An IVP with 'dummy' singularityAn IVP with singularity having continuous solutionsNon-linear IVP questionUniqueness/Existence ODE questionShow following IVP has solutions.Prove that $c_1 phi_1 + c_2 phi_2$ solves the IVPDo we have uniqueness of solutions to the IVP $dotx(t) = x(t)^2, x(0) = 0?$Unique solutions with given inital value problem for ODEThe applications of Picard's theoremProof a sufficient condition to indicate the maximal interval of existence of an IVP 's solution (Wintner 's theorem)Global existence of IVP on $mathbb R$An IVP with singularity having continuous solutions
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An IVP with 'dummy' singularity
An IVP with singularity having continuous solutionsNon-linear IVP questionUniqueness/Existence ODE questionShow following IVP has solutions.Prove that $c_1 phi_1 + c_2 phi_2$ solves the IVPDo we have uniqueness of solutions to the IVP $dotx(t) = x(t)^2, x(0) = 0?$Unique solutions with given inital value problem for ODEThe applications of Picard's theoremProof a sufficient condition to indicate the maximal interval of existence of an IVP 's solution (Wintner 's theorem)Global existence of IVP on $mathbb R$An IVP with singularity having continuous solutions
$begingroup$
Suppose we are given a first order IVP:
$$ (1-x)y' = (1-x)y, quad y(0) = 1 . $$
If one is asked to find the largest interval $I$ so that the solutions is defined, should one answer $ I = (-infty, 1) $ or $ I = (-infty,infty) = mathbbR $?
I think the answer is $ I = (-infty, 1) $ because anyway $x=1$ is a singularity of the equation so the solution can be defined only in either $(-infty, 1)$ or $(1,infty)$ according to the initial value.
I'd appreciate it to any explanation on this. Thank you.
ordinary-differential-equations analysis
edited Mar 15 at 6:28
Paras Khosla
2,468322
asked Mar 15 at 6:27
Hoil LeeHoil Lee
154
$endgroup$
|
show 1 more comment
$begingroup$
Suppose we are given a first order IVP:
$$ (1-x)y' = (1-x)y, quad y(0) = 1 . $$
If one is asked to find the largest interval $I$ so that the solutions is defined, should one answer $ I = (-infty, 1) $ or $ I = (-infty,infty) = mathbbR $?
I think the answer is $ I = (-infty, 1) $ because anyway $x=1$ is a singularity of the equation so the solution can be defined only in either $(-infty, 1)$ or $(1,infty)$ according to the initial value.
I'd appreciate it to any explanation on this. Thank you.
ordinary-differential-equations analysis
edited Mar 15 at 6:28
Paras Khosla
2,468322
asked Mar 15 at 6:27
Hoil LeeHoil Lee
154
$endgroup$
$begingroup$
Welcome to MSE. The function $y(x)=e^x$ satisfies the ODE for all values of $tinmathbbR$, in particular for $t=1$.
$endgroup$
– user539887
Mar 15 at 7:03
$begingroup$
@user539887 Yes it is, but isn't it also true that the solution to an IVP is well-defined only in the interval without singularity?
$endgroup$
– Hoil Lee
Mar 15 at 7:05
$begingroup$
It depends on the definition. I took the most obvious one: a function satisfies (or, if you prefer, is a solution of) a differential equation if for each value in the domain of the function the differential equation holds. And such is the case for $t=1$: the LHS is $(1-1) e^1=0$ and the RHS is $(1-1) e^1=0$.
$endgroup$
– user539887
Mar 15 at 7:09
$begingroup$
@user539887 Thank you for your kind explanations. I think I understand it for this particular problem. What if we have an IVP that is not well-defined for some values of $x$, e.g. $y′+(tan x) cdot y=cos^2 x, quad y(0)=1$? We cannot evaluate the value of the LHS for $x = (2n+1)pi/2$, but $ y = cos x sin x + cos x $ satisfies the DE. Can we say the solution is well-defined on the whole real line?
