Is d'Alembert's Formula Correct for Neumann Boundary Conditions in PDEs?

[docnet]

Homework Statement

solve this PDE

Relevant Equations

$$\tilde{u}(t,x)=\frac{1}{2}\Big[\tilde{g}(x+t)+\tilde{g}(x-t)\Big]+\frac{1}{2}\int^{x+t}_{x-t}\tilde{h}(y)dy$$

Hi all, I was hoping someone could check whether I computed part (4) correctly, where i find the solution u(t,x) using dAlembert's formula:
$$\boxed{\tilde{u}(t,x)=\frac{1}{2}\Big[\tilde{g}(x+t)+\tilde{g}(x-t)\Big]+\frac{1}{2}\int^{x+t}_{x-t}\tilde{h}(y)dy}$$
Does the graph of the solution look correct or does it look off?

I posted all my work just in case someone feels like looking over them ;)

We consider for an unknown function $u:[0,\infty)\times[0,\infty)\rightarrow \mathbb{R}$ the Neumann initial value problem given by
$$
\begin{cases}
\partial_t^2u-\partial_x^2u=0 & \text{in}\quad (0,\infty)\times (0,\infty) \\
u=g & \text{on}\quad\{t=0\}\times(0,\infty)\\\partial_tu=h & \text{on}\quad \{t=0\}\times (0,\infty) \\
\partial_xu=0 & \text{on}\quad (0,\infty)\times\{x=0\}\end{cases}$$
(1) Use even extension in the spatial variable $x$ to extend the unknown function $u$ and the data $g,h$ to functions $\tilde{u}:[0,\infty)\times \mathbb{R}\rightarrow \mathbb{R}$ and $\tilde{g},\tilde{h}:\mathbb{R}\rightarrow \mathbb{R}$. \\\\We define the even extended functions with the following piece-wise functions
$$\tilde{u}(t,x)=
\begin{cases}
u(t,x) & t\geq 0, x\geq0 \\
u(t,-x) & t\geq 0, x<0
\end{cases}$$$$\ \tilde{g}(x)=
\begin{cases}
g(x) & x\geq0 \\
g(-x) & x<0
\end{cases}$$
$$\tilde{h}(x)=
\begin{cases}
h(x) & x\geq0 \\
h(-x) & x<0
\end{cases}$$
(2) The initial value problem satisfied by the unknown function $\tilde{u}$ with data $\tilde{g}$ and $\tilde{h}$ is
$$\begin{cases}
\partial_t^2\tilde{u}-\partial_x^2\tilde{u}=0 & \text{in}\quad (0,\infty)\times \mathbb{R} \\
\tilde{u}(0,\cdot)=\tilde{g} & \text{on}\quad\{t=0\}\times\mathbb{R}\\\partial_t\tilde{u}(0,\cdot)=\tilde{h} & \text{on}\quad \{t=0\}\times\mathbb{R} \end{cases}$$
The first equation is true because for $x<0$
$$\partial_t^2u(t,x)\Rightarrow \partial_t^2u(t,-x)$$
$$\partial_x^2u(t,x)\Rightarrow \partial_x^2u(t,-x)$$
(3) We solve this problem by
plugging in $u_0=\tilde{g}$ and $u_1=\tilde{h}$ into d'Alembert's formula
$$\boxed{\tilde{u}(t,x)=\frac{1}{2}\Big[\tilde{g}(x+t)+\tilde{g}(x-t)\Big]+\frac{1}{2}\int^{x+t}_{x-t}\tilde{h}(y)dy}$$
(4) To recover $u$ from $\tilde{u}$, We consider the solution formula for the following two cases: $0\leq t\leq x$ and
$0\leq x\leq t$.

Let $t\geq 0, x\geq 0$.
$$1. \quad x+t\geq 0 \Rightarrow \tilde{g}(x+t)=g(x+t)$$
$$2. \quad x-t \Rightarrow x-t\geq 0 \quad \text{or} \quad x-t<0$$
If $x-t\geq 0$ $$\tilde{g}(x-t)\Rightarrow(x-t)$$
$$\int^{x+t}_{x-t}\tilde{h}(y)dy\Rightarrow \int^{x+t}_{x-t}h(y)dy$$
If $x-t<0$
$$\tilde{g}(x-t)=g(t-x)$$
$$\int^{x+t}_{x-t}\tilde{h}(y)dy\Rightarrow \int_{t-x}^{x+t}h(y)dy$$
Hence our solution is given by
$$u(t,x)=\frac{1}{2}\Big[g(x+t)+g(x-t)\Big]+\frac{1}{2}\int^{x+t}_{x-t}h(y)dy \quad \text{for} \quad 0\leq t \leq x $$ $$
u(t,x)=\frac{1}{2}\Big[g(x+t)+g(t-x)\Big]+\frac{1}{2}\int_{t-x}^{x+t}h(y)dy \quad \text{for} \quad 0\leq x \leq t$$
(5) We choose $g(x)=x^2$ and $h(y)=y$ then the solution is given by
$$u(t,x)=x^2+t^2+xt\quad \text{for} \quad 0\leq t \leq x \quad \text{and} \quad 0\leq x \leq t$$
Sketch of the solution for $t=0,1,2,3$

 

[Orodruin]

As you can see from your plot, your solution clearly does not satisfy the boundary condition $u_x(t,0) = 0$. Your problem lies in your assumption about the expression of the integral term in d'Alembert's formula for $t > x$ as you are removing part of the integration instead of using that the integrand is an even function.

