Solve the given first order PDE

[chwala]

Homework Statement

See attached. This question is set by me.

Relevant Equations

pde

Solve the given PDE for $u(x,t)$;

$\dfrac{∂u}{∂t} +10 \dfrac{∂u}{∂x} + 9u = 0$

$u(x,0)= e^{-x}$

$-∞ <x<∞ , t>0$


In my lines i have,

$x_t = 10$

$x(t) = 10t+a$

$a = x(t) - 10t$

also,

$u(x(t),t)= u(x(0),0)e^{-9t}$

note this is from, integrating

$u_t[u(x(t),t] = -9u(x(t),t)$ by use of separation of variables.

...

Therefore,

$u(x,t) = (e^{10t-x} ⋅e^{-9t})=e^{t-x}$

I hope i am getting the flow right... cheers.

 

[Orodruin]

I hope i am getting the flow right... cheers.

The easiest way to check this is to insert your result in the original PDE and initial condition to verify that it is the solution.

 

[Mark44]

I hope i am getting the flow right...

The easiest way to check this is to insert your result in the original PDE and initial condition to verify that it is the solution.

@chwala, what Orodruin said above is something I've also said at least once. If you're working problems at this level it's both essential and easy to do this sanity check on your work. I usually start by making sure the initial condition is satisfied, since that's easy to do. Then I verify that my solution satisfies the original ODE or PDE.

 

[pasmith]

Since the equation is linear with constant coefficients, the solutions will be sums of exponentials. It makes sense, given the boundary condition, to try $u(x,t) = T(t)e^{-x}$ with $T(0) = 1$. This yields $$e^{-x}\left( T' - 10T + 9T\right) = 0$$ so that $T' - T = 0$. Hence $T(t) = e^{t}$ and $$u(x,t) = e^{t-x}.$$

 

[Orodruin]

Since the equation is linear with constant coefficients, the solutions will be sums of exponentials.

The general solution is
$$
e^{-9t} f(x - 10t)
$$
That the result is a simple exponential is a consequence of the particular initial condition also being an exponential. Had the initial condition been $u(x,0) = x^3$, then the result would have been
$$
u(x,t) = e^{-9t} (x - 10t)^3
$$
I am not sure I would call that a sum of exponentials.