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I need to solve a transport equation in the form $$ \frac{\partial f}{\partial t} + a(t) \frac{\partial f}{\partial x} = b f + c$$ with $a(t) = A t$. From a reference book I took a solution $f = -c/b + e^{bt} \Phi(x - At^2/2)$ which contain an arbitrary function $\Phi(x - At^2/2)$. Unfortunately I cannot compose this function for the following conditions:

  • initially $t = 0$ the function is zero everywhere $f(0, x) = 0$, which result in $\Phi = c/b$;
  • at the boundary $x = 0$ the function is zero $f(t, 0) = 0$, which means that $\Phi = e^{-bt} c/b $.

The questions:

  1. how to choose the function $\Phi$ in the example above?
  2. is there a technique/algorithm for finding $\Phi$ which satisfy prescribed initial/boundary conditions?
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If you are imposing that at $x = 0$ the function $f(t,0) = 0$ for all time $t$, then necessarily that the only solution is $\Phi(t,x) = e^{-bt}c/b$. This is because your equation is a transport equation, and by prescribing data on $x = 0$ you have a well-posed evolution equation (when $a \neq 0$) where the "time variable" is $x$ and the "space variable" is $t$. And the unique solution is as given above.

But in general you probably are not looking for such a solution.

I expect what you are looking at is a problem on the first quadrant $x\geq 0$ and $t\geq 0$, with boundary condition prescribed at $x = 0$ when $t \geq 0$ and with initial condition prescribed at $t = 0$ for $x \geq 0$.

Then what we have is that $$f(t,0)|_{t\geq 0} = 0 \implies \frac{c}{b} e^{-bt} = \Phi(-at)|_{t\geq 0}$$ This only gives the data for $\Phi(s)$ when $s \leq 0$. You are still free to prescribe $\Phi(s)$ for $s > 0$ using the initial condition $f(0,x)|_{x\geq 0}$. That is $$ \Phi(s)|_{s\geq 0} = \left(f(0,s) + \frac{c}{b}\right)e^{-bt}$$

Note that when $a = 0$ the boundary condition only tells you that $\Phi(0) = c/b$. It tells you absolutely nothing about $\Phi$ for other values of the argument.

Edit after original question was updated

If you are now prescribing initial-boundary data, where $f(t,0) = 0$ for positive times and $f(0,x) = 0$ for positive $x$, then what we need to do is to see how this translates to conditions on $\Phi$. What you need to pay attention to is the fact that the initial and boundary data can only be used to prescribe the free function on corresponding domains. See below for an illustration.

Now $f(t,x) = - \frac{c}{b} + e^{bt}\Phi(x - At^2 / 2)$. We solve this to get $$ \Phi(x - At^2/2) = e^{-bt}\left(f(t,x) + \frac{c}b \right) $$

Using that if $t > 0$ and $x = 0$, $f(t,x) = 0$. We have that

$$ \Phi(0 - At^2/2) = e^{-bt}\frac{c}b $$

for every $t > 0$. This tell you that, changing variables via $s = At^2/2$, we have that for every $s > 0$ (solving $t = \sqrt{2s/A}$)

$$ \Phi(-s) = e^{-b \sqrt{2s/A}} \frac{c}{b} $$

This specifies your arbitrary function along the negative real line.

Now we use that if $t = 0$ and $x > 0$, $f(t,x) = 0$. Plugging in we have

$$ \Phi(x - 0) = e^{-b\cdot 0} \frac{c}{b} = \frac{c}{b} $$

this means that $\Phi(x)$ for $x$ positive is equal to $c/b$.

In other words, you've almost gotten the answer already! What you need to pay attention to is the domain over which your initial/boundary can be applied to give you information. In this case, the boundary data can only be used to specify $\Phi$ along the negative real line, while the initial data can only be used to specify $\Phi$ along the positive real line.

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you are right, I'm looking for a solution in the first quadrant, although I'm not sure if it is possible to prescribe physically meaningful initial conditions at $t=0$, and therefore I'm looking for a way to use the hint with $a = 0$. – zeliboba May 19 '12 at 16:15
What do you mean by "hint with $a = 0$"? If you don't prescribe initial conditions, then in the first quadrant any function $\Phi$ such that $\Phi(-at)|_{t\geq 0} = \frac{c}{b}e^{-bt}$ (in particular, $\Phi(s)$ is arbitrary for $s > 0$ solves the PDE. You simply do not have enough information to specify $\Phi$ completely. – Willie Wong May 21 '12 at 7:33
by hint I mean the solution of ODE obtained from PDE when $a$ is set to zero, i.e. the physically meaningful solution of ${df}/{dt} = c f + b$ is $f = -c/b (1 - e^{bt})$. I assume it should be also valid for PDE with properly chosen $\Phi$, therefore $\Phi$ should be reduced to $c/b$ when $a$ goes to zero. – zeliboba May 21 '12 at 7:55
Well, by saying "the physically meaningful solution of $f' = cf + b$ is $f = - c/b(1-e^{bt})$", you are throwing away a bunch of other solutions (the equation is time translation invariant, so $-c/b(1-e^{bt - d})$ are all solutions). By "physically meaningful" I presume you mean some set of conditions which you have not mentioned in the question statement. Like I said, if you want to further restrict your solution class you have to provide those additional information. – Willie Wong May 21 '12 at 8:10
Also, you have to be even more careful with this "physically reasonable" business. If you formally take $a \to 0$ you end up with an ODE $f' = bf + c$ which is not compatible with your boundary condition $f(t,0) = 0$ unless $c = 0$. Lastly, I am beginning to wonder if what you are asking about is really how to use the method of characteristics. – Willie Wong May 21 '12 at 8:22

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