Three dimensional Laplace equation with constant Temp. on one face. [Solution not satisfying BC] The 2019 Stack Overflow Developer Survey Results Are InLaplace equation in 3D with numerous Non-Homogeneous BC(s) [Strategy Check]Generalized solution of Laplace equation with unknown boundary conditionsNeed help understanding solution to heat equation with two radiating endsSolution to one-dimensional Wave Equation with Method of Characteristics3D Homogenous Laplace equation with integral boundary conditionsEvaluating Fourier coefficients to complete a Laplace equation solutionTwo fluids flowing perpendicular in thermal contact with a Wall [Help to mathematically model]Evaluating Coefficients for a Fourier Series when Exponential terms are present [Approach needed]Handling Boundary functions to determine Fourier coefficientsTwo-dimensional Laplace equation with weird Robin BCLaplace equation in 3D with numerous Non-Homogeneous BC(s) [Strategy Check]
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Three dimensional Laplace equation with constant Temp. on one face. [Solution not satisfying BC]
The 2019 Stack Overflow Developer Survey Results Are InLaplace equation in 3D with numerous Non-Homogeneous BC(s) [Strategy Check]Generalized solution of Laplace equation with unknown boundary conditionsNeed help understanding solution to heat equation with two radiating endsSolution to one-dimensional Wave Equation with Method of Characteristics3D Homogenous Laplace equation with integral boundary conditionsEvaluating Fourier coefficients to complete a Laplace equation solutionTwo fluids flowing perpendicular in thermal contact with a Wall [Help to mathematically model]Evaluating Coefficients for a Fourier Series when Exponential terms are present [Approach needed]Handling Boundary functions to determine Fourier coefficientsTwo-dimensional Laplace equation with weird Robin BCLaplace equation in 3D with numerous Non-Homogeneous BC(s) [Strategy Check]
$begingroup$
The governing differential equation is
$$nabla^2 T=0 tag A$$
The boundary conditions for this problem are as foll0ws:
$$T(0,y,z)=T_hi tag 1A$$
$$T(L,y,z) = T(x,0,z) = T(x,l,z) = T(x,y,0)= T(x,y,mu) = 0 tag 1B$$
I need to solve for the distribution $T(x,y,z)$. I include my attempt here.
Solution attempt
Homogeneous Dirichlet type B.C. on $y$ and $z$ faces allow us to write the following form of preliminary temperature distribution:
$$T(x,y,z) = sum_n,m=1^inftysinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg)T_nm(x) tag 1C$$
We substitute $(1mathrmC)$ in $mathrmA$ and apply the orthogonality properties of $sinbigg(fracnpi ylbigg)$ and $sinbigg(fracmpi zmubigg)$ to obtain the following:
$$fracmathrmd^2T_nm(x)mathrmdx^2 - gamma_nm^2 T_nm(x) = 0 tag 1D$$
where $gamma_nm = sqrt(fracnpil)^2+(fracmpimu)^2$
The general solution of $(1mathrmD)$ is of the form:
$$T_nm(x) = A_nme^gamma x + B_nme^-gamma x tag 1E$$
$A_nm,B_nm$ are the unknown Fourier coefficients that need to be determined. Applying $T(L,y,z) = 0$ on $(1mathrmD)$ and using $(1mathrmE)$, we arrive at:
$$A_nme^gamma L + B_nme^-gamma L = 0$$
$$bfB_nm = - A_nme^2gamma L tag 1F$$
$$T_nm(x) = A_nm(e^gamma x - e^2gamma L - gamma x) tag 1G$$
Using the B.C. $(1mathrmA)$:
$$T_hi = sum_n,m=1^inftysinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg)A_nm(1 - e^2gamma L) tag 1H$$
Multiplying both sides of $(1mathrmH)$ by $int_0^lsinbigg(frackpi ylbigg)mathrmdy$ and $int_0^musinbigg(fracjpi zmubigg)mathrmdz$ and using the principle of orthogonality, we arrive at:
$$T_hiint_0^l int_0^mu sinbigg(frackpi ylbigg)sinbigg(fracjpi zmubigg) mathrmdymathrmdz = fraclmu4 A_kj (1 - e^2gamma L) tag 1I$$
Solving the definite integrals involved , we arrive at (for any arbitrary integer $n$ and $m$):
$$bfA_nm = frac4 T_hinmpi^2(1-e^2gamma L) (1-cos(npi))(1-cos(mpi)) tag 1J$$
$$T(x,y,z) = sum_n,m=1^infty(A_nme^gamma x + B_nme^-gamma x)sinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg) tag 1K$$
On coding this solution in MATLAB. and substituting $x=0$, I find that the answer is not equal to $T_hi$.
For parameter values $L=0.5,l=0.5,mu=0.05,T_hi=50$. When i evaluate for $x=0,y=l/2,z=mu /2$, the answer should be $T=50$, but it evaluates to $96$ using the first four terms in the series. The series surely converges.
Is there something wrong, in the way, I am doing the problem ?Any help is greatly appreciated.
An observation
When I take $z=mu /2, y=l/2$ and $x=0$, and consider only odd values for $n$ and $m$, the solution can be written as:
$$Tbigg(0,fracl2,fracmu2bigg) = frac16 T_hipi^2underbracebigg[1 + frac19 + frac125 + frac149 + ........bigg]_pi^2 /8 = 2T_hi tag 1L$$
So, is there something wrong with my analytical solution ?
proof-verification pde problem-solving heat-equation laplacian
asked Mar 23 at 4:08
Indrasis MitraIndrasis Mitra
50111
$endgroup$
add a comment |
$begingroup$
The governing differential equation is
$$nabla^2 T=0 tag A$$
The boundary conditions for this problem are as foll0ws:
$$T(0,y,z)=T_hi tag 1A$$
$$T(L,y,z) = T(x,0,z) = T(x,l,z) = T(x,y,0)= T(x,y,mu) = 0 tag 1B$$
I need to solve for the distribution $T(x,y,z)$. I include my attempt here.
