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Showing a Solution to a PDE Satisfies Boundary Conditions
Find solutions of a PDE (by substitution?)Checking a solution of a PDEExistence and uniqueness of solution of a PDEFirst Order PDE, How to Deal With This Boundary Condition?Solving PDE with Boundary ConditionsHow to solve this Quasi-linear PDE?IVP Wave Equation $u_tt = 4u_xx + sin(ct)cos(x)$ (PDE)Energy argument to prove that the solution to a PDE is uniquePerturbation of the boundary conditions in PDETrouble in Checking PDE Boundary Solution Problem
$begingroup$
Consider the PDE $$u_xx-4u_xy-5u_yy=0$$ with boundary conditions $u(x,0)=g_0(x) textand u_y(x,0)=g_1(x) textfor xinmathbbR.$ It is given the general solution of the PDE is $$u(x,y)=F(x-y)+G(-5x-y).$$
The solution was derived as follows.
beginalign
u(x,0)=g_o&implies F(x)+G(-5x)=g_0(x) (1) \
u_y(x,o)=g_1(x)&implies -F'(x)-G'(-5x)=g_1(x) (2)\
endalign
Differentiating $(2)$ w.r.t $x$ gives $$F'(x)-5G'(-5x)=g_0(x) (3).$$
Adding $(2)$ and $(3)$: $$-6G'(-5x)=g_1(x)+g'_0(x).$$ Solving this gives $$G(-5x)=frac56int g_1(x) dx+frac56g_0(x)+C_1.$$ Putting this into $(1)$, gives $$F(x)=frac16g_0(x)-frac56int g_1(x) dx-C_1.$$
Hence the solution is $$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^-5x-y g_1(w) dw.$$
My question is, how can I verify this solution?
My attempt:
I have tried to show that $u(x,0)=g_0(x)$. Thus $$u(x,0)=frac16g_0(x)+frac56g_0(x)-frac56int_0^x g_1(w) dw+frac56int_0^-5x g_1(w) dw.$$ But this doesn't seem to work. Thank you in advanced.
integration proof-verification pde
asked 20 hours ago
BellBell
165316
$endgroup$
add a comment |
$begingroup$
Consider the PDE $$u_xx-4u_xy-5u_yy=0$$ with boundary conditions $u(x,0)=g_0(x) textand u_y(x,0)=g_1(x) textfor xinmathbbR.$ It is given the general solution of the PDE is $$u(x,y)=F(x-y)+G(-5x-y).$$
The solution was derived as follows.
beginalign
u(x,0)=g_o&implies F(x)+G(-5x)=g_0(x) (1) \
u_y(x,o)=g_1(x)&implies -F'(x)-G'(-5x)=g_1(x) (2)\
endalign
Differentiating $(2)$ w.r.t $x$ gives $$F'(x)-5G'(-5x)=g_0(x) (3).$$
Adding $(2)$ and $(3)$: $$-6G'(-5x)=g_1(x)+g'_0(x).$$ Solving this gives $$G(-5x)=frac56int g_1(x) dx+frac56g_0(x)+C_1.$$ Putting this into $(1)$, gives $$F(x)=frac16g_0(x)-frac56int g_1(x) dx-C_1.$$
Hence the solution is $$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^-5x-y g_1(w) dw.$$
My question is, how can I verify this solution?
My attempt:
I have tried to show that $u(x,0)=g_0(x)$. Thus $$u(x,0)=frac16g_0(x)+frac56g_0(x)-frac56int_0^x g_1(w) dw+frac56int_0^-5x g_1(w) dw.$$ But this doesn't seem to work. Thank you in advanced.
integration proof-verification pde
asked 20 hours ago
BellBell
165316
$endgroup$
$begingroup$
I recommend editing into your question the argument you used to derive this solution.
$endgroup$
– J.G.
