Convexity and concavity in a textbook figure of the fundamental solution of the heat equation $u_t - u_xx = 0$

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My textbook has this image of the fundamental solution of the heat equation $u_t - u_xx = 0$ for $0.25 le t le 10$:



enter image description here



It says that, when we deduce from the PDE that at locations where $u_xx < 0$ (so $u$ is convex in $x$) the solution decreases in time at fixed $x$ whereas, when $u_xx > 0$ (so $u$ is concave) the solution increases in time. The points of inflection, where $u_xx$ changes sign, separate regions where the solution is increasing in time from those where it is decreasing.



Convex vs. Concave:



enter image description here



Am I thinking about this incorrectly, or does this description not match the figure provided? For constant $x$ "beneath" the line (when $u$ is convex in $x$), we see that the solution decreases till about $t = 1$ and then actually increases. And, for constant $x$ "above" the line (when $u$ is concave in $x$), the solution actually decreases in time till about $t = 1$ and only then seems to increase.



Hmm, I'm not sure if I'm interpreting this incorrectly, or if this is an error by the author?




real-analysis geometry pde convex-analysis mathematical-modeling




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  • 1




    Kindly add the PDE also so that it is easy to determine whether the claims are true or false.
    – Math Lover
    Aug 3 at 21:44










  • @MathLover Oops, sorry. I have added it now.
    – Wyuw
    Aug 3 at 21:48














up vote
1
down vote

favorite












My textbook has this image of the fundamental solution of the heat equation $u_t - u_xx = 0$ for $0.25 le t le 10$:



enter image description here



It says that, when we deduce from the PDE that at locations where $u_xx < 0$ (so $u$ is convex in $x$) the solution decreases in time at fixed $x$ whereas, when $u_xx > 0$ (so $u$ is concave) the solution increases in time. The points of inflection, where $u_xx$ changes sign, separate regions where the solution is increasing in time from those where it is decreasing.



Convex vs. Concave:



enter image description here



Am I thinking about this incorrectly, or does this description not match the figure provided? For constant $x$ "beneath" the line (when $u$ is convex in $x$), we see that the solution decreases till about $t = 1$ and then actually increases. And, for constant $x$ "above" the line (when $u$ is concave in $x$), the solution actually decreases in time till about $t = 1$ and only then seems to increase.



Hmm, I'm not sure if I'm interpreting this incorrectly, or if this is an error by the author?




real-analysis geometry pde convex-analysis mathematical-modeling




share|cite|improve this question

















  • 1




    Kindly add the PDE also so that it is easy to determine whether the claims are true or false.
    – Math Lover
    Aug 3 at 21:44










  • @MathLover Oops, sorry. I have added it now.
    – Wyuw
    Aug 3 at 21:48












up vote
1
down vote

favorite









up vote
1
down vote

favorite











My textbook has this image of the fundamental solution of the heat equation $u_t - u_xx = 0$ for $0.25 le t le 10$:



enter image description here



It says that, when we deduce from the PDE that at locations where $u_xx < 0$ (so $u$ is convex in $x$) the solution decreases in time at fixed $x$ whereas, when $u_xx > 0$ (so $u$ is concave) the solution increases in time. The points of inflection, where $u_xx$ changes sign, separate regions where the solution is increasing in time from those where it is decreasing.



Convex vs. Concave:



enter image description here



Am I thinking about this incorrectly, or does this description not match the figure provided? For constant $x$ "beneath" the line (when $u$ is convex in $x$), we see that the solution decreases till about $t = 1$ and then actually increases. And, for constant $x$ "above" the line (when $u$ is concave in $x$), the solution actually decreases in time till about $t = 1$ and only then seems to increase.



Hmm, I'm not sure if I'm interpreting this incorrectly, or if this is an error by the author?




real-analysis geometry pde convex-analysis mathematical-modeling




share|cite|improve this question













My textbook has this image of the fundamental solution of the heat equation $u_t - u_xx = 0$ for $0.25 le t le 10$:



enter image description here



It says that, when we deduce from the PDE that at locations where $u_xx < 0$ (so $u$ is convex in $x$) the solution decreases in time at fixed $x$ whereas, when $u_xx > 0$ (so $u$ is concave) the solution increases in time. The points of inflection, where $u_xx$ changes sign, separate regions where the solution is increasing in time from those where it is decreasing.



Convex vs. Concave:



enter image description here



Am I thinking about this incorrectly, or does this description not match the figure provided? For constant $x$ "beneath" the line (when $u$ is convex in $x$), we see that the solution decreases till about $t = 1$ and then actually increases. And, for constant $x$ "above" the line (when $u$ is concave in $x$), the solution actually decreases in time till about $t = 1$ and only then seems to increase.