Solution attempt
Homogeneous Dirichlet type B.C. on $y$ and $z$ faces allow us to write the following form of preliminary temperature distribution:
$$T(x,y,z) = sum_n,m=1^inftysinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg)T_nm(x) tag 1C$$
We substitute $(1mathrmC)$ in $mathrmA$ and apply the orthogonality properties of $sinbigg(fracnpi ylbigg)$ and $sinbigg(fracmpi zmubigg)$ to obtain the following:
$$fracmathrmd^2T_nm(x)mathrmdx^2 - gamma_nm^2 T_nm(x) = 0 tag 1D$$
where $gamma_nm = sqrt(fracnpil)^2+(fracmpimu)^2$
The general solution of $(1mathrmD)$ is of the form:
$$T_nm(x) = A_nme^gamma x + B_nme^-gamma x tag 1E$$
$A_nm,B_nm$ are the unknown Fourier coefficients that need to be determined. Applying $T(L,y,z) = 0$ on $(1mathrmD)$ and using $(1mathrmE)$, we arrive at:
$$A_nme^gamma L + B_nme^-gamma L = 0$$
$$bfB_nm = - A_nme^2gamma L tag 1F$$
$$T_nm(x) = A_nm(e^gamma x - e^2gamma L - gamma x) tag 1G$$
Using the B.C. $(1mathrmA)$:
$$T_hi = sum_n,m=1^inftysinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg)A_nm(1 - e^2gamma L) tag 1H$$
Multiplying both sides of $(1mathrmH)$ by $int_0^lsinbigg(frackpi ylbigg)mathrmdy$ and $int_0^musinbigg(fracjpi zmubigg)mathrmdz$ and using the principle of orthogonality, we arrive at:
$$T_hiint_0^l int_0^mu sinbigg(frackpi ylbigg)sinbigg(fracjpi zmubigg) mathrmdymathrmdz = fraclmu4 A_kj (1 - e^2gamma L) tag 1I$$
Solving the definite integrals involved , we arrive at (for any arbitrary integer $n$ and $m$):
$$bfA_nm = frac4 T_hinmpi^2(1-e^2gamma L) (1-cos(npi))(1-cos(mpi)) tag 1J$$
$$T(x,y,z) = sum_n,m=1^infty(A_nme^gamma x + B_nme^-gamma x)sinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg) tag 1K$$
On coding this solution in MATLAB. and substituting $x=0$, I find that the answer is not equal to $T_hi$.
For parameter values $L=0.5,l=0.5,mu=0.05,T_hi=50$. When i evaluate for $x=0,y=l/2,z=mu /2$, the answer should be $T=50$, but it evaluates to $96$ using the first four terms in the series. The series surely converges.
Is there something wrong, in the way, I am doing the problem ?Any help is greatly appreciated.
An observation
When I take $z=mu /2, y=l/2$ and $x=0$, and consider only odd values for $n$ and $m$, the solution can be written as:
$$Tbigg(0,fracl2,fracmu2bigg) = frac16 T_hipi^2underbracebigg[1 + frac19 + frac125 + frac149 + ........bigg]_pi^2 /8 = 2T_hi tag 1L$$
So, is there something wrong with my analytical solution ?
proof-verification pde problem-solving heat-equation laplacian
asked Mar 23 at 4:08
Indrasis MitraIndrasis Mitra
50111
$endgroup$
add a comment |
$begingroup$
The governing differential equation is
$$nabla^2 T=0 tag A$$
The boundary conditions for this problem are as foll0ws:
$$T(0,y,z)=T_hi tag 1A$$
$$T(L,y,z) = T(x,0,z) = T(x,l,z) = T(x,y,0)= T(x,y,mu) = 0 tag 1B$$
I need to solve for the distribution $T(x,y,z)$. I include my attempt here.
Solution attempt
Homogeneous Dirichlet type B.C. on $y$ and $z$ faces allow us to write the following form of preliminary temperature distribution:
$$T(x,y,z) = sum_n,m=1^inftysinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg)T_nm(x) tag 1C$$
We substitute $(1mathrmC)$ in $mathrmA$ and apply the orthogonality properties of $sinbigg(fracnpi ylbigg)$ and $sinbigg(fracmpi zmubigg)$ to obtain the following:
$$fracmathrmd^2T_nm(x)mathrmdx^2 - gamma_nm^2 T_nm(x) = 0 tag 1D$$
where $gamma_nm = sqrt(fracnpil)^2+(fracmpimu)^2$
The general solution of $(1mathrmD)$ is of the form:
$$T_nm(x) = A_nme^gamma x + B_nme^-gamma x tag 1E$$
$A_nm,B_nm$ are the unknown Fourier coefficients that need to be determined. Applying $T(L,y,z) = 0$ on $(1mathrmD)$ and using $(1mathrmE)$, we arrive at:
$$A_nme^gamma L + B_nme^-gamma L = 0$$
$$bfB_nm = - A_nme^2gamma L tag 1F$$
$$T_nm(x) = A_nm(e^gamma x - e^2gamma L - gamma x) tag 1G$$
Using the B.C. $(1mathrmA)$:
$$T_hi = sum_n,m=1^inftysinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg)A_nm(1 - e^2gamma L) tag 1H$$
Multiplying both sides of $(1mathrmH)$ by $int_0^lsinbigg(frackpi ylbigg)mathrmdy$ and $int_0^musinbigg(fracjpi zmubigg)mathrmdz$ and using the principle of orthogonality, we arrive at:
$$T_hiint_0^l int_0^mu sinbigg(frackpi ylbigg)sinbigg(fracjpi zmubigg) mathrmdymathrmdz = fraclmu4 A_kj (1 - e^2gamma L) tag 1I$$
Solving the definite integrals involved , we arrive at (for any arbitrary integer $n$ and $m$):
$$bfA_nm = frac4 T_hinmpi^2(1-e^2gamma L) (1-cos(npi))(1-cos(mpi)) tag 1J$$
$$T(x,y,z) = sum_n,m=1^infty(A_nme^gamma x + B_nme^-gamma x)sinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg) tag 1K$$
On coding this solution in MATLAB. and substituting $x=0$, I find that the answer is not equal to $T_hi$.
For parameter values $L=0.5,l=0.5,mu=0.05,T_hi=50$. When i evaluate for $x=0,y=l/2,z=mu /2$, the answer should be $T=50$, but it evaluates to $96$ using the first four terms in the series. The series surely converges.
