Solution of Laplace eq with Robbins boundary condition

[mario99]

Yes, but I'm confused in how I can evaluate u(x,y)

The integral F_k requires me to input u(0,y) and integrating. But u(0,y) also contains u(0,y).

[imath]F(y)[/imath] depends on [imath]u(0,y)[/imath]. And since [imath]u(0,y)[/imath] is unknown function, you cannot evaluate. You can only write the general solution of the PDE. If they want you to evaluate [imath]F_k[/imath], they must give the value of [imath]u(0,y)[/imath].

For example, if they tell you that:

[imath]u(0,y) = 5y[/imath]

Then,

you can solve [imath]F_k[/imath] and any other constants that depend on [imath]F_k[/imath].

It is not the end of the world. You can play around by choosing any value for [imath]u(0,y)[/imath] and solve [imath]F_k[/imath] if you want. Who will stop you?

 

[shreddinglicks]

I see what you mean. Thanks, this is really helpful.

I understand how you broke up the problem using the superposition principle.

If I use the superposition principle could I also solve a transient problem like below?

 

[mario99]

I see what you mean. Thanks, this is really helpful.

I understand how you broke up the problem using the superposition principle.

If I use the superposition principle could I also solve a transient problem like below?

View attachment 38699

The first equation is a two dimensional heat equation and it will always come with homogeneous boundary conditions in rectangular coordinate. Let me guess, you have made up that problem to have nonhomogeneous boundary conditions, right?

The last equation is a valid two dimensional heat equation and it can be solved normally by separation of variables. The second and third equations are just Laplace equations and they can be solved the same way we did with the OP problem no matter what are the boundary conditions.

The one dimensional heat equation can have nonhomogeneous boundary conditions. It can even have a time-dependent source function. The heat equation with higher dimensions in other coordinate systems can have nonhomogeneous boundary conditions.

Partial Differential Equations become more interesting when the domain is [imath]-\infty < x < \infty[/imath]. Most of the time, you will have no choice but to solve them by either Fourier or Laplace transform.

 

[mario99]

If you are looking for a three dimensional fun challenging Laplace equation in rectangular coordinate system, try this:

[imath]\displaystyle \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = 0[/imath]

[imath]u(0, y, z) = 0[/imath]
[imath]u(a, y, z) = 0[/imath]
[imath]u(x, 0, z) = 0[/imath]
[imath]u(x, b, z) = 0[/imath]
[imath]u(x, y, 0) = f(x,y)[/imath]
[imath]u(x, y, c) = 0[/imath]

[imath]0 < x < a[/imath]
[imath]0 < y < b[/imath]
[imath]0 < z < c[/imath]

----------------------------------------------

If you are looking for a two dimensional fun challenging Heat equation in rectangular coordinate system, try this:

[imath]\displaystyle k\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right) = \frac{\partial u}{\partial t}[/imath]

[imath]\displaystyle \frac{\partial u}{\partial x}\bigg|_{x=0} = 0, \ \ \ \ \ \frac{\partial u}{\partial x}\bigg|_{x=1} = 0[/imath]

[imath]\displaystyle \frac{\partial u}{\partial y}\bigg|_{y=0} = 0, \ \ \ \ \ \frac{\partial u}{\partial y}\bigg|_{y=1} = 0[/imath]

[imath]\displaystyle u(x,y,0) = f(x,y)[/imath]

[imath]\displaystyle 0 < x < 1[/imath]
[imath]\displaystyle 0 < y < 1[/imath]
[imath]\displaystyle t > 0[/imath]

 

[shreddinglicks]

I will try those. I will also try the problem I made up.

Thanks, for all your help so far. You've really helped me.

 

[mario99]

I will try those. I will also try the problem I made up.

Thanks, for all your help so far. You've really helped me.

I have never seen a two dimensional heat equation that has nonhomogeneous boundary conditions in rectangular coordinate system. But by playing around I could force the boundaries to be homogenous. It is similar to your idea of the superposition principle, but an easier version.

Let [imath]u(x,y,t) = g(x,y) + v(x,y,t)[/imath].

After substituting this into the original differential equation (first equation) and applying the boundary conditions, you will get these two equations:

[imath]\displaystyle \frac{\partial^2 g}{\partial x^2} + \frac{\partial^2 g}{\partial y^2} = 0[/imath]

[imath]g(0,y) = f(y)[/imath]
[imath]g(1,y) = 0[/imath]
[imath]g(x,0) = 0[/imath]
[imath]g(x,1) = g(x)[/imath]

[imath]0 < x < 1[/imath]
[imath]0 < y < 1[/imath]

-----------------------------------------

[imath]\displaystyle \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = \frac{\partial v}{\partial t}[/imath]

[imath]v(0,y,t) = 0[/imath]
[imath]v(1,y,t) = 0[/imath]
[imath]v(x,0,t) = 0[/imath]
[imath]v(x,1,t) = 0[/imath]
[imath]v(x,y,0) = h(x,y) - g(x,y) = m(x,y)[/imath]

[imath]0 < x < 1[/imath]
[imath]0 < y < 1[/imath]
[imath]t > 0[/imath]

----------------------------

The first equation is just a Laplace equation that can be solved by the same method we have used in the OP problem. And the second equation is the normal two dimensional Heat equation that we know and it can solve by separation of variables.

 

[shreddinglicks]

I attempted a very similar problem combining all that we have discussed.

 

[shreddinglicks]

 

[shreddinglicks]

 

[mario99]

Your step:

[imath]c_4 = -c_3\tan(\lambda b)[/imath] which is meant to be [imath]c_4 = -c_3\tanh(\lambda b)[/imath]

It is not wrong to do that, but I do not recommend it. I prefer to leave the hyperbolic solution alone and to find the constants from the solution of the superposition principle[imath]\displaystyle \ \rightarrow \Psi_2(x,y) = \sum_{n=1}^{\infty} \cdots[/imath]

When you solved the Heat equation, you wrote the second boundary condition as [imath]\displaystyle \frac{du(b,y,t)}{dy} = 0[/imath]. I am sure you meant to write [imath]\displaystyle \frac{du(x,b,t)}{dy} = 0[/imath].

You write the eigenvalues as:

[imath]\displaystyle \lambda = \frac{n\pi}{a} + \frac{\pi}{2a}[/imath]

when they are simply:

[imath]\displaystyle \lambda = \frac{(2n - 1)\pi}{2a}[/imath]

I will show you how to do it. You have found the first solution as:

[imath]X(x) = c_2\sin(\lambda x)[/imath]

The second boundary condition is [imath]X'(a) = 0[/imath].

Then

[imath]0 = c_2\lambda\cos(\lambda a)[/imath]

We know that [imath]c_2 \neq 0 \ \text{and} \ \lambda \neq 0[/imath], so it must be [imath]\cos(\lambda a) = 0[/imath]

The cosine function is zero when [imath]\displaystyle \lambda a = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2},\cdots \frac{k\pi}{2} \ \ \ \ \ (\lambda a > 0)[/imath]

It is obvious that [imath]k[/imath] is an odd integer, so [imath]k = 2n - 1[/imath].

Then

[imath]\displaystyle \lambda a = \frac{(2n - 1)\pi}{2}[/imath]

Or

[imath]\displaystyle \lambda = \frac{(2n - 1)\pi}{2a}[/imath]

Or

[imath]\displaystyle \lambda_n = \frac{(2n - 1)\pi}{2a}, \ \ \ \ \ n =1,2,3,\cdots[/imath]

I have not checked everything in details, but I think that you have solved the problem perfectly!