Change of variables (PDE)

Often a partial differential equation can be reduced to a simpler form with a known solution by a suitable change of variables.

Change of variable for integral equations is discussed in Integration by substitution.

The article below discusses change of variable for PDEs in two ways:

1. By example
2. By giving the theory of the method

Technique explained by example

For example the following simplified form of the Black–Scholes PDE

${\frac {\partial V}{\partial t}}+{\frac {1}{2}}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}+S{\frac {\partial V}{\partial S}}-V=0.$

is reducible to the Heat equation

${\frac {\partial u}{\partial \tau }}={\frac {\partial ^{2}u}{\partial x^{2}}}$

by the change of variables:

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

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

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

$v(x,\tau )=\exp(-(1/2)x-(9/4)\tau )u(x,\tau )$

in these steps:

• Replace $V(S,t)$ by $v(x(S),\tau (t))$ and apply the chain rule to get

${\frac {1}{2}}(-2v(s,\tau )+2{\frac {\partial \tau }{\partial t}}{\frac {\partial v}{\partial \tau }}+S\left(\left(2{\frac {\partial x}{\partial S}}+S{\frac {\partial ^{2}x}{\partial S^{2}}}\right){\frac {\partial v}{\partial x}}+S\left({\frac {\partial x}{\partial S}}\right)^{2}{\frac {\partial ^{2}v}{\partial x^{2}}}\right)=0$

• Replace $x(S)$ and $\tau (t)$ by $\ln(S)$ and ${\frac {1}{2}}(T-t)$ to get

${\frac {1}{2}}(-2v(\ln(S),{\frac {1}{2}}(T-t))-{\frac {\partial v(\ln(S),{\frac {1}{2}}(T-t))}{\partial \tau }}+{\frac {\partial v(\ln(S),{\frac {1}{2}}(T-t))}{\partial x}}+{\frac {\partial ^{2}v(\ln(S),{\frac {1}{2}}(T-t))}{\partial x}}$

• Replace $\ln(S)$ and ${\frac {1}{2}}(T-t)$ by $x(S)$ and $\tau (t)$ and divide both sides by ${\frac {1}{2}}$ to get

$-2v-{\frac {\partial v}{\partial \tau }}+{\frac {\partial v}{\partial x}}+{\frac {\partial ^{2}v}{\partial x^{2}}}=0$

• Replace $v(x,\tau )$ by $\exp(-(1/2)x-(9/4)\tau )u(x,\tau )$ and divide through by $-\exp(-(1/2)x-(9/4)\tau )u(x,\tau )$ to yield the heat equation.

Advice on the application of change of variable to PDEs is given by mathematician J. Michael Steele:

"There is nothing particularly difficult about changing variables and transforming one equation to another, but there is an element of tedium and complexity that slows us down. There is no universal remedy for this molasses effect, but the calculations do seem to go more quickly if one follows a well-defined plan. If we know that $V(S,t)$ satisfies an equation (like the Black-Scholes equation) we are guaranteed that we can make good use of the equation in the derivation of the equation for a new function $v(x,t)$ defined in terms of the old if we write the old V as a function of the new v and write the new t and x as functions of the old t and S. This order of things puts everything in the direct line of fire of the chain rule; the partial derivatives ${\frac {\partial V}{\partial t}}$, ${\frac {\partial V}{\partial S}}$ and ${\frac {\partial ^{2}V}{\partial S^{2}}}$are easy to compute and at the end, the original equation stands ready for immediate use."

Technique in general

Suppose that we have a function $u(x,t)$ and a change of variables $x_{1},x_{2}$ such that there exist functions $a(x,t),b(x,t)$ such that

$x_{1}=a(x,t)$

$x_{2}=b(x,t)$

and functions $e(x_{1},x_{2}),f(x_{1},x_{2})$ such that

$x=e(x_{1},x_{2})$

$t=f(x_{1},x_{2})$

and furthermore such that

$x_{1}=a(e(x_{1},x_{2}),f(x_{1},x_{2}))$

$x_{2}=b(e(x_{1},x_{2}),f(x_{1},x_{2}))$

and

$x=e(a(x,t),b(x,t))$

$t=f(a(x,t),b(x,t))$

Suppose ${\mathcal {L}}$ is a differential operator such that

${\mathcal {L}}u(x,t)=0$

Then it is also the case that

${\mathcal {L}}v(x_{1},x_{2})=0$

where

$v(x_{1},x_{2})=u(e(x_{1},x_{2}),f(x_{1},x_{2}))$

and we operate as follows to go from ${\mathcal {L}}u(x,t)=0$ to ${\mathcal {L}}v(x_{1},x_{2})=0$:

• Apply the chain rule to ${\mathcal {L}}v(x_{1}(x,t),x_{2}(x,t))=0$ and expand out giving equation $e_{1}$.
• Substitute $a(x,t)$ for $x_{1}(x,t)$ and $b(x,t)$ for $x_{2}(x,t)$ in $e_{1}$ and expand out giving equation $e_{2}$.
• Replace occurrences of $x$ by $e(x_{1},x_{2})$ and $t$ by $f(x_{1},x_{2})$ to yield ${\mathcal {L}}v(x_{1},x_{2})=0$, which will be free of $x$ and $t$.