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$x\in\mathbb R^2$, $a\in\mathbb R^2$, $v=(1,1)$

We know: $v\cdot \nabla f(x)=g(a\cdot x)+v\cdot x$.

1) What is $f(x)$ in terms of $g$?

2) What is $f(x)$ if $f(x_1,x_2)=h(x_1)+h(x_2)$?


The PDE form is $f_{x_1}+f_{x_2}=g(a\cdot x)+x_1+x_2$

Let $h(x)=g(ax)+x_1+x_2$ then this is a quasilinear pde form:

$f_{x_1}+f_{x_2}=h(x_1,x_2)$.

It seems like this is equivalent to the following parametric form:

$dx_1/dt=1$, $dx_2/dt=1$, $df/dt=h$.

Therefore we have $x_1=t+c_1$, $x_2=t+c_2$, $f=\int h dt$.

So $f=\int g(a_1(t+c_1)+a_2(t+c_2))+2t+c_1+c_2dt$;

$f=t^2+(c_1+c_2)t+c_3+\frac{1}{a_1+a_2}G((a_1+a_2)t+a_1c_1+a_2c_2)$

But I think I need the $f$ in terms of $x_1$ and $x_2$?

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    $\begingroup$ Where is this PDE from? What PDEs do you know how to solve? Do you know the method of characteristics? $\endgroup$
    – user9464
    Oct 27, 2019 at 22:51
  • $\begingroup$ Jack didn't ask about the "form", but where it is from. $\endgroup$
    – amsmath
    Oct 27, 2019 at 23:09
  • $\begingroup$ "Therefore we have...?" Of course not. $f =\int h$. I get $f(t+c,t+d) = \int g((a_1+a_2)s + a_1c+a_2d)\,ds$. $\endgroup$
    – amsmath
    Oct 27, 2019 at 23:15

1 Answer 1

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(1) The equation $v\cdot \nabla f(x) = g(a\cdot x) + v\cdot x$ rewrites as $$ f_{,1} + f_{,2} = g(a_1 x_1 + a_2 x_2) + x_1 + x_2 \, . $$ This is a non-homogeneous advection equation (linear transport equation), which can be solved by using the method of characteristics. Following this method, we have the set of parametrized curves $x_1(t) = t+c_1$, $x_2(t) = t+c_2$, which are straight lines with slope one in the plane $x\in\Bbb R^2$. Along these curves, we have $$f(t) = t^2 + Ct + c_3 + \int_0^t g(A\tau+C')\,\text d\tau\, ,$$ where $A = a_1+a_2$, $C = c_1+c_2$ and $C' = a_1c_1+a_2c_2$. Assuming $c_3=F(c_2)$, the variable $t$ is eliminated by substitution of $c_2 = x_2-t$ and $t=x_1-c_1$ (i.e., $c_3=F(x_2-x_1+c_1)$). Similar substitutions are made in the case $c_3=F(c_1)$.

(2) The second part uses abusive notation. Thus, let us define $\varphi$ such that $f(x_1,x_2) = \varphi(x_1) + \varphi(x_2)$. The PDE becomes $$ g(a\cdot x) = \psi(x_1) + \psi(x_2) $$ with $\psi = \varphi' - \text{id}$, which means that $\psi(x_1) = g(a_1 x_1) - \psi(0)$ and $\psi(x_2) = g(a_2 x_2) - \psi(0)$. Since this must be true for all $x$, we must have $a_1 = a_2 =\alpha$. Now we are left with $g(\alpha (x_1+x_2))= \psi(x_1) + \psi(x_2)$, which implies that $\psi(x_1)=\tfrac1{1+\beta} g(\alpha (1+\beta) x_1)$ by setting $x_1=\beta x_2$. The only possibility to have an expression of $\psi$ which does not depend on $\beta$ is that $g$ is a linear function, and we have $\psi(x_1)= \alpha g( x_1)$.

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  • $\begingroup$ Thank you for your answer! For part (1), I think $f(x)$ is a function with two variables? In the end you got a $f(t)$ which seems like a function with only one variable. I think I am wrong but I am not sure. Basically I want to get $f(x_1,x_2)$ in the term of $x_1$ and $x_2$ not in term of t $\endgroup$
    – High GPA
    Oct 29, 2019 at 17:38
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    $\begingroup$ @HighGPA answer edited. $\endgroup$
    – EditPiAf
    Oct 29, 2019 at 19:13

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