Can some one help me solve the following PDE with the given intial and boundary conditions? $\gamma t\frac{\partial^{2}f}{\partial x^{2}}=t\frac{\partial f}{\partial t}-\alpha f$

Initial condition: $f(x,t=0)=0$

Outer Boundary Condition: $f(x\rightarrow\infty,t)=0$

Inner Boundary Condition: $\left.\frac{\partial f}{\partial x}\right|_{x=0}=c\left.\frac{\partial f}{\partial t}\right|_{x=0}-1$

I am only interested in the solution at the inner boundary. i.e.


Other information:




My Skill level: I have solved similar heat equation problems using Laplace transforms but none for the heat equation with the term $\alpha\frac {f}{t}$ Any help is greatly appreciated!

EDIT I have solved the problem using COMSOL and it is giving me results which are close to my physical problem. However I am interested in getting an analytic solution.

  • $\begingroup$ Well, any solution to this will necessarily be weak, since the equation doesn't even make sense at $t=0$. Can you write down the appropriate weak formulation? $\endgroup$
    – Ian
    Commented Oct 24, 2014 at 22:19
  • $\begingroup$ Thanks Ian I am seeking the solution for t>0 $\endgroup$ Commented Oct 24, 2014 at 23:53
  • $\begingroup$ Exact solution when c=0 $f\left(x=0,t>0\right)=\frac{\Gamma\left(1-\alpha\right)}{\Gamma\left(3/2-\alpha\right)}\sqrt{\gamma t}$ $\endgroup$ Commented Oct 25, 2014 at 0:04
  • $\begingroup$ Exact solution when c=0,$\alpha=0$ $f\left(x=0,t>0\right)=\frac{2}{\sqrt{\pi}}\sqrt{\gamma t}$ $\endgroup$ Commented Oct 25, 2014 at 0:06
  • $\begingroup$ First, in order to avoid the ambiguity at $t=0$, I suggest that the equation be written as : $$t \gamma\frac{\partial^{2}f}{\partial x^{2}}=t \frac{\partial f}{\partial t}-\alpha f$$ Second, the condition $$\left.\frac{\partial f}{\partial x}\right|_{x=0}=c\frac{\partial f}{\partial t}-1$$ is impossible because the term on the left is not function of $x$, while the term on the right is function of $x$, because $\frac{\partial f}{\partial t}$ is function of $x$ $\endgroup$
    – JJacquelin
    Commented Oct 25, 2014 at 8:06

1 Answer 1


Assume $\alpha\neq0$ for the key cases:

Of course use separation of variables:

Let $f(x,t)=X(x)T(t)$ ,

Then $\gamma tX''(x)T(t)=tX(x)T'(t)-\alpha X(x)T(t)$

$\gamma tX''(x)T(t)=X(x)(tT'(t)-\alpha T(t))$

$\dfrac{tT'(t)-\alpha T(t)}{\gamma tT(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=\dfrac{\alpha}{t}-\gamma s^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)t^\alpha e^{-\gamma ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore f(x,t)=\int_0^\infty C_1(s)t^\alpha e^{-\gamma ts^2}\sin xs~ds+\int_0^\infty C_2'(s)t^\alpha e^{-\gamma ts^2}\cos xs~ds$

Which is automatically satisfied the condition $f(x,0)=0$

$\left.\dfrac{\partial f}{\partial x}\right|_{x=0}=c\left.\dfrac{\partial f}{\partial t}\right|_{x=0}-1$ :

$\int_0^\infty sC_1(s)t^\alpha e^{-\gamma ts^2}~ds=c\alpha\int_0^\infty C_2'(s)t^{\alpha-1}e^{-\gamma ts^2}~ds-c\gamma\int_0^\infty s^2C_2'(s)t^\alpha e^{-\gamma ts^2}~ds-1$

$\int_0^\infty sC_1(s)t^\alpha e^{-\gamma ts^2}~ds+c\gamma\int_0^\infty s^2C_2'(s)t^\alpha e^{-\gamma ts^2}~ds-c\alpha\int_0^\infty t^{\alpha-1}e^{-\gamma ts^2}~d(C_2(s))=-1$

$\int_0^\infty sC_1(s)t^\alpha e^{-\gamma ts^2}~ds+c\gamma\int_0^\infty s^2C_2'(s)t^\alpha e^{-\gamma ts^2}~ds-c\alpha\left[C_2(s)t^{\alpha-1}e^{-\gamma ts^2}\right]_0^\infty+c\alpha\int_0^\infty C_2(s)~d(t^{\alpha-1}e^{-\gamma ts^2})=-1$

$\int_0^\infty sC_1(s)t^\alpha e^{-\gamma ts^2}~ds+c\gamma\int_0^\infty s^2C_2'(s)t^\alpha e^{-\gamma ts^2}~ds-2c\alpha\gamma\int_0^\infty sC_2(s)t^\alpha e^{-\gamma ts^2}~ds+c\alpha C_2(0)=-1$

$t^\alpha\int_0^\infty(sC_1(s)+c\gamma s^2C_2'(s)-2c\alpha\gamma sC_2(s))e^{-\gamma ts^2}~ds=-c\alpha C_2(0)-1$

$\int_0^\infty(sC_1(s)+c\gamma s^2C_2'(s)-2c\alpha\gamma sC_2(s))e^{-\gamma ts^2}~ds=-(c\alpha C_2(0)+1)t^{-\alpha}$

$sC_1(s)+c\gamma s^2C_2'(s)-2c\alpha\gamma sC_2(s)=-\dfrac{2\gamma^\alpha(c\alpha C_2(0)+1)s^{2\alpha-1}}{\Gamma(\alpha)}$ (according to http://en.wikipedia.org/wiki/Gaussian_integral#Integrals_of_similar_form)

$C_1(s)=-c\gamma sC_2'(s)+2c\alpha\gamma C_2(s)-\dfrac{2\gamma^\alpha(c\alpha C_2(0)+1)s^{2\alpha-2}}{\Gamma(\alpha)}$

$\therefore f(x,t)=\int_0^\infty t^\alpha e^{-\gamma ts^2}(C_2'(s)\cos xs-(c\gamma sC_2'(s)-2c\alpha\gamma C_2(s))\sin xs)~ds-\dfrac{2\gamma^\alpha(c\alpha C_2(0)+1)}{\Gamma(\alpha)}\int_0^\infty t^\alpha s^{2\alpha-2}e^{-\gamma ts^2}\sin xs~ds$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .