Could this linear homogeneous 1D transport equation with variable coefficient yield a solution that can be...
Note: this question was edited to reflect an evolving insight on part of the author. However, the original question still stands (although it has hopefully been formulated more clearly now).
For a physics problem that interests me, I am looking for a solution to the following linear homogeneous 1D transport equation with variable coefficient:
$$
f^{(1,0)}(t,x) - (g_1(t),h_1(x) + g_2(t),h_2(x));f^{(0,1)}(t,x) = 0
$$
In this PDE, all functions and variables are real-valued. $x$ has the meaning of a spatial coordinate and $t$ has the meaning of time. Superscripts between parentheses denote order of differentiation w.r.t. respective arguments. While the two functions of time $g_1(t)$ and $g_2(t)$ can best be considered unspecified1, both functions of space $h_1(x)$ and $h_2(x)$ are given by simple algebraic expressions:
- $h_1(x)=+x(x^gamma-1)$
$h_2(x)=-x(x-1)$,
where $gamma$ is a positive real number. Since this is a transport equation, I have learned that the method of characteristics can be applied to transform a PDE into an ODE. As far as I have understood, I can write for the characteristic $X(t)$:
$$
frac{d}{dt};f(X(t),t) = 0
$$
$$
frac{dX}{dt} = - g_1(t),h_1(X(t)) - g_2(t),h_2(X(t))
$$
$$
f(t,x) = F(X^*(t,x))
$$
here, $X^*(t,x)$ is a solution to the characteristic equation above subject to initial conditions $X(0) = X^*(0,x) = x$. $F(x)$ is an unknown function.
Of course a transport equation will transport any signal we give it. Clearly though, an initial signal equal to e.g. sine of $x$ will not remain a sine of $x$ for long. My question is therefore: given $mathbf{h_1(x)}$ and $mathbf{h_2(x)}$ (but not given $mathbf{g_1(t)}$ and $mathbf{g_2(t)}$), can we say anything about the functional form that $mathbf{F(x)}$ must have for it to be transportable analytically?
There are a few constraints that $f(t,x)$ should adhere to:
- The domain of $f(t,x)$ is $[0,infty)$, $(0,1]$ and its codomain is $[0,infty)$.
- $f(t,1)=0$
$lim_{x downarrow 0} f(t,x) = +infty$.
$f(t,x)$ is a monotonically decreasing function of $x$.- $lim_{x downarrow 0} f^{(0,1)}(t,x) = -infty$
- $lim_{x uparrow 1} f^{(0,1)}(t,x) = -infty$
In short, the inverse function of $f(t,x)$ w.r.t. its second argument could look like a Gaussian or a Lorentzian. In fact, the PDE considered here was derived from a much more horrible looking PDE formulated in terms of this inverse, as well as some other considerations.
1 They each depend on their own set of nonlinear evolution equations which I'm quite sure do not have a closed-form solution or approximation. To pull these equations into the context of this question would in all likelihood not improve it, I feel.
calculus real-analysis differential-equations multivariable-calculus pde
add a comment |
Note: this question was edited to reflect an evolving insight on part of the author. However, the original question still stands (although it has hopefully been formulated more clearly now).
For a physics problem that interests me, I am looking for a solution to the following linear homogeneous 1D transport equation with variable coefficient:
$$
f^{(1,0)}(t,x) - (g_1(t),h_1(x) + g_2(t),h_2(x));f^{(0,1)}(t,x) = 0
$$
In this PDE, all functions and variables are real-valued. $x$ has the meaning of a spatial coordinate and $t$ has the meaning of time. Superscripts between parentheses denote order of differentiation w.r.t. respective arguments. While the two functions of time $g_1(t)$ and $g_2(t)$ can best be considered unspecified1, both functions of space $h_1(x)$ and $h_2(x)$ are given by simple algebraic expressions:
- $h_1(x)=+x(x^gamma-1)$
$h_2(x)=-x(x-1)$,
where $gamma$ is a positive real number. Since this is a transport equation, I have learned that the method of characteristics can be applied to transform a PDE into an ODE. As far as I have understood, I can write for the characteristic $X(t)$:
$$
frac{d}{dt};f(X(t),t) = 0
$$
$$
frac{dX}{dt} = - g_1(t),h_1(X(t)) - g_2(t),h_2(X(t))
$$
$$
f(t,x) = F(X^*(t,x))
$$
here, $X^*(t,x)$ is a solution to the characteristic equation above subject to initial conditions $X(0) = X^*(0,x) = x$. $F(x)$ is an unknown function.