Is there something wrong, in the way, I am doing the problem ?Any help is greatly appreciated.
An observation
When I take $z=mu /2, y=l/2$ and $x=0$, and consider only odd values for $n$ and $m$, the solution can be written as:
$$Tbigg(0,fracl2,fracmu2bigg) = frac16 T_hipi^2underbracebigg[1 + frac19 + frac125 + frac149 + ........bigg]_pi^2 /8 = 2T_hi tag 1L$$
So, is there something wrong with my analytical solution ?
proof-verification pde problem-solving heat-equation laplacian
asked Mar 23 at 4:08
Indrasis MitraIndrasis Mitra
50111
$endgroup$
The governing differential equation is
$$nabla^2 T=0 tag A$$
The boundary conditions for this problem are as foll0ws:
$$T(0,y,z)=T_hi tag 1A$$
$$T(L,y,z) = T(x,0,z) = T(x,l,z) = T(x,y,0)= T(x,y,mu) = 0 tag 1B$$
I need to solve for the distribution $T(x,y,z)$. I include my attempt here.
Solution attempt
Homogeneous Dirichlet type B.C. on $y$ and $z$ faces allow us to write the following form of preliminary temperature distribution:
$$T(x,y,z) = sum_n,m=1^inftysinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg)T_nm(x) tag 1C$$
We substitute $(1mathrmC)$ in $mathrmA$ and apply the orthogonality properties of $sinbigg(fracnpi ylbigg)$ and $sinbigg(fracmpi zmubigg)$ to obtain the following:
$$fracmathrmd^2T_nm(x)mathrmdx^2 - gamma_nm^2 T_nm(x) = 0 tag 1D$$
where $gamma_nm = sqrt(fracnpil)^2+(fracmpimu)^2$
The general solution of $(1mathrmD)$ is of the form:
$$T_nm(x) = A_nme^gamma x + B_nme^-gamma x tag 1E$$
$A_nm,B_nm$ are the unknown Fourier coefficients that need to be determined. Applying $T(L,y,z) = 0$ on $(1mathrmD)$ and using $(1mathrmE)$, we arrive at:
$$A_nme^gamma L + B_nme^-gamma L = 0$$
$$bfB_nm = - A_nme^2gamma L tag 1F$$
$$T_nm(x) = A_nm(e^gamma x - e^2gamma L - gamma x) tag 1G$$
Using the B.C. $(1mathrmA)$:
$$T_hi = sum_n,m=1^inftysinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg)A_nm(1 - e^2gamma L) tag 1H$$
Multiplying both sides of $(1mathrmH)$ by $int_0^lsinbigg(frackpi ylbigg)mathrmdy$ and $int_0^musinbigg(fracjpi zmubigg)mathrmdz$ and using the principle of orthogonality, we arrive at:
$$T_hiint_0^l int_0^mu sinbigg(frackpi ylbigg)sinbigg(fracjpi zmubigg) mathrmdymathrmdz = fraclmu4 A_kj (1 - e^2gamma L) tag 1I$$
Solving the definite integrals involved , we arrive at (for any arbitrary integer $n$ and $m$):
$$bfA_nm = frac4 T_hinmpi^2(1-e^2gamma L) (1-cos(npi))(1-cos(mpi)) tag 1J$$
$$T(x,y,z) = sum_n,m=1^infty(A_nme^gamma x + B_nme^-gamma x)sinbigg(fracnpi ylbigg)sinbigg(fracmpi zmubigg) tag 1K$$
On coding this solution in MATLAB. and substituting $x=0$, I find that the answer is not equal to $T_hi$.
For parameter values $L=0.5,l=0.5,mu=0.05,T_hi=50$. When i evaluate for $x=0,y=l/2,z=mu /2$, the answer should be $T=50$, but it evaluates to $96$ using the first four terms in the series. The series surely converges.
Is there something wrong, in the way, I am doing the problem ?Any help is greatly appreciated.
An observation
When I take $z=mu /2, y=l/2$ and $x=0$, and consider only odd values for $n$ and $m$, the solution can be written as:
$$Tbigg(0,fracl2,fracmu2bigg) = frac16 T_hipi^2underbracebigg[1 + frac19 + frac125 + frac149 + ........bigg]_pi^2 /8 = 2T_hi tag 1L$$
So, is there something wrong with my analytical solution ?
proof-verification pde problem-solving heat-equation laplacian
proof-verification pde problem-solving heat-equation laplacian
asked Mar 23 at 4:08
Indrasis MitraIndrasis Mitra
50111
asked Mar 23 at 4:08
Indrasis MitraIndrasis Mitra
50111
edited Mar 23 at 14:00
Indrasis Mitra
asked Mar 23 at 4:08
Indrasis MitraIndrasis Mitra
50111
asked Mar 23 at 4:08
Indrasis MitraIndrasis Mitra
50111
asked Mar 23 at 4:08
Indrasis MitraIndrasis Mitra
50111
50111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Are you sure you typed it in correctly? I checked your solution against mine and they're the same. I'll leave my answer below in case it helps anyways.
Often times I find it better to start from the beginning when I'm stuck to remove doubt of errors made in assumptions.
begincases
beginalign
Delta u(x,y,z) &= 0 \
u(0,y,z) &= u_0 \
uvert_partialOmegabackslashx=0 &= 0.
endalign
endcases
You know about separation of variables already, so we may take the eigenfunctions in the $y$ and $z$ directions as $sin fracnpi yell$ and $sin fracmpi zmu$. The equation in $x$ is known as $a_mne^gamma_mnx + b_mne^-gamma_mnx.$ Thus $u$ is the resulting infinite linear combination
$$
u(x,y,z) = sum_m,n geq 1left(a_mne^gamma_mnx + b_mne^-gamma_mnxright)sinfracnpi yellsinfracmpi zmu.
$$
The boundary conditions on $y$ and $z$ are already satisfied so we look towards those for $x$.
