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I don't know if my answer can help you because it is a particular solution and will probably need a not standard analysis......

I have found doing some calculation that the metric:

$$g(t)=\frac{dx^2+dy^2}{e^{-4t}-x^2-y^2}$$

satisfies $\frac{dg(t)}{dt}=-2Ric(t)$

Where

$$\frac{dg(t)}{dt}=\frac{4e^{-4t}(dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}\ {\rm and} \ Ric(t)=\frac{-2e^{-4t}( dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}$$

this should be a solution to the Ricci flow on $R^2$.

Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{-4\lambda t}-x^{2 \lambda}-y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$.

Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, when $t$ tends to $\infty$ and $- \infty$, but I want that the metric $g(t)$  remain positive.

a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $-2$

b) While for $t \rightarrow - \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $-2$.

For $t=0$,

I get $g_0= \frac{dx^2+dy^2}{1-x^2-y^2}$ and with a simple calculations, I found that the 1-parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$.

Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{-4}{1-r^2}$.

Now If I make an analysis of singularities (always for $(1-r^2)>0$), I found that if $r$ tends to $1-$, $g_0$ tends to $\infty$ and $R$ tends $- \infty$.

This soliton should be a steady soliton, it has negative curvature and it is not bunded from below.

I have found doing some calculation that the metric:

$$g(t)=\frac{dx^2+dy^2}{e^{-4t}-x^2-y^2}$$

satisfies $\frac{dg(t)}{dt}=-2Ric(t)$

Where

$$\frac{dg(t)}{dt}=\frac{4e^{-4t}(dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}\ {\rm and} \ Ric(t)=\frac{-2e^{-4t}( dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}$$

this should be a solution to the Ricci flow on $R^2$.

Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{-4\lambda t}-x^{2 \lambda}-y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$.

Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, when $t$ tends to $\infty$ and $- \infty$, but I want that the metric $g(t)$  remain positive.

a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $-2$

b) While for $t \rightarrow - \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $-2$.

For $t=0$,

I get $g_0= \frac{dx^2+dy^2}{1-x^2-y^2}$ and with a simple calculations, I found that the 1-parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$.

Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{-4}{1-r^2}$.

Now If I make an analysis of singularities (always for $(1-r^2)>0$), I found that if $r$ tends to $1-$, $g_0$ tends to $\infty$ and $R$ tends $- \infty$.

This soliton should be a steady soliton, it has negative curvature and it is not bunded from below.

I don't know if my answer can help you because it is a particular solution and will probably need a not standard analysis (I'm working.) indeed I take this opportunity, responding to you, to ask a question ......

I have found doing some calculation that the metric:

$$g(t)=\frac{dx^2+dy^2}{e^{-4t}-x^2-y^2}$$

satisfies $\frac{dg(t)}{dt}=-2Ric(t)$

Where

$$\frac{dg(t)}{dt}=\frac{4e^{-4t}(dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}\ {\rm and} \ Ric(t)=\frac{-2e^{-4t}( dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}$$

this should be a solution to the Ricci flow on $R^2$.

Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{-4\lambda t}-x^{2 \lambda}-y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$.

Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, when $t$ tends to $\infty$ and $- \infty$, but I want that the metric $g(t)$  remain positive.

a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $-2$

b) While for $t \rightarrow - \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $-2$.

For $t=0$,

I get $g_0= \frac{dx^2+dy^2}{1-x^2-y^2}$ and with a simple calculations, I found that the 1-parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$.

Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{-4}{1-r^2}$.

Now If I make an analysis of singularities (always for $(1-r^2)>0$), I found that if $r$ tends to $1-$, $g_0$ tends to $\infty$ and $R$ tends $- \infty$.

This soliton should be a steady soliton, it has negative curvature and it is not bunded from below.

My question is: What else can I say? Which other analysis can I do?

@Onil90:Why are you looking for an example of steady Ricci soliton with nonpositive Ricci curvature?

I don't know if my answer can help you because it is a particular solution and will probably need a not standard analysis......

