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Request: PDE clarification - Partial Differential Equations for scientist and engineers Lesson 6 Question 1.
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can someone just double check my working for me please? the correct solution according forTaken From
Partial Differential Equations for scientist and engineers Lesson 6 Question 1.
the question reads as following
Solve the initial Boundary problem
$$PDE:~~u_t = alpha^2 u_xx ~ textfor 0 < x < 1$$
$$BC:~~u(0,t) = 1 ~~&~~ u_x(1,t)+hu(1,t)=1$$
$$IC:~~u(x,0)=sin(pi x)+x$$
so we have a homogeneous PDE with non-homogeneous BC and so we wish to transform it into one with homogeneous BC.
to do so we look for a solution of the type
$$u(x,t) = g_1(t)[1-x/L] + fracg_1(t)+Lg_2(t)1+Lh[x/L]+v(x,t)$$
from the question we see the BC are fixed points this gives us
$$u(x,t) = 1[1-x/1] + frac1+11+1h[x/1]+v(x,t)$$
$$u(x,t) = 1-x + frac2x1+h+v(x,t)$$
where we can denote our steady state equation to be
$$s(x,t) = 1-x + frac2x1+h$$
so our new problem should be of the form
$$PDE:~~v_t = alpha^2 v_xx-s_t ~ textfor 0 < x < 1$$
because $s(x,t)$ is Indept' of t we get
$$PDE:~~v_t = alpha^2 v_xx ~ textfor 0 < x < 1$$
$$BC:~~v(0,t) = 0 ~~&~~ v_x(1,t)+hv(1,t)=0$$
$$IC:~~v(x,0)=sin(pi x)+x - s(x,0) = sin(pi x)+2x -1- frac2x1+h = overlinephi(x)$$
which will have solution
$$v(x,t) = sum_n=1^infty A_n e^-(npialpha)^2tsinnpi x$$
where
$$A_n = 2 int_0^1 overlinephi(x)sinn pi x~dx$$
$$A_n = 2 int_0^1 (sin(pi x)+2x -1- frac2x1+h)sinn pi x~dx$$
$$A_n = 2 left[int_0^1 sin(pi x)sinn pi x ~dx+2int_0^1xsinn pi x~dx - int_0^1 sinn pi x ~dx- frac21+h int_0^1 xsinn pi x ~dx right]$$
$$A_n = 2 left[frac12 delta_1j +2int_0^1xsinn pi x~dx - int_0^1 sinn pi x ~dx- frac21+h int_0^1 xsinn pi x ~dx right]$$
which takes me down an avenue in which im wrong given that they state the solution to be
$$ u(x,t) = x+e^(-pi alpha)^2 sin(pi x) $$
pde
asked Jul 29 at 8:08
Vaas
335213
add a comment |Â
up vote
0
down vote
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can someone just double check my working for me please? the correct solution according forTaken From
Partial Differential Equations for scientist and engineers Lesson 6 Question 1.
the question reads as following
Solve the initial Boundary problem
$$PDE:~~u_t = alpha^2 u_xx ~ textfor 0 < x < 1$$
$$BC:~~u(0,t) = 1 ~~&~~ u_x(1,t)+hu(1,t)=1$$
$$IC:~~u(x,0)=sin(pi x)+x$$
so we have a homogeneous PDE with non-homogeneous BC and so we wish to transform it into one with homogeneous BC.
