Recent content by Gallo

Also it can also be said that ##g^{-t}(x)## is the flow generated by ##-a(x)##, correct?

It seems you are asking too much. The solution is uniquely determined by the boundary conditions. Once found a solution you can check if it is consistent with the value you have at ##(x_i,y_i)##.

Hello I would like to check my reasoning about solutions of first order PDE. I've spell out (almost) all details. I'll consider the following problem: find ##u=u(t,x)## s.t. : $$ \partial_t u(t,x) + a(x) \cdot \nabla u(x) =0 \qquad \qquad u(0,x) = u_0(x)$$ say with smooth coefficient and...

Gotcha! $$\frac{d h(t)}{d t} =\frac{d}{ d t}\int_0^t f(t-\tau) g(\tau) d \tau $$ and using Leibniz integral rule $$ \frac{d h(t)}{d t} = f(0)g(t) + \int_0^t \frac{d f(t-\tau)}{ d t} g(\tau) d \tau $$ The laplace transform of which is $$ s H(s) = G(s) f(0) + G(s)(s F(s) - f(0) ) = s F(s)G(s)...

Prelude Consider the convolution h(t) of two function f(t) and g(t): $$h(t) = f(t) \ast g(t)=\int_0^t f(t-\tau) g(\tau) d \tau$$ then we know that by the properties of convolution $$\frac{d h(t)}{d t} = \frac{d f(t)}{d t} \ast g(t) = f(t) \ast \frac{d g(t)}{d t}$$ Intermezzo We also know that...

Thanks, that's what I thought!

Thanks, following your hint this is I would go: if ##y_{\pm}## solve ##y'_{\pm}(x)+a(x)*y_{\pm}(x)=0## in ## \pm x> 0## then ##y_{\pm} = C_{\pm} \, e^{-A_{\pm}(x)} ## and a solution is given by y(x) = y_{-}(x) + H(x) (u_{+}(x) - u_{-}(x)) where A_{+}(x) = \int_{0^{+}}^{x} a(z) dz...

I'm not sure, what you meant...I would get some kind of integral equation ..

I'm trying to solve the following equation (even if I'm not sure if it's well posed) \partial_{x} \, y(x) + a(x)\, y(x) = \delta(x) with ##\quad \lim_{x \rightarrow \pm \infty}y(x) = 0## It would be a classical first order ODE If it were not for the boundary conditions and the Dirac...

A great quotation from a great man