Converting a PDE into the heat equation

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

1 Answer

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!