$endgroup$
– Hoil Lee
Mar 15 at 7:30
$begingroup$
Actually this was my original question: math.stackexchange.com/questions/3147887/… and I was trying to simplify the DE to have a singularity but also to have a regular solution.
$endgroup$
– Hoil Lee
Mar 15 at 7:34
|
show 1 more comment
$begingroup$
Suppose we are given a first order IVP:
$$ (1-x)y' = (1-x)y, quad y(0) = 1 . $$
If one is asked to find the largest interval $I$ so that the solutions is defined, should one answer $ I = (-infty, 1) $ or $ I = (-infty,infty) = mathbbR $?
I think the answer is $ I = (-infty, 1) $ because anyway $x=1$ is a singularity of the equation so the solution can be defined only in either $(-infty, 1)$ or $(1,infty)$ according to the initial value.
I'd appreciate it to any explanation on this. Thank you.
ordinary-differential-equations analysis
edited Mar 15 at 6:28
Paras Khosla
2,468322
asked Mar 15 at 6:27
Hoil LeeHoil Lee
154
$endgroup$
Suppose we are given a first order IVP:
$$ (1-x)y' = (1-x)y, quad y(0) = 1 . $$
If one is asked to find the largest interval $I$ so that the solutions is defined, should one answer $ I = (-infty, 1) $ or $ I = (-infty,infty) = mathbbR $?
I think the answer is $ I = (-infty, 1) $ because anyway $x=1$ is a singularity of the equation so the solution can be defined only in either $(-infty, 1)$ or $(1,infty)$ according to the initial value.
I'd appreciate it to any explanation on this. Thank you.
ordinary-differential-equations analysis
ordinary-differential-equations analysis
edited Mar 15 at 6:28
Paras Khosla
2,468322
asked Mar 15 at 6:27
Hoil LeeHoil Lee
154
edited Mar 15 at 6:28
Paras Khosla
2,468322
asked Mar 15 at 6:27
Hoil LeeHoil Lee
154
edited Mar 15 at 6:28
Paras Khosla
2,468322
edited Mar 15 at 6:28
Paras Khosla
2,468322
edited Mar 15 at 6:28
Paras Khosla
2,468322
2,468322
asked Mar 15 at 6:27
Hoil LeeHoil Lee
154
asked Mar 15 at 6:27
Hoil LeeHoil Lee
154
asked Mar 15 at 6:27
Hoil LeeHoil Lee
154
154
$begingroup$
Welcome to MSE. The function $y(x)=e^x$ satisfies the ODE for all values of $tinmathbbR$, in particular for $t=1$.
$endgroup$
– user539887
Mar 15 at 7:03
$begingroup$
@user539887 Yes it is, but isn't it also true that the solution to an IVP is well-defined only in the interval without singularity?
$endgroup$
– Hoil Lee
Mar 15 at 7:05
$begingroup$
It depends on the definition. I took the most obvious one: a function satisfies (or, if you prefer, is a solution of) a differential equation if for each value in the domain of the function the differential equation holds. And such is the case for $t=1$: the LHS is $(1-1) e^1=0$ and the RHS is $(1-1) e^1=0$.
$endgroup$
– user539887
Mar 15 at 7:09
$begingroup$
@user539887 Thank you for your kind explanations. I think I understand it for this particular problem. What if we have an IVP that is not well-defined for some values of $x$, e.g. $y′+(tan x) cdot y=cos^2 x, quad y(0)=1$? We cannot evaluate the value of the LHS for $x = (2n+1)pi/2$, but $ y = cos x sin x + cos x $ satisfies the DE. Can we say the solution is well-defined on the whole real line?
$endgroup$
– Hoil Lee
Mar 15 at 7:30
$begingroup$
Actually this was my original question: math.stackexchange.com/questions/3147887/… and I was trying to simplify the DE to have a singularity but also to have a regular solution.