begincases
beginalign
u_0 &= sum_m,n geq 1(a_mn + b_mn)sinfracnpi yellsinfracmpi zmu \
0 &= sum_m,n geq 1 left(a_mne^gamma_mnL + b_mne^-gamma_mnLright)sinfracnpi yellsinfracmpi zmu
endalign
endcases
Note that using orthogonality is not multiplying each side by the respective integrals; it is multiplying each side by the eigenfunctions with dummy indices and then integrating. Continuing with this in mind,
begincases
beginalign
int_0^ell int_0^mu u_0 sinfracnpi yell sinfracmpi zmu , mathrmdz ,mathrmdy &= fracmuell4(a_mn + b_mn) \
0 &= a_mne^gamma_mnL + b_mne^-gamma_mnL
endalign
endcases
The first integral is easily found and we have the system of equations
$$
beginpmatrix
1 & 1 \
e^gamma_mnL & e^-gamma_mnL
endpmatrix
beginpmatrix
a_mn \
b_mn
endpmatrix
=
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mnu_0 \
0
endpmatrix.
$$
Inverting to solve for $a_mn$ and $b_mn$ then,
beginalign
beginpmatrix
a_mn \
b_mn
endpmatrix
&=
-fracoperatornamecsch(gamma_mnL)2
beginpmatrix
e^-gamma_mnL & -1 \
-e^gamma_mnL & 1
endpmatrix
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mn \
0
endpmatrix
\
&= -frac2u_0pi^2fracbig((-1)^m - 1big) big((-1)^n - 1big)sinh(gamma_mnL) mn
beginpmatrix
e^-gamma_mnL \
-e^gamma_mnL
endpmatrix.
endalign
Thus $u$ is written
beginequation
u(x,y,z) = -frac4u_0pi^2 sum_m,n geq 1 Gamma_mn fracsinhbig(gamma_mn(x-L)big)sinh(gamma_mnL) sinfracn pi yell sin fracmpi zmu
endequation
with
$$
Gamma_mn = fracbig((-1)^m - 1big) big((-1)^n - 1big)mn.
$$
$hskip 1 in$ 
Mathematica code - it should be copy-pastable - enjoy.
ClearAll["Global`*"]
[Gamma]mn = Sqrt[((n [Pi])/[ScriptL])^2 + ((m [Pi])/[Mu])^2];
(* u0 not specified yet *)
L = 1;
[ScriptL] = 1;
[Mu] = 1;
(* Keep q low (~10-15 or so) as the number of terms grows as q^2 *)
(* i.e. higher q gives more accuracy but takes longer to run in effect *)
q = 5;
(*The coefficients amn and bmn amn,bmn=a,b/.Solve[a+b[Equal]4/
[Pi]^2(((-1)^m-1)((-1)^n-1))/(m n)u0,a Exp[[Gamma]mn L]+b Exp[-
[Gamma]mn L][Equal]0,a,b][[1]];*)
u[x_, y_,
z_] = (-4 u0)/[Pi]^2 Sum[
Sum[(((-1)^m - 1) ((-1)^n - 1))/(m n) Sinh[[Gamma]mn (x - L)]/
Sinh[[Gamma]mn L] Sin[n [Pi] y/[ScriptL]] Sin[
m [Pi] z/[Mu]], n, 1, q], m, 1, q];
Plot3D[u[0, y, z]/u0, y, 0, [ScriptL], z, 0, [Mu],
PlotLabel ->
"Laplace's Equation Solutionnat x=0 withn[ScriptL]=1, [Mu]=1,
25 terms",
AxesLabel -> "y", "z", "!(*FractionBox[(u), (u0)])",
Boxed -> False]
(* Something extra - it's computationally expensive (3d), at least for my
computer, so let it run for a bit *)
Manipulate[
ContourPlot[u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL],
ColorFunction -> "DarkRainbow"], z, 0, [Mu]]
(* alt+.to quit if it takes too long & reduce q *)
ContourPlot3D[
u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL], z, 0, [Mu],
Mesh -> False, ColorFunction -> "Rainbow",
PlotLabel -> "Contours of Temperature", AxesLabel -> "x", "y", "z"]
answered Mar 24 at 4:26
AEngineerAEngineer
1,6521318
$endgroup$
$begingroup$
Thanks for such a detailed and well-explained answer. It seems the answer i arrived at is same with what you get . Maybe I did some mistake while writing the MATLAB script. Saying that would it be possible for you to post your Mathematica script here ? Also, your note about multiplying with respective eigen functions was helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 5:46
$begingroup$
Sure, glad I could help. I already deleted the code but I could post just the solution and the plot if you'd like.
$endgroup$
– AEngineer
Mar 24 at 5:58
$begingroup$
That would be really helpful. Appreciate it.
$endgroup$
– Indrasis Mitra
Mar 24 at 6:01
$begingroup$
Thanks for the Mathematica script. Some really cool graphics included. I was hoping you could take a look at this problem which I have been trying recently, especially its SP3 part. Any guidance will be really helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 16:12
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Are you sure you typed it in correctly? I checked your solution against mine and they're the same. I'll leave my answer below in case it helps anyways.
Often times I find it better to start from the beginning when I'm stuck to remove doubt of errors made in assumptions.
begincases
beginalign
Delta u(x,y,z) &= 0 \
u(0,y,z) &= u_0 \
uvert_partialOmegabackslashx=0 &= 0.
endalign
endcases
You know about separation of variables already, so we may take the eigenfunctions in the $y$ and $z$ directions as $sin fracnpi yell$ and $sin fracmpi zmu$. The equation in $x$ is known as $a_mne^gamma_mnx + b_mne^-gamma_mnx.$ Thus $u$ is the resulting infinite linear combination
$$
u(x,y,z) = sum_m,n geq 1left(a_mne^gamma_mnx + b_mne^-gamma_mnxright)sinfracnpi yellsinfracmpi zmu.
$$
The boundary conditions on $y$ and $z$ are already satisfied so we look towards those for $x$.
