I confess that I am not following you here: the support of $\delta$ (understood as a measure) is ${0}$. The same stays true if you view $\delta$ as a Schwartz distribution. Your comment doesn't even make sense: how could the support of something be any set of a family of sets? – Alex M. Jul 16 '16 at 08:26
  • @AlexM. $\delta$ is not supported at one point but on a neighborhood of one point. this is why $C^\infty_c$ (with their topology) functions are useful, because by continuity we can identify the value on the neighborhood with the value at the point. but when extending it outside $C^\infty_c$ you'll have many problems if you don't think to it as being supported on an open containing $0$. – reuns Jul 16 '16 at 13:34
  • @user1952009: With all due respect for your very valuable contributions on MSE, this time you've got it all wrong, and (strangely) you are also persevering in pushing things in that wrong direction. Even when viewing $\delta$ as a distribution, its support is still ${0}$, being the complement of the largest open set on which $\delta$ vanishes (and this largest open set is obviously $\Bbb R \setminus {0}$). – Alex M. Jul 16 '16 at 13:54
  • well, let's talk about $\delta'$ : will you say to your students that $\delta'(f)$ depends only on $f$ at $x= 0$, or that $f'(0)$ depends on the values on a neighborhood of $x=0$ ? That's why I say that identifying ${0}$ with $\lim \sup_{\epsilon \to 0}]-\epsilon,\epsilon[$ is really the point of the $C^\infty_c$ topology @AlexM. – reuns Jul 16 '16 at 14:00

2 Answers2

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\partiald{\,\mathrm{f}\pars{x,t}}{t} = \alpha\,\partiald[2]{\,\mathrm{f}\pars{x,t}}{x} - \beta\,\partiald{\,\mathrm{f}\pars{x,t}}{x}\ +\mu_{a}\,\mathrm{P}_{a}\pars{t}\delta\pars{x - 1} + \mu_{b}\,\mathrm{P}_{b}\pars{t}\delta\pars{x - N + 1}}$.

Note that \begin{align} &\pars{\partiald{}{t} - {\beta^{2} \over 4\alpha}} \,\mathrm{f}\pars{x,t} = \alpha\,\pars{% \partiald{}{x} - {\beta \over 2\alpha}}^{2}\,\mathrm{f}\pars{x,t} + \mu_{a}\,\mathrm{P}_{a}\pars{t}\delta\pars{x - 1} + \mu_{b}\,\mathrm{P}_{b}\pars{t}\delta\pars{x - N + 1}\end{align}