19 hours ago
$begingroup$
I have just updated :)
$endgroup$
– Bell
19 hours ago
add a comment |
$begingroup$
Consider the PDE $$u_xx-4u_xy-5u_yy=0$$ with boundary conditions $u(x,0)=g_0(x) textand u_y(x,0)=g_1(x) textfor xinmathbbR.$ It is given the general solution of the PDE is $$u(x,y)=F(x-y)+G(-5x-y).$$
The solution was derived as follows.
beginalign
u(x,0)=g_o&implies F(x)+G(-5x)=g_0(x) (1) \
u_y(x,o)=g_1(x)&implies -F'(x)-G'(-5x)=g_1(x) (2)\
endalign
Differentiating $(2)$ w.r.t $x$ gives $$F'(x)-5G'(-5x)=g_0(x) (3).$$
Adding $(2)$ and $(3)$: $$-6G'(-5x)=g_1(x)+g'_0(x).$$ Solving this gives $$G(-5x)=frac56int g_1(x) dx+frac56g_0(x)+C_1.$$ Putting this into $(1)$, gives $$F(x)=frac16g_0(x)-frac56int g_1(x) dx-C_1.$$
Hence the solution is $$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^-5x-y g_1(w) dw.$$
My question is, how can I verify this solution?
My attempt:
I have tried to show that $u(x,0)=g_0(x)$. Thus $$u(x,0)=frac16g_0(x)+frac56g_0(x)-frac56int_0^x g_1(w) dw+frac56int_0^-5x g_1(w) dw.$$ But this doesn't seem to work. Thank you in advanced.
integration proof-verification pde
asked 20 hours ago
BellBell
165316
$endgroup$
Consider the PDE $$u_xx-4u_xy-5u_yy=0$$ with boundary conditions $u(x,0)=g_0(x) textand u_y(x,0)=g_1(x) textfor xinmathbbR.$ It is given the general solution of the PDE is $$u(x,y)=F(x-y)+G(-5x-y).$$
The solution was derived as follows.
beginalign
u(x,0)=g_o&implies F(x)+G(-5x)=g_0(x) (1) \
u_y(x,o)=g_1(x)&implies -F'(x)-G'(-5x)=g_1(x) (2)\
endalign
Differentiating $(2)$ w.r.t $x$ gives $$F'(x)-5G'(-5x)=g_0(x) (3).$$
Adding $(2)$ and $(3)$: $$-6G'(-5x)=g_1(x)+g'_0(x).$$ Solving this gives $$G(-5x)=frac56int g_1(x) dx+frac56g_0(x)+C_1.$$ Putting this into $(1)$, gives $$F(x)=frac16g_0(x)-frac56int g_1(x) dx-C_1.$$
Hence the solution is $$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^-5x-y g_1(w) dw.$$
My question is, how can I verify this solution?
My attempt:
I have tried to show that $u(x,0)=g_0(x)$. Thus $$u(x,0)=frac16g_0(x)+frac56g_0(x)-frac56int_0^x g_1(w) dw+frac56int_0^-5x g_1(w) dw.$$ But this doesn't seem to work. Thank you in advanced.
integration proof-verification pde
integration proof-verification pde
asked 20 hours ago
BellBell
165316
asked 20 hours ago
BellBell
165316
edited 15 hours ago
Bell
asked 20 hours ago
BellBell
165316
asked 20 hours ago
BellBell
165316
asked 20 hours ago
BellBell
165316
165316
$begingroup$
I recommend editing into your question the argument you used to derive this solution.
$endgroup$
– J.G.
19 hours ago
$begingroup$
I have just updated :)
$endgroup$
– Bell
19 hours ago
add a comment |
$begingroup$
I recommend editing into your question the argument you used to derive this solution.
$endgroup$
– J.G.
19 hours ago
$begingroup$
I have just updated :)
$endgroup$
– Bell
19 hours ago
$begingroup$
I recommend editing into your question the argument you used to derive this solution.
$endgroup$
– J.G.
19 hours ago
$begingroup$
I recommend editing into your question the argument you used to derive this solution.
$endgroup$
– J.G.
19 hours ago
$begingroup$
I have just updated :)
$endgroup$
– Bell
19 hours ago
$begingroup$
I have just updated :)
$endgroup$
– Bell
19 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The last integral of your solution is wrong. It should read:
$$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^frac5x+y5 g_1(w) dw.$$
I am assuming your mistake came from the missing bounds on the integral in the derivation. When you go from
$$-6G'(-5x)=g_1(x)+g'_0(x).$$ to $$G(-5x)=frac56int_colorred0^colorredx g_1(colorredw) dcolorredw+frac56g_0(x)+C_1.$$
Regarding your question if the solution satisfies the PDE+BC: Basically it does by construction. If you really want to double check, simply insert the solution into the PDE and the BC. To do this you will need the chain rule and also this: https://en.m.wikipedia.org/wiki/Leibniz_integral_rule to handle the derivativesof the integrals.
answered 18 hours ago
maxmilgrammaxmilgram
7357
$endgroup$
$begingroup$
Can't believe I missed that. Thank you. Also, could you please edit how it satisfies $u_y(x,0)=g_1(x)$ and the PDE?
