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I don't have much knowledge in solving PDEs other than ones that I'm given in class. I was attempting a physics problem in which I came upon this PDE which is slightly different to the regular wave equation which is well known in how to solve: $$\frac{\partial y}{\partial t} = \frac{1}{x^3}\frac{\partial}{\partial x}\left(x^3\frac{\partial y}{\partial x}\right).$$ I have these boundary conditions of $y (t = 0, x) = y_0$ and $y(t, x=X) = 0$. Could anyone assist me in finding a general solution for $y(t, x)$?

Physics_Learner
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    You can separate variables: $y(x,t)=X(x)T(t)$. Substitute into the equation to find $\frac{T'}{T}=\frac{\partial_x(x^3X')}{x^3X}$. So we have $T'/T=\text{const}:=-k^2$ meaning $T(t)$ are exponential or sinusoidal, and $-k^2x^2X=3xX'+x^2X''$. After the transformations $\zeta=X/x$ and $\chi=kx$, we find Bessel's equation of order one, thus $\zeta=J_1$ and finally $X(x)=x^{-1}J_1(kx)$. – Sal Jun 18 '21 at 22:25
  • Here $k$ is an unknown constant right? Then how would we know what $J_1 (kx)$ is? – Physics_Learner Jun 18 '21 at 22:42
  • $J_1$ is a Bessel function of the first kind. $k$ is an unknown constant. Over what region are you solving the equation? Specifically, is the origin included in this region? – Sal Jun 18 '21 at 23:28

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Looks similar to the diffusion equation with spherical symmetry. We can separate variables: let

$$y(x,t)=X(x)T(t)$$

Substituting into the differential equation we find

$$\frac{T'}{T}=\frac{\partial_x(x^3X')}{x^3X}=\text{const}:=-k^2$$

Here $k$ is our separation constant. The equation for $T(t)$ is simple, yielding

$$ T(t)=A e^{-k^2t} $$

Where the integration constant is $A$. The equation for $X(x)$ is

$$ -k^2x^2X=3xX'+x^2X'' $$ After the transformations $\zeta=X/x$ and $\chi=kx$, we find Bessel's equation of order one, so $\zeta=B J_1+C Y_1$ thus

$$ X(x)=x^{-1}\left(B J_1(kx)+C Y_1(kx) \right) $$

Where $J_1$ is a Bessel function of the first kind and $Y_1$ is a Bessel function of the second kind and $B$ and $C$ are integration constants. Requiring $y$ be finite at the origin sets $C=0$, and the general solution is

$$ y(x,t)=\sum_k B_k e^{-k^2t} x^{-1} J_1(kx) $$

Now the initial and boundary conditions. With $y(t,x_0)=0$ we find

$$ 0=\sum_k B_k e^{-k^2t} x_0^{-1} J_1(kx_0) $$

To hold for all $t$, we must have $J_1(kx_0)=0$, so $kx_0$ are roots of $J_1$. We can rewrite the sum with an integer index by denoting the $n$th root of $J_1$ as $\lambda_n$

$$ y(x,t)=\sum_{n=1}^\infty B_n e^{-\lambda_n^2t/x_0^2} x^{-1} J_1(\lambda_nx/x_0) $$

The $B_n$ are found using the initial condition $y(x,0)=y_0(x)$

$$ y_0=\sum_{n=1}^\infty B_n x^{-1} J_1(\lambda_nx/x_0) $$

$$ x^2 y_0 =\sum_{n=1}^\infty B_n x J_1(\lambda_nx/x_0) $$

Exploiting the orthogonality relation of the Bessel functions we have

$$ B_n=\frac{2x_0}{J_2(\lambda_n)^2}\int\limits_0^1 d\eta \ \eta^2 J_1(\eta \lambda_n) y_0(\eta x_0) $$

Sal
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