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I have a task that looks like $$ \frac{\partial u}{\partial t} = 2\frac{\partial^2 u}{\partial x^2} + \frac{\partial u}{\partial x} + u + 3e^{t},\, u\left(x, 0\right)=\sin{x}$$

Please help, thanks :)

Pavel Naberezhnev
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  • Just a thought. Set $v(x,s)=\mathcal{L}\Big(u(x,t)\Big)$ where $\mathcal{L}$ denotes the Laplace transform. If you apply $\mathcal{L}$ to your PDE and rearrange terms you get $$2v_{xx}+v_{x}+(1-s)v=-\sin(x)+\frac{3}{1-s}$$ where $s>1$. The differential of this transformed PDE only depends on $x$ and can be solved using standard techniques by treating $s$ as a constant. Once you have $v(x,s)$ you can using $\mathcal{L}^{-1}$ to retrieve $u(x,t)$. – Matthew H. Jan 17 '21 at 06:14
  • For example, the PDE $$u_{xx}+e^yu=0$$ has general solution $$u(x,y)=f(y)\cos\Big(e^{y/2}x\Big)+g(y)\sin\Big(e^{y/2}x\Big)$$ by treating $y$ as constant. Here $f,g$ are arbitrary functions. – Matthew H. Jan 17 '21 at 06:14
  • Thanks! :)))))) – Pavel Naberezhnev Jan 17 '21 at 09:27

2 Answers2

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We continue Matthew Pilling solution: Using Laplace transform method we get ode $$2v_{xx}+v_{x}+(1-s)v=-\sin(x)+\frac{3}{1-s}$$ Particular solution is $$v=\frac{s\sin x+\sin x+\cos x}{s^2+2s+2}+\frac{3}{(s-1)^2}$$ Final solution of Cauchy problem is $$u(x,t)=\mathcal{L}^{-1}v=e^{-t}\sin(x+t)+3te^t$$

Aleksas Domarkas
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Some details: $$\mathcal{L}^{-1}\left(\ \frac{3}{(s-1)^2}\right)=3te^t$$ $$\mathcal{L}^{-1}\left(\ \frac{1}{s^2+2s+2}\right)=e^{-t}\sin t$$ $$\mathcal{L}^{-1}\left(\ \frac{s}{s^2+2s+2}\right)=e^{-t}(\cos t-\sin t)$$

Aleksas Domarkas
  • 2,401