$endgroup$
– Hoil Lee
Mar 15 at 7:34
|
show 1 more comment
$begingroup$
Welcome to MSE. The function $y(x)=e^x$ satisfies the ODE for all values of $tinmathbbR$, in particular for $t=1$.
$endgroup$
– user539887
Mar 15 at 7:03
$begingroup$
@user539887 Yes it is, but isn't it also true that the solution to an IVP is well-defined only in the interval without singularity?
$endgroup$
– Hoil Lee
Mar 15 at 7:05
$begingroup$
It depends on the definition. I took the most obvious one: a function satisfies (or, if you prefer, is a solution of) a differential equation if for each value in the domain of the function the differential equation holds. And such is the case for $t=1$: the LHS is $(1-1) e^1=0$ and the RHS is $(1-1) e^1=0$.
$endgroup$
– user539887
Mar 15 at 7:09
$begingroup$
@user539887 Thank you for your kind explanations. I think I understand it for this particular problem. What if we have an IVP that is not well-defined for some values of $x$, e.g. $y′+(tan x) cdot y=cos^2 x, quad y(0)=1$? We cannot evaluate the value of the LHS for $x = (2n+1)pi/2$, but $ y = cos x sin x + cos x $ satisfies the DE. Can we say the solution is well-defined on the whole real line?
$endgroup$
– Hoil Lee
Mar 15 at 7:30
$begingroup$
Actually this was my original question: math.stackexchange.com/questions/3147887/… and I was trying to simplify the DE to have a singularity but also to have a regular solution.
$endgroup$
– Hoil Lee
Mar 15 at 7:34
$begingroup$
Welcome to MSE. The function $y(x)=e^x$ satisfies the ODE for all values of $tinmathbbR$, in particular for $t=1$.
$endgroup$
– user539887
Mar 15 at 7:03
$begingroup$
Welcome to MSE. The function $y(x)=e^x$ satisfies the ODE for all values of $tinmathbbR$, in particular for $t=1$.
$endgroup$
– user539887
Mar 15 at 7:03
$begingroup$
@user539887 Yes it is, but isn't it also true that the solution to an IVP is well-defined only in the interval without singularity?
$endgroup$
– Hoil Lee
Mar 15 at 7:05
$begingroup$
@user539887 Yes it is, but isn't it also true that the solution to an IVP is well-defined only in the interval without singularity?
$endgroup$
– Hoil Lee
Mar 15 at 7:05
$begingroup$
It depends on the definition. I took the most obvious one: a function satisfies (or, if you prefer, is a solution of) a differential equation if for each value in the domain of the function the differential equation holds. And such is the case for $t=1$: the LHS is $(1-1) e^1=0$ and the RHS is $(1-1) e^1=0$.
$endgroup$
– user539887
Mar 15 at 7:09
$begingroup$
It depends on the definition. I took the most obvious one: a function satisfies (or, if you prefer, is a solution of) a differential equation if for each value in the domain of the function the differential equation holds. And such is the case for $t=1$: the LHS is $(1-1) e^1=0$ and the RHS is $(1-1) e^1=0$.
$endgroup$
– user539887
Mar 15 at 7:09
$begingroup$
@user539887 Thank you for your kind explanations. I think I understand it for this particular problem. What if we have an IVP that is not well-defined for some values of $x$, e.g. $y′+(tan x) cdot y=cos^2 x, quad y(0)=1$? We cannot evaluate the value of the LHS for $x = (2n+1)pi/2$, but $ y = cos x sin x + cos x $ satisfies the DE. Can we say the solution is well-defined on the whole real line?
