# Solve first order differential equation in two dimensions

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

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 $$x_2$$?

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

(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 $$\psi(x_1)= \alpha g( x_1)$$.
• 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 Oct 29, 2019 at 17:38