begincases
beginalign
u_0 &= sum_m,n geq 1(a_mn + b_mn)sinfracnpi yellsinfracmpi zmu \
0 &= sum_m,n geq 1 left(a_mne^gamma_mnL + b_mne^-gamma_mnLright)sinfracnpi yellsinfracmpi zmu
endalign
endcases
Note that using orthogonality is not multiplying each side by the respective integrals; it is multiplying each side by the eigenfunctions with dummy indices and then integrating. Continuing with this in mind,
begincases
beginalign
int_0^ell int_0^mu u_0 sinfracnpi yell sinfracmpi zmu , mathrmdz ,mathrmdy &= fracmuell4(a_mn + b_mn) \
0 &= a_mne^gamma_mnL + b_mne^-gamma_mnL
endalign
endcases
The first integral is easily found and we have the system of equations
$$
beginpmatrix
1 & 1 \
e^gamma_mnL & e^-gamma_mnL
endpmatrix
beginpmatrix
a_mn \
b_mn
endpmatrix
=
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mnu_0 \
0
endpmatrix.
$$
Inverting to solve for $a_mn$ and $b_mn$ then,
beginalign
beginpmatrix
a_mn \
b_mn
endpmatrix
&=
-fracoperatornamecsch(gamma_mnL)2
beginpmatrix
e^-gamma_mnL & -1 \
-e^gamma_mnL & 1
endpmatrix
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mn \
0
endpmatrix
\
&= -frac2u_0pi^2fracbig((-1)^m - 1big) big((-1)^n - 1big)sinh(gamma_mnL) mn
beginpmatrix
e^-gamma_mnL \
-e^gamma_mnL
endpmatrix.
endalign
Thus $u$ is written
beginequation
u(x,y,z) = -frac4u_0pi^2 sum_m,n geq 1 Gamma_mn fracsinhbig(gamma_mn(x-L)big)sinh(gamma_mnL) sinfracn pi yell sin fracmpi zmu
endequation
with
$$
Gamma_mn = fracbig((-1)^m - 1big) big((-1)^n - 1big)mn.
$$
$hskip 1 in$
Mathematica code - it should be copy-pastable - enjoy.
ClearAll["Global`*"]
[Gamma]mn = Sqrt[((n [Pi])/[ScriptL])^2 + ((m [Pi])/[Mu])^2];
(* u0 not specified yet *)
L = 1;
[ScriptL] = 1;
[Mu] = 1;
(* Keep q low (~10-15 or so) as the number of terms grows as q^2 *)
(* i.e. higher q gives more accuracy but takes longer to run in effect *)
q = 5;
(*The coefficients amn and bmn amn,bmn=a,b/.Solve[a+b[Equal]4/
[Pi]^2(((-1)^m-1)((-1)^n-1))/(m n)u0,a Exp[[Gamma]mn L]+b Exp[-
[Gamma]mn L][Equal]0,a,b][[1]];*)
u[x_, y_,
z_] = (-4 u0)/[Pi]^2 Sum[
Sum[(((-1)^m - 1) ((-1)^n - 1))/(m n) Sinh[[Gamma]mn (x - L)]/
Sinh[[Gamma]mn L] Sin[n [Pi] y/[ScriptL]] Sin[
m [Pi] z/[Mu]], n, 1, q], m, 1, q];
Plot3D[u[0, y, z]/u0, y, 0, [ScriptL], z, 0, [Mu],
PlotLabel ->
"Laplace's Equation Solutionnat x=0 withn[ScriptL]=1, [Mu]=1,
25 terms",
AxesLabel -> "y", "z", "!(*FractionBox[(u), (u0)])",
Boxed -> False]
(* Something extra - it's computationally expensive (3d), at least for my
computer, so let it run for a bit *)
Manipulate[
ContourPlot[u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL],
ColorFunction -> "DarkRainbow"], z, 0, [Mu]]
(* alt+.to quit if it takes too long & reduce q *)
ContourPlot3D[
u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL], z, 0, [Mu],
Mesh -> False, ColorFunction -> "Rainbow",
PlotLabel -> "Contours of Temperature", AxesLabel -> "x", "y", "z"]
answered Mar 24 at 4:26
AEngineerAEngineer
1,6521318
$endgroup$
$begingroup$
Thanks for such a detailed and well-explained answer. It seems the answer i arrived at is same with what you get . Maybe I did some mistake while writing the MATLAB script. Saying that would it be possible for you to post your Mathematica script here ? Also, your note about multiplying with respective eigen functions was helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 5:46
$begingroup$
Sure, glad I could help. I already deleted the code but I could post just the solution and the plot if you'd like.
$endgroup$
– AEngineer
Mar 24 at 5:58
$begingroup$
That would be really helpful. Appreciate it.
$endgroup$
– Indrasis Mitra
Mar 24 at 6:01
$begingroup$
Thanks for the Mathematica script. Some really cool graphics included. I was hoping you could take a look at this problem which I have been trying recently, especially its SP3 part. Any guidance will be really helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 16:12
add a comment |
$begingroup$
Are you sure you typed it in correctly? I checked your solution against mine and they're the same. I'll leave my answer below in case it helps anyways.
Often times I find it better to start from the beginning when I'm stuck to remove doubt of errors made in assumptions.
begincases
beginalign
Delta u(x,y,z) &= 0 \
u(0,y,z) &= u_0 \
uvert_partialOmegabackslashx=0 &= 0.
endalign
endcases
You know about separation of variables already, so we may take the eigenfunctions in the $y$ and $z$ directions as $sin fracnpi yell$ and $sin fracmpi zmu$. The equation in $x$ is known as $a_mne^gamma_mnx + b_mne^-gamma_mnx.$ Thus $u$ is the resulting infinite linear combination
$$
u(x,y,z) = sum_m,n geq 1left(a_mne^gamma_mnx + b_mne^-gamma_mnxright)sinfracnpi yellsinfracmpi zmu.
$$
The boundary conditions on $y$ and $z$ are already satisfied so we look towards those for $x$.
begincases
beginalign
u_0 &= sum_m,n geq 1(a_mn + b_mn)sinfracnpi yellsinfracmpi zmu \
0 &= sum_m,n geq 1 left(a_mne^gamma_mnL + b_mne^-gamma_mnLright)sinfracnpi yellsinfracmpi zmu
endalign
endcases
Note that using orthogonality is not multiplying each side by the respective integrals; it is multiplying each side by the eigenfunctions with dummy indices and then integrating. Continuing with this in mind,
begincases
beginalign
int_0^ell int_0^mu u_0 sinfracnpi yell sinfracmpi zmu , mathrmdz ,mathrmdy &= fracmuell4(a_mn + b_mn) \
0 &= a_mne^gamma_mnL + b_mne^-gamma_mnL
endalign
endcases
The first integral is easily found and we have the system of equations
$$
beginpmatrix
1 & 1 \
e^gamma_mnL & e^-gamma_mnL
endpmatrix
beginpmatrix
a_mn \
b_mn
endpmatrix
=
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mnu_0 \
0
endpmatrix.