$endgroup$
– Hoil Lee
Mar 15 at 7:30
$begingroup$
@user539887 Thank you for your kind explanations. I think I understand it for this particular problem. What if we have an IVP that is not well-defined for some values of $x$, e.g. $y′+(tan x) cdot y=cos^2 x, quad y(0)=1$? We cannot evaluate the value of the LHS for $x = (2n+1)pi/2$, but $ y = cos x sin x + cos x $ satisfies the DE. Can we say the solution is well-defined on the whole real line?
$endgroup$
– Hoil Lee
Mar 15 at 7:30
$begingroup$
Actually this was my original question: math.stackexchange.com/questions/3147887/… and I was trying to simplify the DE to have a singularity but also to have a regular solution.
$endgroup$
– Hoil Lee
Mar 15 at 7:34
$begingroup$
Actually this was my original question: math.stackexchange.com/questions/3147887/… and I was trying to simplify the DE to have a singularity but also to have a regular solution.
$endgroup$
– Hoil Lee
Mar 15 at 7:34
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Consider the function $$f(a,b,c,x):=begincasesacdot e^x & x<1\b & x=1\ccdot e^x & x>1,endcases$$ and suppose $F(x)=f(a,b,c,x)$ for some $a,b,cinBbb R.$ Here are some facts about $F$ that I leave to you to verify:
- $F:Bbb RtoBbb R.$
$F$ is continuous(ly differentiable) on $Bbb R$ if and only if $a=b=c.$
$y=F(x)$ is a solution to the IVP if and only if $a=1.$
So, we certainly don't have uniqueness outside the interval $(-infty,1).$ The only everywhere-differentiable solution is $y=e^x,$ and all solutions agree with it on $(-infty,1),$ but (unless we impose further restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:32
Cameron BuieCameron Buie
85.9k772161
$endgroup$
$begingroup$
If $a=b=c$ does not hold then $F'(1)$ is undefined. And the result of multiplying zero by something undefined is undefined, not zero. So, under the natural interpretation of the term "solution", such an $F$ is not a solution. When one admits solutions satisfying the equation a.e., then yes, there are many solutions, but I doubt if that was the OP's intention.
$endgroup$
– user539887
Mar 17 at 11:37
$begingroup$
I see your point, but I'm not sure I agree. Consider the IVP $xy'+2y=0,$ $y(1)=1.$ Would you say it has no solution? Would you say that $y=frac1x^2$ is a solution, even though $y'(0)$ is undefined?
$endgroup$
– Cameron Buie
Mar 17 at 13:25
$begingroup$
In this case I would say that $y=frac1x^2$ is a solution defined on $(0,infty)$.
$endgroup$
– user539887
Mar 17 at 13:35
$begingroup$
I would agree that it is a solution, though it's defined on more than just $(0,infty).$ In fact, for any constant $c,$ I would say that $$f(x)=begincasesfrac1x^2 & x>0\fraccx^2 & x<0endcases$$ is a solution with the same domain of definition. Of course, they all agree on $(0,infty),$ which means that our solutions have uniquely-defined behavior on $(0,infty),$ but that doesn't mean they don't exist anywhere else, nor do they fail to be solutions simply by virtue of failing to be differentiable everywhere. Would you agree?
$endgroup$
– Cameron Buie
Mar 17 at 13:50
$begingroup$
I remember a discussion somewhere on MSE about a similar thing. The OP asked about why a solution of an IVP cannot be defined on a disjoint union of two intervals. The most upvoted answer (with which I fully agree) was that then one would lose the uniqueness. I am sorry that I cannot find it.