Of course a transport equation will transport any signal we give it. Clearly though, an initial signal equal to e.g. sine of $x$ will not remain a sine of $x$ for long. My question is therefore: given $mathbf{h_1(x)}$ and $mathbf{h_2(x)}$ (but not given $mathbf{g_1(t)}$ and $mathbf{g_2(t)}$), can we say anything about the functional form that $mathbf{F(x)}$ must have for it to be transportable analytically?
There are a few constraints that $f(t,x)$ should adhere to:
- The domain of $f(t,x)$ is $[0,infty)$, $(0,1]$ and its codomain is $[0,infty)$.
- $f(t,1)=0$
$lim_{x downarrow 0} f(t,x) = +infty$.
$f(t,x)$ is a monotonically decreasing function of $x$.- $lim_{x downarrow 0} f^{(0,1)}(t,x) = -infty$
- $lim_{x uparrow 1} f^{(0,1)}(t,x) = -infty$
In short, the inverse function of $f(t,x)$ w.r.t. its second argument could look like a Gaussian or a Lorentzian. In fact, the PDE considered here was derived from a much more horrible looking PDE formulated in terms of this inverse, as well as some other considerations.
1 They each depend on their own set of nonlinear evolution equations which I'm quite sure do not have a closed-form solution or approximation. To pull these equations into the context of this question would in all likelihood not improve it, I feel.
calculus real-analysis differential-equations multivariable-calculus pde
What's non-linear about this equation? You have $frac{partial f}{partial t} = a(x,t)frac{partial f}{partial x}$
– Dylan
Nov 30 '18 at 12:17
You are correct. I have changed the title and text accordingly.
– Casper Pranger
Nov 30 '18 at 13:28
Thanks to the remark of @Dylan that this is not a non-linear equation at all, I now realize that its classification would be 'one-dimensional transport equation with variable coefficient'. I am currently looking into applying the method of characteristics, though so far I can not say that it is very helpful to me. If I find anything concrete, I will be sure to update the question.
– Casper Pranger
Nov 30 '18 at 17:40
add a comment |
Note: this question was edited to reflect an evolving insight on part of the author. However, the original question still stands (although it has hopefully been formulated more clearly now).
For a physics problem that interests me, I am looking for a solution to the following linear homogeneous 1D transport equation with variable coefficient:
$$
f^{(1,0)}(t,x) - (g_1(t),h_1(x) + g_2(t),h_2(x));f^{(0,1)}(t,x) = 0
$$
In this PDE, all functions and variables are real-valued. $x$ has the meaning of a spatial coordinate and $t$ has the meaning of time. Superscripts between parentheses denote order of differentiation w.r.t. respective arguments. While the two functions of time $g_1(t)$ and $g_2(t)$ can best be considered unspecified1, both functions of space $h_1(x)$ and $h_2(x)$ are given by simple algebraic expressions:
- $h_1(x)=+x(x^gamma-1)$
$h_2(x)=-x(x-1)$,
where $gamma$ is a positive real number. Since this is a transport equation, I have learned that the method of characteristics can be applied to transform a PDE into an ODE. As far as I have understood, I can write for the characteristic $X(t)$:
$$
frac{d}{dt};f(X(t),t) = 0
$$
$$
frac{dX}{dt} = - g_1(t),h_1(X(t)) - g_2(t),h_2(X(t))
$$
$$
f(t,x) = F(X^*(t,x))
$$
here, $X^*(t,x)$ is a solution to the characteristic equation above subject to initial conditions $X(0) = X^*(0,x) = x$. $F(x)$ is an unknown function.
