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I'm having trouble figuring out how to correctly apply boundary conditions to the general solution of a PDE. I'm seeking a particular solution $u(x,t)$ for \begin{align} 4u_{x} + u_{t} &= 0 \qquad 0 < x < \infty \\ u(x,0) &= 0\qquad 0 < x <\infty \\ u(0,t)& = te^{-t} \qquad t>0 \end{align} Specifically, I'm not exactly sure how to handle the condition $u(x,0) = 0$. Here's my work:

In general, a solution to the PDE of the form $au_{x} + bu_{t} = 0$ is $u(x, t) = f(bx - at)$. For our PDE, $a=4$, $b=1$ and we have $u(x, t) = f(x - 4t)$.

For the initial condition $u(0, t) = f(-4t) = te^{-t}$. Let $w = -4t$. Then $f(-4t) = f(w) = -\frac{w}{4}e^{\frac{w}{4}}$ which implies that one particular solution of the PDE is \begin{equation} u(x, t) = f(x - 4t) = (t - \frac{x}{4})e^{\frac{x}{4} - t} \end{equation} Next, $u(x,0) = f(x) = 0 = -\frac{x}{4}e^{\frac{x}{4}}$. But this implies $x = 0$, forcing $u(x, t) = te^{-t}$, which clearly doesn't satisfy our original PDE.

Not sure where to go from here. The problem also stipulates that a solution to the equation $4u_{t} + u_{x}=0$ has no solution and asks us to explain why. I'm hoping I'll be able to answer that question once I figure out what I'm doing wrong here.

Any advice on how to correct my error(s) would be greatly appreciated!

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Your calculus is correct.

The general solution is $u=f(x-4t)$

As you found it, the condition $u(0,t)=te^{-t}$ implies $u(x,t)=(t - \frac{x}{4})e^{\frac{x}{4} - t}$

If you do the same calculus with the condition $u(x,0)=0$, you obtain $f(x)=0$ which implies $u(x,t)=0$.

The two conditions are incompatible because they leads to two different solutions $u(x,y)$. As a consequence, there is no function $u(x,y)$ solution of the PDE which satisfy both two conditions : The problem has no solution.

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