to do so we look for a solution of the type
$$u(x,t) = g_1(t)[1-x/L] + fracg_1(t)+Lg_2(t)1+Lh[x/L]+v(x,t)$$
from the question we see the BC are fixed points this gives us
$$u(x,t) = 1[1-x/1] + frac1+11+1h[x/1]+v(x,t)$$
$$u(x,t) = 1-x + frac2x1+h+v(x,t)$$
where we can denote our steady state equation to be
$$s(x,t) = 1-x + frac2x1+h$$
so our new problem should be of the form
$$PDE:~~v_t = alpha^2 v_xx-s_t ~ textfor 0 < x < 1$$
because $s(x,t)$ is Indept' of t we get
$$PDE:~~v_t = alpha^2 v_xx ~ textfor 0 < x < 1$$
$$BC:~~v(0,t) = 0 ~~&~~ v_x(1,t)+hv(1,t)=0$$
$$IC:~~v(x,0)=sin(pi x)+x - s(x,0) = sin(pi x)+2x -1- frac2x1+h = overlinephi(x)$$
which will have solution
$$v(x,t) = sum_n=1^infty A_n e^-(npialpha)^2tsinnpi x$$
where
$$A_n = 2 int_0^1 overlinephi(x)sinn pi x~dx$$
$$A_n = 2 int_0^1 (sin(pi x)+2x -1- frac2x1+h)sinn pi x~dx$$
$$A_n = 2 left[int_0^1 sin(pi x)sinn pi x ~dx+2int_0^1xsinn pi x~dx - int_0^1 sinn pi x ~dx- frac21+h int_0^1 xsinn pi x ~dx right]$$
$$A_n = 2 left[frac12 delta_1j +2int_0^1xsinn pi x~dx - int_0^1 sinn pi x ~dx- frac21+h int_0^1 xsinn pi x ~dx right]$$
which takes me down an avenue in which im wrong given that they state the solution to be
$$ u(x,t) = x+e^(-pi alpha)^2 sin(pi x) $$
pde
asked Jul 29 at 8:08
Vaas
335213
@Xaver Henderson: Do you and your friends from CRUDE really want to see more questions of this kind?
– Christian Blatter
Jul 29 at 13:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
can someone just double check my working for me please? the correct solution according forTaken From
Partial Differential Equations for scientist and engineers Lesson 6 Question 1.
the question reads as following
Solve the initial Boundary problem
$$PDE:~~u_t = alpha^2 u_xx ~ textfor 0 < x < 1$$
$$BC:~~u(0,t) = 1 ~~&~~ u_x(1,t)+hu(1,t)=1$$
$$IC:~~u(x,0)=sin(pi x)+x$$
so we have a homogeneous PDE with non-homogeneous BC and so we wish to transform it into one with homogeneous BC.
to do so we look for a solution of the type
$$u(x,t) = g_1(t)[1-x/L] + fracg_1(t)+Lg_2(t)1+Lh[x/L]+v(x,t)$$
from the question we see the BC are fixed points this gives us
$$u(x,t) = 1[1-x/1] + frac1+11+1h[x/1]+v(x,t)$$
$$u(x,t) = 1-x + frac2x1+h+v(x,t)$$
where we can denote our steady state equation to be
$$s(x,t) = 1-x + frac2x1+h$$
so our new problem should be of the form
$$PDE:~~v_t = alpha^2 v_xx-s_t ~ textfor 0 < x < 1$$
because $s(x,t)$ is Indept' of t we get
$$PDE:~~v_t = alpha^2 v_xx ~ textfor 0 < x < 1$$
$$BC:~~v(0,t) = 0 ~~&~~ v_x(1,t)+hv(1,t)=0$$
$$IC:~~v(x,0)=sin(pi x)+x - s(x,0) = sin(pi x)+2x -1- frac2x1+h = overlinephi(x)$$
which will have solution
$$v(x,t) = sum_n=1^infty A_n e^-(npialpha)^2tsinnpi x$$
where
$$A_n = 2 int_0^1 overlinephi(x)sinn pi x~dx$$
$$A_n = 2 int_0^1 (sin(pi x)+2x -1- frac2x1+h)sinn pi x~dx$$
$$A_n = 2 left[int_0^1 sin(pi x)sinn pi x ~dx+2int_0^1xsinn pi x~dx - int_0^1 sinn pi x ~dx- frac21+h int_0^1 xsinn pi x ~dx right]$$
$$A_n = 2 left[frac12 delta_1j +2int_0^1xsinn pi x~dx - int_0^1 sinn pi x ~dx- frac21+h int_0^1 xsinn pi x ~dx right]$$
which takes me down an avenue in which im wrong given that they state the solution to be
$$ u(x,t) = x+e^(-pi alpha)^2 sin(pi x) $$
pde
asked Jul 29 at 8:08
Vaas
335213
can someone just double check my working for me please? the correct solution according forTaken From
Partial Differential Equations for scientist and engineers Lesson 6 Question 1.