I have found doing some calculation that the metric:

$$g(t)=\frac{dx^2+dy^2}{e^{-4t}-x^2-y^2}$$

satisfies $\frac{dg(t)}{dt}=-2Ric(t)$

Where

$$\frac{dg(t)}{dt}=\frac{4e^{-4t}(dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}\ {\rm and} \ Ric(t)=\frac{-2e^{-4t}( dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}$$

this should be a solution to the Ricci flow on $R^2$.

Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{-4\lambda t}-x^{2 \lambda}-y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$.

Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, when $t$ tends to $\infty$ and $- \infty$, but I want that the metric $g(t)$  remain positive.

a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $-2$

b) While for $t \rightarrow - \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $-2$.

For $t=0$,

I get $g_0= \frac{dx^2+dy^2}{1-x^2-y^2}$ and with a simple calculations, I found that the 1-parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$.

Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{-4}{1-r^2}$.

Now If I make an analysis of singularities (always for $(1-r^2)>0$), I found that if $r$ tends to $1-$, $g_0$ tends to $\infty$ and $R$ tends $- \infty$.

This soliton should be a steady soliton, it has negative curvature and it is not bunded from below.

I don't know if my answer can help you because it is a particular solution and will probably need a not standard analysis (I'm working.) indeed I take this opportunity, responding to you, to ask a question ......

I have found doing some calculation that the metric:

$$g(t)=\frac{dx^2+dy^2}{e^{-4t}-x^2-y^2}$$

satisfies $\frac{dg(t)}{dt}=-2Ric(t)$

Where

$$\frac{dg(t)}{dt}=\frac{4e^{-4t}(dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}\ {\rm and} \ Ric(t)=\frac{-2e^{-4t}( dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}$$

this should be a solution to the Ricci flow on $R^2$.

Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{-4\lambda t}-x^{2 \lambda}-y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$.

Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, when $t$ tends to $\infty$ and $- \infty$, but I want that the metric $g(t)$  remain positive.

a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $-2$

b) While for $t \rightarrow - \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $-2$.

For $t=0$,

I get $g_0= \frac{dx^2+dy^2}{1-x^2-y^2}$ and with a simple calculations, I found that the 1-parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$.

Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{-4}{1-r^2}$.

Now If I make an analysis of singularities (always for $(1-r^2)>0$), I found that if $r$ tends to $1-$, $g_0$ tends to $\infty$ and $R$ tends $- \infty$.

This soliton should be a steady soliton, it has negative curvature and it is not bunded from below.

My question are: What else can I say? Which other analysis can I do?

@Onil90:Why are you looking for an example of steady Ricci soliton with non positive Ricci curvature?

I don't know if my answer can help you because it is a particular solution and will probably need a not standard analysis (I'm working.) indeed I take this opportunity, responding to you, to ask a question ......

I have found doing some calculation that the metric:

$$g(t)=\frac{dx^2+dy^2}{e^{-4t}-x^2-y^2}$$

satisfies $\frac{dg(t)}{dt}=-2Ric(t)$

Where

$$\frac{dg(t)}{dt}=\frac{4e^{-4t}(dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}\ {\rm and} \ Ric(t)=\frac{-2e^{-4t}( dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}$$

this should be a solution to the Ricci flow on $R^2$.

Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{-4\lambda t}-x^{2 \lambda}-y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$.

Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, when $t$ tends to $\infty$ and $- \infty$, but I want that the metric $g(t)$  remain positive.

a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $-2$

b) While for $t \rightarrow - \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $-2$.

For $t=0$,

I get $g_0= \frac{dx^2+dy^2}{1-x^2-y^2}$ and with a simple calculations, I found that the 1-parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$.

Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{-4}{1-r^2}$.

Now If I make an analysis of singularities (always for $(1-r^2)>0$), I found that if $r$ tends to $1-$, $g_0$ tends to $\infty$ and $R$ tends $- \infty$.

This soliton should be a steady soliton, it has negative curvature and it is not bunded from below.

My question is: What else can I say? Which other analysis can I do?

@Onil90:Why are you looking for an example of steady Ricci soliton with nonpositive Ricci curvature?