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I have to solve the following one Steady State Mixed Boundary Value Problem and i need some help with the analytical solution! Considering the Partial Differential Equation: $$ {\nabla }^{2}T=100$$ applied in a rectangle $ \Omega$ where $ \Omega :\left\{0<x<1, 0\le y\le 0.5\right\}$ with the boundary conditions which given below: $$ \left.\begin{array}{r}T\left(0,y\right)=40,\\ T\left(x,0.5\right)=40,\\ T\left(1,y\right)=40,\\ \nabla T\left(x,0\right)=500\end{array}\right\} $$ I tried the Seperation of Variables Technique where: $$ T\left(x,y\right)={X}_{\left(x\right)}{Y}_{\left(y\right)}\to \frac{{X}_{\left(x\right)}^{"}}{{X}_{\left(x\right)}}=-\frac{{Y}_{\left(y\right)}^{"}}{{Y}_{\left(y\right)}}=-\lambda ,\lambda \in \mathbb{R} $$ and obtain the following: $$ \left(A\right):\left\{\begin{array}{l}{X}_{\left(x\right)}^{"}=-\lambda {X}_{\left(x\right)}\\ X\left(1\right)=X\left(0\right)=40\end{array},\left(0<x<1\right)\right. $$ and $$ \left(B\right):\left\{\begin{array}{l}{Y}_{\left(y\right)}^{"}=\lambda {Y}_{\left(y\right)}\\ Y\left(0.5\right)=40\end{array},\left(0\le y\le 0.5\right)\right. $$ From the (A): $ {\lambda }_{n}={n}^{2}{\pi }^{2}\space, and \space {X}_{n}\left(x\right)=sin\left(n\pi x\right),n\in \mathbb{N}.$ And from the (B): $ {Y}_{\left(y\right)}^{"}={n}^{2}{\pi }^{2}\cdot {Y}_{\left(y\right)}\space,\space where:{Y}_{\left(y\right)}=C\sinh\left[n\pi \left(0.5-y\right)\right]+Dcosh\left[n\pi \left(0.5-y\right)\right].$

Are that steps correct? How can i go on? Need to decompose the problem first? And what kind of form my General-Analytical Solution will have? Thank you!

Alex
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  • Are you sure about the right hand side of the differential equation? The separation of variables technique appears to be inconsistent with it. – M. Wind Jul 13 '21 at 21:39
  • Yes it is actually a Poisson equation with g(x,y)=100 and not a Laplace form! Im not sure about the Seperation of Variables and if i treated right! That's the reason i asked, cause doesn't seems to me right! – Alex Jul 14 '21 at 11:07
  • In that case you should use not the product $X(x)*Y(y)$ because it leads to $X''Y +XY'' = 100$ and you are stuck. Better try the sum $X(x) + Y(y)$ as it leads to $X'' + Y'' = 100$. – M. Wind Jul 14 '21 at 15:38
  • Is that possible? I mean i've never saw it before as a ${X}^{"}\space+\space{Y}^{"}=100$. How to deal that? What would i gain from addition and not multiplication? – Alex Jul 14 '21 at 17:49
  • ${X}^{"}\space+\space{Y}^{"}=100\to \frac{{d}^{2}}{d{x}^{2}}X+\frac{{d}^{2}}{d{y}^{2}}Y=100$ – Alex Jul 14 '21 at 17:51
  • $\frac{{d}^{2}}{d{x}^{2}}X=100&\frac{{d}^{2}}{d{y}^{2}}Y=100$ – Alex Jul 14 '21 at 17:52
  • In a rectangle dont we have the dependency of both directions? I mean it is a 2-D problem. – Alex Jul 14 '21 at 18:09
  • I thought that $T(x,y) = Ax^2 + Bx + (50-A)y^2 +Cy +Dxy +E$ might be worth trying. However, I now see that this function can not match the boundary conditions. Sorry ! I steered you in a wrong direction. – M. Wind Jul 15 '21 at 02:57
  • So you can seek solutions of the Poisson equation with the RHS equal to zero, as you did. But you must still add terms to satisfy the original equation with RHS = 100. This can be done by adding $50x^2$ or $50y^2$ to your solution. That should work. – M. Wind Jul 15 '21 at 03:03

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It is important to realize that the differential equation you were given is an inhomogeneous one. This means that the right hand side is unequal to zero.

The proper procedure follows two lines. First of all we seek a (simple) solution to the inhomogeneous case. We can easily verify that the following solution works:

$$T(x,y) = Ax^2 + Bx + (50-A)y^2 + Dy + Exy + F$$

Presumably we won't need all $6$ terms, i.e. most of the parameters will turn out to be zero.

Next we seek solutions to the homogeneous case, i.e. with a RHS equal to zero. This problem can be solved by the Separation of Variables Technique. We assume $T(x,y)=X(x)*Y(y)$. This leads to these solutions:

$$sin(Cx)e^{-Cy}, cos(Cx)e^{-Cy}, sin(Cx)e^{Cy}, cos(Cx)e^{Cy}$$

And also the same expressions with $x$ and $y$ reversed. You can now construct the general solution which is a linear combination of all the homogeneous solutions (with different C's) plus the special solution found for the inhomogeneous case.

The final step is to match the solution to the boundary conditions.

M. Wind
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  • Thank you very much for your detailed hint and explanation! As i understand, finding a special solution of Poisson equation is generaly a difficult problem, in particular when we also have both Dirichlet and Neumann Boundary Conditions! I thought a decomposition of the problem to a homogeneous one and after that to deal with non homogeneous as a compination but seemed to me very complicated! I will try your hint and hope to have some progress! Thank you! – Alex Jul 15 '21 at 16:26
  • @Alex. You are welcome ! You were already on the right track, setting up the solution for the homogeneous case. But the inhomogeneous case (involving the value 100) was somehow missing. I am glad I could point you in the right direction ! – M. Wind Jul 15 '21 at 20:46