Hmm, I'm not sure if I'm interpreting this incorrectly, or if this is an error by the author?





real-analysis geometry pde convex-analysis mathematical-modeling





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 21:47
























asked Aug 3 at 21:23









Wyuw

1036




1036







  • 1




    Kindly add the PDE also so that it is easy to determine whether the claims are true or false.
    – Math Lover
    Aug 3 at 21:44










  • @MathLover Oops, sorry. I have added it now.
    – Wyuw
    Aug 3 at 21:48












  • 1




    Kindly add the PDE also so that it is easy to determine whether the claims are true or false.
    – Math Lover
    Aug 3 at 21:44










  • @MathLover Oops, sorry. I have added it now.
    – Wyuw
    Aug 3 at 21:48







1




1




Kindly add the PDE also so that it is easy to determine whether the claims are true or false.
– Math Lover
Aug 3 at 21:44




Kindly add the PDE also so that it is easy to determine whether the claims are true or false.
– Math Lover
Aug 3 at 21:44












@MathLover Oops, sorry. I have added it now.
– Wyuw
Aug 3 at 21:48




@MathLover Oops, sorry. I have added it now.
– Wyuw
Aug 3 at 21:48










1 Answer
1






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up vote
4
down vote



accepted










The author has accidentally switched the words "concave" and "convex", but if you ignore those two words and read the maths(which is what I did originally), he is correct :) Its nothing more than combining the PDE $u_t = u_xx$ with the fact that $u_t < 0$ means $u$ is decreasing (and similarly for increasing.)



I remember which way round things are by remembering that $x^2$ is convex, and $(x^2)'' > 0$.



I've also attempted to recreate the graph. Note the contour lines, showing the decrease in height at $x=0$.
enter image description here



Not quite the same, but hopefully this helps. From different angles-
enter image description hereenter image description here



Edit - I just found out that Geogebra has an online 3D grapher. It even has a mode for the oldschool red-blue 3D glasses. I recreated the graph again here, and this time its interactive.






share|cite|improve this answer























  • Silly me! Nice images. You are of course correct for the former. But for $u_xx > 0$ ($u$ concave in $x$), doesn't the solution for any fixed $x$ still decrease as $t$ increases? Fix $x = 0$ and we can see that as $t$ increases $u$ decreases?
    – Wyuw
    Aug 3 at 23:56






  • 1




    Oh. Yes, the author has made a typo. I assumed he was correct because the pictures you gave in the OP classify $pm (x-a)^2 + b$ correctly as convex/concave.
    – Calvin Khor
    Aug 4 at 0:04







  • 1




    @Wyuw It is instructive to notice that if you define $tau = -t$, ("reverse time") and let $w(x,tau) = u(x,t)$, then $w$ solves the equation $w_tau = -w_xx$. So you can see the effect of having the wrong sign in the same graph as well. Namely, there is finite time blowup of the solution instead of decay to 0.
    – Calvin Khor
    Aug 4 at 0:08







  • 1




    @Wyuw Yes. the diagrams at the bottom of your post are correct, the author's math symbols are correct, but he has switched the words "concave" and "convex". I will update my answer to reflect this
    – Calvin Khor
    Aug 4 at 0:17







  • 1




    Ok thank you for this information.
    – Wyuw
    Aug 4 at 0:18










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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










The author has accidentally switched the words "concave" and "convex", but if you ignore those two words and read the maths(which is what I did originally), he is correct :) Its nothing more than combining the PDE $u_t = u_xx$ with the fact that $u_t < 0$ means $u$ is decreasing (and similarly for increasing.)



I remember which way round things are by remembering that $x^2$ is convex, and $(x^2)'' > 0$.



I've also attempted to recreate the graph. Note the contour lines, showing the decrease in height at $x=0$.
enter image description here



Not quite the same, but hopefully this helps. From different angles-
enter image description hereenter image description here



Edit - I just found out that Geogebra has an online 3D grapher. It even has a mode for the oldschool red-blue 3D glasses. I recreated the graph again here, and this time its interactive.






share|cite|improve this answer























  • Silly me! Nice images. You are of course correct for the former. But for $u_xx > 0$ ($u$ concave in $x$), doesn't the solution for any fixed $x$ still decrease as $t$ increases? Fix $x = 0$ and we can see that as $t$ increases $u$ decreases?
    – Wyuw
    Aug 3 at 23:56






  • 1




    Oh. Yes, the author has made a typo. I assumed he was correct because the pictures you gave in the OP classify $pm (x-a)^2 + b$ correctly as convex/concave.
    – Calvin Khor
    Aug 4 at 0:04