$$
Inverting to solve for $a_mn$ and $b_mn$ then,
beginalign
beginpmatrix
a_mn \
b_mn
endpmatrix
&=
-fracoperatornamecsch(gamma_mnL)2
beginpmatrix
e^-gamma_mnL & -1 \
-e^gamma_mnL & 1
endpmatrix
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mn \
0
endpmatrix
\
&= -frac2u_0pi^2fracbig((-1)^m - 1big) big((-1)^n - 1big)sinh(gamma_mnL) mn
beginpmatrix
e^-gamma_mnL \
-e^gamma_mnL
endpmatrix.
endalign
Thus $u$ is written
beginequation
u(x,y,z) = -frac4u_0pi^2 sum_m,n geq 1 Gamma_mn fracsinhbig(gamma_mn(x-L)big)sinh(gamma_mnL) sinfracn pi yell sin fracmpi zmu
endequation
with
$$
Gamma_mn = fracbig((-1)^m - 1big) big((-1)^n - 1big)mn.
$$
$hskip 1 in$
Mathematica code - it should be copy-pastable - enjoy.
ClearAll["Global`*"]
[Gamma]mn = Sqrt[((n [Pi])/[ScriptL])^2 + ((m [Pi])/[Mu])^2];
(* u0 not specified yet *)
L = 1;
[ScriptL] = 1;
[Mu] = 1;
(* Keep q low (~10-15 or so) as the number of terms grows as q^2 *)
(* i.e. higher q gives more accuracy but takes longer to run in effect *)
q = 5;
(*The coefficients amn and bmn amn,bmn=a,b/.Solve[a+b[Equal]4/
[Pi]^2(((-1)^m-1)((-1)^n-1))/(m n)u0,a Exp[[Gamma]mn L]+b Exp[-
[Gamma]mn L][Equal]0,a,b][[1]];*)
u[x_, y_,
z_] = (-4 u0)/[Pi]^2 Sum[
Sum[(((-1)^m - 1) ((-1)^n - 1))/(m n) Sinh[[Gamma]mn (x - L)]/
Sinh[[Gamma]mn L] Sin[n [Pi] y/[ScriptL]] Sin[
m [Pi] z/[Mu]], n, 1, q], m, 1, q];
Plot3D[u[0, y, z]/u0, y, 0, [ScriptL], z, 0, [Mu],
PlotLabel ->
"Laplace's Equation Solutionnat x=0 withn[ScriptL]=1, [Mu]=1,
25 terms",
AxesLabel -> "y", "z", "!(*FractionBox[(u), (u0)])",
Boxed -> False]
(* Something extra - it's computationally expensive (3d), at least for my
computer, so let it run for a bit *)
Manipulate[
ContourPlot[u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL],
ColorFunction -> "DarkRainbow"], z, 0, [Mu]]
(* alt+.to quit if it takes too long & reduce q *)
ContourPlot3D[
u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL], z, 0, [Mu],
Mesh -> False, ColorFunction -> "Rainbow",
PlotLabel -> "Contours of Temperature", AxesLabel -> "x", "y", "z"]
answered Mar 24 at 4:26
AEngineerAEngineer
1,6521318
$endgroup$
$begingroup$
Thanks for such a detailed and well-explained answer. It seems the answer i arrived at is same with what you get . Maybe I did some mistake while writing the MATLAB script. Saying that would it be possible for you to post your Mathematica script here ? Also, your note about multiplying with respective eigen functions was helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 5:46
$begingroup$
Sure, glad I could help. I already deleted the code but I could post just the solution and the plot if you'd like.
$endgroup$
– AEngineer
Mar 24 at 5:58
$begingroup$
That would be really helpful. Appreciate it.
$endgroup$
– Indrasis Mitra
Mar 24 at 6:01
$begingroup$
Thanks for the Mathematica script. Some really cool graphics included. I was hoping you could take a look at this problem which I have been trying recently, especially its SP3 part. Any guidance will be really helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 16:12
add a comment |
$begingroup$
Are you sure you typed it in correctly? I checked your solution against mine and they're the same. I'll leave my answer below in case it helps anyways.
Often times I find it better to start from the beginning when I'm stuck to remove doubt of errors made in assumptions.
begincases
beginalign
Delta u(x,y,z) &= 0 \
u(0,y,z) &= u_0 \
uvert_partialOmegabackslashx=0 &= 0.
endalign
endcases
You know about separation of variables already, so we may take the eigenfunctions in the $y$ and $z$ directions as $sin fracnpi yell$ and $sin fracmpi zmu$. The equation in $x$ is known as $a_mne^gamma_mnx + b_mne^-gamma_mnx.$ Thus $u$ is the resulting infinite linear combination
$$
u(x,y,z) = sum_m,n geq 1left(a_mne^gamma_mnx + b_mne^-gamma_mnxright)sinfracnpi yellsinfracmpi zmu.
$$
The boundary conditions on $y$ and $z$ are already satisfied so we look towards those for $x$.
begincases
beginalign
u_0 &= sum_m,n geq 1(a_mn + b_mn)sinfracnpi yellsinfracmpi zmu \
0 &= sum_m,n geq 1 left(a_mne^gamma_mnL + b_mne^-gamma_mnLright)sinfracnpi yellsinfracmpi zmu
endalign
endcases
Note that using orthogonality is not multiplying each side by the respective integrals; it is multiplying each side by the eigenfunctions with dummy indices and then integrating. Continuing with this in mind,
begincases
beginalign
int_0^ell int_0^mu u_0 sinfracnpi yell sinfracmpi zmu , mathrmdz ,mathrmdy &= fracmuell4(a_mn + b_mn) \
0 &= a_mne^gamma_mnL + b_mne^-gamma_mnL
endalign
endcases
The first integral is easily found and we have the system of equations
$$
beginpmatrix
1 & 1 \
e^gamma_mnL & e^-gamma_mnL
endpmatrix
beginpmatrix
a_mn \
b_mn
endpmatrix
=
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mnu_0 \
0
endpmatrix.
