How to simplify the following PDE by changing the variables I am looking forward to solving the following 1 D convection-advection equation by using the changing of variables.
$$ 4r \frac{∂^2T}{∂r^2} - 2γ \frac{∂T}{∂r}= θ \frac{∂T}{∂t}$$
I have tried to solve the equation by changing the variables as follow:
$$ T(r,t) = U(r,t) exp(C_1 r + C_2 t)$$
and find the values of $C_1$ and $C_2$ so that the PDE is simplified but I found this technique fails due to $r$ in the second derivative term. Can this equation be simplified by this technique?
 A: (I'm assuming that $\gamma$ and $\theta$ are constants.)
Instead of changing the dependent variable $T\mapsto U$, change the independent variable $r\mapsto x$. (I'll use subscripts for partial derivatives.) The chain rule says
$$T_r = T_x x_r$$
$$T_{rr} = (T_x)_r x_r + T_x x_{rr}$$
$$= \big((T_x)_x x_r\big)x_r + T_x x_{rr}$$
$$= T_{xx} x_r\!^2 + T_x x_{rr}$$
so your equation becomes
$$4r\big(T_{xx} x_r\!^2 + T_x x_{rr}\big) - 2\gamma\big(T_x x_r\big) = \theta T_t$$
$$= \big(4rx_r\!^2\big)T_{xx} + \big(4rx_{rr} - 2\gamma x_r\big)T_x$$
To simplify this, we want the first derivative to disappear, so
$$4rx_{rr} - 2\gamma x_r = 0$$
$$\frac{1}{x_r}x_{rr} = \frac{\gamma}{2r}$$
integrating with respect to $r$
$$\ln|x_r| = \frac\gamma2\ln|r| + C_1$$
$$x_r = C_2 r^{\gamma/2}$$
integrating again
$$x = \frac{C_2}{\gamma/2+1} r^{\gamma/2+1} + C_3$$
(In the special case $\gamma = -2$, this would instead be a logarithm: $x = C_2\ln|r| + C_3$.)
We can choose $C_3 = 0$ to simplify:
$$x = \frac{C_2}{\gamma/2+1}r^{\gamma/2+1}$$
$$r = \Bigg(\frac{\gamma/2+1}{C_2}x\Bigg)^{1/(\gamma/2+1)}$$
Now the original equation becomes
$$\big(4rx_r\!^2\big)T_{xx} = \theta T_t$$
$$= 4r\big(C_2\!^2r^\gamma\big)T_{xx}$$
$$= 4C_2\!^2r^{\gamma+1}T_{xx}$$
$$= 4C_2\!^2\Bigg(\frac{\gamma/2+1}{C_2}x\Bigg)^{(\gamma+1)/(\gamma/2+1)}T_{xx}$$
$$= \theta C_4x^{(\gamma+1)/(\gamma/2+1)}T_{xx}$$
(I choose the factor of $\theta$ for this new constant $C_4$ to cancel it from the other side of the equation:)
$$C_4x^{(\gamma+1)/(\gamma/2+1)}T_{xx} = T_t$$
Is this the kind of simplification you were looking for?
