In an old exercise, I tried to solve a the following problem
$$ \begin{cases} u_t + xu_x = u\\ u(0,x)=x^3 \end{cases} $$
I solved the equation with the equalities
$$\frac{1}{dt}=\frac{x}{dx}=\frac{u}{du}$$
Treating 1st and 2nd side of the equalities
$$\frac{1}{dt}=\frac{x}{dx} \quad\rightarrow\quad dt=\frac{dx}{x} \quad\rightarrow\quad t+k_1 = \ln|x| + k_2 \quad\rightarrow\quad k_1 - k_2 =\ln|x|-t$$
$$\rightarrow\quad e^{k_1 - k_2} =e^{\ln|x|-t} \quad\rightarrow\quad e^{k_1 - k_2} =e^{\ln|x|}e^{-t} \quad\rightarrow\quad \pm e^{k_1 - k_2} =xe^{-t} $$
Fixing $\zeta$
$$\zeta:=\pm e^{k_1 - k_2} =xe^{-t} $$
A part of general solution yields
$$\Phi(\zeta)=\Phi(xe^{-t}) $$
Treating now 1st and 3rd
$$\frac{1}{dt}=\frac{u}{du} \quad\rightarrow\quad \frac{du}{u}=dt \quad\rightarrow\quad \ln|u|+k_3=t+k_4 \quad\rightarrow\quad \ln|u|=t+k_4-k_3$$
$$\rightarrow\quad u=\pm e^{k_4-k_3}e^t $$
Fixed the constant $K:=\pm e^{k_4-k_3}$, we can say
$$ u=Ke^t $$
Detected the constant in the general solution, we replace it with $\Phi(\zeta)$.
$$ u_1(t,x)=e^t\Phi(xe^{-t}) $$
Then I tried to solve by verifying the equality between 2nd and 3rd side.
$$\frac{x}{dx}=\frac{u}{du} \quad\rightarrow\quad \frac{du}{dx}=\frac{u}{x} \quad\rightarrow\quad \frac{u'(x)}{u(x)}=\frac{1}{x} \quad\rightarrow$$
$$\rightarrow\quad \ln|u(x)|+k_5=\ln|x|+k_6 \quad\rightarrow\quad \ldots \quad\rightarrow\quad u(x)=\pm e^{k_6-k_5}x $$
Fixed as constant $H:=\pm e^{k_6-k_5}$, we can say
$$ u(x)=Hx $$
Detected the constant in the general solution, we replace it with $\Phi(\zeta)$ found in the first phase.
$$ u_2(t,x)=x\Phi(xe^{-t}) $$
So, both $u_1(t,x)$ and $u_2(t,x)$ are suitable for the PDE.
I tried to sum them up and I noticed that $u(t,x)=u_1(t,x)+u_2(t,x)$ also verifies the PDE!
Question 1: Is there a general form (even more general than the sum) to indicate the most general solution for the PDE?
Then:
I also tried to evaluate the initial condition in order to find the particular solution but I fulfilled the request using $u_1(t,x)$ ONLY. I will show you what I have done
$$x^3 = u(0,x)=\Phi(x)$$
$$\zeta:=x \quad\rightarrow\quad x^3=\zeta^3$$
Yielding:
$$ u_1(t,x)=e^t(xe^{-t})^3=x^3 e^{-2t} $$
But I failed in evaluate the initial condition with $u_2(t,x)$.
Question 2: Can anyone help me to find the particular solution using $u(t,x)=u_1(t,x)+u_2(t,x)$? Can anyone show me the correct passages?
Thanks to all in advance.
The sum of these solutions is a solution because both are different forms for the most general solution possible.
Consider $u_1(t,x)=e^t\Phi(xe^{-t})$ and $u_2(t,x)=x\Psi(xe^{-t})$
Now, we can always set $\Phi(xe^{-t})=xe^{-t}\Phi_1(xe^{-t})$ for some $\Phi_1$, continuous and differentiable too. Then,
$u_1(t,x)=e^t\Phi(xe^{-t})=e^txe^{-t}\Phi_1(xe^{-t})=x\Phi_1(xe^{-t})$
So is, $u_1=u_2$ because obviously $\Psi=\Phi_1$
You had the general solution inasmuch you had an expression depending on an arbitrary function (to determine by the initial conditions)
Solving using $u_2(t,x)=x\Psi(xe^{-t})$
$u_2(0,x)=x^3=x\Psi(x)$
$\Psi(x)=x^2$ and
$u_2(t,x)=x(xe^{-t})^2=x^3e^{-2t}$
The same function, as expected, so, no need to sum them.