Solve the PDE $u_{tt}+2u_{xx}+3u_{xt}-u_t-2u=0$
$begingroup$
Solve the PDE $u_{tt}+2u_{xx}+3u_{xt}-u_t-2u=0$
By a change of variables where $beta=2t-x$ and $alpha =x-t$ I have factorised the equation to $$(frac{partial}{partialalpha}-2)(frac{partial}{partialbeta}+1)u=0$$
But I am unsure as to how to solve this PDE.
ordinary-differential-equations pde
asked Mar 11 '16 at 13:17
123454321123454321
34918
$endgroup$
add a comment |
$begingroup$
Solve the PDE $u_{tt}+2u_{xx}+3u_{xt}-u_t-2u=0$
By a change of variables where $beta=2t-x$ and $alpha =x-t$ I have factorised the equation to $$(frac{partial}{partialalpha}-2)(frac{partial}{partialbeta}+1)u=0$$
But I am unsure as to how to solve this PDE.
ordinary-differential-equations pde
asked Mar 11 '16 at 13:17
123454321123454321
34918
$endgroup$
add a comment |
$begingroup$
Solve the PDE $u_{tt}+2u_{xx}+3u_{xt}-u_t-2u=0$
By a change of variables where $beta=2t-x$ and $alpha =x-t$ I have factorised the equation to $$(frac{partial}{partialalpha}-2)(frac{partial}{partialbeta}+1)u=0$$
But I am unsure as to how to solve this PDE.
ordinary-differential-equations pde
asked Mar 11 '16 at 13:17
123454321123454321
34918
$endgroup$
Solve the PDE $u_{tt}+2u_{xx}+3u_{xt}-u_t-2u=0$
By a change of variables where $beta=2t-x$ and $alpha =x-t$ I have factorised the equation to $$(frac{partial}{partialalpha}-2)(frac{partial}{partialbeta}+1)u=0$$
But I am unsure as to how to solve this PDE.
ordinary-differential-equations pde
ordinary-differential-equations pde
asked Mar 11 '16 at 13:17
123454321123454321
34918
asked Mar 11 '16 at 13:17
123454321123454321
34918
asked Mar 11 '16 at 13:17
123454321123454321
34918
asked Mar 11 '16 at 13:17
123454321123454321
34918
asked Mar 11 '16 at 13:17
123454321123454321
34918
34918
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As you have factorized your equation, you can assume that the solution is
$$ u(alpha, beta) = f(alpha) g (beta).$$
We then find that
$$left(frac{d}{dbeta} +1 right) g(beta) = 0 tag{1}$$
or
$$ left(frac{d}{dalpha}-2 right)f(alpha) =0 tag{2}$$
guarantees that $u$ solves the PDE.
So we have reduced the PDE in the set of ODE's (1) and (2). The solution to the latter reads $$f(alpha) = C_1 e^{2 alpha}$$ and the former implies $$g(beta) = C_2 e^{-beta}.$$
So in total (using the linearity), we have
$$ u(alpha,beta) = C_1(beta) e^{2 alpha} + C_2(alpha) e^{-beta}$$
with arbitrary functions $C_1, C_2$ as the solution of the PDE.
answered Mar 11 '16 at 13:39
FabianFabian
20k3774
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$begingroup$
- $2 {{D}_{x}^{2}}+3 {D_t}, {D_x}+{{D}_{t}^{2}}-{D_t}-2=left( {D_x}+{D_t}+1right) , left( 2 {D_x}+{D_t}-2right) $
- solution of $u_x+u_t+u=0$ is $u_1=e^{-t}f(x-t)$
- solution of $2u_x+u_t-2u=0$ is $u_1=e^{2t}g(x-2t)$
- then $u=u_1+u_2=e^{-t}f(x-t)+e^{2t}g(x-2t)$
answered Jun 3 '18 at 18:21
Aleksas DomarkasAleksas Domarkas
1,54817
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2 Answers
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2 Answers
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active
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$begingroup$
As you have factorized your equation, you can assume that the solution is
$$ u(alpha, beta) = f(alpha) g (beta).$$
We then find that
$$left(frac{d}{dbeta} +1 right) g(beta) = 0 tag{1}$$
or
$$ left(frac{d}{dalpha}-2 right)f(alpha) =0 tag{2}$$
guarantees that $u$ solves the PDE.
So we have reduced the PDE in the set of ODE's (1) and (2). The solution to the latter reads $$f(alpha) = C_1 e^{2 alpha}$$ and the former implies $$g(beta) = C_2 e^{-beta}.$$
So in total (using the linearity), we have
$$ u(alpha,beta) = C_1(beta) e^{2 alpha} + C_2(alpha) e^{-beta}$$
with arbitrary functions $C_1, C_2$ as the solution of the PDE.
answered Mar 11 '16 at 13:39
FabianFabian
20k3774
$endgroup$
add a comment |
$begingroup$
As you have factorized your equation, you can assume that the solution is
$$ u(alpha, beta) = f(alpha) g (beta).$$
We then find that
$$left(frac{d}{dbeta} +1 right) g(beta) = 0 tag{1}$$
or
$$ left(frac{d}{dalpha}-2 right)f(alpha) =0 tag{2}$$
guarantees that $u$ solves the PDE.
