Mixing
Laplace transform on a Boundary value problem (ODE)with . Am i attempting correctly?
$begingroup$
From a system of PDEs where i used the following ansatz: $$theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$$. $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$
So, $$theta_w(x,y) = e^{-beta_h x} F'(x) e^{-beta_c y} G'(y)$$
I have the following two third order linear ODEs which have been arrived at after applying separation of variables
begin{eqnarray}
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are
For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$
$lambda_h$, $beta_h$ and $V$ are constants $>0$, while $mu$ is the constant of separation.
I decided to apply the Laplace transform to each ODE and find the functions $F$ and $G$ individually finally obtaining $theta_w$.
Applying Laplace transform
($bar F$ is $mathcal{L}F(x)$)
$$lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2=0$$
Taking constants before each term as $a$, $b$, $c$ and $d$ makes $a=lambda_h$, $b=-2lambda_hbeta_h$, $c=(lambda_hbeta_h-1)beta_h-mu$ and $d={beta_h}^2$, we have
$$a[s^3bar F(s)-s^2 F(0)-sF'(0)-F''(0)]+b[s^2bar F(s)-sF(0)-F'(0)]+c[sbar F(s)-F(0)]+d[bar F(s)]=0$$
I know that the seperation constant $mu$ is contained in $c$ and is not a known quantity and can attain values $<,=,>0$. Ultimately after applying the first 2 b.c. i arrive at:
$$bar F(s)[as^3+bs^2+cs+d]=F'(0)[s+b+abeta_h]$$
I cannot proceed from here, as $F'(0)$ is not known. Also i have doubts whether my approach here to solve two variable separated ODEs with a separation constant using Laplace transform is correct in the first place or not ?
ordinary-differential-equations pde laplace-transform boundary-value-problem
asked Jan 13 at 6:43
Indrasis MitraIndrasis Mitra
5817
$endgroup$
add a comment |
$begingroup$
From a system of PDEs where i used the following ansatz: $$theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$$. $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$
So, $$theta_w(x,y) = e^{-beta_h x} F'(x) e^{-beta_c y} G'(y)$$
I have the following two third order linear ODEs which have been arrived at after applying separation of variables
begin{eqnarray}
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are
For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$
$lambda_h$, $beta_h$ and $V$ are constants $>0$, while $mu$ is the constant of separation.
I decided to apply the Laplace transform to each ODE and find the functions $F$ and $G$ individually finally obtaining $theta_w$.
Applying Laplace transform
($bar F$ is $mathcal{L}F(x)$)
$$lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2=0$$
Taking constants before each term as $a$, $b$, $c$ and $d$ makes $a=lambda_h$, $b=-2lambda_hbeta_h$, $c=(lambda_hbeta_h-1)beta_h-mu$ and $d={beta_h}^2$, we have
$$a[s^3bar F(s)-s^2 F(0)-sF'(0)-F''(0)]+b[s^2bar F(s)-sF(0)-F'(0)]+c[sbar F(s)-F(0)]+d[bar F(s)]=0$$
I know that the seperation constant $mu$ is contained in $c$ and is not a known quantity and can attain values $<,=,>0$. Ultimately after applying the first 2 b.c. i arrive at:
$$bar F(s)[as^3+bs^2+cs+d]=F'(0)[s+b+abeta_h]$$
I cannot proceed from here, as $F'(0)$ is not known. Also i have doubts whether my approach here to solve two variable separated ODEs with a separation constant using Laplace transform is correct in the first place or not ?
ordinary-differential-equations pde laplace-transform boundary-value-problem
asked Jan 13 at 6:43
Indrasis MitraIndrasis Mitra
5817
$endgroup$
$begingroup$
@Christoph any thoughts ?
$endgroup$
– Indrasis Mitra
Jan 13 at 9:05
add a comment |
$begingroup$
From a system of PDEs where i used the following ansatz: $$theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$$. $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$
So, $$theta_w(x,y) = e^{-beta_h x} F'(x) e^{-beta_c y} G'(y)$$
I have the following two third order linear ODEs which have been arrived at after applying separation of variables
begin{eqnarray}
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are
For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$
$lambda_h$, $beta_h$ and $V$ are constants $>0$, while $mu$ is the constant of separation.
I decided to apply the Laplace transform to each ODE and find the functions $F$ and $G$ individually finally obtaining $theta_w$.
