Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have $$u_t = a(x)u_{xx},\; x\geq 0$$ and initial value $$u(0,x)=u_0(x)$$ in the case of $a(x)=x^2$ with a change of variables $y=ln(x)$ it can be translated to the constant coefficient equivalent equation for $v(t,y)$. Is that the only change of variables that is so straightforward or there are other examples for polynomial $a(x)$? I can't seem to find a way to transfer to constant coefficients if $a(x)=x^{\eta}$ for $\eta>0$ and not equal to a particular value of $2$. Is that because I did not look hard enough or it is in fact not trivial to transfer for an arbitrary power?

share|cite|improve this question

1 Answer 1

up vote 2 down vote accepted

I think this question is base on existence of solution of a degenerate pde with change of variables.

This seem to be related with the PDE issues, however, in fact, this is just related with the oDE issues.

If you solve $u_t=a(x)u_{xx}$ by separation of variables, you will get $\begin{cases}\dfrac{T'(t)}{T(t)}=\lambda\\a(x)X''(x)-\lambda X(x)=0\end{cases}$ , the difficulties will naturally focus on $a(x)X''(x)-\lambda X(x)=0$ . When $a(x)=x^2$ , the ODE is in fact belongs to a Cauchy–Euler equation. Applying the change of variables $y=\ln x$ it will be translated to $Y''(y)-Y'(y)-\lambda Y(y)=0$ .

Since it has a fact that when the change of variables of a PDE are only act on the independent variables and the changings of variables are independent between the independent variables, then acting the change of variables on the ODEs obtained from the separation of variables will be equivalent to acting the same change of variables of the PDE and then performing separation of variables. So you can discover that when applying the change of variables $y=\ln x$ on $u_t=x^2u_{xx}$ you will get $u_t=u_{yy}-u_y$ and then performing separation of variables you will get the corresponding result $\begin{cases}\dfrac{T'(t)}{T(t)}=\lambda\\Y''(y)-Y'(y)-\lambda Y(y)=0\end{cases}$ .

However when $a(x)=x^{\eta}$ for $\eta>0$ and not equal to $2$ , then $x^{\eta}X''(x)-\lambda X(x)=0$ no longer belongs to a Cauchy–Euler equation. It relates to an ODE of the form instead. The form of the solution of the ODE will be more complicated. So it is impossible to transform $u_t=x^{\eta}u_{xx}$ to the related constant coefficients linear PDE by applying the change of variables $y=\ln x$ .

In fact solving $a(x)X''(x)-\lambda X(x)=0$ for the general polynomial $a(x)$ is often a complicated act. The following list some special cases of $a(x)$ :

$a(x)$ is a monomial of any degree (this question): related to

$a(x)$ is a quadratic polynomial of complete square: related to

$a(x)$ is a quadratic polynomial of non-complete square: related to Gaussian hypergeometric equation

$a(x)$ is a complete square of quadratic polynomial: related to or

$a(x)$ is a binomial contains one of term that of the degree $2$: related to

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.