$$
Inverting to solve for $a_mn$ and $b_mn$ then,
beginalign
beginpmatrix
a_mn \
b_mn
endpmatrix
&=
-fracoperatornamecsch(gamma_mnL)2
beginpmatrix
e^-gamma_mnL & -1 \
-e^gamma_mnL & 1
endpmatrix
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mn \
0
endpmatrix
\
&= -frac2u_0pi^2fracbig((-1)^m - 1big) big((-1)^n - 1big)sinh(gamma_mnL) mn
beginpmatrix
e^-gamma_mnL \
-e^gamma_mnL
endpmatrix.
endalign
Thus $u$ is written
beginequation
u(x,y,z) = -frac4u_0pi^2 sum_m,n geq 1 Gamma_mn fracsinhbig(gamma_mn(x-L)big)sinh(gamma_mnL) sinfracn pi yell sin fracmpi zmu
endequation
with
$$
Gamma_mn = fracbig((-1)^m - 1big) big((-1)^n - 1big)mn.
$$
$hskip 1 in$
Mathematica code - it should be copy-pastable - enjoy.
ClearAll["Global`*"]
[Gamma]mn = Sqrt[((n [Pi])/[ScriptL])^2 + ((m [Pi])/[Mu])^2];
(* u0 not specified yet *)
L = 1;
[ScriptL] = 1;
[Mu] = 1;
(* Keep q low (~10-15 or so) as the number of terms grows as q^2 *)
(* i.e. higher q gives more accuracy but takes longer to run in effect *)
q = 5;
(*The coefficients amn and bmn amn,bmn=a,b/.Solve[a+b[Equal]4/
[Pi]^2(((-1)^m-1)((-1)^n-1))/(m n)u0,a Exp[[Gamma]mn L]+b Exp[-
[Gamma]mn L][Equal]0,a,b][[1]];*)
u[x_, y_,
z_] = (-4 u0)/[Pi]^2 Sum[
Sum[(((-1)^m - 1) ((-1)^n - 1))/(m n) Sinh[[Gamma]mn (x - L)]/
Sinh[[Gamma]mn L] Sin[n [Pi] y/[ScriptL]] Sin[
m [Pi] z/[Mu]], n, 1, q], m, 1, q];
Plot3D[u[0, y, z]/u0, y, 0, [ScriptL], z, 0, [Mu],
PlotLabel ->
"Laplace's Equation Solutionnat x=0 withn[ScriptL]=1, [Mu]=1,
25 terms",
AxesLabel -> "y", "z", "!(*FractionBox[(u), (u0)])",
Boxed -> False]
(* Something extra - it's computationally expensive (3d), at least for my
computer, so let it run for a bit *)
Manipulate[
ContourPlot[u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL],
ColorFunction -> "DarkRainbow"], z, 0, [Mu]]
(* alt+.to quit if it takes too long & reduce q *)
ContourPlot3D[
u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL], z, 0, [Mu],
Mesh -> False, ColorFunction -> "Rainbow",
PlotLabel -> "Contours of Temperature", AxesLabel -> "x", "y", "z"]
answered Mar 24 at 4:26
AEngineerAEngineer
1,6521318
$endgroup$
Are you sure you typed it in correctly? I checked your solution against mine and they're the same. I'll leave my answer below in case it helps anyways.
Often times I find it better to start from the beginning when I'm stuck to remove doubt of errors made in assumptions.
begincases
beginalign
Delta u(x,y,z) &= 0 \
u(0,y,z) &= u_0 \
uvert_partialOmegabackslashx=0 &= 0.
endalign
endcases
You know about separation of variables already, so we may take the eigenfunctions in the $y$ and $z$ directions as $sin fracnpi yell$ and $sin fracmpi zmu$. The equation in $x$ is known as $a_mne^gamma_mnx + b_mne^-gamma_mnx.$ Thus $u$ is the resulting infinite linear combination
$$
u(x,y,z) = sum_m,n geq 1left(a_mne^gamma_mnx + b_mne^-gamma_mnxright)sinfracnpi yellsinfracmpi zmu.
$$
The boundary conditions on $y$ and $z$ are already satisfied so we look towards those for $x$.
begincases
beginalign
u_0 &= sum_m,n geq 1(a_mn + b_mn)sinfracnpi yellsinfracmpi zmu \
0 &= sum_m,n geq 1 left(a_mne^gamma_mnL + b_mne^-gamma_mnLright)sinfracnpi yellsinfracmpi zmu
endalign
endcases
Note that using orthogonality is not multiplying each side by the respective integrals; it is multiplying each side by the eigenfunctions with dummy indices and then integrating. Continuing with this in mind,
begincases
beginalign
int_0^ell int_0^mu u_0 sinfracnpi yell sinfracmpi zmu , mathrmdz ,mathrmdy &= fracmuell4(a_mn + b_mn) \
0 &= a_mne^gamma_mnL + b_mne^-gamma_mnL
endalign
endcases
The first integral is easily found and we have the system of equations
$$
beginpmatrix
1 & 1 \
e^gamma_mnL & e^-gamma_mnL
endpmatrix
beginpmatrix
a_mn \
b_mn
endpmatrix
=
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mnu_0 \
0
endpmatrix.
$$
Inverting to solve for $a_mn$ and $b_mn$ then,
beginalign
beginpmatrix
a_mn \
b_mn
endpmatrix
&=
-fracoperatornamecsch(gamma_mnL)2
beginpmatrix
e^-gamma_mnL & -1 \
-e^gamma_mnL & 1
endpmatrix
beginpmatrix
frac4big((-1)^m - 1big) big((-1)^n - 1big)pi^2mn \
0
endpmatrix
\
&= -frac2u_0pi^2fracbig((-1)^m - 1big) big((-1)^n - 1big)sinh(gamma_mnL) mn
beginpmatrix
e^-gamma_mnL \
-e^gamma_mnL
endpmatrix.
endalign
Thus $u$ is written
beginequation
u(x,y,z) = -frac4u_0pi^2 sum_m,n geq 1 Gamma_mn fracsinhbig(gamma_mn(x-L)big)sinh(gamma_mnL) sinfracn pi yell sin fracmpi zmu
endequation
with
$$
Gamma_mn = fracbig((-1)^m - 1big) big((-1)^n - 1big)mn.
$$
$hskip 1 in$
Mathematica code - it should be copy-pastable - enjoy.
