How can I solve these PDE analytically for $z$ and $y$?
\begin{eqnarray} 2 \dot{z} &=& \ddot{y}+ y^{-1}, ~~~ (1)\\ z' + \dot{z}' &=& - \dot{y}'. ~~~~~~~~~~ (2) \end{eqnarray}
Where (.) is the time derivative ($\frac{\partial}{\partial t}$ ) and (’) is the derivative with respect to x
: ($\frac{\partial}{\partial x}$).
Here is my trial:
eq1[x_, t_] = 2*D[z[t, x], t] == D[y[x, t], {t, 2}] + y[x, t]^(-1);
eq2[x_, t_] = D[z[t, x], x] + D[D[z[t, x], t], x] == -D[D[y[x, t], t], x];
DSolve[{eq1[x, t], eq2[x, t]}, {z[t, x], y[x, t]}, {t, x}]
But DSolve
dose not give a solution.
Edit
I try to eliminate z
from the second equation, using the first equation. So any help to make these steps?
Integrate (1) to get
z(x, t)
: To do so, I use:Integrate[ D[y[x, t], {t, 2}] + y[x, t]^(-1), y[x, t], t]
But I don't understand the output:
Out[1]=Integrate[Log[y[x, t]], t] + y[x, t]*Derivative[0, 1][y][x, t]
-Then differentiate z(x, t)
with respect to x
to get z’(x,t)
.
Differentiate (1) with respect to
x
to get $\dot{z}’$Then substitute by $z’$ and $\dot{z}’$ in (2)
Edit:2
I tried to solve
$$\frac{\partial}{\partial t} \left(\frac{\dot{y}^2}{2}+\log (y)\right) =2 \dot{y}\ \dot{z}$$
and
$$\frac{\partial (y+z)}{\partial t}+z=\text{C1}(t)$$
In MA by:
eq1[x, t] := D[D[y[x, t], t]^2/2 + Log[y[x, t]], t] -2*D[y[x, t], t]*D[z[x, t], t]
eq2[x, t] := D[y[x, t] + z[x, t], t] + z[x, t] - c
DSolveValue[{eq1[x, t] == eq2[x, t] == 0, y[x, 0] == 0, z[x, 0]
== 0}, {y[x, t], z[x, t]}, {x, t}]
With or without initial conditions, or by using DSolve
, there is no output solution.
z[t, x]
then in the call toDSolve
you writez[x, t]
. Strange that you did not see this error fromDSolve
Also your latex is confusing. This is not how partial derivatives are written in Latex. $\endgroup$Reduce
. I edit the question to make the partial derivatives clear in the latex equations. $\endgroup$z
ory
and make a single equation in one variable. $\endgroup$