I am searching how to solve the PDE below but I can not seem to find a decent example online. My major did not focus much in solving PDEs so I feel very deficient.

I know how to solve for the steady state - but I am wondering if I can solve the following analytically, not numerically.

\begin{gather*} \frac{\partial F(x, t)}{\partial t} = -g(x)F(x, t) -A\frac{\partial F(x, t)}{\partial x}-B\frac{\partial^{2}F(x, t)}{\partial x^{2}} \\ \\ IC: F(x, 0) = 1 \end{gather*}

I thought about doing a Fourier transform but I end up with a convolution in freq space - and that would be hard to handle. I also know that this is an advection - diffusion - convection PDE but its not specific enough to return any nice Google searches.

Any suggestion or advice would be greatly appreciated.

Thank you in advance!

Follow up:

Consider Case 1, where function g(t) is only a function of time.

\begin{gather*} g(t) : = \frac{\alpha \left(\frac{t}{\beta }\right)^{\alpha -1}}{\beta } \\ \\ \frac{\partial F(x, t)}{\partial t} = -\left(\frac{\alpha \left(\frac{t}{\beta }\right)^{\alpha -1}}{\beta }\right)F(x, t) -A\frac{\partial F(x, t)}{\partial x}-B\frac{\partial^{2}F(x, t)}{\partial x^{2}} \\ \\ \frac{\partial\tilde{F}(\omega , t)}{\partial t} = -\left(\frac{\alpha \left(\frac{t}{\beta }\right)^{\alpha -1}}{\beta }\right)\tilde{F}(\omega , t) -A(i2\pi \omega )\tilde{F}(\omega , t)+B(i2\pi \omega )^{2}\tilde{F}(\omega , t) \\ \\ \frac{\partial\tilde{F}(\omega , t)}{\partial t} = \left[-\left(\frac{\alpha \left(\frac{t}{\beta }\right)^{\alpha -1}}{\beta }\right)-A(i2\pi \omega )+B(i2\pi \omega )^{2}\right]\tilde{F}(\omega , t) \\ \\ \mbox{The DE is of the form:} \\ y^{\prime}+ay = 0\mbox{ so when I solve I get: }\ \\ \\ \tilde{F}(\omega , t) = Ce^{\left(\frac{t}{\beta }\right)^{\alpha }+(A(i2\pi \omega )-B(i2\pi \omega )^{2})t} \\ \\ \mbox{-At this point I don't know how to apply the IC but if I assume C=1,} \\ \mbox{and take the inverse Fourier - I get the solution below - which looks good.} \\ \\ F(x, t) = \frac{e^{\left(\frac{t}{\beta }\right)^{\alpha }-\frac{(x+At)^{2}}{4Bt}}}{2\sqrt{\pi }\sqrt{Bt}} \end{gather*}

When I plug in fixed values for all variables - I get the same values but opposite signs. This is somewhat good news but still I am a little confused.

Additional evidence

Here is the Mathematica ourput for the two derivative terms. Note the second derivative becomes positive.


In[2]:= FourierTransform[-A*D[f[t, x], x], x, w, FourierParameters -> {0, -2 Pi}]

Out[2]= -2 i A Pi w FourierTransform[f[t, x], x, w]


In[1]:= FourierTransform[-B*D[f[t, x], x, x], x, w, FourierParameters -> {0, -2 Pi}]

Out[1]= 4 B Pi^2 w^2 FourierTransform[f[t, x], x, w]

I kept everything consistent with Mathematica; therefore, I am confident my signs should be right.

I really hope someone can shed some light into this!!! I am almost depressed about it.

  • 2
    $\begingroup$ Have you tried Laplace transform of $F(x, t)$ for fixed $x$? That should convert the PDE into an ODE. $\endgroup$
    – Budenn
    Jul 23, 2015 at 17:13
  • $\begingroup$ No, I have never used Laplace transforms. $\endgroup$
    – Edv Beq
    Jul 23, 2015 at 17:20
  • 1
    $\begingroup$ By applying Laplace transform w.r.t. $t$, denoting the result of transform of $F(x, t)$ $\tilde{F}(x, s)$ and using the initial condition, the PDE becomes a linear second-order ODE with variable coefficients $s \tilde{F}(x, s) - 1 = -g(x) \tilde{F}(x, s) + A \frac{\partial \tilde{F}(x, s)}{\partial x} + B\frac{\partial^2 \tilde{F}(x, s)}{\partial x^2}$. $\endgroup$
    – Budenn
    Jul 23, 2015 at 17:46
  • $\begingroup$ I see that and that's pretty cool. Thank you. I will make a follow up post if I can solve it this way. $\endgroup$
    – Edv Beq
    Jul 23, 2015 at 18:50
  • 1
    $\begingroup$ Have you tried directly verifying your solution by plugging it in to the equation? $\endgroup$ Jul 24, 2015 at 21:08


You must log in to answer this question.