Reducing a PDE to a dimensionless form with change of variables

Question

I am working through the following example to refresh my memory on how to use the chain rule when changing variables: Change of variables (PDE)

\begin{equation} \begin{split} \frac{\partial{V}}{\partial{t}} + S\frac{\partial{V}}{\partial{S}} + \frac{1}{2}S^2\frac{\partial^2{V}}{\partial{S}^2} &= V \\ \end{split} \end{equation}

by setting:

$V(S,t) = v(x(S),\tau(t))$

$x(S) = \ln(S)$

$\tau(t) = \frac{1}{2}(T-t)$

$v(x,\tau) = e^{-\frac{1}{2}x - \frac{9}{4}\tau}u(x,\tau)$

Using these I managed to reach to the following stage:

\begin{equation} \begin{split} \frac{\partial{v}}{\partial{\tau}}\frac{\partial{\tau}}{\partial{t}} + S \frac{\partial{v}}{\partial{x}}\frac{\partial{x}}{\partial{S}}-v + \mbox{the second order term} = 0 \end{split} \end{equation}

I am unable to follow how to get $\frac{1}{2}S^2\frac{\partial^2{V}}{\partial{S}^2}$ in to the following form: $\frac{1}{2}S^2\left( \frac{\partial^2{x}}{\partial{S^2}}\frac{\partial{v}}{\partial{x}}+ \left(\frac{\partial{x}}{\partial{S}}\right)^2\frac{\partial^2{v}}{\partial{x^2}} \right)$

Could someone please explain how to get the second order term into that form?

score 1 · Accepted Answer · answered Feb 01 '16 at 18:50

1

$$ \partial_S = \frac{\partial x}{\partial S}\partial_x = \frac{1}{S}\partial_x $$ To get second derivitve you apply the above twice so you get $$ \partial_{SS} = \frac{1}{S}\partial_x\left(\frac{1}{S}\partial_x\right) = -\frac{1}{S^3}\frac{\partial S}{\partial x}\partial_x + \frac{1}{S^2}\partial_{xx} $$ Using the first derivarive with $S$ you find that $S^{-1}\partial_xS = 1$ you get $$ \partial_{SS} = -\frac{1}{S^2}\partial_x + \frac{1}{S^2}\partial_{xx} $$

answered Feb 01 '16 at 18:50

Chinny84

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Where did the minus sign come from in the second equation? – Gabor Bakos Feb 01 '16 at 18:56
The full expansion of the derivatives $$\partial_x s^{-1}\partial_x = \partial_x\left(s^{-1}\right) \partial_x + s^{-1}\partial_{xx}$$ the minus comes from taking the derivative of $s^{-1}$. – Chinny84 Feb 01 '16 at 19:00
If you look at the result though there is no minus sign involved with that term. So not sure how to arrive to that solution. – Gabor Bakos Feb 01 '16 at 19:05
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Compute the detivative $$\frac{\partial^2 x}{\partial S^2}$$ – Chinny84 Feb 01 '16 at 19:07
Not a problem. You did the hard part the rest was applying a method with book keeping. – Chinny84 Feb 01 '16 at 19:10
Let us continue this discussion in chat. – Gabor Bakos Feb 01 '16 at 19:43

Reducing a PDE to a dimensionless form with change of variables

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