the question reads as following
Solve the initial Boundary problem
$$PDE:~~u_t = alpha^2 u_xx ~ textfor 0 < x < 1$$
$$BC:~~u(0,t) = 1 ~~&~~ u_x(1,t)+hu(1,t)=1$$
$$IC:~~u(x,0)=sin(pi x)+x$$
so we have a homogeneous PDE with non-homogeneous BC and so we wish to transform it into one with homogeneous BC.
to do so we look for a solution of the type
$$u(x,t) = g_1(t)[1-x/L] + fracg_1(t)+Lg_2(t)1+Lh[x/L]+v(x,t)$$
from the question we see the BC are fixed points this gives us
$$u(x,t) = 1[1-x/1] + frac1+11+1h[x/1]+v(x,t)$$
$$u(x,t) = 1-x + frac2x1+h+v(x,t)$$
where we can denote our steady state equation to be
$$s(x,t) = 1-x + frac2x1+h$$
so our new problem should be of the form
$$PDE:~~v_t = alpha^2 v_xx-s_t ~ textfor 0 < x < 1$$
because $s(x,t)$ is Indept' of t we get
$$PDE:~~v_t = alpha^2 v_xx ~ textfor 0 < x < 1$$
$$BC:~~v(0,t) = 0 ~~&~~ v_x(1,t)+hv(1,t)=0$$
$$IC:~~v(x,0)=sin(pi x)+x - s(x,0) = sin(pi x)+2x -1- frac2x1+h = overlinephi(x)$$
which will have solution
$$v(x,t) = sum_n=1^infty A_n e^-(npialpha)^2tsinnpi x$$
where
$$A_n = 2 int_0^1 overlinephi(x)sinn pi x~dx$$
$$A_n = 2 int_0^1 (sin(pi x)+2x -1- frac2x1+h)sinn pi x~dx$$
$$A_n = 2 left[int_0^1 sin(pi x)sinn pi x ~dx+2int_0^1xsinn pi x~dx - int_0^1 sinn pi x ~dx- frac21+h int_0^1 xsinn pi x ~dx right]$$
$$A_n = 2 left[frac12 delta_1j +2int_0^1xsinn pi x~dx - int_0^1 sinn pi x ~dx- frac21+h int_0^1 xsinn pi x ~dx right]$$
which takes me down an avenue in which im wrong given that they state the solution to be
$$ u(x,t) = x+e^(-pi alpha)^2 sin(pi x) $$
pde
asked Jul 29 at 8:08
Vaas
335213
asked Jul 29 at 8:08
Vaas
335213
asked Jul 29 at 8:08
Vaas
335213
asked Jul 29 at 8:08
Vaas
335213
335213
@Xaver Henderson: Do you and your friends from CRUDE really want to see more questions of this kind?
– Christian Blatter
Jul 29 at 13:09
add a comment |Â
@Xaver Henderson: Do you and your friends from CRUDE really want to see more questions of this kind?
– Christian Blatter
Jul 29 at 13:09
@Xaver Henderson: Do you and your friends from CRUDE really want to see more questions of this kind?
– Christian Blatter
Jul 29 at 13:09
@Xaver Henderson: Do you and your friends from CRUDE really want to see more questions of this kind?
– Christian Blatter
Jul 29 at 13:09
add a comment |Â
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@Xaver Henderson: Do you and your friends from CRUDE really want to see more questions of this kind?
– Christian Blatter
Jul 29 at 13:09