With $\ds{\,\mathrm{F}\pars{x,t} \equiv \exp\pars{-\,{\beta \over 2\alpha}\,x - {\beta^{2} \over 4\alpha}\,t}\,\mathrm{f}\pars{x,t}}$: \begin{align} \partiald{\,\mathrm{F}\pars{x,t}}{t} & = \alpha\,\partiald[2]{\,\mathrm{F}\pars{x,t}}{x} + \,\mathrm{g}_{a}\pars{t}\delta\pars{x - 1} + \,\mathrm{g}_{b}\pars{t}\delta\pars{x - N + 1}\tag{1} \\[4mm] & \mbox{where}\ \left\lbrace\begin{array}{rcl} \ds{\,\mathrm{g}_{a}\pars{t}} & \ds{=} & \ds{\mu_{a}\exp\pars{-\,{\beta \over 2\alpha}}\exp\pars{-\,{\beta^{2} \over 4\alpha}\,t}}\,\mathrm{P}_{a}\pars{t} \\[2mm] \ds{\,\mathrm{g}_{a}\pars{t}} & \ds{=} & \ds{\mu_{b}\exp\pars{-\,{\beta \over 2\alpha}\,\bracks{N - 1}}\exp\pars{-\,{\beta^{2} \over 4\alpha}\,t}}\,\mathrm{P}_{b}\pars{t} \end{array}\right. \end{align}
Now, I'll 'take' Laplace Transform in both sides of $\pars{1}$: \begin{align} -\,\mathrm{F}\pars{x,0} + s\,\hat{\,\mathrm{F}}\pars{x,s} & = \alpha\,\partiald[2]{\hat{\,\mathrm{F}}\pars{x,s}}{x} + \hat{\,\mathrm{g}}_{a}\pars{s}\delta\pars{x - 1} + \hat{\,\mathrm{g}}_{b}\pars{s}\delta\pars{x - N + 1} \end{align} which leads to \begin{align} \pars{\partiald[2]{}{x} - {s \over \alpha}}\hat{\,\mathrm{F}}\pars{x,s} & = -\,{1 \over \alpha}\bracks{\,\mathrm{F}\pars{x,0} + \hat{\,\mathrm{g}}_{a}\pars{s}\delta\pars{x - 1} + \hat{\,\mathrm{g}}_{b}\pars{s}\delta\pars{x - N + 1}} \end{align}
In terms of the Green's Function $\ds{\,\mathrm{G}\pars{s,x,x'}}$ the solution, for $\ds{\hat{\,\mathrm{F}}\pars{x,s}}$, is written as \begin{align} \hat{\,\mathrm{F}}\pars{x,s} & = \varphi\pars{x,s} \\[3mm] & -\,{1 \over \alpha}\int_{-\infty}^{\infty}\,\mathrm{G}\pars{s,x,x'}\bracks{\,\mathrm{F}\pars{x',0} + \hat{\,\mathrm{g}}_{a}\pars{s}\delta\pars{x' - 1} + \hat{\,\mathrm{g}}_{b}\pars{s}\delta\pars{x' - N + 1}}\,\dd x' \\[8mm] & = \varphi\pars{x,s} - {1 \over \alpha}\bracks{\hat{\,\mathrm{g}}_{a}\pars{s}\,\mathrm{G}\pars{s,x,1} + \hat{\,\mathrm{g}}_{b}\pars{s}\,\mathrm{G}\pars{s,x,N - 1}} \\[4mm] & - {1 \over \alpha}\int_{-\infty}^{\infty}\,\mathrm{G}\pars{s,x,x'} \,\mathrm{F}\pars{x',0}\,\dd x' \end{align} $\ds{\varphi\pars{x,s}}$ satisfies $\ds{\pars{\partiald[2]{}{x} - {s \over \alpha}}\varphi\pars{x,s} = 0}$ and the $\ds{x}$-boundary conditions of $\ds{\hat{\,\mathrm{F}}\pars{x,s}}$ ( lets assume, for example, that it occurs at $\ds{x = 0}$ ). $$ \pars{\partiald[2]{}{x} - {s \over \alpha}}{\,\mathrm{G}}\pars{s,x,x'} = \delta\pars{x - x'}\,,\qquad\,\mathrm{G}\pars{s,0,x'} = 0 $$ The general solution is given by $$ \,\mathrm{G}\pars{s,x,x'} = \left\lbrace\begin{array}{lcl} \ds{0} & \mbox{if} & \ds{x < x'} \\[2mm] \ds{-\root{\alpha \over s}\sinh\pars{\root{s \over \alpha}\bracks{x - x'}}} & \mbox{if} & \ds{x > x'} \end{array}\right. $$
Felix Marin
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  • Thank you for your help. I would like to know if I can get the function in terms of $s$ at some specific points in the transient solution without considering the whole solution in terms of $x$. – Imane_07 Jul 16 '16 at 20:27
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I am not sure whether you will find this helpful, but I would attack this problem as follows.

First step

First, I would try to get rid of the differential operators in $x$ by performing a partial Fourier transform, i.e. a Fourier transform only in $x$ and not in $t$. This would convert these differential operators into algebraic monomials.

Concretely, if we use the tilde to denote the partial Fourier transform

$$\tilde f (t,y) = \frac 1 {\sqrt {2 \pi}} \int \limits _{-\infty} ^\infty \Bbb e ^{- \Bbb i y x} f(t,x) \ \Bbb d x$$

(in the sense of distributions), then

$$\widetilde {\frac {\partial f} {\partial x}} (t,y) = - \Bbb i y \tilde f (t,y), \quad \widetilde {\frac {\partial^2 f} {\partial x^2}} (t,y) = - y^2 \tilde f (t,y), \quad \widetilde {\delta_a} (y) = \frac 1 {\sqrt {2 \pi}} \Bbb e ^{- \Bbb i y a}$$