$endgroup$
– Bell
18 hours ago
$begingroup$
Another point, if $$g(z)=-frac16int g_1left(-fracz5right) dz (z=-5x),$$ does this mean that $$g(-5x-y)=frac56int_0^frac5x+y5 g_1(w) dw?$$ If your answer is yes, then I believe this is where I made my error.
$endgroup$
– Bell
15 hours ago
$begingroup$
@Bell: I further updated my answer!
$endgroup$
– maxmilgram
12 hours ago
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The last integral of your solution is wrong. It should read:
$$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^frac5x+y5 g_1(w) dw.$$
I am assuming your mistake came from the missing bounds on the integral in the derivation. When you go from
$$-6G'(-5x)=g_1(x)+g'_0(x).$$ to $$G(-5x)=frac56int_colorred0^colorredx g_1(colorredw) dcolorredw+frac56g_0(x)+C_1.$$
Regarding your question if the solution satisfies the PDE+BC: Basically it does by construction. If you really want to double check, simply insert the solution into the PDE and the BC. To do this you will need the chain rule and also this: https://en.m.wikipedia.org/wiki/Leibniz_integral_rule to handle the derivativesof the integrals.
answered 18 hours ago
maxmilgrammaxmilgram
7357
$endgroup$
$begingroup$
Can't believe I missed that. Thank you. Also, could you please edit how it satisfies $u_y(x,0)=g_1(x)$ and the PDE?
$endgroup$
– Bell
18 hours ago
$begingroup$
Another point, if $$g(z)=-frac16int g_1left(-fracz5right) dz (z=-5x),$$ does this mean that $$g(-5x-y)=frac56int_0^frac5x+y5 g_1(w) dw?$$ If your answer is yes, then I believe this is where I made my error.
$endgroup$
– Bell
15 hours ago
$begingroup$
@Bell: I further updated my answer!
$endgroup$
– maxmilgram
12 hours ago
add a comment |
$begingroup$
The last integral of your solution is wrong. It should read:
$$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^frac5x+y5 g_1(w) dw.$$
I am assuming your mistake came from the missing bounds on the integral in the derivation. When you go from
$$-6G'(-5x)=g_1(x)+g'_0(x).$$ to $$G(-5x)=frac56int_colorred0^colorredx g_1(colorredw) dcolorredw+frac56g_0(x)+C_1.$$
Regarding your question if the solution satisfies the PDE+BC: Basically it does by construction. If you really want to double check, simply insert the solution into the PDE and the BC. To do this you will need the chain rule and also this: https://en.m.wikipedia.org/wiki/Leibniz_integral_rule to handle the derivativesof the integrals.
answered 18 hours ago
maxmilgrammaxmilgram
7357
$endgroup$
$begingroup$
Can't believe I missed that. Thank you. Also, could you please edit how it satisfies $u_y(x,0)=g_1(x)$ and the PDE?
$endgroup$
– Bell
18 hours ago
$begingroup$
Another point, if $$g(z)=-frac16int g_1left(-fracz5right) dz (z=-5x),$$ does this mean that $$g(-5x-y)=frac56int_0^frac5x+y5 g_1(w) dw?$$ If your answer is yes, then I believe this is where I made my error.
$endgroup$
– Bell
15 hours ago
$begingroup$
@Bell: I further updated my answer!