$endgroup$
– user539887
Mar 17 at 13:56
|
show 2 more comments
Your Answer
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
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active
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votes
$begingroup$
Consider the function $$f(a,b,c,x):=begincasesacdot e^x & x<1\b & x=1\ccdot e^x & x>1,endcases$$ and suppose $F(x)=f(a,b,c,x)$ for some $a,b,cinBbb R.$ Here are some facts about $F$ that I leave to you to verify:
- $F:Bbb RtoBbb R.$
$F$ is continuous(ly differentiable) on $Bbb R$ if and only if $a=b=c.$
$y=F(x)$ is a solution to the IVP if and only if $a=1.$
So, we certainly don't have uniqueness outside the interval $(-infty,1).$ The only everywhere-differentiable solution is $y=e^x,$ and all solutions agree with it on $(-infty,1),$ but (unless we impose further restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:32
Cameron BuieCameron Buie
85.9k772161
$endgroup$
$begingroup$
If $a=b=c$ does not hold then $F'(1)$ is undefined. And the result of multiplying zero by something undefined is undefined, not zero. So, under the natural interpretation of the term "solution", such an $F$ is not a solution. When one admits solutions satisfying the equation a.e., then yes, there are many solutions, but I doubt if that was the OP's intention.
$endgroup$
– user539887
Mar 17 at 11:37
$begingroup$
I see your point, but I'm not sure I agree. Consider the IVP $xy'+2y=0,$ $y(1)=1.$ Would you say it has no solution? Would you say that $y=frac1x^2$ is a solution, even though $y'(0)$ is undefined?
$endgroup$
– Cameron Buie
Mar 17 at 13:25
$begingroup$
In this case I would say that $y=frac1x^2$ is a solution defined on $(0,infty)$.
$endgroup$
– user539887
Mar 17 at 13:35
$begingroup$
I would agree that it is a solution, though it's defined on more than just $(0,infty).$ In fact, for any constant $c,$ I would say that $$f(x)=begincasesfrac1x^2 & x>0\fraccx^2 & x<0endcases$$ is a solution with the same domain of definition. Of course, they all agree on $(0,infty),$ which means that our solutions have uniquely-defined behavior on $(0,infty),$ but that doesn't mean they don't exist anywhere else, nor do they fail to be solutions simply by virtue of failing to be differentiable everywhere. Would you agree?
$endgroup$
– Cameron Buie
Mar 17 at 13:50
$begingroup$
I remember a discussion somewhere on MSE about a similar thing. The OP asked about why a solution of an IVP cannot be defined on a disjoint union of two intervals. The most upvoted answer (with which I fully agree) was that then one would lose the uniqueness. I am sorry that I cannot find it.
$endgroup$
– user539887
Mar 17 at 13:56
|
show 2 more comments
$begingroup$
Consider the function $$f(a,b,c,x):=begincasesacdot e^x & x<1\b & x=1\ccdot e^x & x>1,endcases$$ and suppose $F(x)=f(a,b,c,x)$ for some $a,b,cinBbb R.$ Here are some facts about $F$ that I leave to you to verify:
- $F:Bbb RtoBbb R.$
$F$ is continuous(ly differentiable) on $Bbb R$ if and only if $a=b=c.$
$y=F(x)$ is a solution to the IVP if and only if $a=1.$
So, we certainly don't have uniqueness outside the interval $(-infty,1).$ The only everywhere-differentiable solution is $y=e^x,$ and all solutions agree with it on $(-infty,1),$ but (unless we impose further restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:32
Cameron BuieCameron Buie
85.9k772161
$endgroup$
$begingroup$
If $a=b=c$ does not hold then $F'(1)$ is undefined. And the result of multiplying zero by something undefined is undefined, not zero. So, under the natural interpretation of the term "solution", such an $F$ is not a solution. When one admits solutions satisfying the equation a.e., then yes, there are many solutions, but I doubt if that was the OP's intention.
$endgroup$
– user539887
Mar 17 at 11:37
$begingroup$
I see your point, but I'm not sure I agree. Consider the IVP $xy'+2y=0,$ $y(1)=1.$ Would you say it has no solution? Would you say that $y=frac1x^2$ is a solution, even though $y'(0)$ is undefined?
$endgroup$
– Cameron Buie
Mar 17 at 13:25
$begingroup$
In this case I would say that $y=frac1x^2$ is a solution defined on $(0,infty)$.