Applying Laplace transform
($bar F$ is $mathcal{L}F(x)$)
$$lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2=0$$
Taking constants before each term as $a$, $b$, $c$ and $d$ makes $a=lambda_h$, $b=-2lambda_hbeta_h$, $c=(lambda_hbeta_h-1)beta_h-mu$ and $d={beta_h}^2$, we have
$$a[s^3bar F(s)-s^2 F(0)-sF'(0)-F''(0)]+b[s^2bar F(s)-sF(0)-F'(0)]+c[sbar F(s)-F(0)]+d[bar F(s)]=0$$
I know that the seperation constant $mu$ is contained in $c$ and is not a known quantity and can attain values $<,=,>0$. Ultimately after applying the first 2 b.c. i arrive at:
$$bar F(s)[as^3+bs^2+cs+d]=F'(0)[s+b+abeta_h]$$
I cannot proceed from here, as $F'(0)$ is not known. Also i have doubts whether my approach here to solve two variable separated ODEs with a separation constant using Laplace transform is correct in the first place or not ?
ordinary-differential-equations pde laplace-transform boundary-value-problem
asked Jan 13 at 6:43
Indrasis MitraIndrasis Mitra
5817
$endgroup$
From a system of PDEs where i used the following ansatz: $$theta_w(x,y) = e^{-beta_h x} f(x) e^{-beta_c y} g(y)$$. $F(x) := int f(x) , mathrm{d}x$ and $G(y) := int g(y) , mathrm{d}y$
So, $$theta_w(x,y) = e^{-beta_h x} F'(x) e^{-beta_c y} G'(y)$$
I have the following two third order linear ODEs which have been arrived at after applying separation of variables
begin{eqnarray}
lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2 F &=& 0,\
V lambda_c G''' - 2 V lambda_c beta_c G'' + left( (lambda_c beta_c - 1) V beta_c + mu right) G' + V beta_c^2 G &=& 0,
end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are
For $F$:
$$F(0)=0$$
$$frac{F''(0)}{F'(0)}=beta_h$$
$$frac{F''(1)}{F'(1)}=beta_h$$
$lambda_h$, $beta_h$ and $V$ are constants $>0$, while $mu$ is the constant of separation.
I decided to apply the Laplace transform to each ODE and find the functions $F$ and $G$ individually finally obtaining $theta_w$.
Applying Laplace transform
($bar F$ is $mathcal{L}F(x)$)
$$lambda_h F''' - 2 lambda_h beta_h F'' + left( (lambda_h beta_h - 1) beta_h - mu right) F' + beta_h^2=0$$
Taking constants before each term as $a$, $b$, $c$ and $d$ makes $a=lambda_h$, $b=-2lambda_hbeta_h$, $c=(lambda_hbeta_h-1)beta_h-mu$ and $d={beta_h}^2$, we have
$$a[s^3bar F(s)-s^2 F(0)-sF'(0)-F''(0)]+b[s^2bar F(s)-sF(0)-F'(0)]+c[sbar F(s)-F(0)]+d[bar F(s)]=0$$
I know that the seperation constant $mu$ is contained in $c$ and is not a known quantity and can attain values $<,=,>0$. Ultimately after applying the first 2 b.c. i arrive at:
$$bar F(s)[as^3+bs^2+cs+d]=F'(0)[s+b+abeta_h]$$
I cannot proceed from here, as $F'(0)$ is not known. Also i have doubts whether my approach here to solve two variable separated ODEs with a separation constant using Laplace transform is correct in the first place or not ?
ordinary-differential-equations pde laplace-transform boundary-value-problem
ordinary-differential-equations pde laplace-transform boundary-value-problem
asked Jan 13 at 6:43
Indrasis MitraIndrasis Mitra
5817
asked Jan 13 at 6:43
Indrasis MitraIndrasis Mitra
5817
asked Jan 13 at 6:43
Indrasis MitraIndrasis Mitra
5817
asked Jan 13 at 6:43
Indrasis MitraIndrasis Mitra
5817
asked Jan 13 at 6:43
Indrasis MitraIndrasis Mitra
5817
5817
$begingroup$
@Christoph any thoughts ?
$endgroup$
– Indrasis Mitra
Jan 13 at 9:05
add a comment |
$begingroup$
@Christoph any thoughts ?
$endgroup$
– Indrasis Mitra
Jan 13 at 9:05
$begingroup$
@Christoph any thoughts ?
$endgroup$
– Indrasis Mitra
Jan 13 at 9:05
$begingroup$
@Christoph any thoughts ?
