Problem ut−uxx=et in R×(0,+∞)

Question

kalupunangh · Accepted Answer

Step 1
Another approach consists in using Laplace transform. Let

U (x, s)

be the LT of

u (x, t)

. The eq. writes in Laplace domain

s U - u (x, 0) - \frac{\partial^{2} U}{\partial x^{2}} = \frac{1}{s - 1}

Rearranging terms

\frac{\partial^{2} U}{\partial x^{2}} - s U = - \cos (3 x) - \frac{1}{s - 1}

The solution in spatial domain is

U (x, s) = \frac{1}{s + 9} \cos (3 x) + \frac{1}{s (s - 1)}

Back in time and using fraction decomposition

\frac{1}{s} (s - 1) = \frac{1}{s - 1} - \frac{1}{s}

the solution reads

u (x, t) = e^{- 9 t} \cos (3 x) + e^{+ t} - 1

peterpan7117i · Accepted Answer

Step 1  There were some typos, so I post all the computations:  ϕt+ξ2ϕ=2πδ(ξ)et  ϕteξ2t+ξ2eξ2tϕ=2πδ(ξ)et+ξ2t  (ϕeξ2t)t=2πδ(ξ)et+ξ2t  ϕ(t,ξ)eξ2t−ϕ(0,ξ)=2πδ(ξ)1+ξ2(et(1+ξ2)−1)  With general initial condition, we have the equation  ϕ(t,ξ)=ϕ(0,ξ)e−ξ2t+2πδ(ξ)ξ2+1(et−e−ξ2t)  In this case,  ϕ(0,ξ)e−ξ2t=2π12(δ(x−3)e−ξ2t+δ(x+3)e−ξ2t)  yields  F−1(ϕ(0,ξ)e−ξ2t)(x)=12(e−9t−3ix+e−9+3ix)=e−9t12(e−3ix+e3ix)  So the solution is  u(x,t)=(et−1)+e−9t12(e−3ix+e3ix)=(et−1)+e−9tcos⁡(3x)  You can check that this satisfies the PDE:  ut(x,t)=et−9e−9tcos⁡(3x)

alenahelenash · Accepted Answer

Step 1
When dealing with any mathematical problem, often the best way to do it is to break it up into smaller parts that are easier to deal with. So starting with our original linear inhomogeneous IVP,
$(\partial_{t} - \partial_{x}^{2}) u (t, x) = h (t, x)$
$u (0, x) = u_{0} (x)$
$(t, x) \in R_{> 0} \times R$
We consider the simpler problems
1) $(\partial_{t} - \partial_{x}^{2}) u (t, x) = 0$
$u (0, x) = u_{0} (x)$
2) $(\partial_{t} - \partial_{x}^{2}) u (t, x) = h (t, x)$
$u (0, x) = 0$
(And, in order for uniqueness, all problems are coupled with the additional physical constraint that the integral of $| u |^{2}$ along the x axis is finite for all time)
If we have found a solution $u_{1}$ to (1) and a solution $u_{2}$ to (2) then due to the linearity of the heat operator, $u_{1} + u_{2}$ is a solution of (0).
What is nice about this approach is that (1) and (2) are already well studied problems. For the first, we use the fundamental solution of the heat equation, also known as the heat kernel, and for the second, we use the Green's function (see item 14 in the linked table).
In the following, assume $(τ, ξ) \in \geq 0 \times R$
The fundamental solution is the solution $F (t, x; ξ)$ of the homogeneous IVP
$(\partial_{t} - \partial_{x}^{2}) u (t, x) = 0$
$u (0, x) = δ (x - ξ)$
This is a very well known problem, and the solution is
$F (t, x; ξ) = - \frac{(x - ξ)^{2})}{\sqrt{4 π t}}$
The Green's function is the solution $G (t, x; τ, ξ)$ of the inhomogeneous IVP with zero initial conditions,
$(\partial_{t} - \partial_{x}^{2}) u (t, x) = δ (t - τ) δ (x - ξ)$
$u (0, x) = 0$
And again, the solution of this problem is also well known,
$G (t, x; τ, ξ) = \frac{\exp (- \frac{(x - ξ)^{2}}{4 (t - τ)})}{\sqrt{4 π (t - τ)}} = F (t - τ, x; ξ)$
Step 2
Now that these two solutions are known, we can write $u_{1}$ and $u_{2}$ as convolutions,
$u_{1} = F_{x} \times u_{0}$
$u_{2} = G_{(t, x)} \times h$
I.e,
$u_{1} (t, x) = \int_{R} F (t, x; ξ) u_{0} (ξ) d ξ$
$u_{2} (t, x) = \int_{0}^{t} \int_{- \infty}^{\infty} G (t, x; \int, ξ) h (τ, ξ) d ξ d τ$
And,
$u = u_{1} + u_{2}$
In your case with $u_{0} (x) = \cos (3 x), h (t, x) = e^{t}$ , you end up with.
Though often this procedure will not give you closed forms all that easily, it is the best way to deal with these kinds of problems in general, because typically closed forms are just not possible. To get closed forms, one would usually use separation of variables, but I stress that, in general, techniques like separation of variables will just fail.

Problem u_{t}-u_{x x}=e^{t} in \mathbb{R}\times(0,+\infty)

Answered question

Answer & Explanation

New Questions in Analysis