$endgroup$
– user539887
Mar 17 at 13:35
$begingroup$
I would agree that it is a solution, though it's defined on more than just $(0,infty).$ In fact, for any constant $c,$ I would say that $$f(x)=begincasesfrac1x^2 & x>0\fraccx^2 & x<0endcases$$ is a solution with the same domain of definition. Of course, they all agree on $(0,infty),$ which means that our solutions have uniquely-defined behavior on $(0,infty),$ but that doesn't mean they don't exist anywhere else, nor do they fail to be solutions simply by virtue of failing to be differentiable everywhere. Would you agree?
$endgroup$
– Cameron Buie
Mar 17 at 13:50
$begingroup$
I remember a discussion somewhere on MSE about a similar thing. The OP asked about why a solution of an IVP cannot be defined on a disjoint union of two intervals. The most upvoted answer (with which I fully agree) was that then one would lose the uniqueness. I am sorry that I cannot find it.
$endgroup$
– user539887
Mar 17 at 13:56
|
show 2 more comments
$begingroup$
Consider the function $$f(a,b,c,x):=begincasesacdot e^x & x<1\b & x=1\ccdot e^x & x>1,endcases$$ and suppose $F(x)=f(a,b,c,x)$ for some $a,b,cinBbb R.$ Here are some facts about $F$ that I leave to you to verify:
- $F:Bbb RtoBbb R.$
$F$ is continuous(ly differentiable) on $Bbb R$ if and only if $a=b=c.$
$y=F(x)$ is a solution to the IVP if and only if $a=1.$
So, we certainly don't have uniqueness outside the interval $(-infty,1).$ The only everywhere-differentiable solution is $y=e^x,$ and all solutions agree with it on $(-infty,1),$ but (unless we impose further restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:32
Cameron BuieCameron Buie
85.9k772161
$endgroup$
Consider the function $$f(a,b,c,x):=begincasesacdot e^x & x<1\b & x=1\ccdot e^x & x>1,endcases$$ and suppose $F(x)=f(a,b,c,x)$ for some $a,b,cinBbb R.$ Here are some facts about $F$ that I leave to you to verify:
- $F:Bbb RtoBbb R.$
$F$ is continuous(ly differentiable) on $Bbb R$ if and only if $a=b=c.$
$y=F(x)$ is a solution to the IVP if and only if $a=1.$
So, we certainly don't have uniqueness outside the interval $(-infty,1).$ The only everywhere-differentiable solution is $y=e^x,$ and all solutions agree with it on $(-infty,1),$ but (unless we impose further restrictions on our solutions) need not agree with it anywhere else.
answered Mar 16 at 16:32
Cameron BuieCameron Buie
85.9k772161
edited Mar 17 at 13:25
answered Mar 16 at 16:32
Cameron BuieCameron Buie
85.9k772161
answered Mar 16 at 16:32
Cameron BuieCameron Buie
85.9k772161
answered Mar 16 at 16:32
Cameron BuieCameron Buie
85.9k772161
85.9k772161
$begingroup$
If $a=b=c$ does not hold then $F'(1)$ is undefined. And the result of multiplying zero by something undefined is undefined, not zero. So, under the natural interpretation of the term "solution", such an $F$ is not a solution. When one admits solutions satisfying the equation a.e., then yes, there are many solutions, but I doubt if that was the OP's intention.
$endgroup$
– user539887
Mar 17 at 11:37
$begingroup$
I see your point, but I'm not sure I agree. Consider the IVP $xy'+2y=0,$ $y(1)=1.$ Would you say it has no solution? Would you say that $y=frac1x^2$ is a solution, even though $y'(0)$ is undefined?
$endgroup$
– Cameron Buie
Mar 17 at 13:25
$begingroup$
In this case I would say that $y=frac1x^2$ is a solution defined on $(0,infty)$.