  • 1




    @Wyuw It is instructive to notice that if you define $tau = -t$, ("reverse time") and let $w(x,tau) = u(x,t)$, then $w$ solves the equation $w_tau = -w_xx$. So you can see the effect of having the wrong sign in the same graph as well. Namely, there is finite time blowup of the solution instead of decay to 0.
    – Calvin Khor
    Aug 4 at 0:08







  • 1




    @Wyuw Yes. the diagrams at the bottom of your post are correct, the author's math symbols are correct, but he has switched the words "concave" and "convex". I will update my answer to reflect this
    – Calvin Khor
    Aug 4 at 0:17







  • 1




    Ok thank you for this information.
    – Wyuw
    Aug 4 at 0:18














up vote
4
down vote



accepted










The author has accidentally switched the words "concave" and "convex", but if you ignore those two words and read the maths(which is what I did originally), he is correct :) Its nothing more than combining the PDE $u_t = u_xx$ with the fact that $u_t < 0$ means $u$ is decreasing (and similarly for increasing.)



I remember which way round things are by remembering that $x^2$ is convex, and $(x^2)'' > 0$.



I've also attempted to recreate the graph. Note the contour lines, showing the decrease in height at $x=0$.
enter image description here



Not quite the same, but hopefully this helps. From different angles-
enter image description hereenter image description here



Edit - I just found out that Geogebra has an online 3D grapher. It even has a mode for the oldschool red-blue 3D glasses. I recreated the graph again here, and this time its interactive.






share|cite|improve this answer























  • Silly me! Nice images. You are of course correct for the former. But for $u_xx > 0$ ($u$ concave in $x$), doesn't the solution for any fixed $x$ still decrease as $t$ increases? Fix $x = 0$ and we can see that as $t$ increases $u$ decreases?
    – Wyuw
    Aug 3 at 23:56






  • 1




    Oh. Yes, the author has made a typo. I assumed he was correct because the pictures you gave in the OP classify $pm (x-a)^2 + b$ correctly as convex/concave.
    – Calvin Khor
    Aug 4 at 0:04







  • 1




    @Wyuw It is instructive to notice that if you define $tau = -t$, ("reverse time") and let $w(x,tau) = u(x,t)$, then $w$ solves the equation $w_tau = -w_xx$. So you can see the effect of having the wrong sign in the same graph as well. Namely, there is finite time blowup of the solution instead of decay to 0.
    – Calvin Khor
    Aug 4 at 0:08







  • 1




    @Wyuw Yes. the diagrams at the bottom of your post are correct, the author's math symbols are correct, but he has switched the words "concave" and "convex". I will update my answer to reflect this
    – Calvin Khor
    Aug 4 at 0:17







  • 1




    Ok thank you for this information.
    – Wyuw
    Aug 4 at 0:18












up vote
4
down vote



accepted







up vote
4
down vote



accepted






The author has accidentally switched the words "concave" and "convex", but if you ignore those two words and read the maths(which is what I did originally), he is correct :) Its nothing more than combining the PDE $u_t = u_xx$ with the fact that $u_t < 0$ means $u$ is decreasing (and similarly for increasing.)



I remember which way round things are by remembering that $x^2$ is convex, and $(x^2)'' > 0$.



I've also attempted to recreate the graph. Note the contour lines, showing the decrease in height at $x=0$.
enter image description here



Not quite the same, but hopefully this helps. From different angles-
enter image description hereenter image description here



Edit - I just found out that Geogebra has an online 3D grapher. It even has a mode for the oldschool red-blue 3D glasses. I recreated the graph again here, and this time its interactive.






share|cite|improve this answer















The author has accidentally switched the words "concave" and "convex", but if you ignore those two words and read the maths(which is what I did originally), he is correct :) Its nothing more than combining the PDE $u_t = u_xx$ with the fact that $u_t < 0$ means $u$ is decreasing (and similarly for increasing.)



I remember which way round things are by remembering that $x^2$ is convex, and $(x^2)'' > 0$.



I've also attempted to recreate the graph. Note the contour lines, showing the decrease in height at $x=0$.
enter image description here



Not quite the same, but hopefully this helps. From different angles-
enter image description hereenter image description here



Edit - I just found out that Geogebra has an online 3D grapher. It even has a mode for the oldschool red-blue 3D glasses. I recreated the graph again here, and this time its interactive.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Aug 4 at 0:19


























answered Aug 3 at 22:09









Calvin Khor

7,93911032




7,93911032











  • Silly me! Nice images. You are of course correct for the former. But for $u_xx > 0$ ($u$ concave in $x$), doesn't the solution for any fixed $x$ still decrease as $t$ increases? Fix $x = 0$ and we can see that as $t$ increases $u$ decreases?
    – Wyuw
    Aug 3 at 23:56






