Vtyjkyuk

I have the following partial differential equation (where $u = u(x,t)$)

$$a(x) fracpartial upartial t + f fracpartial upartial x = 0$$

I was given that the solution for $u$ is any differentiable function $h$ of the argument

$$t - frac1fint_0 ^x a(alpha) dalpha$$

i.e

$$u(x,t) = h left( t - frac1fint_0 ^x a(alpha) dalpha right)$$

To get this solution, I assumed that the solution was on the form $u = h(t + g(x))$ and sought out to find $g(x)$.

I substituted the form of the solution into the partial differential equation which resulted in (this is the step I am unsure with)

beginalign
a(x) + f fracpartial gpartial x &= 0 \
implies fracpartial gpartial x &= frac-a(x)f \
implies g(x) &= int_0^xfrac-a(alpha)f dalpha
endalign

Hence, my solution matches the given solution.

However, I am unsure if this working out is, in fact, correct. Also, we were given an initial condition $C(x,t) = 0$ at $t=0$. I, however, hade a change of variable to $u$, defining $u$ as

$$u(x,t) = Klog(fracC(x,t)K)$$,

where $K$ is just a constant.

Applying the initial condition given for $C(x,t)$ to $u$ yields $u(x,0) = -infty$.

I did not use this initial condition when finding my solution, though I have a feeling that it plays a fundamental part in the solution.

Thanks for the help.

Consider the more general problem of this PDE !
$$a(x) fracpartial upartial t + b(t) fracpartial upartial x = 0$$
Your problem is the particular case of $b(t)=$constant.
$$ fracpartial ub(t)partial t + fracpartial ua(x)partial x = 0$$
Change of variables :
$begincases
X=int a(x)dx \
T=int b(t)dt
endcases$
$$ fracpartial upartial T + fracpartial upartial X = 0$$
It is well known and elementary to prove that the general solution is :
$$u(X,T)=F(X-T)$$
where $F$ is an arbitrary function.

Coming back to the original variables :
$$u(x,t)=Fleft(int a(x)dx -int b(t)dt right)$$
In the case of $b(t)=f=$constant :
$$u(x,t)=Fleft(int a(x)dx -f:t right)$$
Since $F$ is an arbitrary function they are an infinity of equivalent forms to express the general solution, for example :

$u(x,t)=Gleft(-frac1fint a(x)dx +:t right)quad$ with $G$ an arbitrary function.

or

$u(x,t)=Hleft(c_1frac1fint a(x)dx -c_1:t +c_0right)quad$ with $H$ an arbitrary function.

Or many other equivalent forms.

Note that the arbitrary coefficient of $int a(x)dx$ and $-f:t$ must be the same : $Hleft(c_1int a(x)dx -c_2f:t +c_0right)$ is not solution of the PDE if $c_1neq c_2$.