$endgroup$
– user539887
Mar 17 at 13:35
$begingroup$
I would agree that it is a solution, though it's defined on more than just $(0,infty).$ In fact, for any constant $c,$ I would say that $$f(x)=begincasesfrac1x^2 & x>0\fraccx^2 & x<0endcases$$ is a solution with the same domain of definition. Of course, they all agree on $(0,infty),$ which means that our solutions have uniquely-defined behavior on $(0,infty),$ but that doesn't mean they don't exist anywhere else, nor do they fail to be solutions simply by virtue of failing to be differentiable everywhere. Would you agree?
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– Cameron Buie
Mar 17 at 13:50
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I remember a discussion somewhere on MSE about a similar thing. The OP asked about why a solution of an IVP cannot be defined on a disjoint union of two intervals. The most upvoted answer (with which I fully agree) was that then one would lose the uniqueness. I am sorry that I cannot find it.
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– user539887
Mar 17 at 13:56
|
show 2 more comments
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If $a=b=c$ does not hold then $F'(1)$ is undefined. And the result of multiplying zero by something undefined is undefined, not zero. So, under the natural interpretation of the term "solution", such an $F$ is not a solution. When one admits solutions satisfying the equation a.e., then yes, there are many solutions, but I doubt if that was the OP's intention.
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– user539887
Mar 17 at 11:37
$begingroup$
I see your point, but I'm not sure I agree. Consider the IVP $xy'+2y=0,$ $y(1)=1.$ Would you say it has no solution? Would you say that $y=frac1x^2$ is a solution, even though $y'(0)$ is undefined?
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– Cameron Buie
Mar 17 at 13:25
$begingroup$
In this case I would say that $y=frac1x^2$ is a solution defined on $(0,infty)$.
$endgroup$
– user539887
Mar 17 at 13:35
$begingroup$
I would agree that it is a solution, though it's defined on more than just $(0,infty).$ In fact, for any constant $c,$ I would say that $$f(x)=begincasesfrac1x^2 & x>0\fraccx^2 & x<0endcases$$ is a solution with the same domain of definition. Of course, they all agree on $(0,infty),$ which means that our solutions have uniquely-defined behavior on $(0,infty),$ but that doesn't mean they don't exist anywhere else, nor do they fail to be solutions simply by virtue of failing to be differentiable everywhere. Would you agree?
$endgroup$
– Cameron Buie
Mar 17 at 13:50
$begingroup$
I remember a discussion somewhere on MSE about a similar thing. The OP asked about why a solution of an IVP cannot be defined on a disjoint union of two intervals. The most upvoted answer (with which I fully agree) was that then one would lose the uniqueness. I am sorry that I cannot find it.
$endgroup$
– user539887
Mar 17 at 13:56
$begingroup$
If $a=b=c$ does not hold then $F'(1)$ is undefined. And the result of multiplying zero by something undefined is undefined, not zero. So, under the natural interpretation of the term "solution", such an $F$ is not a solution. When one admits solutions satisfying the equation a.e., then yes, there are many solutions, but I doubt if that was the OP's intention.
$endgroup$
– user539887
Mar 17 at 11:37
$begingroup$
If $a=b=c$ does not hold then $F'(1)$ is undefined. And the result of multiplying zero by something undefined is undefined, not zero. So, under the natural interpretation of the term "solution", such an $F$ is not a solution. When one admits solutions satisfying the equation a.e., then yes, there are many solutions, but I doubt if that was the OP's intention.
$endgroup$
– user539887
Mar 17 at 11:37
$begingroup$
I see your point, but I'm not sure I agree. Consider the IVP $xy'+2y=0,$ $y(1)=1.$ Would you say it has no solution? Would you say that $y=frac1x^2$ is a solution, even though $y'(0)$ is undefined?