  • 1




    Oh. Yes, the author has made a typo. I assumed he was correct because the pictures you gave in the OP classify $pm (x-a)^2 + b$ correctly as convex/concave.
    – Calvin Khor
    Aug 4 at 0:04







  • 1




    @Wyuw It is instructive to notice that if you define $tau = -t$, ("reverse time") and let $w(x,tau) = u(x,t)$, then $w$ solves the equation $w_tau = -w_xx$. So you can see the effect of having the wrong sign in the same graph as well. Namely, there is finite time blowup of the solution instead of decay to 0.
    – Calvin Khor
    Aug 4 at 0:08







  • 1




    @Wyuw Yes. the diagrams at the bottom of your post are correct, the author's math symbols are correct, but he has switched the words "concave" and "convex". I will update my answer to reflect this
    – Calvin Khor
    Aug 4 at 0:17







  • 1




    Ok thank you for this information.
    – Wyuw
    Aug 4 at 0:18
















  • Silly me! Nice images. You are of course correct for the former. But for $u_xx > 0$ ($u$ concave in $x$), doesn't the solution for any fixed $x$ still decrease as $t$ increases? Fix $x = 0$ and we can see that as $t$ increases $u$ decreases?
    – Wyuw
    Aug 3 at 23:56






  • 1




    Oh. Yes, the author has made a typo. I assumed he was correct because the pictures you gave in the OP classify $pm (x-a)^2 + b$ correctly as convex/concave.
    – Calvin Khor
    Aug 4 at 0:04







  • 1




    @Wyuw It is instructive to notice that if you define $tau = -t$, ("reverse time") and let $w(x,tau) = u(x,t)$, then $w$ solves the equation $w_tau = -w_xx$. So you can see the effect of having the wrong sign in the same graph as well. Namely, there is finite time blowup of the solution instead of decay to 0.
    – Calvin Khor
    Aug 4 at 0:08







  • 1




    @Wyuw Yes. the diagrams at the bottom of your post are correct, the author's math symbols are correct, but he has switched the words "concave" and "convex". I will update my answer to reflect this
    – Calvin Khor
    Aug 4 at 0:17







  • 1




    Ok thank you for this information.
    – Wyuw
    Aug 4 at 0:18















Silly me! Nice images. You are of course correct for the former. But for $u_xx > 0$ ($u$ concave in $x$), doesn't the solution for any fixed $x$ still decrease as $t$ increases? Fix $x = 0$ and we can see that as $t$ increases $u$ decreases?
– Wyuw
Aug 3 at 23:56




Silly me! Nice images. You are of course correct for the former. But for $u_xx > 0$ ($u$ concave in $x$), doesn't the solution for any fixed $x$ still decrease as $t$ increases? Fix $x = 0$ and we can see that as $t$ increases $u$ decreases?
– Wyuw
Aug 3 at 23:56




1




1




Oh. Yes, the author has made a typo. I assumed he was correct because the pictures you gave in the OP classify $pm (x-a)^2 + b$ correctly as convex/concave.
– Calvin Khor
Aug 4 at 0:04





Oh. Yes, the author has made a typo. I assumed he was correct because the pictures you gave in the OP classify $pm (x-a)^2 + b$ correctly as convex/concave.
– Calvin Khor
Aug 4 at 0:04





1




1




@Wyuw It is instructive to notice that if you define $tau = -t$, ("reverse time") and let $w(x,tau) = u(x,t)$, then $w$ solves the equation $w_tau = -w_xx$. So you can see the effect of having the wrong sign in the same graph as well. Namely, there is finite time blowup of the solution instead of decay to 0.
– Calvin Khor
Aug 4 at 0:08





@Wyuw It is instructive to notice that if you define $tau = -t$, ("reverse time") and let $w(x,tau) = u(x,t)$, then $w$ solves the equation $w_tau = -w_xx$. So you can see the effect of having the wrong sign in the same graph as well. Namely, there is finite time blowup of the solution instead of decay to 0.
– Calvin Khor
Aug 4 at 0:08





1




1




@Wyuw Yes. the diagrams at the bottom of your post are correct, the author's math symbols are correct, but he has switched the words "concave" and "convex". I will update my answer to reflect this
– Calvin Khor
Aug 4 at 0:17





@Wyuw Yes. the diagrams at the bottom of your post are correct, the author's math symbols are correct, but he has switched the words "concave" and "convex". I will update my answer to reflect this
– Calvin Khor
Aug 4 at 0:17





1




1




Ok thank you for this information.
– Wyuw
Aug 4 at 0:18




Ok thank you for this information.
– Wyuw
Aug 4 at 0:18












 

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