ClearAll["Global`*"]
[Gamma]mn = Sqrt[((n [Pi])/[ScriptL])^2 + ((m [Pi])/[Mu])^2];
(* u0 not specified yet *)
L = 1;
[ScriptL] = 1;
[Mu] = 1;
(* Keep q low (~10-15 or so) as the number of terms grows as q^2 *)
(* i.e. higher q gives more accuracy but takes longer to run in effect *)
q = 5;
(*The coefficients amn and bmn amn,bmn=a,b/.Solve[a+b[Equal]4/
[Pi]^2(((-1)^m-1)((-1)^n-1))/(m n)u0,a Exp[[Gamma]mn L]+b Exp[-
[Gamma]mn L][Equal]0,a,b][[1]];*)
u[x_, y_,
z_] = (-4 u0)/[Pi]^2 Sum[
Sum[(((-1)^m - 1) ((-1)^n - 1))/(m n) Sinh[[Gamma]mn (x - L)]/
Sinh[[Gamma]mn L] Sin[n [Pi] y/[ScriptL]] Sin[
m [Pi] z/[Mu]], n, 1, q], m, 1, q];
Plot3D[u[0, y, z]/u0, y, 0, [ScriptL], z, 0, [Mu],
PlotLabel ->
"Laplace's Equation Solutionnat x=0 withn[ScriptL]=1, [Mu]=1,
25 terms",
AxesLabel -> "y", "z", "!(*FractionBox[(u), (u0)])",
Boxed -> False]
(* Something extra - it's computationally expensive (3d), at least for my
computer, so let it run for a bit *)
Manipulate[
ContourPlot[u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL],
ColorFunction -> "DarkRainbow"], z, 0, [Mu]]
(* alt+.to quit if it takes too long & reduce q *)
ContourPlot3D[
u[x, y, z]/u0, x, 0, L, y, 0, [ScriptL], z, 0, [Mu],
Mesh -> False, ColorFunction -> "Rainbow",
PlotLabel -> "Contours of Temperature", AxesLabel -> "x", "y", "z"]
answered Mar 24 at 4:26
AEngineerAEngineer
1,6521318
edited Mar 24 at 6:58
answered Mar 24 at 4:26
AEngineerAEngineer
1,6521318
answered Mar 24 at 4:26
AEngineerAEngineer
1,6521318
answered Mar 24 at 4:26
AEngineerAEngineer
1,6521318
1,6521318
$begingroup$
Thanks for such a detailed and well-explained answer. It seems the answer i arrived at is same with what you get . Maybe I did some mistake while writing the MATLAB script. Saying that would it be possible for you to post your Mathematica script here ? Also, your note about multiplying with respective eigen functions was helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 5:46
$begingroup$
Sure, glad I could help. I already deleted the code but I could post just the solution and the plot if you'd like.
$endgroup$
– AEngineer
Mar 24 at 5:58
$begingroup$
That would be really helpful. Appreciate it.
$endgroup$
– Indrasis Mitra
Mar 24 at 6:01
$begingroup$
Thanks for the Mathematica script. Some really cool graphics included. I was hoping you could take a look at this problem which I have been trying recently, especially its SP3 part. Any guidance will be really helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 16:12
add a comment |
$begingroup$
Thanks for such a detailed and well-explained answer. It seems the answer i arrived at is same with what you get . Maybe I did some mistake while writing the MATLAB script. Saying that would it be possible for you to post your Mathematica script here ? Also, your note about multiplying with respective eigen functions was helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 5:46
$begingroup$
Sure, glad I could help. I already deleted the code but I could post just the solution and the plot if you'd like.
$endgroup$
– AEngineer
Mar 24 at 5:58
$begingroup$
That would be really helpful. Appreciate it.
$endgroup$
– Indrasis Mitra
Mar 24 at 6:01
$begingroup$
Thanks for the Mathematica script. Some really cool graphics included. I was hoping you could take a look at this problem which I have been trying recently, especially its SP3 part. Any guidance will be really helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 16:12
$begingroup$
Thanks for such a detailed and well-explained answer. It seems the answer i arrived at is same with what you get . Maybe I did some mistake while writing the MATLAB script. Saying that would it be possible for you to post your Mathematica script here ? Also, your note about multiplying with respective eigen functions was helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 5:46
$begingroup$
Thanks for such a detailed and well-explained answer. It seems the answer i arrived at is same with what you get . Maybe I did some mistake while writing the MATLAB script. Saying that would it be possible for you to post your Mathematica script here ? Also, your note about multiplying with respective eigen functions was helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 5:46
$begingroup$
Sure, glad I could help. I already deleted the code but I could post just the solution and the plot if you'd like.
$endgroup$
– AEngineer
Mar 24 at 5:58
$begingroup$
Sure, glad I could help. I already deleted the code but I could post just the solution and the plot if you'd like.
$endgroup$
– AEngineer
Mar 24 at 5:58
$begingroup$
That would be really helpful. Appreciate it.
$endgroup$
– Indrasis Mitra
Mar 24 at 6:01
$begingroup$
That would be really helpful. Appreciate it.
$endgroup$
– Indrasis Mitra
Mar 24 at 6:01
$begingroup$
Thanks for the Mathematica script. Some really cool graphics included. I was hoping you could take a look at this problem which I have been trying recently, especially its SP3 part. Any guidance will be really helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 16:12
$begingroup$
Thanks for the Mathematica script. Some really cool graphics included. I was hoping you could take a look at this problem which I have been trying recently, especially its SP3 part. Any guidance will be really helpful.
$endgroup$
– Indrasis Mitra
Mar 24 at 16:12
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