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Solving PDE by using Laplace Transform

Markov

Member
Feb 1, 2012
149
Given

$\begin{aligned} & {{u}_{t}}={{u}_{xx}},\text{ }x>0,\text{ }t>0 \\
& u(x,0)={{u}_{0}}, \\
& {{u}_{x}}(0,t)=u(0,t).
\end{aligned}
$

I need to apply the Laplace transform to solve it. I'll denote $u(x,s)=\mathcal L(u(x,\cdot))(s),$ so for the first line I have $s\cdot u(x,s)-u(x,0)=\dfrac{\partial^2 u(x,s)}{\partial x^2},$ now here's my problem, when I did this problem my professor told me I can't apply the transform to the condition $u(x,0)$ why? Well after this for the third line I have $\dfrac{\partial u(x,s)}{\partial x}-u(0,s)=0$ (1). So we have to solve $\dfrac{{{\partial }^{2}}u(x,s)}{\partial {{x}^{2}}}-s\cdot u(x,s)=-{{u}_{0}}$ which gives a a solution $u(x,s)=c_1e^{-\sqrt sx}+c_2e^{\sqrt sx}+\dfrac{u_0}s$ (2).

Now do I need to use (2) and (1) to find the constants? And after that I need to find the inverse Laplace transform, so far, is it correct?
Thanks!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Given

$\begin{aligned} & {{u}_{t}}={{u}_{xx}},\text{ }x>0,\text{ }t>0 \\
& u(x,0)={{u}_{0}}, \\
& {{u}_{x}}(0,t)=u(0,t).
\end{aligned}
$

I need to apply the Laplace transform to solve it. I'll denote $u(x,s)=\mathcal L(u(x,\cdot))(s),$ so for the first line I have $s\cdot u(x,s)-u(x,0)=\dfrac{\partial^2 u(x,s)}{\partial x^2},$ now here's my problem, when I did this problem my professor told me I can't apply the transform to the condition $u(x,0)$ why?
You could, actually, but it wouldn't gain you anything. There's no $t$ dependence in that equation anywhere.

Well after this for the third line I have $\dfrac{\partial u(x,s)}{\partial x}-u(0,s)=0$ (1). So we have to solve $\dfrac{{{\partial }^{2}}u(x,s)}{\partial {{x}^{2}}}-s\cdot u(x,s)=-{{u}_{0}}$ which gives a a solution $u(x,s)=c_1e^{-\sqrt sx}+c_2e^{\sqrt sx}+\dfrac{u_0}s$ (2).

Now do I need to use (2) and (1) to find the constants? And after that I need to find the inverse Laplace transform, so far, is it correct?
Thanks!
You can always plug solutions into the DE to verify that they are correct. I would say they are, and yes, you need to do the inverse LT to find the final solution.
 

Markov

Member
Feb 1, 2012
149
Okay but, do I need to find the inverse now for (2) and that's all?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Use the Initial Value Theorem on $u(x,0)=u_{0}$, and (2), to obtain $c_{1}$ and $c_{2}$. Then take the Inverse LT, and you're done.
 

Markov

Member
Feb 1, 2012
149
If I use $u(x,s)$ at $s=0,$ then (2) will give me problems with the third term. :(
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
If I use $u(x,s)$ at $s=0,$ then (2) will give me problems with the third term. :(
But when you use the IVT, you're not taking the limit as $s\to 0$, but as $s\to\infty$. The word initial refers to the time domain, not the frequency domain. Also keep in mind that you're imposing the IVT as a condition. Incidentally, your (1) really ought to be

$$\frac{\partial u(x,s)}{\partial x}\Bigg|_{x=0}=u(0,s).$$
 

Markov

Member
Feb 1, 2012
149
Oh yes, that now makes sense!