$endgroup$
– Indrasis Mitra
Jan 13 at 9:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't believe that the Laplace-transform approach will work well here, since not all conditions are given at the same point.
In general, the homogeneous third-order linear ODE with constant coefficients $a y''' + b y'' + c y' + d y = 0$ has four types of solutions, depending on the zeros of the characteristic polynomial $P(lambda) = a lambda^3 + b lambda^2 + c lambda + d$.
Such a cubic function has always one real zero, and the first distinction happens according to whether the other two zeros are real or complex.
If all zeros are real, then the solution takes one of the forms
begin{equation}
y = C_1 e^{lambda_1 x} + C_2 e^{lambda_2 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x) e^{lambda_1 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x + C_3 x^2) e^{lambda_1 x},
end{equation}
depending on the multiplicity of the zeros.If only one zero is real then the solution will be of the form
begin{equation}
y = C_1 e^{lambda_1 x} + e^{-b_1 x} left( C_2 cos(b_2 x) + C_3 sin(b_2 x) right).
end{equation}
You could verify these solutions and then perhaps you could exclude some types using the given conditions. The configuration of the zeros of the characteristic polynomial $P$ will depend on the value of your separation constant, and you can distinguish these cases according to a discriminant similar as in the quadratic case (https://en.wikipedia.org/wiki/Cubic_function#The_discriminant), it's just more work!
Finally, this distinction should not be made according to the sign of the separation constant $mu$ but rather according to the sign of the discriminant of the characteristic polynomial $P$, which depends on $mu$.
answered Jan 13 at 10:16
ChristophChristoph
58116
$endgroup$
$begingroup$
much appreciation for these points. So if i am understanding correctly, no general $F$ and $G$ expressions are possible and one needs to take individual $lambda_h$ $beta_h$ values, see the behaviour of the discriminant of $P$, arriving at a general expression for F, repeat the same for $G$ and ultimately arrive at a $theta_w$ expression for those particular set of values
$endgroup$
– Indrasis Mitra
Jan 13 at 17:00
$begingroup$
Well, in principle I think that you can derive conditions on the parameters $lambda_h, beta_h$ in order to fall within a specific case, but I guess with third-order polynomials this would become rather complicated. It may be easier the way you described it.
$endgroup$
– Christoph
Jan 13 at 17:38
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071753%2flaplace-transform-on-a-boundary-value-problem-odewith-am-i-attempting-correc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't believe that the Laplace-transform approach will work well here, since not all conditions are given at the same point.
In general, the homogeneous third-order linear ODE with constant coefficients $a y''' + b y'' + c y' + d y = 0$ has four types of solutions, depending on the zeros of the characteristic polynomial $P(lambda) = a lambda^3 + b lambda^2 + c lambda + d$.
Such a cubic function has always one real zero, and the first distinction happens according to whether the other two zeros are real or complex.
If all zeros are real, then the solution takes one of the forms
begin{equation}
y = C_1 e^{lambda_1 x} + C_2 e^{lambda_2 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x) e^{lambda_1 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x + C_3 x^2) e^{lambda_1 x},
end{equation}
depending on the multiplicity of the zeros.If only one zero is real then the solution will be of the form
begin{equation}
y = C_1 e^{lambda_1 x} + e^{-b_1 x} left( C_2 cos(b_2 x) + C_3 sin(b_2 x) right).
end{equation}
You could verify these solutions and then perhaps you could exclude some types using the given conditions. The configuration of the zeros of the characteristic polynomial $P$ will depend on the value of your separation constant, and you can distinguish these cases according to a discriminant similar as in the quadratic case (https://en.wikipedia.org/wiki/Cubic_function#The_discriminant), it's just more work!
Finally, this distinction should not be made according to the sign of the separation constant $mu$ but rather according to the sign of the discriminant of the characteristic polynomial $P$, which depends on $mu$.
answered Jan 13 at 10:16
ChristophChristoph
58116
$endgroup$
$begingroup$
much appreciation for these points. So if i am understanding correctly, no general $F$ and $G$ expressions are possible and one needs to take individual $lambda_h$ $beta_h$ values, see the behaviour of the discriminant of $P$, arriving at a general expression for F, repeat the same for $G$ and ultimately arrive at a $theta_w$ expression for those particular set of values
$endgroup$
– Indrasis Mitra
Jan 13 at 17:00
$begingroup$
Well, in principle I think that you can derive conditions on the parameters $lambda_h, beta_h$ in order to fall within a specific case, but I guess with third-order polynomials this would become rather complicated. It may be easier the way you described it.