 

[docnet]

As you can see from your plot, your solution clearly does not satisfy the boundary condition $u_x(t,0) = 0$. Your problem lies in your assumption about the expression of the integral term in d'Alembert's formula for $t > x$ as you are removing part of the integration instead of using that the integrand is an even function.

Yes! thank you for noticing as always. Here is the corrected version.

We consider for an unknown function $u:[0,\infty)\times[0,\infty)\rightarrow \mathbb{R}$ the Neumann initial value problem given by
$$
\begin{cases}
\partial_t^2u-\partial_x^2u=0 & \text{in}\quad (0,\infty)\times (0,\infty) \\
u=g & \text{on}\quad\{t=0\}\times(0,\infty)\\\partial_tu=h & \text{on}\quad \{t=0\}\times (0,\infty) \\
\partial_xu=0 & \text{on}\quad (0,\infty)\times\{x=0\}\end{cases}$$
(1) Use even extension in the spatial variable $x$ to extend the unknown function $u$ and the data $g,h$ to functions $\tilde{u}:[0,\infty)\times \mathbb{R}\rightarrow \mathbb{R}$ and $\tilde{g},\tilde{h}:\mathbb{R}\rightarrow \mathbb{R}$. \\\\We define the even extended functions with the following piece-wise functions
$$\tilde{u}(t,x)=
\begin{cases}
u(t,x) & t\geq 0, x\geq0 \\
u(t,-x) & t\geq 0, x<0
\end{cases}$$$$\ \tilde{g}(x)=
\begin{cases}
g(x) & x\geq0 \\
g(-x) & x<0
\end{cases}$$
$$\tilde{h}(x)=
\begin{cases}
h(x) & x\geq0 \\
h(-x) & x<0
\end{cases}$$
(2) The initial value problem satisfied by the unknown function $\tilde{u}$ with data $\tilde{g}$ and $\tilde{h}$ is
$$\begin{cases}
\partial_t^2\tilde{u}-\partial_x^2\tilde{u}=0 & \text{in}\quad (0,\infty)\times \mathbb{R} \\
\tilde{u}(0,\cdot)=\tilde{g} & \text{on}\quad\{t=0\}\times\mathbb{R}\\\partial_t\tilde{u}(0,\cdot)=\tilde{h} & \text{on}\quad \{t=0\}\times\mathbb{R} \end{cases}$$
The first equation is true because for $x<0$
$$\partial_t^2u(t,x)\Rightarrow \partial_t^2u(t,-x)$$
$$\partial_x^2u(t,x)\Rightarrow \partial_x^2u(t,-x)$$
(3) We solve this problem by
plugging in $u_0=\tilde{g}$ and $u_1=\tilde{h}$ into d'Alembert's formula
$$\boxed{\tilde{u}(t,x)=\frac{1}{2}\Big[\tilde{g}(x+t)+\tilde{g}(x-t)\Big]+\frac{1}{2}\int^{x+t}_{x-t}\tilde{h}(y)dy}$$
(4) To recover $u$ from $\tilde{u}$, We consider the solution formula for the following two cases: $0\leq t\leq x$ and
$0\leq x\leq t$.
Let $t\geq 0, x\geq 0$.
$$1. \quad x+t\geq 0 \Rightarrow \tilde{g}(x+t)=g(x+t)$$
$$2. \quad x-t \Rightarrow x-t\geq 0 \quad \text{or} \quad x-t<0$$
If $x-t\geq 0$ $$\tilde{g}(x-t)\Rightarrow(x-t)$$
$$\int^{x+t}_{x-t}\tilde{h}(y)dy\Rightarrow \int^{x+t}_{x-t}h(y)dy$$
If $x-t<0$
$$\tilde{g}(x-t)=g(t-x)$$
$$\int^{x+t}_{x-t}\tilde{h}(y)dy\Rightarrow \int_{t-x}^{0}h(y)dy+\int_{x+t}^{0}h(y)dy$$
Hence our solution is given by
$$\boxed{u(t,x)\begin{cases}\frac{1}{2}\big[g(x+t)+g(x-t)\big]+\frac{1}{2}\int^{x+t}_{x-t}h(y)dy \quad \text{for} \quad 0\leq t \leq x \\
\frac{1}{2}\big[g(x+t)+g(t-x)\big]+\frac{1}{2}\int_0^{x+t}h(y)dy+\frac{1}{2}\int_0^{t-x}h(y)dy \quad \text{for} \quad 0\leq x \leq t\end{cases}}$$
(5) To make a sketch we choose an arbitrary polynomial data $g(x)=x^2$ and $h(y)=y$. The corresponding solution is given by
$$\boxed{u(t,x)=\begin{cases}x^2+t^2+xt\quad \text{for} \quad 0\leq t \leq x \quad \\ \frac{3x^2+3t^2}{2} \quad\text{for}\quad 0\leq x \leq t\end{cases}}$$
Here is a sketch of the solution for $0\leq t\leq x$ at $t=0,1,2,3$ (the solution is valid for $t\leq x$)

And here is a sketch of the solution for $0\leq x\leq t$ at $t=0,1,2,3$

 

[Orodruin]

Better, but your graphs are confusing because they are not valid for all the range you are plotting. I suggest you make a single plot for x>0 with the correct selection of the function for different values of x depending on the time t you are plotting for.

 

[docnet]

Better, but your graphs are confusing because they are not valid for all the range you are plotting. I suggest you make a single plot for x>0 with the correct selection of the function for different values of x depending on the time t you are plotting for.

Last attempt:

Here is a graph of $u(t_i,x)$ for $t=0,1,2,3$

thank you.