$endgroup$
– maxmilgram
12 hours ago
add a comment |
$begingroup$
The last integral of your solution is wrong. It should read:
$$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^frac5x+y5 g_1(w) dw.$$
I am assuming your mistake came from the missing bounds on the integral in the derivation. When you go from
$$-6G'(-5x)=g_1(x)+g'_0(x).$$ to $$G(-5x)=frac56int_colorred0^colorredx g_1(colorredw) dcolorredw+frac56g_0(x)+C_1.$$
Regarding your question if the solution satisfies the PDE+BC: Basically it does by construction. If you really want to double check, simply insert the solution into the PDE and the BC. To do this you will need the chain rule and also this: https://en.m.wikipedia.org/wiki/Leibniz_integral_rule to handle the derivativesof the integrals.
answered 18 hours ago
maxmilgrammaxmilgram
7357
$endgroup$
The last integral of your solution is wrong. It should read:
$$u(x,y)=frac16g_0(x-y)+frac56g_0left(frac5x+y5right)-frac56int_0^x-y g_1(w) dw+frac56int_0^frac5x+y5 g_1(w) dw.$$
I am assuming your mistake came from the missing bounds on the integral in the derivation. When you go from
$$-6G'(-5x)=g_1(x)+g'_0(x).$$ to $$G(-5x)=frac56int_colorred0^colorredx g_1(colorredw) dcolorredw+frac56g_0(x)+C_1.$$
Regarding your question if the solution satisfies the PDE+BC: Basically it does by construction. If you really want to double check, simply insert the solution into the PDE and the BC. To do this you will need the chain rule and also this: https://en.m.wikipedia.org/wiki/Leibniz_integral_rule to handle the derivativesof the integrals.
answered 18 hours ago
maxmilgrammaxmilgram
7357
edited 12 hours ago
answered 18 hours ago
maxmilgrammaxmilgram
7357
answered 18 hours ago
maxmilgrammaxmilgram
7357
answered 18 hours ago
maxmilgrammaxmilgram
7357
7357
$begingroup$
Can't believe I missed that. Thank you. Also, could you please edit how it satisfies $u_y(x,0)=g_1(x)$ and the PDE?
$endgroup$
– Bell
18 hours ago
$begingroup$
Another point, if $$g(z)=-frac16int g_1left(-fracz5right) dz (z=-5x),$$ does this mean that $$g(-5x-y)=frac56int_0^frac5x+y5 g_1(w) dw?$$ If your answer is yes, then I believe this is where I made my error.
$endgroup$
– Bell
15 hours ago
$begingroup$
@Bell: I further updated my answer!
$endgroup$
– maxmilgram
12 hours ago
add a comment |
$begingroup$
Can't believe I missed that. Thank you. Also, could you please edit how it satisfies $u_y(x,0)=g_1(x)$ and the PDE?
$endgroup$
– Bell
18 hours ago
$begingroup$
Another point, if $$g(z)=-frac16int g_1left(-fracz5right) dz (z=-5x),$$ does this mean that $$g(-5x-y)=frac56int_0^frac5x+y5 g_1(w) dw?$$ If your answer is yes, then I believe this is where I made my error.
$endgroup$
– Bell
15 hours ago
$begingroup$
@Bell: I further updated my answer!
$endgroup$
– maxmilgram
12 hours ago
$begingroup$
Can't believe I missed that. Thank you. Also, could you please edit how it satisfies $u_y(x,0)=g_1(x)$ and the PDE?
$endgroup$
– Bell
18 hours ago
$begingroup$
Can't believe I missed that. Thank you. Also, could you please edit how it satisfies $u_y(x,0)=g_1(x)$ and the PDE?
$endgroup$
– Bell
18 hours ago
$begingroup$
Another point, if $$g(z)=-frac16int g_1left(-fracz5right) dz (z=-5x),$$ does this mean that $$g(-5x-y)=frac56int_0^frac5x+y5 g_1(w) dw?$$ If your answer is yes, then I believe this is where I made my error.
$endgroup$
– Bell
15 hours ago
$begingroup$
Another point, if $$g(z)=-frac16int g_1left(-fracz5right) dz (z=-5x),$$ does this mean that $$g(-5x-y)=frac56int_0^frac5x+y5 g_1(w) dw?$$ If your answer is yes, then I believe this is where I made my error.
$endgroup$
– Bell
15 hours ago
$begingroup$
@Bell: I further updated my answer!
$endgroup$
– maxmilgram
12 hours ago
$begingroup$
@Bell: I further updated my answer!
$endgroup$
– maxmilgram
12 hours ago
add a comment |
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$begingroup$
I recommend editing into your question the argument you used to derive this solution.
$endgroup$
– J.G.
19 hours ago
$begingroup$
I have just updated :)
$endgroup$
– Bell
19 hours ago