Agohofidov6

2022-01-15

Problem $u}_{t}-{u}_{xx}={e}^{t$ in $\mathbb{R}\times (0,+\mathrm{\infty})$

kalupunangh

Beginner2022-01-16Added 29 answers

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

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

Rearranging terms

$\frac{{\partial}^{2}U}{\partial {x}^{2}}-sU=-\mathrm{cos}\left(3x\right)-\frac{1}{s-1}$

The solution in spatial domain is

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

Back in time and using fraction decomposition

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

the solution reads

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

Another approach consists in using Laplace transform. Let

Rearranging terms

The solution in spatial domain is

Back in time and using fraction decomposition

the solution reads

peterpan7117i

Beginner2022-01-17Added 39 answers

Step 1

There were some typos, so I post all the computations:

$\varphi}_{t}+{\xi}^{2}\varphi =\sqrt{2\pi \delta}\left(\xi \right){e}^{t$

$\varphi}_{t}{e}^{{\xi}^{2}t}+{\xi}^{2}{e}^{{\xi}^{2}t}\varphi =\sqrt{2\pi \delta}\left(\xi \right){e}^{t+{\xi}^{2}t$

$\left(\varphi {e}^{{\xi}^{2}t}\right)t=\sqrt{2\pi \delta}\left(\xi \right){e}^{t+{\xi}^{2}t}$

$\varphi (t,\xi ){e}^{{\xi}^{2}t}-\varphi (0,\xi )=\sqrt{2\pi}\frac{\delta \left(\xi \right)}{1+{\xi}^{2}}({e}^{t(1+{\xi}^{2})}-1)$

With general initial condition, we have the equation

$\varphi (t,\xi )=\varphi (0,\xi ){e}^{-{\xi}^{2}t}+\sqrt{2\pi}\frac{\delta \left(\xi \right)}{{\xi}^{2}+1}({e}^{t}-{e}^{-{\xi}^{2}t})$

In this case,

$\varphi (0,\xi ){e}^{-{\xi}^{2}t}=\sqrt{2\pi}\frac{1}{2}(\delta (x-3){e}^{-{\xi}^{2}t}+\delta (x+3){e}^{-{\xi}^{2}t})$

yields

${F}^{-1}\left(\varphi (0,\xi ){e}^{-{\xi}^{2}t}\right)\left(x\right)=\frac{1}{2}({e}^{-9t-3ix}+{e}^{-9+3ix})={e}^{-9t}\frac{1}{2}({e}^{-3ix}+{e}^{3ix})$

So the solution is

$u(x,t)=({e}^{t}-1)+{e}^{-9t}\frac{1}{2}({e}^{-3ix}+{e}^{3ix})=({e}^{t}-1)+{e}^{-9t}\mathrm{cos}\left(3x\right)$

You can check that this satisfies the PDE:

${u}_{t}(x,t)={e}^{t}-9{e}^{-9t}\mathrm{cos}\left(3x\right)$

There were some typos, so I post all the computations:

With general initial condition, we have the equation

In this case,

yields

So the solution is

You can check that this satisfies the PDE:

alenahelenash

Expert2022-01-24Added 556 answers

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,

We consider the simpler problems

1)

2)

(And, in order for uniqueness, all problems are coupled with the additional physical constraint that the integral of

If we have found a solution

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

The fundamental solution is the solution

This is a very well known problem, and the solution is

The Green's function is the solution

And again, the solution of this problem is also well known,

Step 2

Now that these two solutions are known, we can write

I.e,

And,

In your case 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.

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