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I am interested in solving the PDE

$$\frac{\partial}{\partial t} f(x,t) = c(x,t) \frac{\partial^2}{\partial x^2} f(x,t)$$

The goal is to solve this in $f$, given the boundary conditions $\frac{\partial}{\partial x} f(\pm \infty, t) = 0$ and the initial condition $f(x,0)=g(x)$ for some known $g$.

If $c(x,t)$ was constant, this would be a standard heat equation, with the solution being $f(x,t) = [g \star k_{ct}](x)$, where $k_{\alpha}(x)$ is the heat kernel $\frac{1}{\sqrt{4 \pi \alpha}}e^{-\frac{x^2}{4\alpha}}$ and $\star$ is the convolution operator.

However, since $c$ varies in both $x$ and $t$, I am not sure if an analytical solution exists. If not, then what is a reasonable analytical approximation to the solution $f$?

For example, one idea is to forget the dependency of $c$ on $x,t$... solve the heat equation, and only after that replace c with $c(x,t)$ in the solution form. So the approximate solution would be $\hat{f}(x,t) = \Big([g \star k_{ct}](x) \Big)_{|\,c=c(x,t)}$. Is this a reasonable idea? If not, are there more reasonable ideas for approximating $f$?

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    $\begingroup$ You could take a perturbative approach, especially if $c$ is dominated by a DC component. Then if $c(x,t)=c_0+ \epsilon c_1(x,t)$, then $f(x,t) = f_0(x,t)+ \epsilon f_1(x,t)$. The solution for $f_0$ is the usual, that for $f_1$ will be an inhomogeoneous solution to the constant-coeff. PDE. $\endgroup$ – Ron Gordon Feb 15 '17 at 0:57
  • $\begingroup$ Thank you Ron! Could you please elaborate a bit your answer. What is special about $c_1(x,t)$ (compared to the general form of $c(x,t)$) that makes solving $f_1$ easy? $\endgroup$ – Golabi Feb 15 '17 at 2:14
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    $\begingroup$ Because we have no idea how to solve for general $c(x,t)$, but we can solve for $f_1$ via an inhomogeneous, constant-coefficient PDE. I will illustrate below. $\endgroup$ – Ron Gordon Feb 15 '17 at 2:30
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We have

$$\frac{\partial}{\partial t} f(x,t) = c(x,t) \frac{\partial^2}{\partial x^2} f(x,t)$$

subject to initial boundary conditions on $f$, e.g., $f(x,0)=g(x)$. Write $c(x,t) = c_0 + \epsilon c_1(x,t)$. Then assume that the solution $f$ behaves as $f(x,t) = f_0(x,t)+\epsilon f_1(x,t)$, where $\epsilon^2=0$. Equating coefficients of $\epsilon$, we get two equations:

$$\frac{\partial}{\partial t} f_0(x,t) - c_0 \frac{\partial^2}{\partial x^2} f_0(x,t) = 0 \quad f_0(x,0)=g(x)$$

$$\frac{\partial}{\partial t} f_1(x,t) - c_0 \frac{\partial^2}{\partial x^2} f_1(x,t) = c_1(x,t) \frac{\partial^2}{\partial x^2} f_0(x,t) \quad f_1(x,0)=0$$

The first equation is the typical homogeoneous heat equation with a constant diffusivity. We incorporate the initial boundary condition here.

The second equation is an inhomogeoneous heat equation with a constant diffusivity. Note that the RHS is known. Also we impose zero initial conditions here.

The idea is that you split up the diffusivity term into a constant term and a smaller, non-constant term. The value of $\epsilon$ can be what you need it to be.

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  • $\begingroup$ Thank you Ron for your very helpful answer... the whole argument is now clear except one point. You said "write $c(x,t)=c_0+\epsilon c_1(x,t)$ and assume that the solution $f$ behaves as $f(x,t)=f_0(x,t)+\epsilon f_1(x,t)$". When does your "assume" hold? Specifically, if I knew $c(x,t)=c_0+\epsilon c_1(x,t)$ (call it A1) is a good assumption, when does it impliy that $f(x,t)=f_0(x,t)+\epsilon f_1(x,t)$ (call it A2) is satisfied? The solution $f$ is related to $c$ in a nonlinear way, so it is not quite clear if A1 implies A2. $\endgroup$ – Golabi Feb 15 '17 at 6:47
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    $\begingroup$ @Golabi: the assumption holds in most cases. It really doesn't hold when there is some essential singularity in the equation, which is not the case here. $\endgroup$ – Ron Gordon Feb 15 '17 at 13:28

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