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From Evans' PDE, chapter 2.1.2

Prove $u(x,t)=g(x-tb)+\int_0^t f(x+(s-t)b,s)\,ds, u\in C^1(\mathbb R^n\times[0,\infty))$ solves \begin{cases} u_t+b\cdot Du=f &\text{in }\mathbb R^n\times(0,\infty),\\ u=g &\text{on }\Gamma:=\mathbb R^n\times\{t=0\}. \end{cases} where $b\in \mathbb R^n, g\in C^1(\Gamma), f\in C^0(\mathbb R^n\times(0,\infty))$.

My attempt: Let $P$ be the primitive of $f$, and for brevity write $g'$ instead of $g'(x-tb)$ \begin{align} u_t &= g'\cdot(-b) +\frac{\partial}{\partial t}(P(x,t)-P(x-tb,0)) \\ &= -g'\cdot b+f(x,t)-f(x-tb,0)\cdot(-b)\\ Du &= g' +\frac{\partial}{\partial x}(P(x,t)-P(x-tb,0)) \\ &= g'+f(x,t)-f(x-tb,0)\\ \end{align} \begin{align}\require{cancel} u_t+b\cdot Du &= \cancel{-g'\cdot b}+f(x,t)\cancel{-f(x-tb,0)\cdot(-b)}+\cancel{b\cdot g'}+b\cdot f(x,t)\cancel{-b\cdot f(x-tb,0)}\\ &= f(x,t)+b\cdot f(x,t). \end{align} I was expecting $u_t+b\cdot Du = f(x,t)$, where is the error?

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Remember the Leibniz integral rule. In the scalar case $n=1$, we have $$ u_t(x,t) = -b g'(x-tb) + f(x,t) -b \int_0^t f_x(x+(s-t)b,s)\, ds $$ $$ Du(x,t) = g'(x-tb) + \int_0^t f_x(x+(s-t)b,s)\, ds $$ i.e. $u_t + b\cdot Du = f$, and the boundary condition is satisfied as well. Now it remains to do the same for arbitrary dimension $n$.

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  • $\begingroup$ Many thanks! I have few questions, the integrand in first line is $f_x$ or $f_t$? Your computations are the solution of the problem or we have to write them for dimension $n$? (and if this is the case do you have any hint to start?) Thank you $\endgroup$
    – sound wave
    Commented Jun 23, 2021 at 12:31
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    $\begingroup$ @soundwave It's $f_x$. According to the Leibniz rule, we must compute $$ \partial_t f(X, s) = \frac{\partial X}{\partial t}\, f_X(X, s) $$ where $X=x+(s-t)b$ and $(x,t)\mapsto f_x(x,t)$ is a partial derivative of $f$. You can proceed the same way using the Leibniz rule in higher dimension $n>1$. $\endgroup$
    – EditPiAf
    Commented Jun 24, 2021 at 8:53

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