Solving the homogeneous heat equation using the method of fundamental solutions

Question

I'm trying to solve the PDE $u_t=Du_{xx}$ using the method of fundamental solutions. I've used the ansatz $u(x,t)=t^{\alpha/2}f(x/\sqrt{t})$ and this has given me the following ODE: $$\frac{\alpha}{2}f(\phi)-\frac{1}{2}\phi f'(\phi)=Df''(\phi)$$ where $\phi=x/\sqrt{t}$.

I'm then told to let $\alpha=-1/2$, leaving me with the ODE $$-\frac{1}{4}f(\phi)-\frac{1}{2}\phi f'(\phi)=Df''(\phi)$$

I'm not sure how to solve this though.

Note: $D \in \mathbb{R}_{>0}$

Answer 1

Let's start from the top. Make the ansatz $$ u(x,t) = t^\alpha f\left(\frac{x}{2\sqrt{Dt}}\right) $$ Then after some massaging of the differential equation and substituting $\phi = x/(2\sqrt{Dt})$, you end up with $$ f''(\phi) + 2\phi f'(\phi) - 4\alpha f(\phi) = 0 $$ The form of the differential operator suggests the substitution $f(\phi) = e^{-\phi^2}h(\phi)$, which then gives $$ h''(\phi) - 2\phi h'(\phi) - 2(1+2\alpha)h(\phi) = 0 $$ This is a well-known differential equation called Hermite's equation. These solutions normally blow up extremely quickly for $\phi \rightarrow\pm \infty$, but for certain special values of $\alpha$ it will have polynomial solutions. Those values are $\alpha = -n-1/2, n\in \mathbb N$, and the polynomials are denoted by $H_n(\phi)$. These will be the only solutions for which $f(\phi)$ remains finite for all $\phi$. So, substituting back into the original expression, we find the fundamental solutions are $$ u(x,t) = A_n t^{-n-1/2}H_n\left(\frac{x}{2\sqrt{Dt}}\right)\exp\left(\frac{x^2}{4Dt}\right) $$ for some normalization constant $A_n$.

Answer 2

This answer was for the original version of the question asked

I would divide through by $y=f(\phi)$ first, and integrate w.r.t $y$:

$$\int\frac{y''}{y}dy=C\int 2\phi - 1 dy=Cy(2\phi-1)+G$$

Where $G \in \mathbb{R}$.

Then use a substitution, however I am not sure if the LHS integral has a closed form (exact) solution.

Answer 3

The DE can be rearranged to:

$$Df''(\phi)+\frac{1}{2}\phi f'(\phi)+\frac{1}{4}f(\phi)=0$$

Or in alternate notation:

$$Dy''+\frac{1}{2}\phi y'+\frac{1}{4}y=0$$

This has an Integrating Factor of

$$e^{\int \frac{1}{2D}\phi d\phi}$$

Evaluating the Integral, $I$;

$$I=\int \frac{1}{2D}\phi d\phi=\frac{\phi^2}{4D}+C$$

Hence I.F. is $Ae^{\frac{\phi^2}{4D}}$, where $A=e^C$

Multiplying the DE by the IF, and cancelling the $A$'s:

$$De^{\frac{\phi^2}{4D}}y''+\frac{1}{2}\phi e^{\frac{\phi^2}{4D}}y'+\frac{1}{4}e^{\frac{\phi^2}{4D}}y=0$$

Hence

$$(De^{\frac{\phi^2}{4D}}y')'=-\frac{1}{4}e^{\frac{\phi^2}{4D}}y$$

Integrating w.r.t. $\phi$, and taking terms without $\phi$ outside integral:

$$De^{\frac{\phi^2}{4D}}y'=-\frac{1}{4}y\int e^{\frac{\phi^2}{4D}} d\phi$$

$$\frac{y'}{y}=-\frac{1}{4De^{\frac{\phi^2}{4D}}}\int e^{\frac{\phi^2}{4D}} d\phi$$

Hence

$$\int\frac{y'}{y}dy=-\int[\frac{1}{4De^{\frac{\phi^2}{4D}}}\int e^{\frac{\phi^2}{4D}} d\phi] dy$$

$$\ln y=-[\frac{1}{4De^{\frac{\phi^2}{4D}}}\int e^{\frac{\phi^2}{4D}} d\phi] \int dy=[\frac{y+C}{4De^{\frac{\phi^2}{4D}}}\int e^{\frac{\phi^2}{4D}} d\phi] $$

The Integral has no exact solution afaik.