Of course a transport equation will transport any signal we give it. Clearly though, an initial signal equal to e.g. sine of $x$ will not remain a sine of $x$ for long. My question is therefore: given $mathbf{h_1(x)}$ and $mathbf{h_2(x)}$ (but not given $mathbf{g_1(t)}$ and $mathbf{g_2(t)}$), can we say anything about the functional form that $mathbf{F(x)}$ must have for it to be transportable analytically?
There are a few constraints that $f(t,x)$ should adhere to:
- The domain of $f(t,x)$ is $[0,infty)$, $(0,1]$ and its codomain is $[0,infty)$.
- $f(t,1)=0$
$lim_{x downarrow 0} f(t,x) = +infty$.
$f(t,x)$ is a monotonically decreasing function of $x$.- $lim_{x downarrow 0} f^{(0,1)}(t,x) = -infty$
- $lim_{x uparrow 1} f^{(0,1)}(t,x) = -infty$
In short, the inverse function of $f(t,x)$ w.r.t. its second argument could look like a Gaussian or a Lorentzian. In fact, the PDE considered here was derived from a much more horrible looking PDE formulated in terms of this inverse, as well as some other considerations.
1 They each depend on their own set of nonlinear evolution equations which I'm quite sure do not have a closed-form solution or approximation. To pull these equations into the context of this question would in all likelihood not improve it, I feel.
calculus real-analysis differential-equations multivariable-calculus pde
Note: this question was edited to reflect an evolving insight on part of the author. However, the original question still stands (although it has hopefully been formulated more clearly now).
For a physics problem that interests me, I am looking for a solution to the following linear homogeneous 1D transport equation with variable coefficient:
$$
f^{(1,0)}(t,x) - (g_1(t),h_1(x) + g_2(t),h_2(x));f^{(0,1)}(t,x) = 0
$$
In this PDE, all functions and variables are real-valued. $x$ has the meaning of a spatial coordinate and $t$ has the meaning of time. Superscripts between parentheses denote order of differentiation w.r.t. respective arguments. While the two functions of time $g_1(t)$ and $g_2(t)$ can best be considered unspecified1, both functions of space $h_1(x)$ and $h_2(x)$ are given by simple algebraic expressions:
- $h_1(x)=+x(x^gamma-1)$
$h_2(x)=-x(x-1)$,
where $gamma$ is a positive real number. Since this is a transport equation, I have learned that the method of characteristics can be applied to transform a PDE into an ODE. As far as I have understood, I can write for the characteristic $X(t)$:
$$
frac{d}{dt};f(X(t),t) = 0
$$
$$
frac{dX}{dt} = - g_1(t),h_1(X(t)) - g_2(t),h_2(X(t))
$$
$$
f(t,x) = F(X^*(t,x))
$$
here, $X^*(t,x)$ is a solution to the characteristic equation above subject to initial conditions $X(0) = X^*(0,x) = x$. $F(x)$ is an unknown function.
Of course a transport equation will transport any signal we give it. Clearly though, an initial signal equal to e.g. sine of $x$ will not remain a sine of $x$ for long. My question is therefore: given $mathbf{h_1(x)}$ and $mathbf{h_2(x)}$ (but not given $mathbf{g_1(t)}$ and $mathbf{g_2(t)}$), can we say anything about the functional form that $mathbf{F(x)}$ must have for it to be transportable analytically?
There are a few constraints that $f(t,x)$ should adhere to:
- The domain of $f(t,x)$ is $[0,infty)$, $(0,1]$ and its codomain is $[0,infty)$.
- $f(t,1)=0$
$lim_{x downarrow 0} f(t,x) = +infty$.