So we have reduced the PDE in the set of ODE's (1) and (2). The solution to the latter reads $$f(alpha) = C_1 e^{2 alpha}$$ and the former implies $$g(beta) = C_2 e^{-beta}.$$
So in total (using the linearity), we have
$$ u(alpha,beta) = C_1(beta) e^{2 alpha} + C_2(alpha) e^{-beta}$$
with arbitrary functions $C_1, C_2$ as the solution of the PDE.
answered Mar 11 '16 at 13:39
FabianFabian
20k3774
$endgroup$
add a comment |
$begingroup$
As you have factorized your equation, you can assume that the solution is
$$ u(alpha, beta) = f(alpha) g (beta).$$
We then find that
$$left(frac{d}{dbeta} +1 right) g(beta) = 0 tag{1}$$
or
$$ left(frac{d}{dalpha}-2 right)f(alpha) =0 tag{2}$$
guarantees that $u$ solves the PDE.
So we have reduced the PDE in the set of ODE's (1) and (2). The solution to the latter reads $$f(alpha) = C_1 e^{2 alpha}$$ and the former implies $$g(beta) = C_2 e^{-beta}.$$
So in total (using the linearity), we have
$$ u(alpha,beta) = C_1(beta) e^{2 alpha} + C_2(alpha) e^{-beta}$$
with arbitrary functions $C_1, C_2$ as the solution of the PDE.
answered Mar 11 '16 at 13:39
FabianFabian
20k3774
$endgroup$
As you have factorized your equation, you can assume that the solution is
$$ u(alpha, beta) = f(alpha) g (beta).$$
We then find that
$$left(frac{d}{dbeta} +1 right) g(beta) = 0 tag{1}$$
or
$$ left(frac{d}{dalpha}-2 right)f(alpha) =0 tag{2}$$
guarantees that $u$ solves the PDE.
So we have reduced the PDE in the set of ODE's (1) and (2). The solution to the latter reads $$f(alpha) = C_1 e^{2 alpha}$$ and the former implies $$g(beta) = C_2 e^{-beta}.$$
So in total (using the linearity), we have
$$ u(alpha,beta) = C_1(beta) e^{2 alpha} + C_2(alpha) e^{-beta}$$
with arbitrary functions $C_1, C_2$ as the solution of the PDE.
answered Mar 11 '16 at 13:39
FabianFabian
20k3774
answered Mar 11 '16 at 13:39
FabianFabian
20k3774
answered Mar 11 '16 at 13:39
FabianFabian
20k3774
answered Mar 11 '16 at 13:39
FabianFabian
20k3774
20k3774
add a comment |
add a comment |
$begingroup$
- $2 {{D}_{x}^{2}}+3 {D_t}, {D_x}+{{D}_{t}^{2}}-{D_t}-2=left( {D_x}+{D_t}+1right) , left( 2 {D_x}+{D_t}-2right) $
- solution of $u_x+u_t+u=0$ is $u_1=e^{-t}f(x-t)$
- solution of $2u_x+u_t-2u=0$ is $u_1=e^{2t}g(x-2t)$
- then $u=u_1+u_2=e^{-t}f(x-t)+e^{2t}g(x-2t)$
answered Jun 3 '18 at 18:21
Aleksas DomarkasAleksas Domarkas
1,54817
$endgroup$
add a comment |
$begingroup$
- $2 {{D}_{x}^{2}}+3 {D_t}, {D_x}+{{D}_{t}^{2}}-{D_t}-2=left( {D_x}+{D_t}+1right) , left( 2 {D_x}+{D_t}-2right) $
- solution of $u_x+u_t+u=0$ is $u_1=e^{-t}f(x-t)$
- solution of $2u_x+u_t-2u=0$ is $u_1=e^{2t}g(x-2t)$
- then $u=u_1+u_2=e^{-t}f(x-t)+e^{2t}g(x-2t)$
answered Jun 3 '18 at 18:21
Aleksas DomarkasAleksas Domarkas
1,54817
$endgroup$
add a comment |
$begingroup$
- $2 {{D}_{x}^{2}}+3 {D_t}, {D_x}+{{D}_{t}^{2}}-{D_t}-2=left( {D_x}+{D_t}+1right) , left( 2 {D_x}+{D_t}-2right) $
- solution of $u_x+u_t+u=0$ is $u_1=e^{-t}f(x-t)$
- solution of $2u_x+u_t-2u=0$ is $u_1=e^{2t}g(x-2t)$
- then $u=u_1+u_2=e^{-t}f(x-t)+e^{2t}g(x-2t)$
answered Jun 3 '18 at 18:21
Aleksas DomarkasAleksas Domarkas
1,54817
$endgroup$
- $2 {{D}_{x}^{2}}+3 {D_t}, {D_x}+{{D}_{t}^{2}}-{D_t}-2=left( {D_x}+{D_t}+1right) , left( 2 {D_x}+{D_t}-2right) $
- solution of $u_x+u_t+u=0$ is $u_1=e^{-t}f(x-t)$
- solution of $2u_x+u_t-2u=0$ is $u_1=e^{2t}g(x-2t)$
- then $u=u_1+u_2=e^{-t}f(x-t)+e^{2t}g(x-2t)$
answered Jun 3 '18 at 18:21
Aleksas DomarkasAleksas Domarkas
1,54817
edited Jan 6 at 7:23
answered Jun 3 '18 at 18:21
Aleksas DomarkasAleksas Domarkas
1,54817
answered Jun 3 '18 at 18:21
Aleksas DomarkasAleksas Domarkas
1,54817
answered Jun 3 '18 at 18:21
Aleksas DomarkasAleksas Domarkas
1,54817
1,54817
add a comment |
add a comment |
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