How to solve this PDE equation?

Question

(I never had a PDE course unfortunately, but this problem came up from some physical problem.)

Solve for $g(x,t)$, where: $$\frac{\partial{g(x,t)}}{\partial t} + \frac{\partial}{\partial x}\left(g(x,t)\frac{-x}{a}\right)=0$$ with some initial condition $g(x,0) = f(x)$, and constant $a$.

Mathematica tells me its solution should be $g(x,t)=e^{t/a} f(x e^{t/a})$, but I would like to know how to solve it myself too. How to do it? Or can you point me in the right direction?

Edit: you suggested separating variables $g(x,t)=\psi(x)\phi(t)$. This gives the following: $$\frac{\partial\ln{\phi(t)}}{\partial t} = \frac{\partial}{\partial x} \left(\psi(x) \frac{x}{a}\right)$$ How to continue from here then? The final answer does not look immediately separable to me either...

Answer 1

$$\frac{\partial{g(x,t)}}{\partial t} + \frac{\partial}{\partial x}\left(g(x,t)\frac{-x}{a}\right)=0$$ $$\frac{\partial{g(x,t)}}{\partial t}-\frac{x}{a}\frac{\partial{g(x,t)}}{\partial x}=\frac{1}{a}g(x,t)$$ By luck, the method of separation of variables is perfect in this case, according to the initial condition $g(x,0)=f(x)$. But this is not always so simple.

Alternatively one can use the method of characteristics in order to find not only some particular solutions, but the general solution.

The Charpit-Lagrange system of equations is: https://en.wikipedia.org/wiki/Method_of_characteristics

$$\frac{dt}{1}=\frac{dx}{-x/a}=\frac{dg}{g/a}$$ This easily leads to two characteristic equations : $$\frac{dt}{a}+\frac{dx}{x}=0\qquad\to\qquad xe^{t/a}=c_1$$ $$\frac{dt}{a}-\frac{dg}{g}=0\qquad\to\qquad ge^{-t/a}=c_2$$ The general solution is : $\quad ge^{-t/a}=F(xe^{t/a})\quad$ where $F$ is an arbitrary function. $$g(x,t)=e^{t/a}F(xe^{t/a})$$ The arbitrary function $F$ is no longer arbitrary when the initial condition is specified : $$g(x,0)=f(x)=e^{0/a}F(xe^{0/a})=F(x)$$ This is why this case is so simple $F(x)=f(x)$ and the result is $$g(x,t)=e^{t/a}f(xe^{t/a})$$

Answer 2

I can give you a hint.

Assume that your function $g(x, t)$ can be set to be equal to multiplication of two functions $\psi(x)$ and $\phi(t)$.

Then apply your conditions and see how far you can go with that :)

Answer 3

Let's look for solution in the form $g(x,t)=X(x)T(t)$. We have $X(x)T'(t)-xX'(x)T(t)/a-X(x)T(t)/a=0$. Separating the variables we obtain: $\frac{T'}{T}=\frac{xX'+X}{aX}=k$ (it's equal to some constant because we have two functions with independent variables on different sides). Now you just split PDE into two ODE: $\frac{T'(t)}{T(t)}=k$ and $\frac{xX'(x)+X(x)}{aX(x)}=k$. Integrating the first one we get $\ln(T)=kt+c$, $T(t)=Ce^{kt}$. The second equation can be rewritten as $\frac{xX'(x)}{aX(x)}=k-\frac{1}{a}$ or $\frac{X'(x)}{X(x)}=\frac{ak-1}{x}$. Integrating bot sides we get $\ln X(x)=(ak-1)\ln x+c_1$, $X(x)=C_1x^{ak-1}$. Thus, $g(x,t)=Bx^{ak-1}e^{kt}$ (where $B$ is some constant). Use boundary conditions to find constants $k$ and $B$.