Confused over the solution of partial differential equation $xu_x+u_t=0$

Question

Consider,

$$ \displaystyle x\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t} = 0 $$

with initial values $ t = 0 : \ u(x, 0) = f(x) $ and calculate the solution $ u(x,t) $ of the above Cauchy problem using the method of characteristics.

And here is the solution, I will point out where i am stuck:

1. We parametrise the initial conditions by $\mathbb n:x_0(\mathbb n)=\mathbb n$, $t_0(\mathbb n)=0, u_0(\mathbb n)=f(\mathbb n)$

2. Solve the characteristic equations

   $$ \matrix{ \frac{\partial x(\sigma,\mathbb n)}{\partial \sigma} = x, && x(0,\mathbb n)=n \\ \frac{\partial t(\sigma,\mathbb n)}{\partial \sigma} = 1, && x(0,\mathbb n)=n \\ t(0,\mathbb n)=0: && x(\sigma,\mathbb n)=e^{\sigma}\mathbb n \\ t(\sigma,\mathbb n)=\sigma } $$ This is where i am stuck and confused

   How did they get $x(\sigma,\mathbb n)=e^{\sigma}\mathbb n$? I just cannot see where the $e$ came from, please forgive my stupidity but can someone please tell me how they got this solution? When i integrate i do not get this!

   I will put the rest of the solution so the reader can follow, I am only stuck with the part mentioned though

3. Calculate $\sigma$ and $\mathbb n$ in terms of $x$ and $t$ (coordinate change)

   $$ \sigma = t, \mathbb n = xe^-{t} $$

4. Solve the compatibility condition $ \frac{\partial u}{\partial \sigma} = 0, u(0,\mathbb n) = f(\mathbb n) $.

   Hence, $u(\sigma)=f(\mathbb n)$

5. Hence we have $u(x,t)=f(xe^{-t})$

Answer 1

You're solving the initial value problem: $$ \dfrac{\partial}{\partial \sigma} x(\sigma,n) = x(\sigma,n),\ x(0,n) = n $$ The variable $n$ is not important here: this is just the ordinary differential equation $$ \dfrac{d}{d\sigma} x = x, \ x(0) = n$$

You do remember how to solve constant-coefficient first-order linear equations, don't you?

Answer 2

If you'll abide the strangeness of it, group theory (Lie theory) provides an answer. Use this stretching group: $$ G(x,t,u)=(\lambda x,\lambda^\beta t, \lambda^\alpha u)\lambda_o=1 $$$\lambda_o=1$ is the unit transformation without which no group is complete. Plug these transformed variables into the equation and you get $$ \lambda x \frac{\lambda^\alpha \partial u}{\lambda \partial x}+\frac{\lambda^\alpha \partial u}{\lambda^\beta \partial t}=0 $$Here you can see, for invariance to this group to exist, $\alpha=\alpha -\beta$. Therefore, $\beta =0$ and $\alpha$ can be any value, so we'll leave it as a variable. Since $$ \lambda^\alpha u(x,t)=u(\lambda x,y) $$we can take the partial derivative of both sides w.r.t. $\lambda$ and set $\lambda =\lambda_o =1$ to get $$ \alpha u=xu_x+0u_t $$The characteristic equation is $$ \frac{du}{\alpha u}=\frac{dx}{x}=\frac{dt}{0} $$Two independent integrals are $\frac{u}{x^\alpha}$ and $t$. These are, in fact, stabilizers of the transformation group, and according to Sophus Lie they form an embedded manifold within your solution manifold. Hence, the most general solution to your PDE is to take one stabilizer and set it equal to a function of the other. This results in $$ u=x^\alpha f(t) $$Now we take partial derivatives. $$ u_x=\alpha x^{\alpha -1}f $$ $$ u_t=x^\alpha f_t $$Plugging these back into the PDE, we arrive at $$ \alpha x^\alpha f+x^\alpha f_t=0 $$The x-terms drop out (as they must, else you are in error) and the equation simplifies to $$ \frac{f_t}{f}=-\alpha $$Here comes your exponential term. $$ \frac{d}{dt}ln(f)=-\alpha $$ $$ ln(f)=-\alpha t+C\rightarrow f=Ce^{-\alpha t} $$Since $u=x^\alpha f(t)$, we have arrived at your answer. $$ u=C(xe^{-t})^\alpha $$The only reason this works is because an infinite continuous group (in this case an infinite cyclic group) preserves the structure of any smooth manifold to which it is invariant. An entire family of groups where $x'=\lambda x$, $t'=t$ and $u'=\lambda^\alpha u$ is invariant to your PDE, a very high degree of symmetry.