$f(t,x)$ is a monotonically decreasing function of $x$.- $lim_{x downarrow 0} f^{(0,1)}(t,x) = -infty$
- $lim_{x uparrow 1} f^{(0,1)}(t,x) = -infty$
In short, the inverse function of $f(t,x)$ w.r.t. its second argument could look like a Gaussian or a Lorentzian. In fact, the PDE considered here was derived from a much more horrible looking PDE formulated in terms of this inverse, as well as some other considerations.
1 They each depend on their own set of nonlinear evolution equations which I'm quite sure do not have a closed-form solution or approximation. To pull these equations into the context of this question would in all likelihood not improve it, I feel.
calculus real-analysis differential-equations multivariable-calculus pde
calculus real-analysis differential-equations multivariable-calculus pde
edited Dec 4 '18 at 10:53
asked Nov 30 '18 at 11:17
Casper Pranger
12
12
What's non-linear about this equation? You have $frac{partial f}{partial t} = a(x,t)frac{partial f}{partial x}$
– Dylan
Nov 30 '18 at 12:17
You are correct. I have changed the title and text accordingly.
– Casper Pranger
Nov 30 '18 at 13:28
Thanks to the remark of @Dylan that this is not a non-linear equation at all, I now realize that its classification would be 'one-dimensional transport equation with variable coefficient'. I am currently looking into applying the method of characteristics, though so far I can not say that it is very helpful to me. If I find anything concrete, I will be sure to update the question.
– Casper Pranger
Nov 30 '18 at 17:40
add a comment |
What's non-linear about this equation? You have $frac{partial f}{partial t} = a(x,t)frac{partial f}{partial x}$
– Dylan
Nov 30 '18 at 12:17
You are correct. I have changed the title and text accordingly.
– Casper Pranger
Nov 30 '18 at 13:28
Thanks to the remark of @Dylan that this is not a non-linear equation at all, I now realize that its classification would be 'one-dimensional transport equation with variable coefficient'. I am currently looking into applying the method of characteristics, though so far I can not say that it is very helpful to me. If I find anything concrete, I will be sure to update the question.
– Casper Pranger
Nov 30 '18 at 17:40
What's non-linear about this equation? You have $frac{partial f}{partial t} = a(x,t)frac{partial f}{partial x}$
– Dylan
Nov 30 '18 at 12:17
What's non-linear about this equation? You have $frac{partial f}{partial t} = a(x,t)frac{partial f}{partial x}$
– Dylan
Nov 30 '18 at 12:17
You are correct. I have changed the title and text accordingly.
– Casper Pranger
Nov 30 '18 at 13:28
You are correct. I have changed the title and text accordingly.
– Casper Pranger
Nov 30 '18 at 13:28
Thanks to the remark of @Dylan that this is not a non-linear equation at all, I now realize that its classification would be 'one-dimensional transport equation with variable coefficient'. I am currently looking into applying the method of characteristics, though so far I can not say that it is very helpful to me. If I find anything concrete, I will be sure to update the question.
– Casper Pranger
Nov 30 '18 at 17:40
Thanks to the remark of @Dylan that this is not a non-linear equation at all, I now realize that its classification would be 'one-dimensional transport equation with variable coefficient'. I am currently looking into applying the method of characteristics, though so far I can not say that it is very helpful to me. If I find anything concrete, I will be sure to update the question.
– Casper Pranger
Nov 30 '18 at 17:40
add a comment |
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What's non-linear about this equation? You have $frac{partial f}{partial t} = a(x,t)frac{partial f}{partial x}$
– Dylan
Nov 30 '18 at 12:17
You are correct. I have changed the title and text accordingly.
– Casper Pranger
Nov 30 '18 at 13:28
Thanks to the remark of @Dylan that this is not a non-linear equation at all, I now realize that its classification would be 'one-dimensional transport equation with variable coefficient'. I am currently looking into applying the method of characteristics, though so far I can not say that it is very helpful to me. If I find anything concrete, I will be sure to update the question.
– Casper Pranger
Nov 30 '18 at 17:40