# Solving PDE by using Laplace Transform

#### Markov

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

#### Ackbach

##### Indicium Physicus
Staff member
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
If I use $u(x,s)$ at $s=0,$ then (2) will give me problems with the third term.

#### Ackbach

##### Indicium Physicus
Staff member
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
Oh yes, that now makes sense!