$endgroup$
– Christoph
Jan 13 at 17:38
add a comment |
$begingroup$
I don't believe that the Laplace-transform approach will work well here, since not all conditions are given at the same point.
In general, the homogeneous third-order linear ODE with constant coefficients $a y''' + b y'' + c y' + d y = 0$ has four types of solutions, depending on the zeros of the characteristic polynomial $P(lambda) = a lambda^3 + b lambda^2 + c lambda + d$.
Such a cubic function has always one real zero, and the first distinction happens according to whether the other two zeros are real or complex.
If all zeros are real, then the solution takes one of the forms
begin{equation}
y = C_1 e^{lambda_1 x} + C_2 e^{lambda_2 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x) e^{lambda_1 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x + C_3 x^2) e^{lambda_1 x},
end{equation}
depending on the multiplicity of the zeros.If only one zero is real then the solution will be of the form
begin{equation}
y = C_1 e^{lambda_1 x} + e^{-b_1 x} left( C_2 cos(b_2 x) + C_3 sin(b_2 x) right).
end{equation}
You could verify these solutions and then perhaps you could exclude some types using the given conditions. The configuration of the zeros of the characteristic polynomial $P$ will depend on the value of your separation constant, and you can distinguish these cases according to a discriminant similar as in the quadratic case (https://en.wikipedia.org/wiki/Cubic_function#The_discriminant), it's just more work!
Finally, this distinction should not be made according to the sign of the separation constant $mu$ but rather according to the sign of the discriminant of the characteristic polynomial $P$, which depends on $mu$.
answered Jan 13 at 10:16
ChristophChristoph
58116
$endgroup$
$begingroup$
much appreciation for these points. So if i am understanding correctly, no general $F$ and $G$ expressions are possible and one needs to take individual $lambda_h$ $beta_h$ values, see the behaviour of the discriminant of $P$, arriving at a general expression for F, repeat the same for $G$ and ultimately arrive at a $theta_w$ expression for those particular set of values
$endgroup$
– Indrasis Mitra
Jan 13 at 17:00
$begingroup$
Well, in principle I think that you can derive conditions on the parameters $lambda_h, beta_h$ in order to fall within a specific case, but I guess with third-order polynomials this would become rather complicated. It may be easier the way you described it.
$endgroup$
– Christoph
Jan 13 at 17:38
add a comment |
$begingroup$
I don't believe that the Laplace-transform approach will work well here, since not all conditions are given at the same point.
In general, the homogeneous third-order linear ODE with constant coefficients $a y''' + b y'' + c y' + d y = 0$ has four types of solutions, depending on the zeros of the characteristic polynomial $P(lambda) = a lambda^3 + b lambda^2 + c lambda + d$.
Such a cubic function has always one real zero, and the first distinction happens according to whether the other two zeros are real or complex.
If all zeros are real, then the solution takes one of the forms
begin{equation}
y = C_1 e^{lambda_1 x} + C_2 e^{lambda_2 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x) e^{lambda_1 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x + C_3 x^2) e^{lambda_1 x},
end{equation}
depending on the multiplicity of the zeros.If only one zero is real then the solution will be of the form
begin{equation}
y = C_1 e^{lambda_1 x} + e^{-b_1 x} left( C_2 cos(b_2 x) + C_3 sin(b_2 x) right).
end{equation}
You could verify these solutions and then perhaps you could exclude some types using the given conditions. The configuration of the zeros of the characteristic polynomial $P$ will depend on the value of your separation constant, and you can distinguish these cases according to a discriminant similar as in the quadratic case (https://en.wikipedia.org/wiki/Cubic_function#The_discriminant), it's just more work!
Finally, this distinction should not be made according to the sign of the separation constant $mu$ but rather according to the sign of the discriminant of the characteristic polynomial $P$, which depends on $mu$.
answered Jan 13 at 10:16
ChristophChristoph
58116
$endgroup$
I don't believe that the Laplace-transform approach will work well here, since not all conditions are given at the same point.
In general, the homogeneous third-order linear ODE with constant coefficients $a y''' + b y'' + c y' + d y = 0$ has four types of solutions, depending on the zeros of the characteristic polynomial $P(lambda) = a lambda^3 + b lambda^2 + c lambda + d$.
