# How do I solve a PDE with multiple Dirac functions?

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: $$\frac{dP_a}{dt}=-\mu_a P_a + \lim _{x\rightarrow 0} \left( \alpha \frac{\partial^2 f}{\partial x^2}-\beta \frac{\partial f}{\partial x}\right)$$ $$\frac{dP_b}{dt}=-\mu_b P_b + \lim _{x\rightarrow N} \left( \alpha \frac{\partial^2 f}{\partial x^2}-\beta \frac{\partial f}{\partial x}\right)$$

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.

• 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

## 2 Answers

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.

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

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

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 $$\,\mathrm{G}\pars{s,x,x'} = \left\lbrace\begin{array}{lcl} \ds{0} & \mbox{if} & \ds{x < x'} \\[2mm] \ds{-\root{\alpha \over s}\sinh\pars{\root{s \over \alpha}\bracks{x - x'}}} & \mbox{if} & \ds{x > x'} \end{array}\right.$$

• 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