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

• docnet
In summary, the conversation discusses an unknown function and its Neumann initial value problem, where the function is extended using even extension and solved using d'Alembert's formula. The correct solution is provided using the extended function and a sketch is shown for different values of t and x.
docnet
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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##

Delta2
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
Orodruin said:
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##

Last edited:
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
Orodruin said:
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.

1. What is a PDE?

A PDE, or partial differential equation, is a type of mathematical equation that involves multiple variables and their partial derivatives. It is commonly used to model physical phenomena in fields such as physics, engineering, and economics.

2. What are Neumann boundary conditions?

Neumann boundary conditions are a type of boundary condition used in solving PDEs. They specify the value of the derivative of the solution at the boundary of a domain, rather than the value of the solution itself. In other words, they describe how the solution changes at the boundary.

3. How do I solve a PDE with Neumann boundary conditions?

To solve a PDE with Neumann boundary conditions, you will need to use a specific method or technique, such as the finite difference method or the method of characteristics. These methods involve breaking down the PDE into smaller, simpler equations that can be solved using numerical or analytical techniques.

4. What are some applications of solving PDEs with Neumann boundary conditions?

Solving PDEs with Neumann boundary conditions has many practical applications. For example, it can be used to model heat transfer in a material with a fixed temperature at the boundary, or to simulate the flow of fluids in a pipe with a fixed pressure at the ends. It is also commonly used in financial modeling to predict the behavior of stock prices with fixed boundaries.

5. Are there any limitations to solving PDEs with Neumann boundary conditions?

There are some limitations to using Neumann boundary conditions in PDEs. For example, they may not be applicable in certain situations where the boundary conditions are not well-defined or change over time. In addition, the accuracy of the solution may be affected by the choice of boundary conditions, and it is important to carefully consider the physical meaning and implications of the chosen conditions.

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