(where $\delta_a (x) = \delta (x-a)$), so (using that $\widetilde {\frac {\partial f} {\partial t}} = \frac {\partial \tilde f} {\partial t}$) this Fourier transform would transform the initial equation into

$$\frac {\partial \tilde f} {\partial t} (t,y) = (- \alpha y^2 + \beta \Bbb i y) \tilde f (t,y) + \frac 1 {\sqrt {2 \pi}} \mu_a P_a (t) \Bbb e ^{- \Bbb i y} + \frac 1 {\sqrt {2 \pi}} \mu_b P_b (t) \Bbb e ^{- \Bbb i y (N-1)}$$

which is an inhomogeneous differential equation of order $1$ that you can solve with the standard tool.

Second step

To solve the above equation, first treat the homogeneous case, thus considering the equation

$$\frac {\partial \tilde f} {\partial t} (t,y) = (- \alpha y^2 + \beta \Bbb i y) \tilde f (t,y) .$$

This has the solution

$$\tilde f (t,y) = C \Bbb e ^{(- \alpha y^2 + \beta \Bbb i y) t}$$

where $C>0$ is some arbitrary constant.

Next, treat the inhomogeneous equation by considering $C$ to be a function $C(t)$, and not a constant anymore, so

$$\tilde f (t,y) = C(t) \Bbb e ^{(- \alpha y^2 + \beta \Bbb i y) t} .$$

Replacing this formula for $\tilde f$ into the inhomogeneous equation leads to the new differential equation

$$C'(t) = \Bbb e ^{- (- \alpha y^2 + \beta \Bbb i y) t} \left( \frac 1 {\sqrt {2 \pi}} \mu_a P_a (t) \Bbb e ^{- \Bbb i y} + \frac 1 {\sqrt {2 \pi}} \mu_b P_b (t) \Bbb e ^{- \Bbb i y (N-1)} \right)$$

which can be integrated directly to yield

$$C(t) = \int \limits _{t_0} ^t \Bbb e ^{- (- \alpha y^2 + \beta \Bbb i y) s} \left( \frac 1 {\sqrt {2 \pi}} \mu_a P_a (s) \Bbb e ^{- \Bbb i y} + \frac 1 {\sqrt {2 \pi}} \mu_b P_b (s) \Bbb e ^{- \Bbb i y (N-1)} \right) \Bbb d s$$

with $t_0$ arbitrary in the correct domain of definition required by $f$ (that you do not give, so I shall assume $f$ to be defined on $\Bbb R^2$ - in which case $t_0 \in \Bbb R$). Plugging this back in the formula of $\tilde f (t,y)$ yields

$$\tilde f (t,y) = \Bbb e ^{(- \alpha y^2 + \beta \Bbb i y) t} \ \int \limits _{t_0} ^t \Bbb e ^{- (- \alpha y^2 + \beta \Bbb i y) s} \left( \frac 1 {\sqrt {2 \pi}} \mu_a P_a (s) \Bbb e ^{- \Bbb i y} + \frac 1 {\sqrt {2 \pi}} \mu_b P_b (s) \Bbb e ^{- \Bbb i y (N-1)} \right) \Bbb d s .$$

Third step

To obtain $f$, perform an inverse Fourier transform in the $y$ variable, obtaining

$$f(t,x) = \frac 1 {\sqrt {2 \pi}} \int \limits _{-\infty} ^\infty \Bbb e ^{\Bbb i x y} \tilde f (t,y) \ \Bbb d y = \\ \frac 1 {2 \pi} \int \limits _{-\infty} ^\infty \int \limits _{t_0} ^t e ^{\Bbb i x y} \Bbb e ^{(- \alpha y^2 + \beta \Bbb i y) t} \ \Bbb e ^{- (- \alpha y^2 + \beta \Bbb i y) s} \left( \mu_a P_a (s) \Bbb e ^{- \Bbb i y} + \mu_b P_b (s) \Bbb e ^{- \Bbb i y (N-1)} \right) \Bbb d s \ \Bbb d y.$$

Maybe this can be further slightly simplified; I do not know if it answers your question in the way you were hoping for, but if not it can at least suggest an approach. Not having concrete formulae for $P_a$ and $P_b$ necessarily leaves things in a somewhat unfinished, generic state.

Alex M.
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