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how to solve $u_t + uu_x =f(x,t)$ with initial condition $u(x,0)= \phi(x)$

For the simple case $f(x,t)=1 $ and $\phi(x)=x$, I've tried to proceed like for the homogeneous one. I find the characteristics are parabolae $$X(t)=\frac {t^2}{2} +X(0)(1+t)$$

but the solution won't be constant along these characteristics because of the source term f!

I find that if I define $v(t)=u(X(t),t) \ \ $, $v$ would be solution to $v'=1$ hence $$v(t)=t+k=t+v(0)=t+u(X(0),0)=t+X(0)$$ plugging this into u I get $$ u(X(t),t)=u(\frac {t^2}{2} +X(0)(1+t),t)$$ and then, solving for $X(0)$ in terms of $x$ $$u(x,t)=\frac{1}{1+t}(x-\frac {t^2}{2} )+t$$

Am I missing something here?

And how should I procced for a generic f and $\phi$?

aflous
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1 Answers1

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Your solution is correct. A slightly different way to obtain it is: write $u(x,t)=w(x,t)+t$, so that $w$ satisfies the equation $w_t+(w+t)w_x=0$. The function $w$ is constant along curves $X(t)$ that satisfy $X'(t)=\phi(X(0))+t$. This leads to $$X(t)=X(0)+\phi(X(0))t+\frac{t^2}{2}$$ which is easy to put in implicit form if $\phi$ is a nice function like $\phi(x)=x$. In this case,
$$X(t)=X(0)(1+t)+\frac{t^2}{2}$$ which in implicit form is $$\frac{x-t^2/2}{1+t}=C$$ Thus, $w$ is of the form $$w(x,t) = \Psi\left(\frac{x-t^2/2}{1+t}\right)$$ The initial condition is satisfied with $\Psi(z)=z$, i.e., $$w(x,t) = \frac{x-t^2/2}{1+t} $$ Add $t$ to get $u(x,t)$.

More generally, if $f$ depends only on $t$, you can write it as $f(x,t)=F'(t)$ for some $F$, and let $u(x,t)=w(x,t)+F(t)$ similar to the above. Then $w_t+(w+F(t))w_x=0$. The function $w$ is constant along curves $X(t)$ that satisfy $X'(t)=\phi(X(0))+F(t)$. This leads to $$X(t)=X(0)+\phi(X(0))t+\int_0^t F(s)\,ds$$ ... and so on.

I don't know how to handle $f$ that depends on $x$.

Another version of this problem, where $f$ is a function of $u$, is considered in the paper Exact solution of generalized inviscid Burgers' equation where some more-or-less explicit formulas are derived for the solution. (One must be able to integrate, invert, and integrate again...)

Post No Bulls
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