$endgroup$
– Cameron Buie
Mar 17 at 13:25
$begingroup$
I see your point, but I'm not sure I agree. Consider the IVP $xy'+2y=0,$ $y(1)=1.$ Would you say it has no solution? Would you say that $y=frac1x^2$ is a solution, even though $y'(0)$ is undefined?
$endgroup$
– Cameron Buie
Mar 17 at 13:25
$begingroup$
In this case I would say that $y=frac1x^2$ is a solution defined on $(0,infty)$.
$endgroup$
– user539887
Mar 17 at 13:35
$begingroup$
In this case I would say that $y=frac1x^2$ is a solution defined on $(0,infty)$.
$endgroup$
– user539887
Mar 17 at 13:35
$begingroup$
I would agree that it is a solution, though it's defined on more than just $(0,infty).$ In fact, for any constant $c,$ I would say that $$f(x)=begincasesfrac1x^2 & x>0\fraccx^2 & x<0endcases$$ is a solution with the same domain of definition. Of course, they all agree on $(0,infty),$ which means that our solutions have uniquely-defined behavior on $(0,infty),$ but that doesn't mean they don't exist anywhere else, nor do they fail to be solutions simply by virtue of failing to be differentiable everywhere. Would you agree?
$endgroup$
– Cameron Buie
Mar 17 at 13:50
$begingroup$
I would agree that it is a solution, though it's defined on more than just $(0,infty).$ In fact, for any constant $c,$ I would say that $$f(x)=begincasesfrac1x^2 & x>0\fraccx^2 & x<0endcases$$ is a solution with the same domain of definition. Of course, they all agree on $(0,infty),$ which means that our solutions have uniquely-defined behavior on $(0,infty),$ but that doesn't mean they don't exist anywhere else, nor do they fail to be solutions simply by virtue of failing to be differentiable everywhere. Would you agree?
$endgroup$
– Cameron Buie
Mar 17 at 13:50
$begingroup$
I remember a discussion somewhere on MSE about a similar thing. The OP asked about why a solution of an IVP cannot be defined on a disjoint union of two intervals. The most upvoted answer (with which I fully agree) was that then one would lose the uniqueness. I am sorry that I cannot find it.
$endgroup$
– user539887
Mar 17 at 13:56
$begingroup$
I remember a discussion somewhere on MSE about a similar thing. The OP asked about why a solution of an IVP cannot be defined on a disjoint union of two intervals. The most upvoted answer (with which I fully agree) was that then one would lose the uniqueness. I am sorry that I cannot find it.
$endgroup$
– user539887
Mar 17 at 13:56
|
show 2 more comments
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$begingroup$
Welcome to MSE. The function $y(x)=e^x$ satisfies the ODE for all values of $tinmathbbR$, in particular for $t=1$.
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– user539887
Mar 15 at 7:03
$begingroup$
@user539887 Yes it is, but isn't it also true that the solution to an IVP is well-defined only in the interval without singularity?
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– Hoil Lee
Mar 15 at 7:05
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It depends on the definition. I took the most obvious one: a function satisfies (or, if you prefer, is a solution of) a differential equation if for each value in the domain of the function the differential equation holds. And such is the case for $t=1$: the LHS is $(1-1) e^1=0$ and the RHS is $(1-1) e^1=0$.
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– user539887
Mar 15 at 7:09
$begingroup$
@user539887 Thank you for your kind explanations. I think I understand it for this particular problem. What if we have an IVP that is not well-defined for some values of $x$, e.g. $y′+(tan x) cdot y=cos^2 x, quad y(0)=1$? We cannot evaluate the value of the LHS for $x = (2n+1)pi/2$, but $ y = cos x sin x + cos x $ satisfies the DE. Can we say the solution is well-defined on the whole real line?
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– Hoil Lee
Mar 15 at 7:30
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Actually this was my original question: math.stackexchange.com/questions/3147887/… and I was trying to simplify the DE to have a singularity but also to have a regular solution.
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– Hoil Lee
Mar 15 at 7:34