Such a cubic function has always one real zero, and the first distinction happens according to whether the other two zeros are real or complex.
If all zeros are real, then the solution takes one of the forms
begin{equation}
y = C_1 e^{lambda_1 x} + C_2 e^{lambda_2 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x) e^{lambda_1 x} + C_3 e^{lambda_3 x}, quad y = (C_1 + C_2 x + C_3 x^2) e^{lambda_1 x},
end{equation}
depending on the multiplicity of the zeros.If only one zero is real then the solution will be of the form
begin{equation}
y = C_1 e^{lambda_1 x} + e^{-b_1 x} left( C_2 cos(b_2 x) + C_3 sin(b_2 x) right).
end{equation}
You could verify these solutions and then perhaps you could exclude some types using the given conditions. The configuration of the zeros of the characteristic polynomial $P$ will depend on the value of your separation constant, and you can distinguish these cases according to a discriminant similar as in the quadratic case (https://en.wikipedia.org/wiki/Cubic_function#The_discriminant), it's just more work!
Finally, this distinction should not be made according to the sign of the separation constant $mu$ but rather according to the sign of the discriminant of the characteristic polynomial $P$, which depends on $mu$.
answered Jan 13 at 10:16
ChristophChristoph
58116
edited Jan 13 at 10:49
answered Jan 13 at 10:16
ChristophChristoph
58116
answered Jan 13 at 10:16
ChristophChristoph
58116
answered Jan 13 at 10:16
ChristophChristoph
58116
58116
$begingroup$
much appreciation for these points. So if i am understanding correctly, no general $F$ and $G$ expressions are possible and one needs to take individual $lambda_h$ $beta_h$ values, see the behaviour of the discriminant of $P$, arriving at a general expression for F, repeat the same for $G$ and ultimately arrive at a $theta_w$ expression for those particular set of values
$endgroup$
– Indrasis Mitra
Jan 13 at 17:00
$begingroup$
Well, in principle I think that you can derive conditions on the parameters $lambda_h, beta_h$ in order to fall within a specific case, but I guess with third-order polynomials this would become rather complicated. It may be easier the way you described it.
$endgroup$
– Christoph
Jan 13 at 17:38
add a comment |
$begingroup$
much appreciation for these points. So if i am understanding correctly, no general $F$ and $G$ expressions are possible and one needs to take individual $lambda_h$ $beta_h$ values, see the behaviour of the discriminant of $P$, arriving at a general expression for F, repeat the same for $G$ and ultimately arrive at a $theta_w$ expression for those particular set of values
$endgroup$
– Indrasis Mitra
Jan 13 at 17:00
$begingroup$
Well, in principle I think that you can derive conditions on the parameters $lambda_h, beta_h$ in order to fall within a specific case, but I guess with third-order polynomials this would become rather complicated. It may be easier the way you described it.
$endgroup$
– Christoph
Jan 13 at 17:38
$begingroup$
much appreciation for these points. So if i am understanding correctly, no general $F$ and $G$ expressions are possible and one needs to take individual $lambda_h$ $beta_h$ values, see the behaviour of the discriminant of $P$, arriving at a general expression for F, repeat the same for $G$ and ultimately arrive at a $theta_w$ expression for those particular set of values
$endgroup$
– Indrasis Mitra
Jan 13 at 17:00
$begingroup$
much appreciation for these points. So if i am understanding correctly, no general $F$ and $G$ expressions are possible and one needs to take individual $lambda_h$ $beta_h$ values, see the behaviour of the discriminant of $P$, arriving at a general expression for F, repeat the same for $G$ and ultimately arrive at a $theta_w$ expression for those particular set of values
$endgroup$
– Indrasis Mitra
Jan 13 at 17:00
$begingroup$
Well, in principle I think that you can derive conditions on the parameters $lambda_h, beta_h$ in order to fall within a specific case, but I guess with third-order polynomials this would become rather complicated. It may be easier the way you described it.
$endgroup$
– Christoph
Jan 13 at 17:38
$begingroup$
Well, in principle I think that you can derive conditions on the parameters $lambda_h, beta_h$ in order to fall within a specific case, but I guess with third-order polynomials this would become rather complicated. It may be easier the way you described it.
$endgroup$
– Christoph
Jan 13 at 17:38
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071753%2flaplace-transform-on-a-boundary-value-problem-odewith-am-i-attempting-correc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
@Christoph any thoughts ?
$endgroup$
– Indrasis Mitra
Jan 13 at 9:05