Converting a PDE into the heat equation

Question

I am looking at a PDE ${v_t}\left( {t,y} \right) = 4y{v_{yy}}\left( {t,y} \right) + 2{v_y}\left( {t,y} \right)$ defined for $\left( {t,y} \right) \in (\pi ,\infty ) \times (0,\infty )$. I think I should be able to transform this into the heat equation, but I am unable to think up of a transformation that would let me get rid of the y in $4y{v_{yy}}\left( {t,y} \right)$, without introducing more variables in the coefficients. The $\pi$ in the domain suggests that the transformation for the first variable at least would involve sin. Any suggestions for what kind of transformation I should perhaps try? Grateful for a nudge in any direction actually, I am stuck

Answer 1

There are two things in this equation that separate it from a heat equation:

1. The coefficient in front of $v_{yy}$ depending on $y$. The usual heat equation has constant coefficients.
2. The term $v_y$. The heat equation does not have any first order spatial derivatives.

We can deal with each thing separately.

To deal with the coefficient: changing coefficients in differential equations can usually be done by rescaling. We will define a new function $w$ such that $$w(t,y)=v(\lambda t, \mu y),$$ for $\lambda$ and $\mu$ to be determined. First we compute the relevant time and space derivatives for this new function:

• $w_t=\lambda v_t$.
• $w_y=\mu v_y$.
• $w_{yy}=\mu v_{yy}$.

Now we plug these into your equation to obtain $$w_{t} = \frac{\lambda}{\mu^2}4yw_{yy} + \frac{\lambda}{\mu}2w_{y},$$ so we can choose $\mu=\lambda=4y$, and the equation becomes $$w_{t} = w_{yy} + 2w_{y}.$$ Much nicer!

To deal with the first spatial derivative: The equation we just arrived can be written as $$w_{t} - 2w_{y} = w_{yy}.$$ You might have seen this type of equation before, it's a transport equation with diffusion. The left hand side is often called a transport operator, which causes the solution to travel in time. If the equation were just $$w_{t} - 2w_{y} = 0,$$ then the solution would simply transport the initial data $w(0,y)$ horizontally.

This suggests a way to deal with the extra term: we can change the spatial coordinates again to make the solution move in time, and this should counteract the transport term. I.e. let's define $$u(t,y) = w(t, y+ct)$$ and choose $c$ later. We compute derivatives again:

• $u_{t} = w_t + cw_y$.
• $u_{y} = w_y$.
• $u_{yy} = w_{yy}$.

If we pick $c=-2$ and plug these into the equation, we get $$u_{t} = u_{yy},$$ our beloved heat equation.

I hope this helps!